Jupiter and mercury motions analysis (revised) (2)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
Jupiter And Mercury Motions Analysis (Revised) (2)
The Author Authorized To Be Used By
Mr. Gerges Francis Tawdrous
A Student–Physics Department- Physics
& Mathematics Faculty –
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Dr. Budochkina, Svetlana Aleksandrovna
Associate Professor (Mathematical Analysis
and Theory of Functions Department)
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Phone +201022532292
E-Mail: mrwaheid@gmail.com
Curriculum Vitae http://vixra.org/abs/1902.0044
Phone +7 (495) 952-35-83
E-Mail: budochkina-sa@rudn.ru, sbudotchkina@yandex.ru
Website
http://web-local.rudn.ru/web-local/prep/rj/index.php?id=2944&p=19024
The Assumption Of S. Virgin Mary -Written in Cairo –Egypt –8th
September 2021
Abstract
Paper Argument
- Jupiter Motion analysis disproves Newton Theory of The Sun Mass Gravity and
Planet Motion Independency Concept.
- Because
- Jupiter Motion depends on Mercury and Mars Cycles Periods
- This dependency proves that,
- The solar planets motions be integrated into One Unified Motion
- Means,
- The solar planets be similar to gears move in one machine they do their motions in
comparison with each other, or
- The solar planets be similar to train carriages they move together One Unified
General Motion
- The solar system distances be similar to railways be built by energy and the planets
be similar to trains move on these railways
- Jupiter Motion analysis disproves Newton theory in 2 basic points which are
- (1) The Planet Motion Can't be Independent from the other planets motions
- (2) The Planet Motion Be done by the same force by which Planet Matter is
created, Because the motion features be effective in Planet Matter Dimensions.

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Paper Conclusion
- The Solar System Distances Be Created In A Network Form And Based On One
Geometrical Design.
- Means
- The distances be distributed based on geometrical design in comparison with other
distances and no a single distance be created independent from the other distances
Please scan the figure (ORCID)

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Paper Contents
Subject Page No.
1- Introduction 4
2- Methodology 7
3- Jupiter Motion Data Analysis 10
3-1 Preface 11
3-2 Mercury, Mars and Jupiter Motions Comparison 13
3-3 Mercury, Mars and Jupiter Motions Analysis 21
3-4 Earth and Pluto Motions Analysis 42
4-Mercury Motion Analysis 57
4-1 Preface 58
4-2 The Theory basic Ideas 59
4-3 Light Beam (1.16 mkm/s) Existence Proves Discussion 61
4-4 Jupiter Orbital Circumference Analysis 100
5-The Solar System Distances Be Created In A Network Form 118
5-1 Preface 119
5-2 The Continuum effect Through the Solar System Distances 121
5-3 The Solar System Distances Distribution 126
5-4 The Solar System Distances Dependency On One Another 133
6- The Solar Planets Motions Depend On Different Rates of Time 135
6-1 Preface 136
6-2 Earth And Pluto Rate Of Time 137
6-3 Mercury Motion Rate Of Time 146
6-4 Saturn Motion Rate Of Time 149
6-5 Uranus And Jupiter Rate Of Time 151
7- Planets Unified General Motion 152
7-1 The Planets Unified General Motion Description 153
7-2 The Solar Planets Motions Are Complementary One Another 154
7-3 The Solar Planets Move An Unified Motion (A Team Motion) 159
8- The Sun Age Historical Proves 164
8-1 The Sun Circles The Earth 165
8-2 The Rate (1.0725) 166
8-3 The Sun Diameter Analysis 170
8-4 The Sun Rays Analysis 173
9- Mercury Jupiter Distance Analysis 178
9-1 The Main Idea 179
9-2 Venus Jupiter Distance Analysis 184
9-3 Jupiter and Pluto Motions For The Distance 720.7 mkm 192
9-4 Planets Motions Rates of Time 195
9-5 Planets Motions Interaction 200
9-6 Mars Motion For The Distance 720.7 mkm 215
10- The Solar Planets Motions Analysis 227
10-1 Preface 228
10- 2 The Equal Distances Method 229
10-3 The Inner And Outer Planets Motions Be Distinguished 235
Appendix No.1 The Solar System Equal Distances List 244
References and Biography 246

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1-Introdcution
- Jupiter Orbital Period =4331 days = 2π x 687 days (Mars Orbital Period)
- This data is one noticeable data I have asked about it many times before, because
no clear answer for the question (Why?)
- What's the relationship between Jupiter motion and mars motion based on which
this relationship be produced…
- Why Jupiter Orbital Period = Mars Orbital Period x 2π?
- The answer usually be (Pure Coincidences of numbers)!
- But
- The Question Is So Deep Than The Superficial Answer.
- The question is produced based on a completely different vision of the solar
system motion contradicts and disproves Newton Vision inherited by his theory of
the sun mass gravity.
- The solar planets don't move by the sun mass gravity, the idea is an imaginary one
- The Solar Planet Motion can't be independent from other planets motions
- Simply
- The solar system motion facts contradict Newton Theory of The Sun mass gravity
and the planet motion independency Concept
- The ideas simply are mistaken
- The relationship between Jupiter orbital period and mars orbital period discovered
a deep interaction of the planets motions which create it
- What does this paper try to do?
- We analyze 3 planets motions which are (Mercury – Mars – Jupiter) to prove that
the 3 planets move relative to one another as gears in one machine. Their motions
be integrated together into one unified general motion disproves the planet motion
independency concept and Newton Theory of The sun mass gravity.
- Shortly
- The Nature Of Motion Is Different

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- Newton Was Mistaken
- How the motion be different from what Newton had described?
- Because
- The 3 planets depend on each other data in their motion. let's see examples in
following
- Example no. (1)
- Jupiter orbital period =4331 mkm. But Jupiter moves during (687 day = Mars
orbital period) a distance = 778.6 mkm = Jupiter orbital distance
- Example no. (2)
- Jupiter (13.1 km/s) moves during 24.6 hours (Mars rotation period) a distance =
1.16 mkm
- But
- 4900 mkm (Jupiter Orbital Circumference) = 1.16 mkm x 4224
- Where
- 4222.6 hours = Mercury Day Period (very near to 4224)
- If the data be produced by pure coincidences of numbers, we would to consider
tenths of data as pure coincidences results – no clear logic behind such thinking
- The straight logic tells that, the 3 planets motions have interaction with one
another and this interaction causes the planets motions data to be created taking
into consideration the other planets motions data and based on that no
independency of motion can be supposed…
- I try to show that, the 3 planets motions data analysis gives clearly another vision
for the solar system contradicting Newton Theory of the Sun mass gravity.
- The next question is
- Why and how these 3 planets motions be integrated in one unified motion or create
between one another motions interaction?

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- This question we should answer in this paper discussion
- Let's take a look on the paper contents
- Point no.(3) Jupiter Motions Data Analysis
- In this point we analyze Mercury, Mars And Jupiter Motions Data
- Point no.(4) The Solar System Distances Be Created In A Network Form
- In this point we try to prove that the distances be created in a network form
- Point no.(5) Mercury Motion Analysis
- Mercury Motion provides another reading for the planets motions data.
- Means
- The data which we have discussed in Jupiter Motion Data Analysis we will repeat
its discussion by using another reading…

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2-Methodology
- I use the planets data analysis to discover the planets motions laws
- This method provides good comparisons between The Solar Planets Motions Data
and the theories which try to explain These Planets Motions.
- The planets data analysis is so useful method because it protects greatly from the
imagination.
- By thinking approach we can find suitable tests for the solar system motion
theories, and based on that, let's test Newton Theory of the sun mass gravity.
- The sun mass gravity tells us that,
- The solar planet moves depending on the sun mass gravity independently from all
other planets motions. By that Each Planet Motion Data is almost created
independent from all other planets motions data. this idea we understood simply
from the planet motion independency concept.
- We don't need to discuss the independency concept here but we need to test if the
planets motions data supports it. Because if it doesn't that may refer to another
conclusion tells (Planet Motion Independency Concept Is An Imaginary Idea)
- Let's discuss an example in following:
Example No. (1)
(I)
- 778.6 mkm (Jupiter Orbital Distance) = 1.16 x 671 mkm (Jupiter Venus Distance)
- 720.7 mkm (Jupiter Mercury Distance) = 1.16 x 629 mkm (Jupiter Earth Distance)
- 550.7 mkm (Jupiter Mars Distance) =1.16 x π x149.6 mkm (Earth orbital distance)
- (Max error 1%)
- Shortly
- The Distances Between Jupiter On One Side And Mercury, Venus, Earth And
Mars On The Other Side, All These Distances Are Rated With The Rate (1.16)

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Example No. (1)
(II)
- 778.6 mkm (Jupiter Orbital Distance) =1.072 x 720.7mkm (Jupiter Mercury Distance)
- 720.7 mkm (Jupiter Mercury Distance) =1.072 x 671 mkm (Jupiter Venus Distance)
- 671 mkm (Jupiter Venus Distance) =1.072 x 629 mkm (Jupiter Earth Distance)
- 629 mkm (Jupiter Earth Distance) = (1.072)2
x 550.7 mkm (Jupiter Mars Distance)
- The previous data tries to explain one feature of Jupiter Motion.
- We deal with the same 5 distances which are
- Jupiter Orbital Distance = 778.6 mkm
- Jupiter Mercury Distance = 720.7 mkm
- Jupiter Venus Distance = 671 mkm
- Jupiter Earth Distance = 629 mkm
- Jupiter Mars Distance = 550.7 mkm
- These 5 distances are rated with each other based on the rates (1.16) and (1.0725)
- The question is why these 5 distances be rated with (1.0725) and (1.16)?!
- Because these distances are created based on one another by using one geometrical
design.
- (Imagine we a have a square, each dimension in this square be in proportionality
with this square data and by that all dimensions be rated with one another
geometrically)…
- The 5 distances be rated with one another because they are created as dimensions
in one geometrical design.
- I want to say these distances can't be independent distances from one another. they
are created by using one geometrical system…
- The direct conclusion tells that (Planet Motion Can't Be Independent)…
- I try to show that
- Newton Simple Idea is an imaginary one because

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- The solar system is so complex machine, and the simple vision without data
analysis gives us no facts. The solar system is a building of Gold made based on a
complex geometrical design.
- The provided data is an example for hundreds similar data. each part of the solar
system design refers to a complex machine.
- The direct question is (Why Jupiter distances are rated with one another by these 2
rates 1.0725 and 1.16)??
- The answer is not a simple one
- The rate 1.0725 controls 40% of all distances in the solar system (appendix no. 1
provides a list for these distances)
- Why this rate is used between 40% of all distances? This question still needs
answer but the using is a fact can't be disproved.
- Planets data analysis put planet motion data in comparison with the theory tri to
explain it, by that, we can understand much better both the planet motion data and
the theory which tries to explain it.

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3- Jupiter Motion Data Analysis
3-1 Preface
3-2 Mercury, Mars and Jupiter Motions Comparison
3-3 Mercury, Mars and Jupiter Motions Analysis
3-4 Earth and Pluto Motions Analysis

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3-1 Preface
- Jupiter Motion Data Analysis disproves Newton Theory of The Sun Mass Gravity..
- Because
- Jupiter Motion Data Analysis Proves The Solar Planets Motions Be Integrated Into
One Unified Motion.
- The unified motion can't be created of planets move by the sun mass gravity but be
created by planets move by one force uses them as players in comparisons with
one another.
- Means,
- The force which created the solar planets matters forces them to move as puppets
on a puppets theater relative to one another.
- Let's use one data as example to explain this meaning as clear as possible
Example no. (1)
4331 solar days = 24.6 hours x 175.94 = 687 solar days x 2π
- Where
- 4331 solar days = Jupiter Orbital Period
- 175.94 solar days = Mercury Day Period
- 687 solar days = Mars Orbital Period
- 24.6 hours = Mars Rotation Period
- This data is the main data in this point discussion.
- Let's ask why the 3 cycles periods of time be created in proportionality with one
another?
- We don't understand the clear meaning behind this data!
- Because, for example
- 175.94 solar days = Mercury Day Period, But what's the relationship with (24.6 h)
here? how to explain this equality between data and why they are found?
- Many other questions we need to answer in our try to explain this data,

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- But
- The usual answer be (Pure Coincidence Of Numbers)
- And
- We shouldn't occupy ourselves by fighting with the pure coincidence claim, on the
contrary, we should kill this claim at once by a direct shoot because it prevents us
from a very important understanding for the solar system motion.
- As a result, we ask the direct question
- Does this data depend on planets real motions or be created by pure
coincidence of numbers?
- The answer be in the next point directly, where we bring the 3 planets motions data
and create comparisons between them to prove that These 3 Planets Motions
Can't Be Done Independently From One Another.
- Based on that,
- We analyze the three planets motions in the next point aiming to prove that the 3
planets motions can't be independent from one another.

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3-2 Mercury, Mars and Jupiter Motions Comparison
I-Data
(No. 1) (Mercury and Mars Motions)
- Mercury moves during its day period (4222.6 h.) a distance = 720.7 mkm
- Mars moves during (2802 hours) a distance = 243 mkm
- Mercury moves during (1407.6 hours) a distance = 243 mkm (error 1%)
- Mars moves during 346.6 d. a distance =720.7 mkm (Mercury Jupiter Dis.)
- Mercury moves during 346.6 d. a distance =1419 mkm (with 1433 error 1%)
- Mars moves during 687 d. a distance = 1433 mkm (Mars orbit. Circum)
- Mercury moves during 687 d. a distance =2815mkm (Mercury Uranus Dis.)
- Mars moves during 4331d. a distance = 9010mkm (Saturn orbit. Circum)
- Mercury moves during 4331d. a distance = 2815 mkm x 2π
- Mars moves during 224.7 d. a distance = π x 149.6 mkm (Earth orb. Dis)
- Mercury moves during 224.7 d. a distance = 920 mkm (with 929 error 1%)
- Mars moves during 365.25 d. a distance = π x 243 mkm (0.5%)
- Mercury moves during 365.25 d. a distance = 2 x 748 mkm
- Mars moves during 5040s. a distance= 121464 km= Saturn Diameter (+1%)
- Mercury moves during 5040s. a distance= 238896 km= 2 Saturn Diameters (-1%)
Where
1407.6 h = Mercury Rotation Period 2802 h = Venus Day Period
346.6 days = the nodal year 224.7 days = Venus orbital period
687 days = Mars Orbital Period
4331 days = Jupiter orbital period = 2π x 687 days

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(No. 2) (Mars And Jupiter Motions)
- Mars moves during 636.7 days a distance =1325 mkm= (Venus Saturn Dis.)
- Jupiter moves during 636.7 days a distance =720.7 mkm
- Mars moves during 687 d. a distance= 1433 mkm= Mars Orbital Circumference
- Jupiter moves during 687 d. a distance= 778.6 mkm= Jupiter Orbital Distance
- Mars moves during 4331d. a distance= 9010 mkm= Saturn orbital circumference
- Jupiter moves during 4331 d. a distance= 4900 mkm= Jupiter Orbital Circum.
- Mars moves during 778.6 d. a distance = 1622 mkm = Uranus Neptune Dis.
- Jupiter moves during 778.6 d a distance= 881 mkm =58 days x 15.19 mkm.
- ( 15.19 mkm = the 9 solar planets motions distances total per a solar day).
- Notice
- 2.082 mkm / day x 778.6 days = 1622 mkm = 1.1318 mkm /day x 1433 days
- Where
- 2.082 mkm /day = Mars Velocity Per A Solar Day
- 4.095 mkm /day = Mercury Velocity Per A Solar Day
- 1.13184 mkm /day = Jupiter Velocity Per A Solar Day

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(No. 3) (Mercury And Jupiter Motions)
- Mercury moves during 687 d. a distance= 2815 mkm= Mercury Uranus Distance
- Mercury moves during 1433 d. a distance =5848 mkm= Mercury Pluto Dis.
- Jupiter moves during 1433 d a distance =1622 mkm=Uranus Neptune Dis.
- Mercury moves during 4331d. a distance= 2815 mkm x 2π
- Mercury moves during 4222.6 h a distance = 720.7 mkm
- Jupiter moves during 4222.6 h a distance = 200 mkm = 629 mkm/π
- Mercury moves during 550.7 d a distance = π x 720.7 mkm (0.4%)
- Jupiter moves during 550.7 d a distance = 629 mkm (error 1%)
Please Remember,
Data Group No. (1)
- Mars moves during 4331 x π d a distance= 28255 mkm= Neptune orbital circumference
Data Group No. (2)
Data Group No. (3)

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II-Discussion
- The planets move defined distances in defined periods of time.
- The Defined Distance means a distance is known in the solar system, as to be any
planet orbital distance or a distance between any 2 planets.
- The Defined Period Of Time means a period of any planet cycle, as 365.25 days
(Earth orbital period) or any planet orbital period. Or any planet rotation period or
any planet day period. All these periods are defined periods of time
- The argument tells that
- If the solar planets move their motions independently from one another, based on
that, the planets motions during defined periods of time should pass (random)
distances. Because these Planets are independent in their motion from one
another. (For example) Mercury motion depends on its orbital period (88 days) and
doesn't interest neither for Mars orbital period (687 days) nor for Jupiter orbital
period (4331 days) and by that, Mercury during these periods (687 days and 4331
days) should move Random distances (not be one of the solar system distances).
- The data disproves Planet Motion Independency Concept. because the (3)
planets move defined distances in defined periods of time. These planets motions
aren't independent from one another. These motions are done based on One
Geometrical Design. And these motions be similar to Chess Board Pieces
Motions. Each Motion is calculated geometrically and be obligatory and takes into
consideration the other planets motions and their effects.

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- Why Did We Create The Comparisons Between The 3 Planets Motions?
- We have found an interesting relationship between the 3 planets cycles periods,
let's remember this relationship in following…
- Where
- We have referred to this data in the point preface, Then we have asked why the 3
planets cycles periods be in proportionality with one another? Is there a real reason
behind or it's found by pure coincidence of number? And to find a definitive
answer for the "Pure Coincidences" claim, we have compared between the 3
planets motions to prove that these planets motions can't be independent from one
another.
- And
- The previous planets motions comparisons prove that their motions can't be
created independently from one another.
- So,
- The simple puzzle of the cycles periods be supported now by huge number of the 3
planets motions data which shows that some great force be behind the 3 planets
and forces them to move (in proportionality) with one another, and forces each
planet motion to take into consideration the other planets motions data!
- The question is
- Why Each Planet Motions Data Be In Proportionality With The Other
Planets Motions Data?

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- In the next point (3-3) we analyze the 3 planets motions data in details…
- But
- There are 2 features of motion we have to refer to them because we should use in
our discussion, these 2 features are
- (1st
Feature)
- Each planet motion uses a different rate of time from the other planets motions
rates of time… that means, the planets motions don't use one rate of time but
different rates of time which is an important feature of the solar planets motions.
- (2nd
Feature)
- The solar system distances values be used as periods of time by the solar planets
motions depending on their motions rates of time.
- For example
- 1 hour of Pluto motion =4.61 hours of the Earth moon motion
- This is the rate of time between Pluto and The Moon Motions.
- But 50% of all distances in the solar system be rated based on the rate (4.61)
- That's happened because these distances be used as periods of time
- For example
- Mercury moves during its rotation period (1407.6 h) a distance = 243 mkm (1%)
- Venus moves during its day period (2802 h) x 2 a distance = 706 mkm
- Where
- 243 days = Venus Rotation Period
- 706 days = π x 224.7 days (= Venus Orbital Period)
- These 2 data is an example of similar hundreds, where usually the distances be
used as periods of time by the planets motions.

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Discussion
- The previous 2 features are basic features used in the solar planets motions
- Many explanations can be provided to answer why or how these 2 features be used
by the planets motions.
- But
- We will analyze (The 3 planets Motions Data),
- That means, we follow the data and don't try to explain it till we have a complete
vision about it… means, we don't create our own ideas to explain how the data
works because the data is a treasure we shouldn't destroy it by our ideas, we should
analyze the data as deep as possible and there in the depth we can see the basic
understanding behind it.
- I want to say,
- These features of motions are working features because of that they be explained
through hundreds of data as a result for their using frequently. Now the reason
behind these features using is one more question we don't deal here because we try
to know if really these features be used by the planets motions data or not …
- I mean,
- We should move step by step, Because
- Newton told us that, The Solar Planet moves by The Sun Mass Gravity Force
independently in its motion from all other solar planets motions. this is the basic
theory based on which the solar planets motions be described and understood
- And, Here
- We create comparisons between the 3 planets motions data and analyze them.
- By this analysis and comparisons we should prove or disprove Newton Theory.
And If we disprove Newton theory we have to provide an alternative description
for the solar planets motions
- Because of that,

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- I have provided the 2 new features because we use them in the planets motions
comparisons and analysis. And by the analysis we should prove these 2 features
are used really in the planets motions and based on that we provide a new vision
for the solar planets motions.
- Why this vision be proved?
- Because it depends on the 3 planets motions data
- Shortly
- We provide a description for the planets motions under the analysis form, where
this description be created based on the planets motions real data and disproves
Newton Theory and its description.
- Because of that, the question (why these features be used?) will be discussed later
in this paper but not in our analysis and comparisons discussion.
- Notice
- Because the solar system distances be used as periods of time and the planets use
different rates of time among themselves. As a result for this fact, the solar system
distances be created in a network form which should be discussed and proved in
(Point no.5)

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3-3 Mercury, Mars and Jupiter Motions Analysis
I –Data
(1)
- Where

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Discussion
Equation no.(1)
- 4331 days = Jupiter Orbital Period, and 175.94 days = Mercury Day Period, and
- 687 days = Mars Orbital Period , and 24.6 hours = Mars Rotation Period
- This is the data which we try to understand in this discussion point. Our analysis
started in the point preface (3-1) and then we compared the 3 planets motions data
in point (3-2) to prove that the 3 planets motions can't be independent from one
another,
- Now , Let's try to understand this data as possible in following:
- The data tells that
- This conclusion is a try to understand the equation no. (1).
- Based on this conclusion we have 3 rates which are (24.6, 3.9 and 2π) should be
found between these 3 planets motions data
- Because
- The idea tells, the planets motions be interacted with one another based on these 3
rates of time.
- We still don't understand how planets motions rates of time (if found) effect on
these planets motions. but we have 3 rates should be effective on the 3 planets
motions data and because of that we should test each rate individually
- Let's do that in following:
1 hour of Mercury Motion = 24.6 hours of Jupiter Motion
And
1 hour of Mercury Motion = 3.9 hours of Mars Motion
And
1 hour of Mars Motion = 2π hours of Jupiter Motion

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3-3-1 The Motions Rates Of Time
- The idea tells us
- 1 hour of Mercury motion = 24.6 hours of Jupiter motion
- Let's test this rule in following
(Test No.1) (Part no.1)
- 671 h (of Mercury Motion) = 16506.6 h =687.8 solar days of Jupiter motion
- Let's analyze this data in following:
- 687 days x 24 h =24.6 h (Mars rotation period) x 671
- And
(I)
- Mercury moves during (671 hours) a distance = 2 x 57.24 mkm (1)
- (Mercury orbital distance 57.9 mkm has error 1% with 57.24 mkm)
But
- Mercury rotation period (1407.6 h) = Mars rotation period (24.6 h) x 57.2 (2)
(II)
- Jupiter moves during (687 days) a distance = 778.6 mkm
- (Jupiter orbital distance =778.6 mkm)

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- 671 h (of Mercury Motion) = 2616.9 h = 109 solar days of Mars motion
- We analyze this data also in following:
(I)
- Mercury moves during (671 hours) a distance = 2 x 57.24 mkm (1)
- (Mercury orbital distance 57.9 mkm has error 1% with 57.24 mkm)
But
- Mercury rotation period (1407.6 h) = Mars rotation period (24.6 h) x 57.2 (2)
(II)
- Mars moves during (109 days) a distance = 227.9 mkm
- (Mars orbital distance =227.9 mkm)

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Discussion
- Shortly
- 671 hours of Mercury Motion = 687 days of Jupiter motion = 109 days of
Mars motion
- But
- During 671 hours Mercury moves a distance 57.2 mkm= 2 x Mercury Orbital
Distance (error 1%)
- And
- During 687 days Jupiter moves a distance 778.6 mkm = Jupiter Orbital
Distance
- And
- During 109 days Mars moves a distance 227.9 mkm = Mars Orbital Distance
- How to understand this data?
- Can we answer Why? the data is incredible rich and priceless
- Why these motions be done based on this data?!
- For more complex discussion
(1)
- Mercury rotation period = Mars rotation period x 57.2
- But
- Mercury moves during 671 hours a distance = 2 x 57.2 mkm
(2)
- Mercury day period 4222.6 hours = Mars day period 24.7 hours x 171.7
- But
- Mercury Mars Distance = 170 mkm (error 1% with 171.7)
- It's incredible puzzled data
- How to explain it?!!

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- But
- We should see this data as clear as possible
- The 3 planets create 3 different rates of time for their motions and these rates are
- 1 h Mercury Motion = 24.6 h Jupiter Motion = 3.9 h Mars Motion
- As a result
- 671 h Mercury Motion = 687 day Jupiter Motion = 109 days Mars Motion
- But
- During 671 hours Mercury moves a distance = its orbital distance x 2 (error 1%)
- During 687 days Jupiter moves a distance = its orbital distance
- During 109 days Mars moves a distance = its orbital distance
- Why?
- 1 h of Mercury Motion = 24.6 h of Jupiter Motion
- But
- 24.6 hours = Mars Rotation Period (be used by Jupiter not by Mars!)
- Also
- 687 days = Mars orbital period (be used by Jupiter not by Mars)
- Why?
- Let's provide more data for better vision
- (17.4 deg /7 deg) = (10747 days /4331 days) = (24.1/9.7) =(24.6/9.9) =2.48
- 7 degrees = Mercury Orbital Inclination
- 17.4 degrees = The inner planets orbital inclinations total
- 10747 days = Saturn Orbital Period
- 4331 days = Jupiter Orbital Period
- 24.1 km/s = Mars Velocity
- 9.7 km/s = Saturn Velocity
- 9.9 hours = Jupiter Rotation Period
- 2.5 d degrees = Saturn Orbital Inclination

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Discussion (Continued)
- We have one period of time seen in 3 different forms, which is
- 671 hours for Mercury Motion
- 687 days for Jupiter Motion
- 109 days for Mars Motion
- These 3 periods of time are equal
- Jupiter moves during its period a distance = Its Orbital Distance
- Mars moves during its period a distance = Its Orbital Distance
- But
- Mercury moves during its period a distance = 2 x Its Orbital Distance (error 1%)
- Why Mercury passes (double) of its orbital distance?
- We even don't know why each planet moves a distance = its orbital distance during
this period of time?
- And the second question is (Why Mercury pass a double value of its orbital
distance?)
- Let's analyze more data for some better understanding
- More Data Analysis
- Mercury Day Period =4222.6 hours which is less than (176 solar days) with a
value (1.4 h =5040 seconds)
- During 5040 seconds Mercury (47.4 km/s) moves a distance = 2 x 119448 km =
Saturn diameter (= 120536 km) (error -1%)
- But
- During 5040 seconds Mars (24.1 km/s) moves a distance =121464 km = Saturn
diameter (= 120536 km) (error +1%)
- Again the data tells when Mars moves (one value) during the same period of time
Mercury moves (double value)!
- During 5040 seconds Jupiter (13.1 km/s) moves a distance =66024 km !!

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- Let's now see Jupiter motion distance
- During 5040 seconds Jupiter (13.1 km/s) moves a distance =66024 km !!
- 120536 km (Saturn diameter) = 66024 km x 1.8 (error 1.4%)
- And
- What's the rate (1.8)??
- 2.082 mkm =1.16 mkm x 1.8
- Where
- 2.082 mkm = Mars Motion Distance During A Solar Day
- 1.16 mkm = Jupiter Motion Distance During Mars Rotation period (24.6h)
- The rate (1.8) be used between Saturn diameter (1205.6km) and Jupiter motion
distance 66024 km because both Mercury and Mars motion distances be relate to
Saturn diameter and I thought that may be this distance (120536 km) is the target
of their motions.
- Notice
- 5040 seconds = 88 seconds x 57.2
- (where 57.2 = Mercury Rotation Period / Mars Rotation Period)
- We know, Mercury day period = 4222.6 hours =175.94 solar days
- but
- Mercury Orbital Period = 88 days
- We have asked before,
- Why Mercury day period= 2 x Mercury Orbital Period?
- Means,
- Our 3 planets motions data discussion can answer the old question why mercury
day period = 2 x Mercury orbital period
- Because
- Mercury moves during the period (double it orbital distance) but Jupiter and Mars
move distances equal (their orbital distances)

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- The double value is the reason behind
- But we have to analyze it as deep as possible
- Our key is the equation (5040 seconds = 88 seconds x 57.2)
- But
- This equation we can't discuss here, because the period 88 seconds is seen by us as
88 days and this rate of time (1 second = 1 day) is the rate of time between light
and planet motions
- The fact is that, the planet motion depends on a light beam motion that because the
matter is created out of light and moves with its parent light by using different rate
of time. This story we should discuss through Mercury Motion Data Analysis
where Mercury Data provides another and different reading for the planets motions
data
- Means,
- The data which we have discussed about the 3 planets, this all data has another
reading and we should discuss this same data once again by a new vision… this
process we can do only by Mercury motion analysis.
- Shortly
- The question (Why Mercury day period be = 2 Mercury Orbital Periods?) this
question we should discuss in Mercury Motion Data Analysis (Point No. 4)
- Notice (3)
- Neptune orbital inclination = 1.8 degrees! And
- 3.1 deg (Jupiter axial tilt) – 1.3 deg (Jupiter orbital inclination) = 1.8 deg

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- Because (1 Hour Of Mercury Motion =24.6 Hours Of Jupiter Motion)
- 175.94 solar days of Mercury Motion = 4331 solar days of Jupiter motion
- But
- Mercury moves during its day period (175.94 days) a distance = 720.7 mkm
- And
- Jupiter moves during it orbital period (4331 days) a distance = 4900 mkm
- Where
- 4900 mkm = 6.8 x 720.7 mkm
- 88 solar days of Mercury Motion = 2164.8 solar days of Jupiter motion
- But
- Mercury moves during (88 days) a distance = 360 mkm
- And
- Jupiter moves during (2164.8 days) a distance = 2450 mkm
- Where
- 360 mkm =Mercury Orbital Circumference
- 2450 mkm = π x 778.6 mkm (Jupiter Orbital Distance)
- 1407.6 h of Mercury Motion = 34627 h = 1442.8 solar days of Jupiter motion.
- But
- Mercury moves during its rotation period (1407.6h) a distance = 243mkm (1%)
- And
- Jupiter moves during (1442.8 solar days) a distance = 1622.7 mkm (0.6%)
- Where
- 243 days =Venus rotation period
- 1622.7 mkm = Neptune Uranus Distance

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More Discussion
- Let's summarize the data shortly
(1)
- Mercury moves during (175.94 days) a distance =720.7 mkm and Jupiter moves
during 4331 days a distance =4900 mkm
- Both passed distances are related to its planet motion because
- 720.7 mkm = Mercury Jupiter Distance
- 4900 mkm = Jupiter Orbital Distance
(2)
- Mercury moves during (88 days) a distance =360 mkm and Jupiter moves during
2165.5 days a distance = 2450 mkm = π x 778.6 mkm
- Both passed distances are related to its planet motion because
- 360 mkm = Mercury Orbital Circumference
- π x 778.6 mkm = Jupiter Orbital Distance
- 88 days = Mercury orbital period
(3)
- Mercury moves during (1407.6 h) a distance =243 mkm (1%)
- And Jupiter moves during (1407.6 h x 24.6=1442.8 days) a distance = 1622.7 mkm
- Both passed distances are NOT related to its planet motion because
- 243 days = Venus Rotation Period
- 1622.7 mkm = Uranus Neptune Distance
- In Brief
- We Should Return To The Argument Of The Discussion In Point No. (3-2)
- Because
- If these planets motions be independent from one another, the distances should be
Random Distances
- Similar to that,

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- If these rates if time are invented and not used by the planets real motions, the
passed distances should be random and not defined distances in the solar system
- Some question can be raised here tells that
- The distance 243 mkm isn't a distance but it's used as a period of time (243 Solar
days = Venus Rotation Period)?
- But
- Mars moves during Venus day period (116.75 solar days) a distance =243 mkm,
- The passed distance by Mars proves that, this distance be mentioned by the solar
system geometrical design because the same distance be passed by 2 planets
during (defined periods of time which are 1407.6 hours and 116.75 days)
- (1407.6 h = Mercury rotation period and 116.75 = the nodal year)
- That tells the distance is mentioned geometrically
- But
- Because the second planet is Mars
- That tells the distance 243 mkm isn't only designed geometrically but also
effective in our 3 planets motions data analysis..
- That makes the distance 243 mkm is similar to any defined distance (or even more
effective than the others)

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3-3-2 The Rate of time 24.6
I - Data
(1)
778.6 mkm x 24.6 =19153.6 = 2π x 3030 mkm (= Uranus Pluto Distance) (0.6%)
(2)
720.7 mkm x 24.6 =17730 = 2π x 2815 mkm (Mercury Uranus Distance)
(3)
671 mkm x 24.6 = 16506.6 = 2π x 2644 mkm (Mars Uranus Distance) (0.7%)
(4)
629 mkm x 24.6 = 15473.4 = π x 4900 mkm (Jupiter Orbital Circumference)
(5)
550.7 mkm x 24.6 = 13547.2 = π x 4331 mkm (Jupiter Orbital Period =4331 days)
(6)
2094 mkm x 24.6 = 51513 = 2π2
x 2644 mkm (Mars Uranus Distance) (1.3%)
(7)
3716.5 mkm x 24.6 = 91426 = (Pluto orbital period 90560 days error 1%)
(8)
5127 mkm x 24.6 =126124= 3.9 x 32000 mkm (=37100 mkm- 4900 mkm)
778.6 mkm = Jupiter Orbital Distance
720.7 mkm = Jupiter Mercury Distance
671 mkm = Jupiter Venus Distance
629 mkm = Jupiter Earth Distance
550.7 mkm = Jupiter Mars Distance
2094 mkm = Jupiter Uranus Distance
3716.5 mkm = Jupiter Neptune Distance
5127 mkm = Jupiter Pluto Distance
4900 mkm = Jupiter Orbital Circumference
37100 mkm = Pluto Orbital Circumference

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II – Discussion
- The previous data uses all distances between Jupiter and the other planets, and
shows that these distances be rated with (24.6) with other defined distances in the
solar system.
- Why?
- We can move step by step for better vision
- As we have seen,
- The planets move depend on the rate (24.6) and we have concluded that because
the planets move (defined distances) in (defined periods of time) which use the
rate (24.6) between them based on the rule (1 hour of Mercury Motion = 24.6
hours of Jupiter Motion)
- As A Result
- The planets move defined distances during defined periods of time use the rate
(24.6) between them.
- And we know that the distances be used as periods of time by the planets motions
- As a result
- The distances be rated based on the rate (24.6) as seen in the previous data
- I want to say that,
- The distances in the solar system are defined by some very complex method, based
on that the previous distances has a complex explanation.
- Means
- This discussion is so simple and can't explain clearly how these distances be
created or distributed … let's use one example for better understanding
- 720.7 mkm (Mercury Jupiter Distance) x 24.6 = 2π x 2815 mkm (Mercury
Uranus Distance)
- Both distances are Mercury distances, the rate (24.6) is found between 2 distances
of Mercury to Jupiter and Uranus
- Why?

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- I have no answer,
- I want to say, the understood general idea tells that (there's a rate of time 24.6
between Mercury and Jupiter Motions)
- And
- The distances be used as periods of time
- As result
- The rate 24.6 which controls the periods of time be used also as distances
distribution rate
- This is the general idea
- But each distance has its own story how to be created based on this rate (24.6)
- This idea is complex because
- Many other rates of time be used for the planets motions, and as a result, the
distance which is distributed based on the rate (24.6) will be redistributed based on
another rate for example (3.9)
- By that the distances be created in some a network form
- But
- This network can't be discovered clearly unless we discovered all planets motions
rates of time and form them in one general map
- This map which is designed by the planets rates of time will be the indicator for
the distances network map
- Let's discuss the second rate of time (3.9) which is found between Mars and
Mercury, where
- 1 Hour Of Mercury Motion Be = 3.91 Hours Of Mars Motion

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3-3-3 The Rate of time 3.91
I - Data
(1)
687 days (Mars orbital period) = 175.94 days (Mercury Day Period) x 3.91
(2)
227.9 mkm (Mars Orbital Distance) = 57.9 mkm (Mercury Orbital Distance ) x 3.91
(error 0.7%)
(3)
671 mkm (Venus Jupiter Distance) = 170 mkm (Mercury Mars Distance ) x 3.91
(error 1%)
(4)
360 mkm (Mercury Orbital Circumference) =91.7 mkm (Mercury Earth Distance ) x 3.91
(5)
2815 mkm (Mercury Uranus Distance) =720 mkm (Mercury Jupiter Distance ) x 3.91
(6)
2π x 2764 mkm (Venus Uranus Distance) = (4437.2 mkm Mercury Neptune Distance)
x 3.91
(7)
1407.6 mkm = 360 mkm (Mercury Orbital Circumference) x 3.91
(1407.6 hours = Mercury rotation period)
(8)
5378 mkm = 1375.6 mkm (Mercury Saturn Distance) x 3.91
Where
5378 hours =224.7 days = Venus Orbital Period

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II – Discussion
- Equations no (1 and 2) summarizes the idea clearly
- Because
- (227.9 mkm/ 57.9 mkm) = (687 days /175.94 days) =3.91 (error 0.8%)
- Where
- 227.9 mkm =Mars Orbital Distance
- 57.9 mkm =Mercury Orbital Distance
- 687 days = Mars Orbital Period
- 175.94 days = Mercury Day Period
- This data gives a clear reference for the rate (3.91) as a basic one between Mercury
and Mars Data
- If this data be found alone – no one can consider it be created resulting from pure
coincidence of numbers – but it must be created based on a geometrical reason
- The point is that,
- The basic data gives more extension for this one and proves that it be created as a
basic player in the planets motion – let's remember the basic data
- 4331 days (Jupiter Orbital Period) = 2π x 687 days ( Mars Orbital Period) =
24.6 x 175.94 days (Mercury day period) - And
- 24.6 hours = Mars Rotation Period - Simply
- The rate (3.91) is the rate of time between Mercury and Mars motions and based
on this rate the solar system distances be distributed and created…
- For that reason the rate (3.91) is found deeply and widely in the solar system
distances distribution as be seen in the data
- I want to say
- The discussion is so limited, Because many basic rules of the planets motions be
unknown for us – because of that the discussion can't reach to the data depth which
is the very rich point of discussion.
- The situation is complex because

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- Newton theory of The Sun Mass Gravity and planet motion independency is an
imaginary idea and completely mistaken
- The fact is that, the sun is created after all solar planets creation and motion, and
even the sun is created depending on the solar planets motions!
- I want to say
- We try to wake up from the imaginary dreams. We see the facts with difficulties
and can't see clearly the world around, but our situation now is better that when we
followed Newton theory because it's completely imaginary idea
- We now understand some of the planets motions facts (not all of them), but in
Newton Theory no facts be provided but some imaginary ideas
- Now, Let's ask, why not this paper provides an imaginary idea also? as Newton
one? why this paper provides facts?
- Because We Analyze The Planets Motions Data
- If someone asked Newton why the rate (24.6) be used frequently between the solar
system distances? Newton would answer telling that (Because of Pure Coincidence
of numbers!)
- Why did he tell this answer? Because he has no another explanation
- That means,
- Hundreds of Planets Motions Data will be considered as (Pure Coincidences of
numbers) and as a result we would have to remove Planets Motions Data and get
ride of them because no available explanations for any of them
- That's the point I fight for
- Because
- The theory be acceptable without reference for planets motions data. means, the
theory be acceptable spite its contradiction with planets motions data and by that
we lost our method to test the theory and know if it's a correct one or built on an
imaginary idea -
- Let's discuss equation no. (3)

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Equation no. (3)
671 mkm (Venus Jupiter Distance) = 170 mkm (Mercury Mars Distance ) x 3.91
- The equation is understood
- One question only we need to discuss let's provide it immediately
- (4222.6 h Mercury Day Period) = 24.6 h (Mars rotation period) =171.7
- (1407.6 h Mercury Rotation Period) = 24.6 h (Mars rotation period) =57.2
- But
- 170 mkm = Mercury Mars Distance (error 1%)
- 57.9 mkm = Mercury Orbital Distance (error 1%)
- Why the distances be used as rates in these 2 data?
- There's A Geometrical Machine Behind This Data, but how to discover it?

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3-3-4 The Rate of time (2π)
I - Data
(1)
4331 days (Jupiter Orbital Period) = 687 days (Mars Orbital Period) x 2π
(2)
550.7 mkm (Jupiter Mars Distance) x 2 x 24.6 = 4331 mkm x 2π
(3)
4495.1 mkm (Neptune orbital distance) = 720.7 mkm (Mercury Jupiter distance) x 2π
II – Discussion
- No need to insert many data depends on the rate (2π), because
- It's the basic rate in the solar planets motions and tenths or even hundreds of data
depends on the rate (2π).
- And the solar system distances distribution equation depends on this rate, let's see
this equation in following
1.16 mkm/s x (2) x 86400 seconds =(2) x 100733 mkm = (37100 mkm – 4900
mkm) x 2π = 28255 mkm + (2) x 86400 mkm
- Where
- 100733 mkm = The 9 Planets Orbital Circumferences Total
- 37100 mkm = Pluto Orbital Circumference
- 4900 mkm = Jupiter Orbital Circumference
- This equation we should discuss later, where this is the main equation of the solar
system distances distribution (and the solar system energy distribution because
Distance = Energy)
- The rate (2π) creates a very important effect on this e equation discussion
- Shortly
- This equation discussion gives a new meaning for the rate (2π) but we have to wait
because this meaning depends on another reading for the data (be discussed in
point no.4 Mercury Motion Analysis

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- A Conclusion
- Many distances be distributed based on the rate (2π), and that tells the rate of time
between Jupiter and Mars motions is a general rate used in the solar system and
that tells it's not a rate between 2 planets, instead the 2 planets use one general rate
in the solar system
- i.e.
- We have 2 passengers decided to leave their own car and use the Metro
- It’s a general rate in the solar system and its effect beyond Jupiter and Mars
motions range
- As a result
- In Kepler (3rd
Law) (The distance3
= The Period2
x Constant)
- In this law, if we use the solar planets motions data, the constant will be (25)
- But
- Mars Axial Tilt =25.2 degrees
- Means,
- Mars axial tilt be used in kepler law as a constant for all planets because the rate
between Jupiter and Mars is (2π) which is a general rate in the solar system

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3-4 Earth and Pluto Motions Analysis
(Point No. 1) (Preface)
- We analyze Earth and Pluto Motions here because of 2 basic reasons:
- (1) The rate of time between Earth and Pluto = 2π
- (2) The rate of time between Pluto and The Earth moon = 4.61
- We remember that
- 1 hour of Mars Motion = 2π hours of Jupiter Motion
- And also
- 1 hour of Earth Motion = 2π hours of Pluto Motion
- By that we have one rate of time be used for 2 groups of planets motions
- That shows the basic reason behind the analysis of Pluto and Earth motion in out
discussion.
- The second reason is the rate (4.61) between Pluto and The Earth Moon Motions,
where this rate (4.61) effects and distribute 50% of all distances in the solar system
- That shows the rate (4.61) has a massive effect
- But
- The basic point is that,
- The rate (2π) between Jupiter and mars is a powerful rate and effects greatly on the
solar planets motions…
- Now this same rate be used again by Earth and Pluto
- And this rate using produced the rate (4.61) which effects on 50% of all distances
of the solar system
- All these reasons pushes to provide a deep analysis between Earth and Pluto
motions…

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(Point No. 2) (Earth, Its Moon And Pluto Rates Of Time)
(Earth And Pluto Rate of Time)
(A)
- Earth (29.8 km/s) moves during a solar day (24 h) a distance = 2574720 km
- Pluto (4.7 km/s) moves during its day (153.3) a distance = 2593836 km
- The moon displacements total during (29.53 days) = 2598693 km
- Pluto and the moon motion distances be equal but they are different from Earth
motion distance with (1%).
(B)
- (Earth velocity / Pluto velocity) = 2π (error 1%)
- (Pluto Day period/ earth day period) = 2π (error 1.7%)
- Pluto orbital distance 5906 mkm = Earth orbital circumference 940 mkm x 2π
- Pluto orbital period 90560 days = Earth Cycle (1461 days) x 2π3
(C)
- 1 hour of Earth motion= 29.53 hours of the moon motion (explained in discussion)
(Pluto And The Moon Rate Of Time)
(1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
(2)
Tan (12.2) x 708.7 hours = 153.3 hours
(3)
Tan (13.17) x 655.7 hours = 153.3 hours
(4)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
(5)
413600 km cos (10.96 deg) = 406000 km
(6)
(10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)

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II- Discussion
- How to define the rate of time between 2 planets motions
- Let's remember how we have defined the 3 planets rates of time, we have used the
following equation
- 4331 days =175.94 x 24.6 = 2π x 687 days
- That means, the rate of time be defined as a rate between the planets cycles periods
- But
- Te difficulty is to know which cycle period be used. For example in this equation
if we use Mercury orbital period (88) in place of its day period (175.94) the
produced rate will not be working because it's not be used frequently as the day
period. That means, the planet uses its data for specific direction…
- For example, 1 hour of Mercury motion = 9.18 hours of Pluto motion. this rate of
time be produced as a rate between Mercury rotation period and Pluto day period.
Here Mercury doesn't use its day period with Pluto but used it with Jupiter. It's te
planet direction of motion…
- How can we discover that? the rate be used frequently in its direction –for example
– Mercury moves during its day period a distance = 720.7 mkm =Mercury Jupiter
distance, this data tells Mercury uses its day period with Jupiter motion interaction
- Let's return to Earth and Pluto Discussion
- 1 Hour Of Earth Motion = 2π Hours Of Pluto Motion
- How to know that?
- 153 hours (Pluto Day Period) = 24 hours(Earth day period) x 2π (error 1%)
- And
- Earth velocity 29.8 km/s = Pluto velocity 4.7 km/s x 2π (error 1.7%)
- The errors don't effect negatively because the planets motions distances are
different with rates (1% and 2%), by that we consider the errors found for
geometrical necessities.

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- The question should be
- How the rate (2π) be transported as a rate between Jupiter and mars to be a rate
between Earth and Pluto?
- Let's see the following data:
- Venus Velocity = Uranus Velocity x 5.1
- Mars Velocity = Pluto Velocity x 5.1
- The Moon Velocity = Neptune Velocity x 5.1
- That means,
- The rate (5.1) is a continuous force moves through the solar planets motions and
effects on planets according to its geometrical design effect
- We clearly deal with a system of data and a geometrical mechanism found behind
this data.
- The transportation process for the rate (2π) is done between Mars and Pluto
Motions.
- That may explain the deep interaction between Pluto and the Earth Moon Motions.
- The moon data analysis will show that the relationship with Pluto motion is so
strong one.
-
- Notice
- The 3 planets velocities total =112.2 km/sec
- (The 3 planets are Mercury 47.4 km/s + Venus 35 km/s + Earth 29.8 km/s)
- 112.2 km/sec = 24.1 km/s (Mars Velocity) x 4.61 (error 1%)
- But
- 1 hour of Pluto motion be = 4.61 hours of the moon motion
- That tells some complex machine be found behind the transportation of the rate
(2π)
- Specially because
- Mars orbital distance (227.9 mkm) x (2π) = 1433.5 mkm = Saturn orbital distance

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- That, again tells that, there's a geometrical reason for Mercury to move during
(5040 seconds) a distance = 2 Saturn diameter and Mars moves during this same
period (5040 seconds) a distance = Saturn diameter
- The definition of the distance to be = Saturn diameter is a choice based on a
geometrical reason.
- Why?
- Because the period 5040 seconds is engraved period inside Mercury day Period,
and Mercury day needs (5040 seconds) to be (176 solar days)
- The geometrical system created this period (5040 s) inside Mercury Day Period
because of its massive effect on the planets motions
- That tells Saturn diameter has a massive effect on planets motions, I don't catch yet
its effect, I can only add some data supports this idea
- More Data
- Pluto (4.7 km/s) moves during (51118 seconds) a distance = 2 x 120536 km
(Saturn diameter = 120536 km)
- (Uranus diameter = 51118 km)
- And
- Pluto (4.7 km/s) moves during (2 x 120536 seconds) a distance = 1.13184 mkm
- (1.13184 mkm = Jupiter Motion Distance During A Solar Day)
- Jupiter moves during 4495.1 days a distance =5127 mkm (Jupiter Pluto Distance)
- (4495.1 mkm = Neptune Orbital Distance = π x Saturn Orbital Distance)
- Now let's discuss the moon and Pluto motions data

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II- Discussion (Continued)
- In following we analyze the moon and Pluto motion data
- Notice
- No pure coincidences of number can be claimed here because Pluto data be in
proportionality with the Earth and its Moon. Means each planet motion be
interacted based on its position
- The question we try to solve is (How This Data Be Created?)
Equation no. (C)
1 hour of Earth motion = 29.53 hours of the moon motion.
- Because
- The Earth Day Period = 24 hours
- The moon Day Period = 708.7 hours
- And
- Pluto Day Period = 24 hours
- The moon Day Period = 708.7 hours
- Because of that
- 1 hour of Pluto motion = 4.61 hours of the moon motion.
- We can see that, the machine works with consistency, for example
o 1 hour of Earth motion = 2π hours of Pluto motion
o 1 hour of Pluto motion = 4.61 hours of the moon motion
o And
o 1 hour of Earth motion = 2π x 4.61 = 29.53 hours of the moon motion
o The rates of time are created based on each planet motion data and be in full
harmony as we see. That tells we deal with a geometrical mechanism works
consistency…
Let's discuss more data in following …

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Discussion (continued)
Equation no. (1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
- Where
- 406000 km = Pluto motion distance per a solar day
- 88000 km = The moon displacement per a solar day
- 708.7 hours = The Moon Day Period
- 153.3 hours = Pluto Day Period
- Again the rates of time causes (2 Planets All Data) to be in proportionality with
one another. in following equations we see that Pluto and the moon motions data
be in full proportionality which is very interesting data.
Equation no. (2)
Tan (12.2) x 708.7 hours = 153.3 hours
- Where
- The angle 12.2 degrees = 13.177 deg – 0.98562 deg
- 13.177 degrees = The Moon Motion Degrees Per A Solar Day
- 0.98562 degrees = The Earth Motion Degrees Per A Solar Day
- Equation no. (2) shows that the moon and Pluto days periods are created based on
each other geometrically. The fact can't be denied. Pluto day period (153.3 h) is
created as a function in the moon day period (708.7 h).
Equation no. (3)
Tan (13.17) x 655.7 hours = 153.3 hours
- Where
- 655.7 hours = The Moon Rotation Period.
- 153.3 hours = Pluto Day Period and (153.3- ) = Pluto Rotation Period

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- Equation no. (3) shows, Pluto rotation period is created depending on the moon
rotation period based on the angle (13.177 deg).
- The data of the 3 equations are strong and prove the idea clearly, because it's A
Directed Data. Means, this data be created based on geometrical mechanism be
found between Pluto and the moon motions. For that reason, the equations use (all)
data of the 2 planets. That shows the data is directed and can't be created by any
random process. It's data be created based on geometrical reasons and rules.
- The proportionality in data between Earth and its moon on one side and Pluto on
the other side shows that a geometrical mechanism must be found behind them.
Equation no. (4)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
- 708.7 hours = The moon Day period
- 655.7 hours = The Moon Rotation Period
- 10.7 hours = Saturn Day Period
- 9.9 hours = Jupiter Day Period
- The rate (4.61) we have defined, where (1 hour of Pluto motion =4.6 hours of the
moon motion).
- Equation no.(4) tells that this rate (4.61) is known by Jupiter and Saturn motions.
- That tells, the rate (4.61) is transported among the solar planets and that explains
how this rate be created between the moon motion and Pluto motion.

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Equation no. (5)
413600 km cos (10.96 deg) = 406000 km
- Where
- 406000 km = Pluto Motion Distance During A Solar Day
- Equation no. (5) shows that, the motions interaction between the moon and Pluto is
a deep interaction. And effects on the planet motion basic data. Because
- The moon displacements total during 29.53 days =2598693 km = 2π x 413600 km
- The distance (413600 km) should be the moon apogee orbital radius if the moon
apogee orbital circumference = the moon displacements total during 29.53 days.
- This is the fact which should be performed
- But
- The moon apogee orbital circumference be (2550973 km) less than the moon
displacements total during (29.53 days) with 2%. This fact arises the question
Why? the moon apogee circumference should be 2598693 km (surely). But for
some geometrical miracle this distance became =2550973 km!
- The radius 413600 km be changed into 406000 km and this change be done based
on the equation no. (5). The angle (10.96) caused this change. We have discussed
this angle before deeply because it's the main angle in the moon orbit geometrical
structure (we will review a short notice about it later). The angle (10.96 deg)
causes the radius 413600 km to be 406000 km – why?
Equation no. (6)
(10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)
- The angle 10.96 deg be created based on the rate between Pluto Day Period
(153.3h) and Earth Day Period (24 h). this rate be used between the angle (10.96)
which causes to decrease the radius from 413600 km to 406000 km, and the angle
1.7 deg which is the moon daily motion angle in the moon orbital equation (θ1= θ0
+ 1.7 deg). (The moon orbital equation be in my previous paper)

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Notice (1)
- The moon displacements total 2598693 km should be the moon apogee orbital
circumference, by that, the moon apogee orbital radius should be =413600 km
- But
- It's not true, the moon apogee orbital radius =406000 km
- Why? because of the angle (10.96 degrees)
- Let's remember this angle in following:
o The moon orbital perigee radius =363000 km
o The moon orbital apogee radius = 406000 km
o If we draw these 2 radiuses in 2 circles as in the figure
o The tangent DB = 181843 km, also
o The perigee orbital circumference = 2.28 mkm
o (181843 km /2.28 mkm) = 0.08
o 137 degrees x 0.08 = (10.96 degrees)
o As we remember that 137 degrees = 95.1 deg x 1.44 deg where
o 95.1 deg =90 deg + 5.1 deg (the moon orbital inclination)
o 1.44 deg = the moon orbit regression degrees per month.
o We have supposed that the angle 137 degrees is created based on the moon
orbital inclination (degrees) and the moon orbit regression angle monthly.
And this angle (137 deg) is the basic angle in the moon orbital geometrical
structure (hypothesis).
o The calculations of the tangent DB = 181843 km aimed to explain one basic
idea which is, the moon orbit is built geometrically to create the rate (0.08)
by which the moon orbital inclination and regression will create the angle
(10.96 deg).
o Let's explain that more clearly

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o The moon orbit inclination and regression controls the moon orbital motion.
this is a hypothesis. Now, the moon orbital motion creates the moon orbit
based on a geometrical design, the moon orbital motion uses the moon orbit
geometrical design to create the angle (10.96 deg) from the basic one (137
deg). by that the original angle is (137 deg) which is created by the moon
inclination and regression unification, and the born angle is (10.96 deg)
which is the motor of the machine.
o What we need to refer to is that, the moon orbit geometrical design is used
in this process. That's why the distance 413600 be decreased and became
406000 km. By that, the main player is the angle (10.96 deg).
- This explanation answers the question,
- Why Pluto Motion Distance During A Solar Day = The Moon Apogee Orbital
Radius=406000 km?

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Notice
- The previous data still have another reading let's summarize it in following:
o Pluto day period is a very long day period in comparison with the outer
planets days periods, Why?
o We notice also that, Pluto Moves During Its Day Period A Distance =
Earth Motion Distance During Earth Day Period. That means, Pluto Day
Period be extending to perform this requirement.
o i.e.
o Pluto Day Period be extending to be 153.3 h to enable Pluto to move a
distance = Earth motion distance during Earth day period.
o Because of that,
o Pluto day period (and rotation period) be created as functions in the Earth
moon day and rotation periods as we have seen in the data discussion.
o i.e.
o The interaction between Pluto motion on one side with the Earth and its
moon motions on the other side caused the following results:
o (1st
Result) the moon apogee orbital radius be decreased from 413600 km to
be 406000 km
o (2nd
Result) Pluto Day Period be extending to be =153.3 days.
Notice
- Uranus effect is the main force behind the changes for the 2 planets data.
- Uranus effect on Pluto is proved by Planet 8 days Cycle where Uranus uses Pluto
day period in its motion.
- Uranus effect on the moon is proved by the fact, Uranus orbital circumference =
19.2 Earth orbital circumference. The moon moves a cycle called Metonic cycle
and extends for 19 years. Metonic Cycle is caused almost by Uranus effect on the
moon motion.

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The Rate of time 4.61
I- Data
778.6 mkm (Jupiter Orbital Distance) = 4.61 x 170 mkm
550.7mkm (Jupiter Mars Distance) = 4.61 x 119.7 mkm
2094 mkm (Jupiter Uranus Distance) = 4.61 x 2 x 227.9 mkm
2872.5 mkm (Uranus Orbital Distance) = 4.61 x 629 mkm
3033.5 mkm (Uranus Pluto Distance) =4.61 x 654.9 mkm
5127 mkm (Jupiter Pluto Distance) = 4.61 x 2 x 550.7 mkm
5906 mkm (Pluto Orbital Distance) =4.61 x 1284 mkm
1375.6 mkm (Mercury Saturn Distance) =4.61 x 2 x 149.6 mkm
3062 mkm (Saturn Neptune Distance) =4.61 x 671 mkm
4345.5 mkm (Earth Neptune Distance) =4.61 x 940 mkm
4267.2 mkm (Mars Neptune Distance) =4.61 x 929 mkm
Where
170 mkm = Mercury Mars Distance
119.7 mkm = Venus Mars Distance
227.9 mkm = Mars Orbital Distance
654.9 mkm = Jupiter Saturn Distance
1284 mkm = Earth Saturn Distance
149.6 mkm = Earth Orbital Distance
940 mkm = Earth Orbital Circumference
929 mkm = Earth Jupiter Distance when the 2 planets be on the sun 2 sides
(Max error 1%)

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II- Discussion
- 1 hour of Pluto motion =4.61 hours of the moon motion
- We have discussed that deeply before
- But
- The previous distances are (22 distances)
- And
- The solar system distances total be (45 distances)
- That means,
- These distances in the pervious data be considered as 50% of all distances in the
solar system. that shows the rate (4.61) between Pluto and the moon motions be so
important and interesting rate which controls around 50% of all distances
- I have no explanation why this is done or how the 2 planets rate of time can effect
so greatly on the solar system distances distribution and planets motions data
- What we need to do is that, to prove the following results:
o The solar system distances are created in a network form and not as
individual or independent distances
o The solar system distances be created based on a map uses the planets
motions rates of time to define and distribute the distances
o The data clearly provides another vision different from Newton Theory of
the sun mass gravity… we should see this data frequently and has a time to
think deeply in the question (how this data be created?)
o It's another vision…The eyes which see one distance (Mars Orbital Distance
for example) and think this distance be created by effect of the sun mass
gravity on Mars Mass! we try to tell them (please look at the other distances
in the same time) and then we will discover that (Mars orbital distance) is a

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part of some great machine be controlled by one geometrical design which
distribute the distances based on its geometrical rules
o It's one system of data and one machine mechanism which causes to create
and distribute the distances and because of that they be created in a network
form
o I try to show that, the solar planets motions data contradict the solar system
description inherited from Newton Theory of the sun mass gravity
o The features which we have concluded aren't so wide but they provide an
extending vision contradicts clearly and strongly Newton theory
o We have to adopt different rates of time for the planets motions which is a
strange feature from Newton theory
o We have to accept the distances using as periods of time and this feature
also is strange one
o And then
o We have to consider the solar system distances as distance be created in a
network form and has one geometrical design.
o The machine has different strange features are unknown by Newton Theory
of the sun mass gravity
o There are many other features we have to discover to provide a complete
description for the solar system motion.
- Notice
- Jupiter Data still need more analysis we should do that but with Mercury Motion
analysis because Mercury provides one more reading for the solar planets motions
data.

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4-Mercury Motion Analysis
4-1 Preface
4-2 The Theory basic Ideas
4-3 The Light Beam (1.16 mkm/s) Proves Discussion
4-4 Jupiter Orbital Circumference Analysis

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4-1 Preface
- Jupiter motion data analysis shows that, Jupiter Motion depends on Mercury and
Mars Cycles Periods because there's one great machine causes the 3 planets to
move in comparison with one another.
- And
- What's This Great Machine?
- The Fact is that,
- Matter is created out of light beam and the born matter doesn't separate from its
light parent but move together in one unified motion by using different rates of
time.
- Means,
- The planets are created out of one light beam which causes their motions
- The planets be similar ships on a sea and the sea is light motion network
- That caused the great difference between Newton Theory and the planets motions
data and facts. While Newton saw the planets are independent rigid bodies move
by the sun mass effect on them, the planets be in fact ships sail over a sea of light
motions which control and define the planets motions range.
- Mercury Data tells this story in details and explained clearly how the planet moves
accompanying with a light beam motion.
- Let's provide the complete theory of Planet motion depending on light motion and
then we should discuss The 3planets motions data (again) from Mercury Motion
Point of View to see that how the light beam creates this data we have discussed
before

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4-2 The Theory Basic Ideas
(I)
- The theory hypothesis
- The solar system be created out of one light beam its velocity =1.16 mkm/sec
-
- The hypothesis proves depends on planets motions data
- Because
- The Planets Matters are created out of energy, And
- The Planets Distances are created out of energy
- This energy is provided by one light beam its velocity (1.16 mkm/sec)
- As a result
- This light beam motion feature be registered in the planets dimensions and
distances…
- That's why I suppose there's a light beam its velocity =1.16 mkm/sec
(II)
- Planet and light motions rate of time is
- 1 Second Of Light Motion = 1 Solar Day Of Planet Motion
(one second of light motion energy causes planet to move for one solar day)
- The planets divide this general rate of time among their motions and by that small
rates of time be used for the planets motions (as we have examples of these small
rates of time)
- The planets motions unify their small rates of time to produce the general one.
- By that, the solar group works used as a great clock, its input is 1 second of light
motion and its output is 1 solar day of a planet motion.
(III)
- Light supposed velocity (1.16mkm/sec) started from Mercury its journey to Jupiter
directly and then Jupiter energy is sent to Pluto directly (1st
trajectory)
- There's one more trajectory started from Mercury also to Pluto directly

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- We analyze all motions to see as possible how they are done
- Saturn data be a parallel player found with the light motion data ….
- In Mercury Motion data analysis we should remember the question
- (why Mercury day period = 2 Mercury orbital periods?)

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4-3 The Light Beam (1.16 mkm/s) Proves Discussion
I - Data
(1)
1.16 mkm/s x 86400 s = 100733 mkm
And
0.3 mkm/s x 86400 s = 25920 mkm
(2)
1.16 mkm/s x 500 s = 580 mkm
1.16 mkm/s x 580 s = 671 mkm (Venus Jupiter Distance)
1.16 mkm/s x 671 s = 778.6 mkm (Jupiter Orbital Distance)
1.16 mkm/s x 629 s = 720.7 mkm (error 1%) (Mercury Jupiter Distance)
1.16 mkm/s x 5127 s = 5906 mkm (error 0.7%) (Pluto Orbital Distance)
(3)
1.16 mkm/s x 4224 seconds = 4900 mkm
(4)
1.16 mkm per second x 2 x 86400 seconds =2 x 100733 mkm = (37100 mkm – 4900
mkm) x 2π = 28255 mkm + 2 x 86400 mkm
(5)
1.16 mkm/s x 5040 seconds = 5848 mkm (Mercury Pluto Distance)
And
5848 mkm (Mercury Pluto Distance) – 4900 mkm = 948 mkm
(6)
1.16 mkm/s x 4900 seconds = 5678.1 mkm (Mars Pluto Distance)
And
1.16 mkm/s x 5687.1 seconds = 6585.4 mkm
(7)
193 seconds x 0.3mkm/sec = 57.9 mkm (Mercury Orbital Distance)
196.5 seconds x 1.16 mkm/sec = 227.9 mkm (Mars Orbital Distance)
(8)
5040 sec = 88 sec x 57.27 (this data answers why Mercury day = double its period)

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II- Discussion
Equation no. (1)
1.16 mkm/s x 86400 s = 100733 mkm
And
0.3 mkm/s x 86400 s = 25920 mkm
- Light known velocity (0.3 mkm/s) travels during a solar day (86400 s) a distance =
25920 mkm.
- Because it's a light motion we see this value (25920 mkm) as a period of time
(25920 years = the Precession Cycle)
- Also
- The distance 25920 mkm be = The Planets Motions Distances Total During A
Period =1461 days (where 1461 days =365 +365 +365 +366)
- Because the period 1461 days is (Earth Cycle) we consider that the distance 25920
mkm be mentioned by planets motions total
- Notice, the distances total added the Earth moon motion distance to the 9 planets
motions distance to create the total per a solar day (17.75 mkm) where the moon
motion daily distance be considered = Earth motion daily distance because they
aren't separated from one another through their motions.
- The light supposed velocity (1.16 mkm/s) travels during a solar day (86400 s) a
distance = 100733 mkm = The 9 Planets Orbital Circumferences Total
- Equation no. (1) tries to prove that, Light Motion Uses The Solar Day Period As
Its Basic Period Of Motion.
- Light known velocity and light supposed velocity both use the solar day as the
basic period of time. Because of that, both light beams passes essential distances in
the solar system geometry.
- The next discussions will prove that these distances (25920 mkm and 100733
mkm) are essential distances in the solar system geometry.

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- Notice
- The rule tells
- (Light Motion For 1 Second = Planet Motion For 1 Solar Day)
- This rule which is a supposed one and we should prove it through the paper
discussions…
- This rule tells that,
- The light motion depends on the solar day as a period of time as similar to its
dependency on the second as a period of time.
- That means,
- These 2 periods of time (the second and the solar day) are the 2 basic columns for
light and planet motions accompaniment.
- And by that, light motion doesn't use the solar day as similar to any other planet
day period. On the contrary, the solar day is the basic period of time for light
motion (For both velocities, the known and supposed).

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Equation no. (2)
1.16 mkm/s x 500 s = 580 mkm
1.16 mkm/s x 580 s = 671 mkm (Venus Jupiter Distance)
1.16 mkm/s x 671 s = 778.6 mkm (Jupiter Orbital Distance)
1.16 mkm/s x 629 s = 720.7 mkm (error 1%) (Mercury Jupiter Distance)
1.16 mkm/s x 5127 s = 5906 mkm (error 0.7%) (Pluto Orbital Distance)
- The data show specific features of motions which are:
- The light supposed velocity (1.16 mkm/s) uses the distances values as a periods of
time by the rate (1 mkm = 1 second).
- Based on that, the distances be created depends on one another and no distance be
created independent from the other distances. That means, the distances be created
in a network form of distances and not as independent distances from one another.
- The data shows how the solar system distances be created in a network form
- In details
- Light supposed velocity (1.16 mkm/s) uses the period (500 sec) to pass a distance
580 mkm, and
- Light supposed velocity (1.16 mkm/s) uses the distance (580 mkm) as a period of
time (580 sec) to pass a distance 671 mkm (= Venus Jupiter Distance), and
time (671 sec) to pass a distance 778.6 mkm (=Jupiter Orbital Distance), and
time (629 sec) to pass a distance 720.7 mkm (=Jupiter Mercury Distance), …etc
- We have discussed this data before but now we have another reading for this same
data
- Notice
- (The Distances Network Theory Be Discussed In Point No.5)

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The Motion 1st
Trajectory
- The Trajectory Summary
- Light supposed velocity (1.16 mkm/s) travels starting from Mercury to Jupiter,
creating Jupiter orbital circumference
- And, then
- Light supposed velocity (1.16 mkm/s) travels from Jupiter to Pluto (Directly) and
doesn't touch any other planet in the way.
- The motion trajectory is built on 2 basic equations and many other additional data.
because
- Mercury Motion uses light supposed velocity (1.16 mkm/s) because of that,
different data of Mercury motion be created based on this light supposed velocity
(1.16 mkm/s) energy
- Shortly
- The motion 1st
trajectory is defined based on Equation no. (3) and (4) as basic
equations. And through discussion many other data be added
Notice
Equation no. (4) is the solar system geometry main equation and it's the paper main
one we discuss it along the paper,

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Equation no. (3)
1.16 mkm/s x 4224 seconds = 4900 mkm
- Where
- Equation no. (3) tells, the light supposed velocity (1.16 mkm/s) travels during
4224 seconds a distance = 4900 mkm = Jupiter Orbital Circumference.
- Now, let's suppose 1 second of light motion = 1 hour of Mercury Motion
- based on that, 4224 seconds be equivalent to 4224 hours (but 4222.6 hours =
Mercury Day Period).
- Equation no. (3) provides 3 braches of data! let's summarize them in following:
- (1st
Branch) Light supposed velocity (1.16 mkm/s) moves in (4224 s) a distance
=4900 mkm but Mercury moves during (4222.6 h) a distance = 720.7 mkm
(=Mercury Jupiter Distance) and we suppose that 1 second of light motion = 1
hour of Mercury Motion. Because of that we need to find some connection
between 4900 mkm and 720.7 mkm to create a comparison between light motion
and planet motion.
- The comparison is complex because, I suppose the rate (2π) distinguish between
light and planet motions. By that, if light moves 4900 mkm, a planet motion
should be 778.6 mkm but Mercury motion distance only be 720.7 mkm, where,
- 778.6 mkm (Jupiter orbital distance) =1.0725 x 720.7 mkm (Mercury Jupiter
Distance)
- The rate (1.0725) creates a negative effect on the rule of the value (2π). although
the rate (1.0725) be used between 40% of all distances in the solar system. But
because it be found here creates more complexity for the discussion and causes
some negative effect for the comparison between light and planet motions. because
always in other data we have found that the value (2π) distinguish clearly between
light and planet motions but in our equation the rate (1.0725) causes some
difficulty for this comparison vision.

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- Notice
- The distance 720.7 mkm (Mercury Jupiter Distance) be discussed in point no. (9)
of this paper.
- (2nd
Branch) Mercury Day Period Creation.
- We have supposed that, 1 second of light motion be equivalent to 1 hour of
Mercury Motion. that means the period (4224 s ) be equivalent to (4224 h) and
Mercury Day Period = (4222.6 hours).
- This supposition suggests that, Light supposed velocity (1.16 mkm/) motion causes
to create Mercury Day Period be (= 4222.6 hours).
- The data is accurate and clear
- NOW
- The data tells that…
- Mercury is the source of the light supposed velocity (1.16 mkm/s)
- The light supposed velocity (1.16 mkm/s) started from Mercury and sent to Jupiter,
creating the distance (4900 mkm = Jupiter Orbital Circumference).
- This motion depends on Mercury Day Period.
- As a result, Mercury Day Period be defined by light supposed velocity (1.16
mkm/s) motion effect on Mercury Motion.
- i.e.
- Mercury Day Period = 4222.6 hours And
- 4900 mkm= 1.16 mkm/s x 4224 seconds
- We suppose that, 1 hour of Mercury Motion = 1 second of light motion… means
- The difference (4224 -4222.6 = 1.4) is created by the light supposed velocity (1.16
mkm/s) motion effect. that means,
- The light motion effected on Mercury Motion and caused Mercury day period to
be 4222.6 hours and less than 4224 hours with 1.4 hour = 5040 seconds
- Why? Because

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- Light supposed velocity (1.16 mkm/s) travels during (5040 s) a distance =5848
mkm = Mercury Pluto Distance
- The light supposed velocity (1.16 mkm/s) created a connection between Mercury
and Pluto motions by creating the period 5040 seconds as a difference between
Mercury Day period (4222.6 h) and 176 solar days (=4224 hours).
- We can't discover the geometrical machine behind this data. But the data clearly
shows that, the light supposed velocity (1.16 mkm/s) travels during 5040 seconds a
distance = 5848 mkm.
- We have 3 information we got based on this analysis which are:
- (1st
) Mercury Is The Origin Point Of The Light Supposed Velocity (1.16 mkm/s)
- (2nd
) light supposed velocity (1.16 mkm/s) motion created Jupiter orbital
circumference 4900 mkm.
- (3rd
) Mercury Day Period be (4222.6 h) and less than (4224 h) with (5040 s)
because of light supposed velocity (1.16 mkm/s) effect on Mercury motion, to
create a connection between Mercury and Pluto motions.
- Notice
- The accompaniment of The Light supposed velocity and Mercury Motions caused
Mercury day Period to be (4222.6 hours) to create a connection between Mercury
and Pluto based on the distance (5848 mkm = Mercury Pluto Distance)
- Is There any Proof for this conclusion?
- Yes, There's a proof in Mercury Motion data,
- Let's analyze Mercury Motion Data in following:

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Mercury Motion Data
- 5040 seconds = 84 minutes
- Mercury (4.095 mkm/ daily) moves during 84 solar days a distance = 344 mkm
- And
- Mercury (4.095 mkm/ daily) moves during 346.6 solar days a distance = 1419.5 mkm
- And
- Mercury (4.095 mkm/ daily) moves during 1419.5 solar days a distance = 5813 mkm
- But
- Mercury (4.095 mkm/ daily) moves during 1433.5 solar days a distance = 5870 mkm
Pluto Orbital Distance
- NASA Planetary Fact sheet had defined Pluto orbital distance as 5870 mkm
- And
- From 5 years NASA had modified this distance to be 5906 mkm
- In the theoretical calculations both distances are required!
- Let's explain that in following
o 5906 mkm – 57.9 mkm (Mercury orbital distance) = 5848 mkm
o This distance 5848 mkm accurately = 1.16 mkm/s x 5040 seconds
But
o Mercury motion during 1419.5 days passes a distance =5813 mkm = 5870
mkm – 57.9 mkm - And
o Mercury motion during 1433.5 days passes a distance =5870 mkm (Pluto
orbital distance)
Shortly
o Light motion depends on the distance (5906 mkm as Pluto orbital distance)
but Mercury motion depends on the distance (5870 mkm as Pluto orbital
distance).
o That tells both distances are found geometrically but also tells some
difference be found between light and planet motions.

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Mercury Motion Data Analysis
- The previous data shows that,
- Mercury uses light motion strategy. By that, Mercury moves a distance and then
uses this distance as a period of time to move another distance. By that the
distances be created in a network form.
- We don't know how a planet moves with low velocity can use distances values as
periods of time. The data needs more analysis to discover the rule on which this
using is done.
- Mercury motion aims to pass a distance = 5848 mkm= Mercury Pluto Distance,
this distance is the distinguish one in the 1st
trajectory of motion and also the 2nd
trajectory of motion. Many data be created in proportionality with this distance
which proves some geometrical machine be found behind these motions.
- Mercury moves the distance (5848 mkm) following light motion
- Mercury uses the rate (1 mkm= 1 solar day)
- Notice
- Jupiter Orbital Circumference (4900 mkm) is a very rich distance of discussions
and arguments. Because of that,
- Jupiter orbital circumference analysis be discussed in a separated point no. (4-4) .
- The idea is that, the light supposed velocity (1.16 mkm/s) creates a connection
between Mercury and Pluto by using Jupiter orbital circumference (4900). That
means, Jupiter orbital circumference be used as a central point in the light
trajectory of motion from Mercury to Pluto, because of that, the light supposed
velocity (1.16 mkm/s) created Jupiter orbital circumference (4900 mkm) by
different geometrical features which be required for this task. As a result we find
Jupiter orbital circumference is so complex distance and filled of questions.

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The Conclusions
- The data tells,
- Mercury is the origin point of the light supposed velocity (1.16 mkm/s) motion.
that means, the light starts its motion from Mercury point. (A Conclusion No. 1)
- The light supposed velocity (1.16 mkm/s) passes the distance (4900 mkm = Jupiter
orbital circumference) in a period (4224 seconds). By that the light connects
between Mercury motion and Jupiter motion. (A Conclusion No. 2)
- Mercury Day Period be less than (4224 hours) by 5040 seconds because light
supposed velocity (1.16 mkm/s) travels during 5040s a distance = (5848 mkm) =
(Mercury Pluto Distance)
- Light supposed velocity (1.16 mkm/s) uses (5040s) to pass the distance (5848
mkm) and create a connection between Mercury motion and Pluto motion. (A
Conclusion No. 3)
- Notice
- 5848 mkm (Mercury Pluto Distance) –4900 mkm (Jupiter orbital circumference) =
946.5 mkm
- Where
- 940 mkm = Earth Orbital Circumference (error 0.7%)
- The value (946.5 mkm) is very important we will discuss in 2nd
Trajectory of
motion because it's related to Saturn motion data.

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Mercury Motion Data Analysis (Continued)
- The previous data analysis concluded that, Mercury is the source of light supposed
velocity (1.16 mkm/s).
- So, let's ask,
- Is Mercury Motion data in harmony with light supposed velocity (1.16 mkm/s)?
(I)
- Light suppose velocity (1.16 mkm/s) needs only 50 seconds to pass Mercury
orbital distance (57.9 mkm).
- Can this (50 s) be a data belong to Mercury Motion?
- Mercury (47.4 km/s) moves during 50.3 s a distance = 2390 km (Pluto diameter)
- Mercury (47.4 km/s) moves during 50.3 m x 60 s a distance = 142984 km (Jupiter
diameter)
- Mercury Venus Distance = 50.3 mkm ( 1mkm = 1 second)
- Mercury Neptune Distance (4437 mkm) = 50.3 mkm x 88
- (Mercury Orbital Period =88 days)
- Mercury Pluto Distance (5848 mkm) = 50.3 mkm x 116.2
- (57.9 mkm Mercury orbital distance x2 = 115.8 mkm)
- The data shows that, the period 50.3 seconds is a value be used frequently by
Mercury motion data. and mercury moves during (50.3 s) a distance = Pluto
diameter and moves during (50.3 m) a distance = Jupiter diameter, showing that
the 2 planets are the same planets mentioned by the 1st
trajectory motion.
- The basic difficulty is that, the light motion creates geometrical effects which we
can't discover because we don’t know neither their rule nor their effects. We have
only some data be transported from a point to another and that creates difficulties
for the analysis process.

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(II)
- Mercury Orbital Distance =57.9 mkm
- Mercury rotation period =58.65 solar days.
- The rate (1 mkm = 1 day) be used frequently in different motions data.
- The 2 values (57.9 mkm and 58.65 mkm) have an error (1%)
- Means, the 2 values (57.9 mkm and 58.66 days) can be created by the light
supposed velocity (1.16 mkm/s) motion during 50 seconds
- I try to show that, Mercury motion different data be in proportionality with light
supposed velocity (1.16 mkm/s) motion. which proves that Mercury is the origin
point of the light supposed velocity (1.16 mkm/s) motion.
- Because not only the period (5040 mkm) can be used or the period 4222.6 h
(which us equivalent to 4224 s) but also the period (50s)
- Notice
- 153.3 hours = Pluto day period
- 151.7 hours = 0.99 x 153.3 h (Pluto day period)
- But
- 176 mkm = 1.16 mkm/s x 151.7 seconds
- Where
- Mercury Day period =175.94 solar days (if 1 day = 1 mkm, the 2 values be
equivalent)

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- (3rd
Branch) Saturn Motion Effect.
- Saturn (0.838 mkm / daily) moves during = 5848 days, a distance = 4900 mkm
- This data shows Saturn motion effect on Mercury motion data
- And
- We have to discuss this data in separated point because Saturn creates one more
reading for Mercury Motion Data.
- That means, the data be used by light motion for reasons but the same data be used
again by planet motion for different reason.

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Equation no. (4)
1.16 mkm per second x 2 x 86400 seconds =2 x 100733 mkm = (37100 mkm –
4900 mkm) x 2π = 28255 mkm + 2 x 86400 mkm = 202318 mkm (error 1%)
- Where
- 100733 mkm = The 9 Solar Planets Orbital Circumferences Total
- 28255 mkm = Neptune orbital Circumference
- Equation no. (4) is the paper basic equation. Because it explains The Basic Energy
Trajectory Through The Solar System…
- First, We remember that we have discussed this equation before but here we
discuss it through another approach which shows a different reading for the data
- Also we notice that,
- The equation uses the rate (2) because of Neptune orbital circumference. Where
the equation 3 parts can be used without the rate (2), But the last part (28255 mkm
+ 2 x 86400 mkm) needs the rate (2) because of Neptune orbital circumference
(28255 mkm). That explains why the equation uses the rate (2).
- Now let's divide the equation into 4 Parts.
(Part No. 1)
- 1.16 mkm per second x 2 x 86400 seconds = 2 x 100733 mkm
- Part no. (1) tells us a known information which is
- Light supposed velocity (1.16 mkm/s) travels during one solar day (86400s) a
distance = 100733 mkm = The 9 Solar Planets Orbital Circumferences Total
- The rate (2) is used for Neptune Orbital Circumference.
- This part of equation tells that, light motion depends on the solar day period. This
idea we have discussed in equation no. (1) of this discussion.
- Also we know that

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- Light known velocity (0.3 mkm/s) travels during (86400 s) a distance = 25920
mkm
- Where
- 25920 mkm = 17.75 mkm x 1461 days
- 17.75 mkm = the planets distances total during a solar day in addition to the Earth
moon motion distance during a solar day, and
- 1461 days = Earth Cycle (365 +365 +365 +366)
- The data tries to prove that, the light motion depends on the solar day period.
- Notice
- (25920 mkm x 2) =37100 mkm x 1.392
- Where
- The sun diameter =1.392 mkm
- The data uses the rate (2) because this rate be used in equation no. (4) which
expresses the solar system basic energy trajectory.
- That means the value (25920 mkm x 2) refers to the total energy according to light
known velocity (0.3 mkm/s) range of motion.
- The data tells the value (25920 mkm x 2) expresses really the total energy because
37100 km (Pluto orbital circumference) is the greatest orbit
- That means, when the total energy be expresses by the value (2 x 100733 mkm)
this energy be transported to the light known velocity based on the same rate
because it's one machine.
- The data proves the energy unification through the solar system spite of using 2
different velocities of light.

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(Part No. 3)
- 2 x 100733 mkm = (37100 mkm – 4900 mkm) x 2π =
- This part is very important because
- The data shows the energy motion (or light motion) started from 4900 mkm
- That means,
- To create Pluto orbital circumference 37100 mkm, the light starts from the point
4900 mkm and measures a distance = 100733 mkm by geometrical calculations.
- We have 3 points here
- The start point which is (4900 mkm =Jupiter orbital circumference)
- The direction of motion (toward Pluto)
- The measurement unit is 100733 mkm (defined by light supposed velocity motion
during a solar day)
- The data tells, Pluto orbital circumference is created depending on (4900 mkm )
Jupiter orbital circumference by using the light motion.
- We remember
- The distance 4900 mkm be passed by light supposed velocity (1.16 mkm/s)
depends on the period (4224 s) which is defined in comparison with Mercury day
Period (4222.6 hours)
- As A Result, we conclude that,
- Light supposed velocity (1.16 mkm/s) starts from Mercury to Jupiter and continues
from Jupiter to Pluto. that shows one clear trajectory of motion from Mercury to
Jupiter to Pluto.
- That means, the motion from Mercury to Jupiter considers Mercury is the head of
the inner planets and by that the connection between Mercury and Jupiter express
the connection between Jupiter and all inner planets.
- And
- The connection between Jupiter and Pluto doesn't contain any other planet. It’s one
directed motion from Jupiter to Pluto immediately

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- Shortly
- Jupiter and Pluto orbital circumferences be created by light supposed velocity
(1.16 mkm/s) motion depends on Mercury motion data by creating different steps
of motions to connect between Mercury and Pluto
- This data confirms that the distances be created based on network form and
according to one geometrical design.
Notice
- (2π) This rate distinguish between light motion and planet motion as we have
discussed before.
- The rate (2π) distinguishes here clearly between the planet and light motions.
where the difference (37100 mkm – 4900 mkm) refer to the planets motions where
the value 2 x 100733 mkm refers to the light motion and the rate (2π) distinguish
between the 2 motions.
- Venus equation which we will study with the part no. (4) of this equation uses also
the rate (2π) between the energy source and the planet received energy. we should
refer to this meaning in its discussion.
- And
- The data gives a new meaning for the rate (2π).
- It's the rate between light motion and planet motion or
- It's the rate between the giver energy motion and the receiver energy motion
- That tells Mars energy be provided by Jupiter motion and Pluto energy be
provided by Earth motion.

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(Part No. 4)
- 2 x 100733 mkm = 28255 mkm + 2 x 86400 mkm
- The total energy be reflected from Pluto to Neptune. The distances behave
perfectly as light beams.
- Neptune had seized 14% of the total energy and the rest of energy be reflected
toward the inner planets into 2 trajectories of energy each has (86400 mkm)
- We suppose that (Distance = Energy)
- The energy be sent from Neptune to Venus and Earth, let's see their equations in
following
- 2 x 86400 mkm = (28255 mkm – 680 mkm) x 2π (Venus Equation)
- Where
- 28255 mkm = Neptune Orbital Circumference
- 680 mkm = Venus Orbital Circumference
- Earth equation is perfectly similar to Venus equation where between both the error
is less than 1%
- Also we see the rate (2π) which distinguish between light and planet motion as we
have seen in our equation no. (3) (or it distinguish between the source of energy
and the planet received energy).
- Part (4) of the equation tells that, Neptune had seized 14% of the total energy and
the rest of energy is sent to Venus and Earth into equal trajectories of energy each
has (86400 mkm).
- That means, the inner planets energy be received from Neptune. And because of
that, the inner planets distance be controlled by Neptune orbital distance (4495.1
mkm) as the data shows that clearly.
- 4495.1 mkm (Neptune orbital distance) = 78.3 mkm x 57.9 mkm
- = 108.2 mkm x 41.4 mkm
- = (149.6 mkm x 120 mkm)/4

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- Where
- 57.9 mkm = Mercury Orbital Distance
- 108.2 mkm = Venus Orbital Distance
- 149.6 mkm = Earth Orbital Distance 41.4 mkm = Earth Venus Distance
- 78.3 mkm = Earth Mars Distance 120 mkm = Venus Mars Distance
- Now let's ask
- Why Neptune had seized 14% of the total energy but Pluto seized nothing?
- Because
- Mars Migration Theory tells us…
- Pluto was The Mercury Moon and had migrated with Mars Migration.
- Neptune, in that time, was occupied the point (5906 mkm = Pluto current orbital
distance). And
- Pluto be migrated and thrown to the end of the solar group, flying along way and
had collided with Neptune at the point (5906 mkm).
- Pluto had pushed Neptune and put it out of Pluto (influence area!), for that reason
Pluto eccentricity distance = 1410 mkm= Pluto Neptune Distance
- The collision between Pluto and Neptune produced Pluto moons and the Kuiper
belt.
- Neptune be forced to be more near to the sun with orbital distance =4495.1 mkm
and without orbit.
- Neptune be out of the distances network (the railways) generally and needed
energy to built its orbit
- That explains why Bode Law Couldn't predict Neptune orbital distance, because
it's created (additionally) on the original geometrical design to repair the planets
migration negative effects on the solar system motion. means, it's a distance not
designed geometrically with the original design. So the equations depend on the
original design couldn't predict Neptune orbital distance.
- Please review Mars Migration Theory (Please See The References).

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Notice
- 202318 mkm = (449197 km)2
- Where
- 449197 km =Jupiter Circumference
- Accurately
- (202318 mkm) / (449197 km)2
= (361/360)
- What's (361 degrees)?
- The cycle is 360 degrees but the Earth moon orbital regresses 19 degrees per year
and during 19 years moves Metonic Cycle (19 years) (=361 degrees).
- The moon orbit regression effects on many of the planets motions data and
because of that the usual rate (360) be changed into (361).

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The Motion 1st
Trajectory (A Summary)
- Mercury is the origin point of this motion trajectory
- The energy be transported from Mercury to Jupiter by the light supposed velocity
(1.16 mkm/s) motion and by this energy Jupiter orbital circumference be created
- The energy be transported from Jupiter to Pluto by the light supposed velocity
(1.16 mkm/s)
- The energy be reflected from Pluto to Neptune and
- The energy be reflected from Neptune to the inner planets
- The Motion 1st
Trajectory Shows the energy trajectory from Mercury to
Jupiter to Pluto, which creates the solar system basic energy trajectory.

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The Motion 2nd
Trajectory
- Light supposed velocity (1.16 mkm/s) travels starting from Mercury to Pluto
directly
- In this trajectory of motion the light doesn’t use a central point as did with Jupiter
orbital circumference. Instead the light travels from Mercury to Pluto directly.
- As a result, this Trajectory of motion be formed based on the following data,
- Light supposed velocity (1.16 mkm/s) travels during (5040 seconds) a distance =
(5848 mkm= Mercury Pluto Distance)
-
- This trajectory of motion we have studied deeply. Because the period (5040 s) is
distinguish Mercury day period as we have discussed.
- But this trajectory of motion still have more important data
- This trajectory of motion uses one equation only which is no. (5), let's discuss it in
following…

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Equation no. (5)
5848 mkm (Mercury Pluto Distance) – 4900 mkm = 948 mkm
- Where
- 5848 mkm = Mercury Pluto Distance
- 948 mkm = Earth orbital circumference (940 mkm) error (1%)
- Equation no. (5) shows Earth orbital circumference be taken into account with the
light supposed velocity motion for the 2 distances (5848 mkm and 4900 mkm)
- But
- Why the distance (948 mkm) be important here?
- The Earth moon daily displacement (=88000 km), and if the moon passes Earth
orbital circumference by using its displacement (88000 km), the moon needs a
period =10747 days to pass a distance = (945.5 mkm)
- Where 10747 days = Saturn Orbital Period
- The important information isn't (Saturn orbital period) but its (Saturn data),
because Saturn data be used frequently in the motions trajectories data between
Mercury and Pluto, for light and planet trajectories of motions.
- i.e.
- Through different data analysis we reach to Saturn data. this time we found (10747
days = Saturn orbital period), and in previous discussion we have found 3 data be
related Saturn motion defined by the light and mercury motions by using the
period (5040 seconds) which are:
o Saturn moves during (5848 days) a distance = 4900 mkm
o Mercury moves during (5040 seconds) a distance = 2 x Saturn diameters
o Mercury moves during 1433.5 days a distance = 5870 mkm = Pluto orbital
distance (old definition) or near to (5848 mkm = Mercury Pluto Distance)
Where (1433.5mkm = Saturn Orbital Distance)

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- Why (5848 mkm) – (4900 mkm) = 948 mkm (Earth Orbital Circumference)?
(error 1%)
- It's A Geometrical Distribution For Distances,
- But not only,
- Because this distribution contains data (10747 days) be used by Saturn and the
Earth moon motions…
- This data we should discuss with Saturn motion data because Saturn motion is
related to the light known velocity (0.3mkm/s) motion.
- The period 10747 days also be used by the moon to pass a distance = the distance
be passed by light known velocity motion during a solar day.

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The Motion 3rd
Trajectory
- Light supposed velocity (1.16 mkm/s) uses 2 different trajectories of motions (no.
1 and no. 2) to pass a distance =5848 mkm (Mercury Pluto Distance)
- And these motions be done by light velocity
- The first trajectory, the light travels from Mercury to Jupiter and then to Pluto
- The second trajectory, the light travels from Mercury to Pluto Directly
- The third trajectory expresses a motion from Mercury to Pluto (5848 mkm) but by
using mercury motion (Mercury velocity per a solar day =4.095 mkm)
- We have seen this data before, let's remember them:
o Mercury moves during 84 days a distance = 344 mkm
o Mercury moves during 346.6 days a distance = 1419 mkm
o Mercury moves during 1433.5 days a distance = 5848 mkm
o Where
o 5848 mkm = (Mercury Pluto distance)
o 1433.5 mkm = Saturn orbital distance
o 346.6 days = the nodal year.
- This trajectory of motion refers (frequently) to Saturn motion data among mercury
motion data. we don't know why that's happening…
- We can conclude that, the planet (Mercury) follows the light motion but with
different rate of time.
- Because this behavior is described by the matter creation theory. Where the theory
tells, the matter is born out of light but the born matter doesn't separate from its
light parent. On the contrary, they move together in one unified motion but by
using different rates of time.

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- That's happening here. Mercury moves with light suppose velocity (1.16 mkm/s)
by using a different rate of time.
- Notice
- Equation no. (5) shows that Saturn data effects on Mercury motion data
- The fact is that, Saturn motion data be accompanying for Mercury motion data
along the way. We found Saturn motion data frequently in Mercury motion data
analysis.

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Equation no. (6)
1.16 mkm/s x 4900 seconds = 5678.1 mkm (Mars Pluto Distance)
And
- Where
- 4900 mkm =Jupiter Orbital Circumference
- Light supposed velocity (1.16 mkm/sec) travels during 4900 seconds a distance =
5678.1 mkm (Mars Pluto Distance).
- We remember that,
- The light supposed velocity (1.16 mkm/sec) depends on the period (4224 sec) to
pass the distance 4900 mkm.
- And depends on the period (5040 second to pass) the distance 5848 mkm (Mercury
Pluto distance)
- And we have asked why light supposed velocity (1.16 mkm/sec) created 2
trajectories between Mercury and Pluto positions? Or in more clear words why
light supposed velocity creates Jupiter orbital circumference (4900 mkm)by using
(4224 seconds)?
- Equation no. (6) gives an answer.
- Because light supposed velocity creates a connection between Mars and Pluto
positions with Mercury.
- Mars migration theory tells, Pluto was the Mercury moon and Mars was the next
planet to Mercury. And the 2 planets were migrated from their original positions.
- Light supposed velocity (1.16 mkm/sec) creates a connection between Mercury
and Pluto through the period 5040 second and another connection between Pluto
and Mars through the distance (4900 mkm)

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- Notice
- Light supposed velocity (1.16 mkm/sec) travels during 5687.1 seconds a distance
= 6585.4 mkm
- By the rate (1 mkm = 1 day) this distance 6858.4 mkm be = 6858.4 solar days
- Where (Saros Cycle =6858.4 solar days)
- Please remember
- Light known velocity (0.3 mkm/sec) moves during a solar day (86400 s) a distance
=25920 mkm,
- Where we see this distance (25920 mkm) as a period of time (25920 years = The
Procession Cycle)
- I try to show that, the period of Sarod Cycle (6585.39 days) is similar to the period
(25920 years), both be created by light motion behavior, which proves it's usual a
behavior in the solar system
- But that creates one more connection between Mars motion and the Earth moon
motion because Saros is a cycle produced by the moon motion effect.

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Equation no. (7)
193 seconds x 0.3mkm/sec = 57.9 mkm (Mercury Orbital Distance)
196.5 seconds x 1.16 mkm/sec = 227.9 mkm (Mars Orbital Distance)
- Equation no.(7) tries to explain why Mercury and Mars Motions have interaction
as we have proved in Jupiter motion analysis
- The reason is that, the 2 planets orbital periods are defined based on equal periods
approximately (193 seconds and 196.5 seconds the difference 2%), by using the 2
light velocities
- That shows the period (193 sec or 196.5 sec) are the reason of the 2planets motions
integration or interaction
- But
- As we have seen Jupiter motion also be integrated with Mercury and Mars motions
- Can this period (196.5sec) effect on Jupiter motion also?
- Yes
- 8 Jupiter days = 8 x 9.9 x 3600 = 285120 seconds = π x 90756.5 seconds
- Pluto orbital period =90560 days
- 90756.5 seconds - 90560 seconds = 196.5 seconds
- Where
- Light motion uses any period (of 1 solar day) as (1 second)

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The Motion Trajectories Summary
- As we have discussed in the previous analysis,
- There are 3 motions started from Mercury moving toward Pluto, and these motions
are:
o Light supposed velocity (1.16 mkm/) motion during (5040 s) to pass (5848
mkm = Mercury Pluto Distance)
o Light motion from Mercury to Jupiter by using (4224s)
o 1.16 mkm/s x4224 seconds =4900 mkm (= Jupiter orbital circumference),
where Mercury uses Jupiter orbital circumference as its central point to
connect with Pluto.
o Mercury Motion depends on (5040) to pass in three steps toward Pluto by
the data
o 4.095 mkm x 84 days =344 mkm
o 4.095 mkm x 346.6 days =1419 mkm
o 4.095 mkm x 1433.5 days =5870 mkm (very near to 5848 mkm)
o Notice
o Light supposed velocity (1.16 mkm/s) motion depends on the rate (2π) but
Planet motion depends on (R). So if Mercury want to connects with Jupiter,
light motion doesn't use the distance (778.6 mkm Jupiter orbital distance)
but use (4900 mkm =Jupiter orbital circumference)
o 778.6 mkm =1.0725 x 720.7 mkm (Mercury Jupiter Distance)

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Equation no. (8)
5040 sec = 88 sec x 57.27
5040 seconds are required for Mercury Day Period to be 176 solar days
88 days = Mercury Orbital Period
57.27 =Mercury Rotation Period / Mars Rotation Period
Let's remember our question
Why Mercury Day Period = 2 Mercury Orbital Periods?
Where
- 4331 days = Jupiter Orbital Period, and 175.94 days = Mercury Day Period, and
- 687 days = Mars Orbital Period , and 24.6 hours = Mars Rotation Period
- Based on this data we have concluded the following rates of time
- And
- 671 hours of Mercury Motion = 687 days of Jupiter motion = 109 days of
Mars motion
- The we have discovered that,
- During 671 hours Mercury moves a distance 57.2 mkm= 2 x Mercury Orbital
Distance (error 1%)
- And
- During 687 days Jupiter moves a distance 778.6 mkm = Jupiter Orbital
Distance
1 hour of Mercury Motion = 24.6 hours of Jupiter Motion
And
1 hour of Mercury Motion = 3.9 hours of Mars Motion
And
1 hour of Mars Motion = 2π hours of Jupiter Motion

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- And
- During 109 days Mars moves a distance 227.9 mkm = Mars Orbital Distance
- The question was
- Why Mercury moves distance = (2 x Mercury Orbital Distance), But Mars And
Jupiter Move Distances Only Equal Each Planet Orbital Distance?
- Then
- We have found, 5040 s are required for Mercury day to be =176 solar days
- And
- During 5040 seconds Mercury (47.4 km/s) moves a distance = 2 x 119448 km =
Saturn diameter (= 120536 km) (error -1%)
- But
- During 5040 seconds Mars (24.1 km/s) moves a distance =121464 km = Saturn
diameter (= 120536 km) (error +1%)
- This data gives one more reference that, the (one value for Mars) produces (2
values for Mercury)!
- We have a reason to find the rate (2) in data and because Mercury day = (2)
Mercury orbital period –some interesting feeling pushed to dig under this data may
we find answer why Mercury Day Period = 2 Mercury Orbital Period
- Let's return to the equation
Equation no. (8) (continued)
5040 sec = 88 sec x 57.27
- Light motion for 1 second causes a planet motion for 1 solar day,
- Because of this rule we couldn't discuss this equation in Jupiter Motion analysis
- Light moves 88 seconds and caused Mercury to move 88 solar day
- That means, this equation is Mercury orbital revolution equation… because of that
the rate (57.2) be produced…. Where 57.9 mkm = Mercury orbital distance (error
1%)

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- Let's remember this data
- Mercury rotation period 1407.6 h / Mars rotation period 24.6 h = 57.2
- And
- Mercury day period 4222.6 h / Mars rotation period 24.6 h = 171.7
- Both values (57.2 and 171.7) are Mercury distances with the same error (1%)
- (57.9 mkm= Mercury orbital distance, and 170 mkm = Mercury Mars Distance)
- How Mercury orbital period 88 days can produce Mercury day period?
- 57.2 =2 x 28.6 (Neptune axial tilt =28.3 degrees error 1%)
- It's Neptune effect on Mercury Motion
- Why Neptune??
- Because
- Neptune Orbital Distance 4495.1 mkm = π x Saturn Orbital Distance 1433.5 mkm
- Notice
- Neptune orbital period 59800 days = Mercury day period 175.94 days x 339
- But
- Venus rotation period 5832.5 hours = Uranus rotation period x 339
- It's the machine by which the rate (5.1) be transported, let's remember it here
o Venus Velocity = Uranus Velocity x 5.1
o Mars Velocity = Pluto Velocity x 5.1
o The Moon Velocity = Neptune Velocity x 5.1
- The machine by which the rate (2π) be transported and used as the time rate
between Mars And Jupiter. And be used again as the time rate Between Earth
And Pluto Motions.
- That tells why
- Mercury moves during its rotation period a distance 243mkm (error 1%) and
- Mars moves during Venus day period a distance 243mkm (error 1%)

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-
The General Discussion
- I have tried to arrange the data in best form to show that, the data be created based
on a geometrical design for a light beam motion by which the solar planets and
their distances be created.
- The paper hypothesis gives a chance for better explanation….
- The hypothesis tells a light beam its velocity (1.16 mkm/sec) caused the solar
system creation and motion, and by effect of the light motion the distances be
created in a network form.
- The hypothesis shows itself so complex because we need to prove (at first) that
there's a light beam its velocity =1.16 mkm/s
- By the planets motions data arrangement I have tried to show that the planets
motions data be created depends on a light beam its velocity =1.16 mkm/s and if
this light beam isn't found there's no chance to create the solar planets motions data
and their distances as their values. This discussion is done in so deep details where
Jupiter motion data and its distances with the inner planets and with Pluto, this all
data will be simply removed if there's no light beam its velocity =1.16 mkm/s.
- The data still have more support for the hypothesis but needs at first to discover
the geometrical design based on which the planets motions data be created.
- Although, this try is a strong one, but many basic questions are still left behind, we
have to face them as a part of our effort searching for the truth
- (A Question) what's the relationship between Planet motion distance and a
distance be found between 2 planets? If these 2 distances be equal why does this
give any geometrical meaning?
- For example
- Light supposed velocity (1.16 mkm/s) travels during 5040 seconds a distance =
5848 mkm = Mercury Pluto Distance
- What's the relationship between the distance be passed by light motion (5848
mkm) and Mercury Pluto Distance (5848 mkm)?

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- If these 2 distances are equal why this is important?
- My answer for this question is a simple one ( I Don't Know How The Machine
Works).
- For example
- Earth moves during a solar day (24 h) a distance = 2574720 km
- Pluto moves during its day period (153.3 h) a distance = 2593836 km
- These 3 distances are equal with max error only (1%)
- Can these equal distances effect on any planet motion data? yes
- As we discuss in (The Planets Motions Depend On Different Rates Of Time)
the three planets motion data be created in full proportionality with one another.
can this data be in proportionality because of the distances equality or the distances
equality be a result of the data proportionality, I can't give a decision.
- The fact is that, the planets data be in proportionality and the distances are equal.
- To prove this point of view let's bring the data here for better understanding…
- More Data
Group no. (1) (Pluto And Earth Rate Of Time)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ earth day period) = 6.37 (0.8%)
Group no. (2) (Pluto And The moon Rate Of Time)
(1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
(2)
Tan (12.2) x 708.7 hours = 153.3 hours
(3)
Tan (13.17) x 655.7 hours = 153.3 hours

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- Where
- 406000 km = Pluto Motion Distance during a solar day
- 88000 km = The moon displacement daily
- And
- And
- 153.3 hours = Also (153.3-) Pluto Rotation Period
- The data confirms the point of view, the equal distances effects on the planets
motion data be caused it to be in proportionality with one another. why or how?
We still search for the machine geometrical details.
- The discussion is limited and can't receives its power, as a creature crippled and
can't walk! Why? because we don't know the used geometrical rules based on
which the solar system be created and moving.
- For example
- The equal distances (as we have seen in Earth, its moon and Pluto data) is a
method used widely in the solar system. That means, all planets use this method.
All planets create equal distances with one another and based on these equal
distances the planets motions data be in proportionality with one another.
- Let's use one more example:
- Mercury moves during its rotation period (58.65 days) a distance = 243 mkm(1%)
- Mars moves during Venus day period ( 116.75 days) a distance = 243 mkm
- These are the equal distances, where is the proportionality of data?

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- (Mercury Day Period/ Venus Day Period) =(Mercury orbital period/ Mercury
rotation period)
- The data shows that, Venus Day period be created in proportionality with Mercury
cycles periods. Can that be a result for the equal distances?
- The equal distances is a method used thousands of times in the solar system, where
many planets motions distances be passed in defined periods of time be equal one
another. we don't know what effect be done from this distances equality and by
that a great part of the machine be out of our sight…
- Shortly,
- We have to depend on some conclusions (and even guess) to conclude how the
solar group be created and moving
- The other option is the imaginary ideas. As Newton idea of the sun mass gravity,
where he imagined that the sun is the planets motions reason while the fact is that
the sun is created after all planets creation and motion. and in fact the sun rays be
created by using the planets motions energies total. means, based on the planets
motions the sun be created, and that tells the planets can be found without the sun
but the sun can't be found without the planets. Newton idea is completely
mistaken.
- Our situation is not so better. The point is that, the geometrical rules book is inside
the light nature, and we can't catch how the geometrical effects be done but we can
discover them in every where.
- Let's summarize the machine basic difficulties
o Planet moves side by side with a light motion by that the planet motion
refers to light motion features but we can't catch the velocity which creates
these features
o The equal distances method is the method caused the planets motions. this
method works geometrically.

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o That means, this method works as a waterwheel or as a lever. A tool works
by a geometrical effect. But we don't understand what benefit be produced
by the equality of distances. This equality causes the planets motions but
how?
o We notice also that,
Jupiter motion depends on light supposed velocity (1.16 mkm/s) and
all distances around Jupiter be defined based on this velocity (1.16
mkm/) but
Saturn motion depends on the light (0.3 mkm/s) where different data
belong to Saturn depends on light known velocity.
The Earth moon motion be in proportionality with both light beam
velocities! The data confirms this meaning
10921 km x 27.3 days =300000 km (error 0.5%) the equation tells if
the moon rotates around its axis one time per solar day as Earth does,
the moon would pass by its circumference during 27.32 days (the
moon orbital period) a distance =300000 km = light known velocity
(0.3 mkm/s) motion distance during 1 second.
43000 km x 27.3 days =1160000 km (error 1.3%) the equation tells if
the moon is a point without circumference and moves from perigee
point to apogee in a solar day and in the next day moves from apogee
to perigee … during the moon orbital period (27.3 days)the moon
would move a distance =1.16 mkm
Notice (43000 km = the distance between perigee and apogee points)
10921 km x 86400s =940 mkm, the equation tells if Earth revolves
around the sun a complete revolution (940 mkm) in one solar day
only (86400 s), the moon circumference (10921 km) will equal a
distance of Earth motion during 1 second period of time. (notice, the
geometrical design can create the effect without planet real motion)

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4-4 Jupiter Orbital Circumference Analysis
I-Data
Group No. (1)
(A)
1.16 mkm/s x 4224 seconds = 4900 mkm (Light Motion Reading)
(B)
(449197 km / 27.78) = (4900 mkm /0.3 mkm)
(C)
1.16 mkm/s x 4224 seconds = 4900 mkm (Planet Motion Reading)
(D)
4900 mkm = 360 mkm +680 mkm +940 mkm +1433 mkm +1433 mkm
(E)
1.1318 mkm/s x 4222.6 days = 4780 mkm
(F)
4900 mkm = 449197 km x 10921 km
But
4846 mkm =1.392 mkm x 3475 km
(G)
9.7 km/s x 142984 seconds = 1386945 km (notice 1.392 mkm = the sun diameter)
(H)
4900 mkm = 0.384 mkm x 12756 km =0.406 mkm x 12104 km
(I) (please remember)
(37100 mkm – 4900 mkm) x 2π =28255mkm + 2 x 86400 mkm =2 x 1.16 mkm/s x 86400 s
Group No. (2)
(J)
(2 x The Sun Mass) = Jupiter Mass x 2094

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II-Discussion
- Jupiter orbital circumference is a very rich distance for discussion and analysis,
also it provides surprised using of data.
- There are 3 basic features of this distance makes it very specific distance, let's refer
to them in following:
- (1) The Distance 4900 mkm Is Defined By Many Planets Motions
- (2) The Distance 4900 mkm be in proportionality with The Sun Diameter.
- (3) The Distance 4900 Is Defined also by Light Motion Also.
- Let's analyze the distance 4900 mkm in following…
Equation No. (A)
1.16 mkm/s x 4224 seconds = 4900 mkm (Light Motion Reading)
- We have studied this equation deeply in the previous discussion.
- This equation is provided here to remember its meaning only because we need it in
the following discussion
- Equation no. (A) tells,
- Light supposed velocity (1.16 mkm/s) travels during the period 4224 seconds a
distance = 4900 mkm = Jupiter orbital circumference
- We consider that, 1 second of light motion be = 1 hour of Mercury motion and by
that the period (4224 s) be = (4224 hours)
- This analysis lead us to conclude that light supposed velocity (1.16 mkm/s) motion
effects on Mercury motion and causes Mercury day period to be 4222.6 hours and
decreased it from (4224 h) by (5040 seconds) because the light supposed velocity
(1.16 mkm/s) travels during 5040 seconds a distance = 5848 mkm (= Mercury
Pluto Distance). And by that the light supposed velocity creates a connection
between Mercury and Pluto motions.
- This is a summary for this equation which we have discussed deeply before in the
light supposed velocity proves discussion.

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Equation No. (B)
(449197 km / 27.78) = (4900 mkm /0.3 mkm) (error 1%)
- Where
- 449197 km = Jupiter Circumference
- 0.3 mkm/sec = light known velocity (300000 km /s)
- 27.78 km/s = The Moon Velocity.
- Equation no. (B) tells, the period the Earth moon needs to move a distance =
449197 km = Jupiter circumference = 16330 sec
- Light known velocity (0.3 mkm/) needs this same period of time (16330 sec) to
pass a distance =4900 mkm (= Jupiter orbital circumference)
- i.e.
- during the moon passes Jupiter circumference, the light known velocity (0.3
mkm/s) passes Jupiter orbital circumference! Why?
- Equation no. (f) causes more puzzles for this discussion, where (4900 mkm =
449197 km x 10921 km), it tells Jupiter orbital circumference be defined as a
multiplication between the moon circumference and Jupiter circumference!
- Geometrically we considered Jupiter orbital circumference (4900 mkm) be built
based on the 2 measurements (Jupiter circumference and the moon circumference!)
- This idea be supported by the fact
- Jupiter (13.1 km/s) moves during 10921 sec a distance = 142984 km =Jupiter
diameter.
- This data gives a feeling, the geometrical structure of Jupiter orbital circumference
depended on Jupiter circumference and the moon circumference…. But
- What's the relation of this geometrical structure with light known velocity? How
light known velocity (0.3 mkm/s) be in proportionality with this same data?!
because light known velocity (0.3 mkm/s) should be an independent velocity and
has no proportionality with the planets motions velocities!

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- Can we understand this data?!
- Jupiter orbital circumference be created by a motion of light supposed velocity
(1.16 mkm/s) during the period (4224 seconds). But the distance (4900 mkm) be in
proportionality with light known velocity (0.3 mkm/s) in comparison with planets
motions velocities…
- Shortly why (449197 km / 27.78) = (4900 mkm /0.3 mkm)?
- Or
- (light velocity 0.3 mkm/ the moon velocity 27.78 km) = (4900 mkm /449197 km)?
- The rate (0.3 mkm /27.78) = 10800
- And
- The rate (4900 mkm/449197 km) =10921 (the moon circumference =10921 km)
- The difference between 10800 and 10921 is 1%
Notice
- Why the moon velocity be =27.78 km/s? because the moon moves daily a
distance = Earth motion distance daily because they aren't separated from one
another. means the moon daily motion distance =2.574 mkm but this distance be
effected by the rate 1.0725. (we have referred to that frequently before). We don't
know why the distance 2.574 mkm be effected by the rate (1.0725) but 40% of all
distances in the solar system be effected by this same rate 1.0725 and appendix
no.1provides a list for these distances. By that the moon distance 2.574 mkm be
=2.4 mkm daily which =27.78 km/sec.
Notice
- Earth (29.8 km/s) moves during (16330 sec) a distance = 4 x Saturn diameter (1%)
- Saturn (9.7 km/s) moves during (16330 sec) a distance = Uranus Circumference (1.3%)
- Neptune (5.4 km/s) moves during (16330 sec) a distance = 88000 km (the moon
daily displacement).

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Equation No. (C)
1.16 mkm/s x 4224 seconds = 4900 mkm (Planet Motion Reading)
- This equation we know perfectly
- Light supposed velocity (1.16 mkm/s) travels during 4224 seconds a distance =
4900 mkm = Jupiter Orbital Circumference
- But
- This equation has another reading, let's try with it in following
(1)
o Jupiter (13.1 km/s) moves during 24.6 h a distance = 1.16 mkm.
o Where (24.6 hours = Mars rotation period)
(2)
o Jupiter orbital circumference (4900 mkm) be divided on the distance 1.16
mkm and the rate 4224 (or 4222.6).
o Means, if each 1 hour of Mercury Day period be = 24.6 hours, based on that,
Jupiter during the period (4222.6) x (24.6 h) moves a distance =4900 mkm =
Jupiter orbital circumference
o (4331 days x 24 h = 4224 x 24.6h) where 4331 days = Jupiter orbital period
o Can that refer to some interaction of motions among the 3 planets (Mercury,
Mars and Jupiter)? Of course yes, this fact is proved in the planets unified
motion discussion in this paper (point no.6 of this paper) where huge
number of data shows clearly the motions interactions among the 3 planets.
(3)
o Notice
o Light supposed velocity (1.16 mkm/sec)moves during 4900 seconds a
distance = 5684 mkm (Mars Pluto Distance).
o By that,

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o Light supposed velocity (1.16 mkm/sec) creates ac connection between
Mercury, Jupiter and Mars motions, where Mars be connected with Mercury
through Pluto.
o The result is that,
o A strong connection be found between Mercury, Mars and Jupiter. And this
connection is seen clearly in the light motion data.
o But
o Can this connection be a real one? Is there really an interaction of motion
between Mercury, Mars and Jupiter?
o This question is answered in point no. (6) of this paper (The Solar Planets
move one unified general motion), where this point provides tenths of data
proves that, Mercury, Mars and Jupiter Motions can't be considered as 3
independent planets motions, On the contrary, The Data shows clearly that,
they move together one unified general motion which is proved strongly by
tenths of these planets motions data.
(4)
o I want to say that,
o We have 2 pictures for the same scene. As if we use 2 covers for the same
place. One cover is put on another. the 2 covers are equal in length. But they
are different in nature. Similar to that, we have equal distances (4900 mkm)
be passed by (light motion) and by (planet motion). the motions create
interactions in data. while we don't see the light motion but we see only the
planet motion. but we discover the interactions which are done as a result
for light motion effect on planet motion.
(5)
o The real data causes even more confusion, because light moves 1.16 mkm in
one second. but Jupiter moves 1.16 mkm in 24.6 hours, by that the simple
interactions which depend on (1 second) should be explained in some

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complex form based on the period (24.6 h) and then we have to ask (why
Jupiter uses Mars rotation period 24.6 h?). Because we know that some
relationship is found between Mercury Day Period and Jupiter orbital
circumference. And now we have a relationship between Jupiter orbital
circumference (4900 mkm) and Mars rotation period (24.6 h) through the
distance 1.16 mkm. That simply creates a relationship (or a proportionality)
between Mercury day period and Mars rotation period! (why?)
o Notice/
o (Mercury Day period 4222.6h) / (Mars rotation period 24.6 h) =170 (error 1%)
o Mercury Mars Distance = 170 mkm
o But
o (1407.6 h/ 24.6 h) = 2 x ( (4900 mkm) /(170 mkm)), where
o 1407.6 h = Mercury Rotation Period
o 24.6 h = Mars Rotation Period
o 4900 mkm = Jupiter Orbital Circumference
o I wish the argument logic is clear.
o Jupiter moves a distance 1.16 mkm during a period (24.6 h =Mars Rotation
Period) because (1st
, Jupiter motion and Jupiter orbital circumference 4900
mkm have a relationship with Mercury Day Period). And (2nd
, Mercury
motion has a relationship with Mars Motion as the previous data refers).
We deal with a machine of gears, the main motor is the light motion and the planets
are gears, because of that, their data be in proportionality with one another according
to the motions interactions.
Notice
Mars moves during 13.37 hours a distance = 1.16 mkm
And 2.082 mkm(Mars motion distance per a solar day) =1.16 mkm x 1.8
(but 1.3 deg Jupiter orbital inclination – 3.1 deg Jupiter axial tilt) =1.8 deg (Neptune
orbital inclination)

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Equation No. (D)
4900 mkm = 360 mkm +680 mkm +940 mkm +1433 mkm +1433 mkm
(Error 1%)
- Where
- 360 mkm = Mercury Orbital Circumference
- 940 mkm = Earth Orbital Circumference
- 1433 mkm = Mars Orbital Circumference
- The data considers that, the Earth Moon Orbital Circumference around the sun
=1433 mkm = Mars Orbital Circumference.
- If we accept this idea, the data will give a clear meaning. Which tells that (Jupiter
orbital circumference = the inner planets orbital circumferences total)
- Then we have to ask.
Why does Jupiter Orbital Circumference=
= The Inner Planets Orbital Circumferences Total?
- More data can help our analysis,
(I)
- We accept that, Jupiter orbital circumference 4900 mkm be passed by a light
supposed velocity (1.16 mkm/s) in a period = 4224 seconds. And by this motion
the light caused to create Mercury day period (4222.6 h) where we accept that 1
second of light motion be = 1 hour of Mercury motion.
- Venus and Earth Data also be effected by (4224), Let's remember their data
o 4224 mkm = 2π x 671 mkm (Venus Jupiter Distance)
o And
o (5040 sec/4224 sec) = (928 mkm/778.6 mkm)
o Where

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o 778.6 mkm = Jupiter orbital distance
o 928 mkm = Earth Jupiter distance (notice 928 mkm= 778.6 mkm +149.6 mkm)
o 5040 sec = Mercury Day Period be less by 5040 seconds than 176 solar days
(II)
o 778.6 mkm = 1.0725 x 720.7 mkm (Mercury Jupiter Distance)
o 720.7 mkm = 1.0725 x 671 mkm (Venus Jupiter Distance)
o 671 mkm = 1.0725 x 629 mkm (Earth Jupiter Distance)
o 778.6 mkm = Jupiter Orbital Distance
(III)
o 1.16 mkm /sec x 500 sec = 580 mkm
o 1.16 mkm /sec x 580 sec = 671 mkm (Venus Jupiter Distance)
o 1.16 mkm /sec x 671 sec = 778.6 mkm (Jupiter Orbital Distance)
o 1.16 mkm /sec x 629 sec = 720.7 mkm (Mercury Jupiter Distance) (1%)
- The data shows, the distances between Jupiter and the inner planets are defined
based on Geometrical Calculations, because of that all distances are rated with
(1.0725) and (1.16).
- Means, these planets motions are not independent motions, on the contrary they
are (complementary motions) and be created based on One Geometrical Design.
- The previous data shows a complex geometrical machine behind. More analysis
provides even more complex interactions for example
- (778.6 mkm/ 629 mkm) = (720.7 mkm/580 mkm)
- These distances also we have used in the previous data. that means, the distances
between Jupiter and the inner planets be created by some geometrical system
makes these distances be in proportionality to one another mutually.
- Means,
- Not just, Jupiter orbital circumference be = the inner planets orbital circumferences
total, but also all used distances be created in proportionality with one another.

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Equation No. (E)
1.1318 mkm/s x 4222.6 days = 4780 mkm
- Where
- 1.1318 mkm /day = Jupiter motion distance during a solar day
- 4222.6 hours = Mercury Day Period
- Equation (E) supposes that 1 hour of Mercury Motion = 1 day of Jupiter motion.
by that, Jupiter uses Mercury Day Period as 4222.6 Solar Days
- During 4222.6 days Jupiter moves 4780 mkm. But during 4331 days Jupiter moves
4900 mkm.
- That means, the period 4222.6 days is incorrect period, because if we want to
measure the distance 4900 mkm we should use the period 4331 days
- But
- (4222.6 days /4331 days) = (654.9 mkm/ 671 mkm)
- Where
- 654.9 mkm = Jupiter Saturn Distance
- 671 mkm = Jupiter Venus Distance
- That tells the rate (4222.6 /4331) is defined geometrically and by that, the distance
4780 mkm which is less than 4900 mkm by 2.5% may refer to the distance 4846
mkm (with error 1.3%)
- Based on that
- Equation no. (E) tells, Jupiter moves during 4222.6 days a distance = (4780 mkm)
= 4846 mkm (the inner planets orbital circumferences total) (error 1.38%)
- Equation no. (E) can refer to that, 1 hour of Mercury Motion be = 24 hours of
Jupiter motion.
- This is the same rate of Uranus where 1 hour of Mercury Motion be =24 hours
of Uranus motion.
- Be we know that, 1 h of Mercury motion = 9.18 h of Pluto motion

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- Based on that
- 1 h of Pluto motion = 2.62 h of Uranus Motion = 2.62 h of Jupiter Motion
- We should discuss that in details in the planets motions depend on different rates
of time (Point no. 7) of this paper.
- The rate 2.62 is so specific for both planets
- Uranus Circumference x 2.62 = Uranus motion distance during its day period
(421056 km)
- And
- Jupiter Circumference x 2.62 = 1.16 mkm (error 1.4%)
- I try to show that, the distance (1.16 mkm) is found again under some new data. it's
not some random number found by chance in Jupiter data. it's a basic distance
based on which Jupiter motion and orbital circumference be designed, because of
that we face this distance (1.16 mkm) in any place we move.
- The negative point is that, we can't discover the geometrical design based on which
this distance 1.16 mkm be defined. For that reason, we meet this distance (1.16
mkm) frequently but couldn't catch the machine which produces it.
- Notice
- Light known velocity (0.3 mkm/s) x 4780s =1433.5mkm (Saturn Orbital Distance)
- Light known velocity (0.3 mkm/s) x 4879s =1461 mkm
- Where
- 4879 km = Mercury Diameter
- 1461 days = Earth Cycle (365 +365 +365 +366 days =1461 days)
- The distance 4780 mkm be used as (4780 seconds) where usually light uses the
rate (1 mkm = 1 second) but the planet uses the rate (1 mkm = 1 day).
- Please notice, again the distance (1461 mkm) be seen by us as a period of time
(1461 days) it's a usual behavior

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Equation No. (F)
4900 mkm = 449197 km x 10921 km
But
4846 mkm =1.392 mkm x 3475 km
- Where
- 449197 km =Jupiter Circumference
- 10921 km = The Moon Circumference
- And
- 1.392 mkm = The Sun Diameter
- 3475 km = The Moon Diameter
- 4846 mkm = The Inner Planets Orbital Circumferences Total
(4846 mkm = 360+680+940+1433+1433)
- Equation (F) tells, the multiplication of the sun diameter and the moon diameter =
accurately the value (4846 mkm)
- But
- The multiplication of the Jupiter Circumference and the moon Circumference =
accurately the value (4900 mkm)
- Can that refer to some relationship between the distance 4900 mkm and the sun
creation?
- Let's move step by step in following:
o Jupiter diameter x π2
= 1411196 km which is different from the sun diameter
1392000 km with the rate (1.38%)
o But
o Saturn (9.7 km/s) moves during 142984 seconds a distance = 1386945 km
(notice 1.392 mkm = the sun diameter and the difference 0.4%)
o Can this data tell (the sun is created in proportionality with Jupiter diameter)

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o We have 2 reasons for this conclusion. Because (1st
) the data refers to a
proportionality between the sun and Jupiter diameter and (2nd
) the data
refers to proportionality between the sun and Jupiter orbital circumference.
That shows 2 basic data be in proportionality with the sun diameter.
o The second player is the Earth moon
o The moon circumference (10921 km) be in proportionality with Jupiter
orbital circumference 4900 mkm based on Jupiter circumference (449197
km).
o And
o Jupiter (13.1 km/s) moves during 10921 sec a distance = 142984 km =
Jupiter diameter.
o I don't need to prove that, this data be created based on geometrical system.
Because the data proves that strongly and clearly. I need to discover this
geometrical system based on which this data be created.
o Shortly
o Jupiter is the basic player found on other side and effects on the sun creation
process. Where this event is happening, Jupiter adopted the Earth moon as
its own son and uses its data as part of Jupiter own data. as a result the earth
moon (small planet) has some chance to effect on the sun creation data!
o Again
o Why We See The Sun Disc =The Moon Disc? Because
o (The sun diameter / the moon diameter) = (Earth orbital distance/ Earth
moon Distance)
o Why The Diameters Rate = The Distances Rate?
o Because all players are created in proportionality with 4900 mkm (Jupiter
orbital circumference).
o The data can't be created by any pure coincidence. The fact can be seen if
we try to answer the question (how the matter is created?)

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Equation No. (H)
4900 mkm = 0.384 mkm x 12756 km =0.406 mkm x 12104 km
- Where
- 0.384 mkm = the moon orbital distance
- 0.406 mkm = the moon orbital Radius At Apogee
- 12756 km = Earth Diameter
- 12104 km = Venus Diameter
- Equation no. (H) shows, the distance 4900 mkm controls the data around the
moon motion.
- Earth and Venus Diameters must be ancient and aren't changed. That tells the
value (4900 mkm) use the 2 planets diameters as references to define the moon
motion range. By that the moon orbital distance be =0.384 mkm depends on Earth
diameter and the moon apogee radius be 0.406 mkm depends on Venus diameter.
- The data shows this meaning but no geometrical mechanism be discovered to
answer how this data be created in proportionality with one another.
- More data can support the same meaning for example
- This data tells that (the apogee orbital radius 406000 km be defined by Venus),
where (406000 km = 3475 km x 116.75), and 3475 km = The Moon Diameter and
116.75 days = Venus Day Period.
- The data gives the meaning but no geometrical mechanism be discovered.
- Shortly
- The distance 4900 mkm controls the sun creation process and the moon
orbital motion definition.
- The distance 4900 mkm uses Earth and Venus diameters as references to
define the moon orbital distance and its apogee radius.

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- Notice
- 1433 mkm =550.7 x 2.5986 mkm
- 2.598 mkm (the moon displacements total during 259.53 days),
- 550.7 mkm= Mars Jupiter Distance,
- 1433 mkm = Saturn Orbital Distance
- This data tells why the data considers the moon orbital circumference round the
sun be = 1433 mkm = Mars orbital circumference and Not be =940 mkm = Earth
orbital circumference.
- The data shows that, the Earth moon is the son of the inner planets. The moon isn't
belonged to the Earth alone where Venus effects greatly on the moon apogee
radius definition on one side where Mars still effect on the moon motion and
causes its orbital circumference around the sun be considered =1433 mkm = Mars
orbital circumference on the other side.
- The geometrical machine based on which this data be created has answers for the
question how this data be created.
Comments
- The distance 4900 mkm defines the moon diameter and its orbital motion range
- And
- The distance 4900 mkm controls the sun creation process and defines the sun
diameter.
- Because of that, the sun and moon data be created in proportionality as we have
discussed before.
- But
- Is the aforementioned data sufficient for this conclusion? Suppose (4900 mkm =
the sun diameter 1.392 mkm x 3475 km the moon diameter) can this data be a
sufficient for this heavy conclusion? Let's discuss the next data to answer this
question.

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Group No. (2)
Equation No. (J)
(2 x The Sun Mass) = Jupiter Mass x 2094
Where
2094 mkm = Jupiter Uranus Distance
- How to make the rate 2094 be = the distance 2094 mkm?
- Geometrically, this process depends on Jupiter motion. because Jupiter diameter
142984 km = (1 million km/7) where Mercury orbital inclination = 7 degrees
- That means, the interaction of motions between Jupiter and Mercury can produce
the distance 1 mkm based on Jupiter diameter.
- By that, the rate 2094 x 1 mkm = 2094 mkm (Jupiter Uranus Distance)
- Why do we need this data?
- I try to prove that, the sun is created by an effect of Jupiter motion. We have seen a
proportionality of data between the sun diameter and the distance 4900 mkm
(Jupiter orbital circumference) and also between the sun diameter and Jupiter
diameter. and this data shows a new proportionality between the sun mass and
Jupiter mass.
- Equation no. (J) shows that, Jupiter Uranus Distance has some role in the sun
creation (or perhaps Jupiter Uranus Motions Interaction does this role)
- Why does The Equation Use 2 Values Of The Sun Mass?
- We remember, Jupiter energy which is sent to Pluto used (2 solar days) where,
light supposed velocity (1.16 mkm/s) travels during one solar day a distance = the
solar planets orbital circumferences total, but Jupiter energy uses 2 solar days
which provides energy = 2 x the solar planets orbital circumferences total
- This is happened because of Neptune orbital circumference as we have discussed
before (equation no. 3). So if the used energy be double value, and the matter is
created of energy, that caused the sun mass be used in double value.

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- Based on different data, Jupiter motion shows its role in the sun creation.
- Jupiter Uranus Distance shows specific role in this process.
- The difficulty is that, We can't discover the geometrical mechanism by which this
process is done or this data be created. the data tells what's happened but doesn't
tell how that's happened. By that the argument provides clear meanings but not
clear mechanism of motions.
- One more important equation we need to remember is the following
- 90000 mkm = 2872.5 mkm (Uranus orbital distance) x (97/3.1)
- Where
- C2
= 90000 mkm (if the period of time be = 1 second)
- 97 deg = 97.8 deg (Uranus Axial Tilt) – 0.8 deg (Uranus orbital inclination)
- 3.1 deg = Jupiter Axial Tilt
- We need this data because the sun rays be created from the value (C2
)
- The data shows that, the interaction between Uranus and Jupiter effects on the sun
creation process.
- This data supports the meaning we get from the analysis of the distance 2094 mkm
(Jupiter Uranus Distance)
- Notice
- The machine depends on layers, So, the rate (97/3.1) uses the value 97
- But, other rates can use different data for example
- (97.8 deg /3.1 deg) x 2872.5 mkm = 90560 mkm
- (where 90560 days = Pluto Orbital Period)
- Also
- (98.6 deg /3.1 deg) x 778.6 mkm x 2 = 49528 mkm
- Where
- 98.6 deg =97.8 deg (Uranus axial tilt) + 0.8 deg (Uranus orbital inclination)
- 778.6 mkm = Jupiter orbital distance and 49528 km = Neptune diameter.

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The Conclusions
- Jupiter orbital circumference 4900 mkm effects greatly on the machine motion
- The distance 4900 mkm defines the moon diameter and its motion range and by
that Jupiter motion effect on the moon motion be defined clearly
- Also
- The distance 4900 mkm defines the sun diameter 1.392 mkm, and that be done
through interaction with Jupiter diameter.
- These effects are found by the energy provided for the sun rays creation process
- This energy is seen through the interaction between Jupiter and Uranus motions.

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5-The Solar System Distances Be Created In A Network Form
5-1 Preface
5-2 The Continuum effect Through the Solar System Distances
5-3 The Solar System Distances Distribution
5-4 The Solar System Distances Dependency On One Another

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5-1 Preface
- The Solar Planets data analysis shows that, The Planets Distances Be Created In A
Network Form.
- Means,
- The solar system distances are considered as one geometrical design or one
geometrical system. There's no one distance defined as independent distance or
created out of this one geometrical design.
- Let's suppose, this geometrical design be a square. So all dimensions in this
square be defined geometrically based on this square angles and data. means no
one dimension can be created independently from this square data because all of
them are parts in the same one geometrical design.
- Suppose we have a triangle its 2 angles are (60 degrees and 80 degrees) what's the
third one? (40 degrees). Can this angle be any thing else than (40 degrees)? NO it's
obligatory value defined by the geometrical basic rules
- Similar to that,
- The Solar System Distances Be Created Based On One Geometrical Design.
- But
- By using this vision we will have surprises. Because this vision tells (for example)
Venus Jupiter Distance be= 671 mkm Because Earth Jupiter distance = 629 mkm
- That creates another vision
- While we consider Venus is a planet and Earth is another planet and they move
independently from one another. We are astonished now with this fact that, their
distances to Jupiter be defined based on one another! How? And Why?
- How (Because The Solar System Distances Be Created In A Network Form)
- Why (this answer will be discussed in Mercury Motion Analysis Point)
- The fact is proved strongly that
- The Solar System Distances Be Created in A Network

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- There Are Many Basic Proves For This Fact which are:
- (1) The Continuum effect Through the Solar System Distances
- (2) The Solar System Distances Distribution
- (3) The Solar System Distances Dependency On One Another
- Let's discuss these proves in their details in following
- Notice
- The idea tells that,
- By energy the distances be created in a network form
- means
- The Distances Be Similar To The Railways And The Planets Be Similar To The
Trains Move On The Railways.
- The important point is that, the distance be created by energy and because of that
it's a defined trajectory of motion. that makes the planet moves in some obligatory
motion because the energy created the distances.
- Means, the distances aren’t common space but defined distances be built for
specific reason. Based on that (Distance = Energy)
- And
- The created distance defined a trajectory of motion
- As a result
- The solar system distances be as a map created by energy

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5-2 The Continuum effect Through the Solar System Distances
Point No. (I)
The Continuum Effect Description
I- Data
(1)
50% of all distances in the solar system be equal one another
(2)
40% of all distances in the solar system be rated one another
Let's provide this data in details in following:
Data Group No. (A)
Why These Distances Are Equal?
(1)
Saturn Orbital Distance = Saturn Uranus Distance
= Mars Orbital Circumference
= Pluto Neptune Distance (error 1.5%)
= Pluto eccentricity Distance (error 1.5%)
= Neptune Orbital Distance/π
= Uranus Orbital Distance /2
= Mercury Jupiter Distance x 2
(2)
Mercury Neptune Distance = Saturn Pluto Distance
Jupiter Pluto Distance = Uranus Neptune Circumference
Earth Neptune Distance = Mercury Saturn Circumference (0.5%)
(3)
Jupiter Mercury Distance = 2 Mercury Orbital Circumference
Jupiter Venus Distance = Venus Orbital Circumference (1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (1.2%)
(Earth and Jupiter at 2 different sides from the sun)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (0.8%)
Pluto Orbital Distance = Earth Orbital Circumference x 2π
(These Equal Distances Be Around 50% Of All Distances In The Solar System)

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Data Group No. (B)
Why These Distances Are NOT Equal?
1. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
2. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
3. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
4. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
5. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (0.6%)
6. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (0.8%)
7. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (0.7%)
8. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (0.7%)
9. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4864
= (0.5 %)
(These Rated Distances Be Around 40% Of All Distances In The Solar System)
Additional Data
(10)
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

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II- Discussion
- Let's suppose that,
- The solar planet moves interpedently from all other planets motions depending on
the sun mass gravity, in this case how these equal distances (50%) and the rated
distances (40%) be created?
- No logic can imagine a different reason for each 2 equal distances? We conclude
logically that (One Reason Caused These Distances To Be Equal)
- Based on that, we don't deal with independent distances instead we deal with one
machine be created by all distances integration. and As A Result, ONE REASON
effects on (the machine) caused (all distances to be effected by this one reason)
- I want to say,
- These 2 groups of data disproves The Planet Motion Independency Concept.
- There's no way to create the equal and rated distances in huge percentage if the
planets motions be independent from one another.
- But
- Why These Distances Are Equal?
- The answer is somehow complex and will be clear in the point no.(10) in which we
should discuss the equal distances method… but here let's provide a short answer
in following:
- The solar planets don't move by the sun mass gravity. But they move by equal
distances method (or at least the outer planets move by this method)
- The equal distances method is a method be used by many planets. Let's explain
how this method be used…. 2 Planets move equal distances in defined periods of
time (as example, Earth moves during its day period a distance =2.574mkm and
Pluto moves during its day period a distance = 2.598 mkm, these 2 distances are
different with 1% only and by that they can be considered equal distances. As a
result for these 2 motions equal distances Earth and Pluto motions data be in
proportionality with one another. this data we should discuss in point no.(6)

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- This feature be used hundreds of time by the solar planets, I guessed that, because
this method is used hundreds or even thousands of times in the solar system, this
method causes the planets motions (but I don't know how yet). The idea is that,
this method (the equal distances) works by a geometrical rule, as a waterwheel or a
lever, these methods work by geometrical rules. Similar to that the equal distances
method caused the planets motions
- As a result for using the equal distances method, 50% of all distances in the solar
system be created equal one another and also 40% of all distances be rated with
one another by the same one rate (1.0725)
- But
- Please note
- The equal distances be created because the solar system uses the equal distances
method as I have explained it before and as we will discuss and prove deeply in
point no. (10) of this paper.
- But
- Why Does The Rate (1.0725) effect On 40% Of All Distances?
- The rate (1.0725) effect on (40% distances) is a proof for a continuous effect
through the distances. There's Some Effect Like A Continuum effects on the
solar system distances!
- How that can be done if the planets move independently from one another? and
- If the distances be created independently from one another how one effect can be
continuous through (40%) of all distances?
- I try to prove, the distances are created in a network or by one geometrical design.
- means,
- the distances be created together in one geometrical system as one machine
because of that one effect can continue through (40%) of all distances
- the explanation of using this rate (1.0725) isn't our question…. Our basic notice is
that this effect has a continuous task through the solar system distances which can't

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be done except if these distances be created based on one geometrical system or in
one Network
- I wish the argument is powerful and proved strongly
- The data gives us a chance to show that, the planet motion can't be independent
from other planets motions. we here don't deal with independent data but with one
system of data.

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5-3 The Solar System Distances Distribution
(1st
Point) The Idea Explanation
- There are 2 basic features we should discuss about the solar planet motion, thee 2
features are:
- (1) The Planets Motions Use Different Rates Of Time
- (2) The Planets Motions Use The Distances As Periods Of Time
- These 2 features we should discuss and prove in this paper discussion
- I need here jut to explain how these features be used to test their effects on the
distances distribution
- There are different rates of time be used by the solar planets, for example:
- 1 hour of Mercury motion = 9.18 hours of Pluto Motion
- And
- 1 hour of Earth motion = 6.37 hours of Pluto Motion
- And also
- 1 hour of Pluto motion = 4.61 hours of The Moon Motion
- We will prove this data in the paper next discussion, but let's take a look on the
second feature in following… (for example)
- Mercury moves during its rotation period a distance = 243 mkm (error 1%)
- Venus moves during 365.25 days a distance =1106 mkm
- If 1mkm= 1 solar day
- 243 mkm will be = 243 solar days (Venus Rotation Period), and
- 1106 mkm will be = 1106 days =2π x 175.94 days (Mercury Day Period)
- These are just examples for hundreds of data be used similar to this way
- I want to say that, the planets motions use the distances values as periods of time
- But
- The first feature tells, different rates of time be used by the planets motions

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- As we have seen (1hour of Mercury motion = 9.18 h of Pluto motion),
- If different rates of time be used how the distances values be used as periods of
time?
- Means,
- 1 mkm be =1 solar day, or equal 9.18 solar days, or equal 6.37 solar days ……etc?
- How to use this feature (the using of distances values as periods of time)??
- Let's use the rate between Pluto motion and the Earth moon motion as example for
our discussion
- Notice
- 1 Hour Of Pluto Motion = 4.61 Hours Of The Moon Motion
- How do I know this rate?
- By their motions distances
- Pluto move during a solar day 406000 km
- and
- the Earth moon displacement per a solar day =88000 km
- (406000 km) = 4.61 x (88000 km)
- As A Result
- 708.7 hours (The Moon Day Period) = 4.61 x 153.3 hours (Pluto Day Period)
- And the rate (4.61) effects on many other data between Pluto and the moon
motions. this data we should discuss deeply in point no. (6) of this paper
- Now
- If the 2 features be used really by the planets motions. the rate (4.61) should be
found in the solar system distances distribution… because the distances be used as
periods of time and (1 hour of Pluto motion = 4.61 hours of the moon motion) and
that necessitates to create some distances based on this rate…
- Let's see the data in following

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(2nd
Point) The Rate (4.61) Effect On The Distances Distribution
I- Data
778.6 mkm (Jupiter Orbital Distance) = 4.61 x 170 mkm
550.7mkm (Jupiter Mars Distance) = 4.61 x 119.7 mkm
2094 mkm (Jupiter Uranus Distance) = 4.61 x 2 x 227.9 mkm
2872.5 mkm (Uranus Orbital Distance) = 4.61 x 629 mkm
3033.5 mkm (Uranus Pluto Distance) =4.61 x 654.9 mkm
5127 mkm (Jupiter Pluto Distance) = 4.61 x 2 x 550.7 mkm
5906 mkm (Pluto Orbital Distance) =4.61 x 1284 mkm
1375.6 mkm (Mercury Saturn Distance) =4.61 x 2 x 149.6 mkm
3062 mkm (Saturn Neptune Distance) =4.61 x 671 mkm
4345.5 mkm (Earth Neptune Distance) =4.61 x 940 mkm
4267.2 mkm (Mars Neptune Distance) =4.61 x 929 mkm
Where
170 mkm = Mercury Mars Distance
119.7 mkm = Venus Mars Distance
227.9 mkm = Mars Orbital Distance
654.9 mkm = Jupiter Saturn Distance
1284 mkm = Earth Saturn Distance
149.6 mkm = Earth Orbital Distance
940 mkm = Earth Orbital Circumference
929 mkm = Earth Jupiter Distance when the 2 planets be on the sun 2 sides
(Max error 1%)

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II- Discussion
- The solar system orbital and internal distances total be (45 distances)
- The previous data shows that (22 distances) are rated with the rate (4.61)
- Means
- 50% of all distances in the solar system be rated with one another based on this
same rate (4.61).
- The data proves the distances distribution depends on the planets motions rates
time.
- The data proves, The Solar System Distances Be Created As A Network.

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(3rd
Point) The Distances Distribution Proof.
I - Data
Equation no. (A)
1.16 mkm per second x (2) x 86400 seconds =(2) x 100733 mkm = (37100 mkm –
4900 mkm) x 2π = 28255 mkm + (2) x 86400 mkm
- Where
- 100733 mkm = The 9 Planets Orbital Circumferences Total
-
- This equation is the paper main equation and the solar system main equation and
the most important one generally
- This discussion we discuss deeply with Mercury Motion analysis because of the
light supposed velocity (1.16 mkm/sec) motion which is found in Part one of the
II- Discussion
- This equation is the paper main equation and the solar system main equation and
the most important one generally
- This discussion we discuss deeply with Mercury Motion analysis because of the
light supposed velocity (1.16 mkm/sec) motion which is found in Part one of the
equation.
- We discuss here only Part (3) and part (4)

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- (Part no. 3)
- (37100 mkm – 4900 mkm) x 2π
- Part no. (3) shows why the equation be a real one and not imaginary equation.
Because the difference in distance between Pluto orbital circumference (37100
mkm) and Jupiter orbital circumference (4900 mkm) creates this equation.
- The equation tells that
- The difference in distances between Pluto orbital circumference (37100 mkm) and
Jupiter orbital circumference (4900 mkm) be used as a radius in a circle and in this
case this circle circumference will be = 2 x 100733 mkm
- Where
- 100733 mkm= The 9 Planets Orbital Circumferences Total
- That shows there are 2 equal distances which are the difference between Pluto and
Jupiter orbital circumferences and the planets orbital circumferences total
- These are 2 equal distances in this equation
- But
- Let's see the part no. (4) of it in following
- (Part no. 4)
- 28255 mkm + 2 x 86400 mkm
- Where
- 28255 mkm =Neptune Orbital Circumference
- 86400 mkm=??
- Let's See The Next Equation
- 2 x 86400 mkm = (28255 mkm – 680 mkm) x 2π (Venus Equation)
- Where
- 680 mkm = Venus orbital circumference
- If we use Earth orbital circumference in place of Venus the error be (1%)

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- How to understand the part no. (4)
- (Part no. 4) (continued)
- 28255 mkm + 2 x 86400 mkm
- We had 2 equal distances which are the (32200 mkm x2π) and (2 x 100733 mkm)
- And now we have a third distance equal them which is
- (28255 mkm+ 2 x 86400 mkm)
- This equation part tells that distances between Neptune and Venus (1st
Trajectory)
and between Neptune and Earth (2nd
Trajectory), these distances + Neptune orbital
circumference be the third distance and = (2 x 100733 mkm)
- I try to show, these distances are distributed in proportionality with one another
- That proves the solar system distances be created in a network form and
distributed geometrically
- This equation will be discussed again wit Mercury Motion analysis. In this
discussion we should see more clear that these distances are distributed based on
geometrical rules.

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5-4 The Solar System Distances Dependency On One Another
I-Data
(1)
1.16 x 580 mkm = 671 mkm (Venus Jupiter Distance)
1.16 x 671 mkm = 778.6 mkm (Jupiter Orbital Distance)
1.16 x 629 mkm = 720.7 mkm (error 1%) (Mercury Jupiter Distance)
1.16 x 542 mkm = 629 mkm (Earth Jupiter Distance)
1.16 x 5127 mkm = 5906 mkm (error 0.7%) (Pluto Orbital Distance)
(2)
1.16 x 778.6 mkm x 2 = 1806 mkm
1.16 x 1806 mkm = 2094 mkm (Jupiter Uranus Distance)
1.16 x 2094 mkm = 2 x 1205 mkm (error 0.8%) (Mars Saturn Distance)
(3)
(1.16)2
x 170 mkm = 227.9 mkm (error 0.4%) (Mars Orbital Distance)
Where 170 mkm = Mercury mars Distance
(5)
1.16 x 1980 mkm = 2296.8 mkm
1.16 x 2296.8 mkm = 2644.6 mkm (error 0.7%) (Mars Uranus Distance)
1.16 x 2644.5 mkm = 3033.5 mkm (error 1%) (Uranus Pluto Distance)

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II-Discussion
- The previous distances are example for similar huge number of distances which
depend on the rate (1.16)
- We should analyze this rate (1.16) later
- But
- We need here to see how the distances be distributed base don this rate (1.16)
- The next data can help our analysis greatly
More Data
- 778.6 mkm (Jupiter Orbital Distance) = 1.0725 x 720.7 mkm (error 1%)
- 720.7 mkm (Jupiter Mercury Distance) = 1.0725 x 671 mkm
- 671 mkm (Jupiter Venus Distance) = 1.0725 x 629 mkm
- 629 mkm (Jupiter Earth Distance) = 1.0725 x 580 mkm (error 1%)
- This data also proves, these distances be created based on a geometrical design.
Because the same distances be rated on (1.0725 and 1.16).
- Please remember 1.16 =(1.077)2
The analysis shows that the solar system distances can't be created interpedently from
one another but created in a network form. And because of that the distances be
distributed based on a geometrical design and as a result the distances be rated with
the rates (1.16) and (1.0725)

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6- The Solar Planets Motions Depend On Different Rates of Time
6-1 Preface
6-2 Earth And Pluto Rate Of Time
6-3 Mercury Motion Rate Of Time
6-4 Saturn Motion Rate Of Time
6-5 Uranus And Jupiter Rate Of Time

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6-1 Preface
- The different rates of time is the logical result for light motion accompanying
planet motion.
- The matter is created out of light but be not separated from its light parent, on the
contrary, they move together one unified motion. but they use different rates of
time for their motions.
- The basic rate of time is (1 second of light motion = 1 solar day of planet motion)
- The solar planets divides this great rate (1s =86400 s) among their motions and
work as a great clock.
- The rate of time between Earth and the sun (1 s = 365.25 s) is produced through
this process.
- But
- How to prove that? for example If
- 1 hour of Mercury motion = 9.18 hours of Pluto motion, can we prove that?
- The rate of time is one data of the planets motion data, because of that we can
discover the rates of time by the planets motions data analysis.
Notice
- Mercury data analysis refers to the rate 1 hour of Mercury motion = 1 second of
light motion and
- 1 hour of Mercury motion = 9.18 hours of Pluto motion

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6-2 Earth And Pluto Rate Of Time
I- Data
Group no. (1) (Earth and Pluto Rate Of Time)
(A)
- Earth (29.8 km/s) moves during a solar day (24 h) a distance = 2574720 km
- Pluto (4.7 km/s) moves during its day (153.3) a distance = 2593836 km
- Pluto and the moon motion distances be equal but they are different from Earth
motion distance with (1%).
(B)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ earth day period) = 6.37 (0.8%)
(C)
- 1 hour of Earth motion= 29.53 hours of the moon motion (explained in discussion)
(1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
(2)
Tan (12.2) x 708.7 hours = 153.3 hours
(3)
Tan (13.17) x 655.7 hours = 153.3 hours
(4)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
(5)
413600 km cos (10.96 deg) = 406000 km
(6)
(10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)

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II- Discussion
- The Rate Of Time Depends On Planets Velocities Rate (A hypothesis)
- But, this rule sometimes be violated and in this case we can use the simple rule
tells (Equal Distances Causes Different Rates Of Time)
- 1 hour of Earth motion = 6.37 hours of Pluto motion
- This rate is concluded simply from Earth and Pluto velocities
Because
- Earth velocity (29.8 km/s) = Pluto velocity (4.7 km/s) x 6.37
- Also
- Pluto Day Period (153.3 h) = Earth Day Period (24 h) x 6.37
- The rate (6.37) which creates the rate of time between Pluto and Earth Motions,
effects greatly on the 2 planets motions data. As we have seen.
- Let's summarize the data idea in following:
- There's a proportionality between Pluto Data on one side and Earth and its moon
data on the other side. No Pure Coincidence can be claimed, because the
connection among the three planets be defined based on their positions to one
another. the question be (How This Data Be Created?)

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Equation no. (C)
1 hour of Earth motion = 29.53 hours of the moon motion.
- Because
- The moon displacements total during the moon day period (29.53 days) = Earth
motion distance during a solar day Period. (error 1%)
- Because of that
- 1 Day Of Earth Motion =29.53 Days Of The Moon Motion.
- We can see that, the machine works with consistency, for example
o 1 hour of Earth motion =6.37 hours of Pluto motion
o 1 hour of Pluto motion = 4.61 hours of the moon motion
o And
o 1 hour of Earth motion = 6.37 x 4.61 = 29.53 hours of the moon motion
o The rates of time are created based on each planet motion data and be in full
harmony as we see. That tells we deal with a geometrical mechanism works
consistency…
- Let's see the rate of time of Pluto and the moon motions

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Discussion (continued)
Equation no. (1)
(406000 km /88000 km) = (708.7 h/ 153.3 h) = 4.61
- Where
- 406000 km = Pluto motion distance per a solar day
- 88000 km = The moon displacement per a solar day
- Based on the rule (Equal distances creates different rates of time).
- This rate of time be concluded because Pluto moves during its day period (153.3 h)
a distance = the moon displacements total during its day period (29.53 days). By
that 708.7 hours of the moon motion = 153.3 hours of Pluto motion. And So, the
rate of time be concluded (1 h of Pluto motion =4.61 h of the moon motion)
- Again the rates of time causes (2 Planets All Data) to be in proportionality with
one another. in following equations we see that Pluto and the moon motions data
be in full proportionality which is very interesting data.
Equation no. (2)
Tan (12.2) x 708.7 hours = 153.3 hours
- Where
- Equation no. (2) shows that the moon and Pluto days periods are created based on
each other geometrically. The fact can't be denied. Pluto day period (153.3 h) is
created as a function in the moon day period (708.7 h).

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Equation no. (3)
Tan (13.17) x 655.7 hours = 153.3 hours
- Where
- 153.3 hours = Pluto Day Period and (153.3- ) = Pluto Rotation Period
- Equation no. (3) shows, Pluto rotation period is created depending on the moon
rotation period based on the angle (13.177 deg).
- The data of the 3 equations are strong and prove the idea clearly, because it's A
Directed Data. Means, this data be created based on geometrical mechanism be
found between Pluto and the moon motions. For that reason, the equations use (all)
data of the 2 planets. That shows the data is directed and can't be created by any
random process. It's data be created based on geometrical reasons and rules.
- The proportionality in data between Earth and its moon on one side and Pluto on
the other side shows that a geometrical mechanism must be found behind them.
Equation no. (4)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
- 708.7 hours = The moon Day period
- 10.7 hours = Saturn Day Period
- 9.9 hours = Jupiter Day Period
- The rate (4.61) we have defined, where (1 hour of Pluto motion =4.6 hours of the
moon motion).
- Equation no.(4) tells that this rate (4.61) is known by Jupiter and Saturn motions.
- That tells, the rate (4.61) is transported among the solar planets and that explains
how this rate be created between the moon motion and Pluto motion.

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Equation no. (5)
413600 km cos (10.96 deg) = 406000 km
- Where
- 406000 km = Pluto Motion Distance During A Solar Day
- Equation no. (5) shows that, the motions interaction between the moon and Pluto is
a deep interaction. And effects on the planet motion basic data. Because
- The moon displacements total during 29.53 days =2598693 km = 2π x 413600 km
- The distance (413600 km) should be the moon apogee orbital radius if the moon
apogee orbital circumference = the moon displacements total during 29.53 days.
- This is the fact which should be performed
- But
- The moon apogee orbital circumference be (2550973 km) less than the moon
displacements total during (29.53 days) with 2%. This fact arises the question
Why? the moon apogee circumference should be 2598693 km (surely). But for
some geometrical miracle this distance became =2550973 km!
- The radius 413600 km be changed into 406000 km and this change be done based
on the equation no. (5). The angle (10.96) caused this change. We have discussed
this angle before deeply because it's the main angle in the moon orbit geometrical
structure (we will review a short notice about it later). The angle (10.96 deg)
causes the radius 413600 km to be 406000 km – why?
Equation no. (6)
(10.96 deg /1.7 deg) = (153.3 h /24 h) (error 1%)
- The angle 10.96 deg be created based on the rate between Pluto Day Period
(153.3h) and Earth Day Period (24 h). this rate be used between the angle (10.96)
which causes to decrease the radius from 413600 km to 406000 km, and the angle
1.7 deg which is the moon daily motion angle in the moon orbital equation (θ1= θ0
+ 1.7 deg). (The moon orbital equation be in my previous paper)

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Notice (1)
- The moon displacements total 2598693 km should be the moon apogee orbital
circumference, by that, the moon apogee orbital radius should be =413600 km
- But
- It's not true, the moon apogee orbital radius =406000 km
- Why? because of the angle (10.96 degrees)
- Let's remember this angle in following:
o The moon orbital perigee radius =363000 km
o The moon orbital apogee radius = 406000 km
o If we draw these 2 radiuses in 2 circles as in the figure
o The tangent DB = 181843 km, also
o The perigee orbital circumference = 2.28 mkm
o (181843 km /2.28 mkm) = 0.08
o 137 degrees x 0.08 = (10.96 degrees)
o As we remember that 137 degrees = 95.1 deg x 1.44 deg where
o 95.1 deg =90 deg + 5.1 deg (the moon orbital inclination)
o 1.44 deg = the moon orbit regression degrees per month.
o We have supposed that the angle 137 degrees is created based on the moon
orbital inclination (degrees) and the moon orbit regression angle monthly.
And this angle (137 deg) is the basic angle in the moon orbital geometrical
structure (hypothesis).
o The calculations of the tangent DB = 181843 km aimed to explain one basic
idea which is, the moon orbit is built geometrically to create the rate (0.08)
by which the moon orbital inclination and regression will create the angle
(10.96 deg).
o Let's explain that more clearly

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o The moon orbit inclination and regression controls the moon orbital motion.
this is a hypothesis. Now, the moon orbital motion creates the moon orbit
based on a geometrical design, the moon orbital motion uses the moon orbit
geometrical design to create the angle (10.96 deg) from the basic one (137
deg). by that the original angle is (137 deg) which is created by the moon
inclination and regression unification, and the born angle is (10.96 deg)
which is the motor of the machine.
o What we need to refer to is that, the moon orbit geometrical design is used
in this process. That's why the distance 413600 be decreased and became
406000 km. By that, the main player is the angle (10.96 deg).
- This explanation answers the question,
- Why Pluto Motion Distance During A Solar Day = The Moon Apogee Orbital
Radius=406000 km?

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Notice (2)
- The previous data still have another reading let's summarize it in following:
o Pluto day period is a very long day period in comparison with the outer
planets days periods, Why?
o We notice also that, Pluto Moves During Its Day Period A Distance =
Earth Motion Distance During Earth Day Period. That means, Pluto Day
Period be extending to perform this requirement.
o i.e.
o Pluto Day Period be extending to be 153.3 h to enable Pluto to move a
distance = Earth motion distance during Earth day period.
o Because of that,
o Pluto day period (and rotation period) be created as functions in the Earth
moon day and rotation periods as we have seen in the data discussion.
o i.e.
o The interaction between Pluto motion on one side with the Earth and its
moon motions on the other side caused the following results:
o (1st
Result) the moon apogee orbital radius be decreased from 413600 km to
be 406000 km
o (2nd
Result) Pluto Day Period be extending to be =153.3 days.
Notice (3)
- Uranus effect is the main force behind the changes for the 2 planets data.
- Uranus effect on Pluto is proved by Planet 8 days Cycle where Uranus uses Pluto
day period in its motion.
- Uranus effect on the moon is proved by the fact, Uranus orbital circumference =
19.2 Earth orbital circumference. The moon moves a cycle called Metonic cycle
and extends for 19 years. Metonic Cycle is caused almost by Uranus effect on the
moon motion.

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6-3 Mercury Motion Rate Of Time
I- Data
(1)
112.2 km/s = 4.7km/s x 23.9
(2)
1407.6 hours =153.3 hours x 9.18
(3)
90560 days = 4222.6 hours x 9.18 x 2.33 (error 2%)
(4)
(2 x 153.3 h x 3600 s) /5040 s =23.9 x 9.18
II- Discussion
Equation (1)
112.2 km/s = 4.7 km/s x 23.9
- Where
- 112.2 km/s = The 3 Planets Velocities Total
- 4.7 km/s = Pluto Velocity
- Let's suppose, the 3 planets velocities (Mercury, Venus and Earth) be unified in
one velocity =112.2 km/s, So this velocity works as one gear and Pluto velocity
works as another gear, move relative to one another.
- By this supposed motion of 4 planets, the rate (1 hour = 23.9 hours) be created and
that causes 1 hours of a planet motion be = 23.9 hours of another planet motion.
- This process has a great effect on the solar planets rates of time because if the
planets motions rates of time depend only on the planets velocities that makes the
max rate is (10.08). Means, 1 hour of a planet motion be = 10.08 hours of another
planet motion. And this is a very narrow range of the rates of time. The 3 planets
velocities total creates an extension for this process and enables the system to
create the rate (1h = 23.9 h).

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Equation (2)
1407.6 hours =153.3 hours x 9.18
- Where
- 1407.6 hours = Mercury Rotation Period
- 153.3 days = Pluto Day Period.
- (9.18) = The rate of time between Mercury and Pluto motions.
- We conclude that mercury and Pluto motions rates of time is (9.18) based on this
equation. Means
- 1 hour of Mercury Motion = (9.18) hours of Pluto Motion.
- By this same method we conclude the rate of time between Pluto and the moon
motions. It was the rate between the moon day period and Pluto day period. But
here the rate (9.18) is between Mercury Rotation Period and Pluto Day Period
(where Pluto day period =153.3 h and Pluto rotation period = -153.3 h)
Equation (3)
90560 days = 4222.6 hours x 9.18 x 2.33 (error 2%)
- Where
- 90560 days = Pluto Orbital Period
- 4222.6 hours = Mercury day period
- 2.33 = ??
- Mercury velocity (47.4 km/s) = 10.08 x Pluto velocity (4.7km/s). if the planets
rates of time depend on their velocities. That means 1 hour of Mercury motion =
10.08 hours of Pluto motion. and in this case 24 hours of Pluto motion = 2.37
hours of Mercury Motion.
- That because the equation uses (Pluto orbital period 90560 solar days) in
comparison with 4222.6 hours (Mercury day period). The equation proves that the
rate (9.18) is the used one between the 2 planets.

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Equation (4)
(2 x 153.3 h x 3600 s) /5040 s =23.9 x 9.18
- Where
- 5040 seconds are required for Mercury day period to be 176 solar days
- 153.3 h = Pluto Day Period.
- 23.9 = the rate between the 3 planets velocities total (112.2 km/s) and Pluto
velocity (4.7 km/s).
- This equation also proves that the rate of time between Mercury and Pluto is (9.18)
- The data proves that, 1 hour or Mercury motion = (9.18) of Pluto motion.
Notice
- Different data of Mercury and Pluto Motions shows that the rate of time between
them is (9.18)
- Because this rate is used in different data in the 2 planets motions data, we
conclude that it's a fact and real one.
- I wish the proves show its power clearly. Because we deal with geometrical data.
We find for example a rate (9.18) between Mercury and Pluto motions data and
this rate be used between them frequently. That happens also with Earth (6.37) and
with the moon (4.61).
- The point is that, the rate of time has a massive effect on planets motion but we
still need to discover this effect clearly.
- It's noticeable to take into consideration the method by which the rate of time
effect on Planet Motion.
Notice
- If Mercury and Pluto motions rate of time be defined based on their velocities
only, The rate will be 10.08 (means 1 h of Mercury motion = 10.08 h of Pluto
motion). the rate (9.18) effect only on Pluto and Mercury motions but no other
planets. For example (Mercury 1 hour =46.1 h of The Earth moon motion), in this
case (46.1 = 10.08 x 4.61) , the data doesn’t use (9.18) but (10.08).

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6-4 Saturn Motion Rate Of Time
I- Data
(1)
Saturn moves daily 838080 km = the Earth moon displacement 88000 km x 9.5
(2)
Saturn diameter 120536 km = Earth diameter 12756 km x 9.5
(3)
Saturn orbital distance 1433.5 mkm = Earth orbital distance 149.6 mkm x 9.5
(4)
Earth axial tilt 23.4 deg = Saturn orbital inclination 2.5 deg x 9.5 (error 1.7%)
(5)
Saturn Orbital Period 10747 days = 365.25 days x 29.53 days
II-Discussion
- Equation no.(1) shows that, the rate of time be produced as a rate between Saturn
motion distance per a solar day and the moon displacement per a solar day (9.5)
and this rate controls different data between Saturn on one side and Earth with it
moon on the other side.
- This feature is similar perfectly to Pluto data proportionality with Earth data.
- And also the data between Pluto and Mercury.
- The data tells that, the rate of time has a massive effect on the planets motions
data. This feature which we see frequently in different planets data.
- The data (1, 2 ,3 and 4) shows that the rate of time (9.5) creates a proportionality
between Saturn and Earth motions data.
- Let's discuss Equation No. (5) in following
Equation no. (5)
Saturn Orbital Period 10747 days = 365.25 days x 29.53 days
- We know that,
- 1 day of Earth Motion = 29.53 days of the moon motion
- And also we know

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- 1 day of the sun motion = 365.25 days of Earth Motion
- Based on that,
- 1 day of the sun motion = 365.25 x 29.53 days of The moon Motion, means
- 1 Day Of The Sun Motion = 10747 Days Of The Moon Motion
- Where
- Saturn Orbital Period =10747 days
- How this data be created?
- More interesting data be in our waiting
- The Earth orbital circumference =940 mkm but the moon daily displacement be
=88000 km, that means
- If the moon passes Earth orbital circumference (940 mkm) by using it daily
displacement (88000 km) the moon needs a period = 10747 days (error 0.6%)
- This data tells, the period 10747 days be defined regardless Saturn motion. or in
better words, the period 10747 days be defined before Saturn Creation.
- The data forces us to guess that, the Earth moon must be created before Saturn
creation. and as a result for the interaction with the moon motion Saturn uses the
period be defined by the moon motion. (also Saturn orbital distance =1433 mkm =
Mars orbital circumference. That tells may some interaction of motions between
Saturn and Mars caused Saturn to use mars orbital circumference as its orbital
distance. And if we put this data beside each other we may conclude that Saturn is
created after Mars migration and the moon creation).
- One more data can help our investigation
- The moon daily motion distance = Earth daily motion distance =2.574 mkm
(because they are not separated from one another)
- The moon distance be effected by the rate 1.0725 and because of that (2.574 mkm
=1.0725 x 2.4 mkm) the moon motion distance daily be =2.4 mkm
- During 10747 days the moon moves (2.4 mkm) a total distance =25920 mkm

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6-5 Uranus Rate Of Time
Notice
- Mercury Uranus Distance = 2815.2 mkm
- Mercury rotation period = 1407.6 hours
- 2 x Mercury rotation periods = 2815.2 hours
- If 1 mkm = 1 hour, So the distance 2815.2 mkm be = 2 Mercury rotation periods
And
- Mercury (47.4 km/s) moves during its rotation period (1407.6 h) a distance = 243
mkm (error 1%)
- 243 solar days = Venus rotation period
- If each 1 mkm be =1 solar days, the distance 240 mkm be = 243 days (error 1%)
Based on that
- Mercury as a player uses (1 mkm = 1 hour) and (1 mkm = 1 day =24 hours)
- That creates the rate (1 hour = 24 hours)
- But between whom? Almost
- Between Mercury and Uranus, means
- 1 hour of Mercury Motion =24 hours of Uranus Motion.
Notice
- 1 h of Pluto motion=2.62 h of Jupiter motion =2.62 h of Uranus motion.
- The rate 2.62 is so important for Jupiter and Uranus because
- Uranus Circumference 160592 km x 2.62 = 421056 km (Uranus Motion
Distance During Its Day Period)
- Jupiter Circumference 449197 km x 2.62 = 1160000 km (Jupiter Motion
Distance During Mars Rotation Period) (error 1.4%)
- This data we have discussed before.

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7- Planets Unified General Motion
7-1 The Planets Unified General Motion Description
7-2 The Solar Planets Motions Are Complementary One Another
7-3 The Solar Planets Move An Unified Motion (A Team Motion)

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7-1 The Planets Unified General Motion Description
- I suppose that (The Planets Motions Create One Unified General Motion)
- This hypothesis tells the planets motions be similar to Train Carriages Motion
- Also It tells the planets motions be similar to Chess Pieces Motions
- The hypothesis discovers One Law controls the solar system motion and data.
- The planets unified general motion forces each planet motion to be complementary
with other planets motions to perform The Unified General Motion. as a result,
The Planet motion be An Obligatory Motion.
- The Solar Planet creates its data to be in consistency with its motion.
- The planet can't change its motion, but can change its data to be in consistency
with its motion. And the planet data be created in consistency with its motion.
- Because the planet motion be complementary with other planets motions.
- The planet data be created complementary with other planets data
- Based On This Vision
- The solar planets data be created complementary to other planets data and based
on that the planets data be created depends on One Geometrical Design.
- And
- The Planets Creation And Motions Data Be Controlled By One Equation Only
Notice
- We discuss again a part of the 3 Planets Motions data because we add more data
and extensive discussion for it (The 3 planets are Mercury, Mars and Jupiter)

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7-2 The Solar Planets Motions Are Complementary One Another
I-Data
(The Interaction between Mercury, Mars and Jupiter)
(No. 1) (Mercury and Mars Motions)
- Mercury moves during its day period (4222.6 h.) a distance = 720.7 mkm
- Mars moves during (2802 hours) a distance = 243 mkm
- Mercury moves during (1407.6 hours) a distance = 243 mkm (error 1%)
- Mars moves during 346.6 d. a distance =720.7 mkm (Mercury Jupiter Dis.)
- Mercury moves during 346.6 d. a distance =1419 mkm (with 1433 error 1%)
- Mars moves during 687 d. a distance = 1433 mkm (Mars orbit. Circum)
- Mercury moves during 687 d. a distance =2815mkm (Mercury Uranus Dis.)
- Mars moves during 4331d. a distance = 9010mkm (Saturn orbit. Circum)
- Mercury moves during 4331d. a distance = 2815 mkm x 2π
- Mars moves during 224.7 d. a distance = π x 149.6 mkm (Earth orb. Dis)
- Mercury moves during 224.7 d. a distance = 920 mkm (with 928 error 1%)
- Mars moves during 365.25 d. a distance = π x 243 mkm (0.5%)
- Mercury moves during 365.25 d. a distance = 2 x 748 mkm
- Mars moves during 5040s. a distance= 121464 km= Saturn Diameter (+1%)
- Mercury moves during 5040s. a distance= 238896 km= 2 Saturn Diameters (-1%)
Where
1407.6 h = Mercury Rotation Period 2802 h = Venus Day Period
346.6 days = the nodal year 224.7 days = Venus orbital period
687 days = Mars Orbital Period
4331 days = Jupiter orbital period = 2π x 687 days

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(No. 2) (Mars And Jupiter Motions)
- Mars moves during 636.7 days a distance =1325 mkm= (Venus Saturn Dis.)
- Jupiter moves during 636.7 days a distance =720.7 mkm
- Mars moves during 778.6 d. a distance = 1622 mkm = Uranus Neptune Dis.
- Jupiter moves during 778.6 d a distance= 881 mkm =58 days x 15.19 mkm.
- ( 15.19 mkm = the 9 solar planets motions distances total per a solar day).
- Notice
- 2.082 mkm / day x 778.6 days = 1622 mkm = 1.1318 mkm /day x 1433 days

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(No. 3) (Mercury And Jupiter Motions)
- Mercury moves during 1433 d. a distance =5848 mkm= Mercury Pluto Dis.
- Jupiter moves during 1433 d a distance =1622 mkm=Uranus Neptune Dis.
- Mercury moves during 4222.6 h a distance = 720.7 mkm
- Jupiter moves during 4222.6 h a distance = 200 mkm = 629 mkm/π
- Mercury moves during 550.7 d a distance = π x 720.7 mkm (0.4%)
- Jupiter moves during 550.7 d a distance = 629 mkm (error 1%)
Please Remember,
Data Group No. (1)
- Mars moves during 4331 x π d a distance= 28255 mkm= Neptune orbital circumference
Data Group No. (2)
Data Group No. (3)

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II-Discussion
- The planets move defined distances in defined periods of time.
- The Defined Distance means a distance is known in the solar system, as to be any
planet orbital distance or a distance between any 2 planets.
- The Defined Periods Of Time means a period of any planet cycle, as 365.25 days
(Earth orbital period) or any planet orbital period. Or any planet rotation period or
any planet day period. All these periods are defined periods of time
- The argument tells that
- If the solar planets move their motions independently from one another, based on
that, the planets motions during defined periods of time should pass (random)
distances. Because these Planets are independent in their motion from one
another. (For example) Mercury motion depends on its orbital period (88 days) and
doesn't interest neither for Mars orbital period (687 days) nor for Jupiter orbital
period (4331 days) and by that, Mercury during these periods (687 days and 4331
days) should move some random distances (not defined in the solar system
distances).
- The data disproves Planet Motion Independency Concept. because the (3)
planets move defined distances in defined periods of time. These planets motions
aren't independent from one another. These motions are done based on One
Geometrical Design. And these motions be are similar to Chess Board Pieces
Motions. Each Motion is calculated geometrically and be obligatory.

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- Notice
- The previous analysis of motions prove the interactions of motions among the
three planets (Mercury, Mars and Jupiter). These interactions we have referred to
them during our analysis for Jupiter orbital circumference (4900 mkm).
- The 2 parts of discussions can provide a clear vision for how the planet motion be
done. Because the planets motions data help us greatly and protect us from any
imaginary vision. Because of that, the motions interaction which we have
concluded based on the planets data analysis concerning Jupiter orbital
circumference be proved here clearly by the 3 planets motions analysis
- It's simply a fact
- If the geometrical machine based on which this data be created be discovered the
reasons will be known to answer how or why these motions have interactions.
- But even before this machine discovery the data shows the motions interactions in
both sides. In the planets data and in the planets motions.

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7-3 The Solar Planets Move An Unified Motion (A Team Motion)
- Let's Summarize The Proof Idea In Following:
- This proof depends on 2 planets motion cycles are discovered later. Which are
(planet 8 days cycles) and (planet 6 days cycles). We study one Cycle only in this
paper because both cycles provide the same argument.
- The Cycle Shows That Different Planets Motions Be Used In The Same One
Cycle To Create One Final Result.
- Means,
- Many planets move relative one another to create one result.
- For Example
- Jupiter moves a distance (A) and Uranus moves a distance (B).
- The different distance between (A) and (B) = Jupiter Diameter
- Now
- The machine uses this behavior frequently, and (the different distance = Jupiter
Diameter) be created frequently while the cycle uses its different periods (8 days,
16 days, 24 days….etc).
- Based on that,
- We can't consider these planets motions are independent from one another because
the different distance be defined by the 2 planets motions. we have to consider
these planets motions as a team motion. They move relative to one another which
proves The Unified General Motion Concept.
- Based on that,
- (Planet 8 and 6 Days Cycle) disproves Planet Independent Motion Concept.
- In following we study Planet 8 Days Cycle

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(Planet 8 Days Cycle)
(I)
- Jupiter (13.1 km/s) moves during its day period (9.9 h) a distance = 466884 km
- But
- 466884 km = 449197 km (Jupiter Circumference) + 17687 km
- Where
- (8 x 17687 km = 141496 km (Jupiter Diameter) (error 1%)
- Based on that, we have concluded that, Jupiter has a cycle of 8 days
- Jupiter (13.1 km/s) moves during 8 Jupiter days (79.2 h) a distance = 3735072 km
- (3735072 km= 8 Jupiter circumferences + 141496 km (Jupiter diameter) (1%)
(II)
- The distance 3735072 km be passed also by Saturn and Neptune with a rate 80%
depends on one another as following:
- Saturn (9.7 km/s) moves during 10 Saturn days (107 h) a distance = 3736440 km
- (10 Saturn Circumferences = 3786750 km, the difference =50310 km = Uranus
Diameter error 1.5%)
- Neptune (5.4 km/s) moves during 12 Neptune days (193.2 h) a distance = 3755808
km
- (24 Neptune Circumferences = 3734323 km, the difference =21485 km = Mars
Circumference (error 0.6 %))
(III)
- Uranus (6.8 km/s) moves during Pluto day period (153.3 h) a distance = 3752784
km
- The distance 3752784 km = Jupiter motion distance during 8 days + 17687 km
- And because
- 17687 km x (8) = 141496 km (Jupiter Diameter) (error 1%)
- That tells another Cycle is found between Uranus and Jupiter based on 8 Pluto
days

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- That means, the distance be passed by Uranus during 8 Pluto days equal the
distance be passed by Jupiter during 64 Jupiter days and equal the distance be
passed by Saturn during 80 Saturn days and equal the distance be passed by
Neptune during 100 Neptune days
Let's see that in following
(1)
Jupiter (13.1 km/s) moves during (64 Jupiter days) a distance =29880756 km
(2)
Saturn (9.7 km/s) moves during (80 Saturn days) a distance =29891520 km
(3)
Neptune (5.4 km/s) moves during (100 Neptune days) a distance =31298400 km
(4)
Uranus (6.8 km/s) moves during (8 Pluto days) a distance =30022272 km
Comments
- Uranus motion distance (30022272 km) – Jupiter motion distance (29880756 km)
= 141496 km (Jupiter Diameter)
- The differences between these distances are less than 1 % (generally) and based on
that we can't consider they are different distances but we have to consider they are
equal distances.
- Although still there are small differences which are found for geometrical reasons
for example the difference between Jupiter and Saturn motions distances =
29880756 km – 29891520 km = 10921 = the moon circumference
- The data shows Planets Motions Dependency, because the different distances are
defined geometrically and that means these aren't 2 different distances of 2 plants
independent motions. On the contrary, the 2 distances are planned geometrically
and the 2 planets are 2 players to perform one different distance.

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The Discussion
- Let's discuss the previous data
(A)
- The outer planets are 5 planets, they consist 2 teams,
- The first team is consisted of Jupiter, Saturn and Neptune, these 3 planets move
based on a cycle (8 days cycle) depends on Jupiter motion with the rate 80%,
- That means
- The distance be passed by Jupiter in 8 Jupiter days be equal the distance be passed
by Saturn in 10 Saturn days and equal the distance be passed by Neptune during 12
Neptune Days
- The (small) difference between these 3 distances have geometrical necessities, as
we have seen in the difference between Jupiter and Saturn motions distances which
= 10921 km = The Earth Moon Circumference
- The moon circumference itself tells that it's a cycle because if it's not a cycle we
would find a part of the moon circumference
(B)
- The second team is Uranus and Pluto….
- Uranus uses Pluto day period (153.3 hours), and by that, Uranus (6.8 km/s) moves
during Pluto day period (153.3 hours) a distance = 3752784 km
- Because
- 3752784 km = Jupiter motion distance during 8 Jupiter days +17687 km
- Because of this data, we have concluded that, these motions depends on (8 days
Cycle), because
- Uranus needs to move during a period (= 8 Pluto days) to cause this value (17687
km) be = (141496 km (Jupiter Diameter) (1%)
- Because of Jupiter diameter we conclude that Uranus has a cycle of (8 Pluto days)
- Based on that

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- Uranus motion distance during 8 Pluto days = Jupiter motion distance during 64
Jupiter days = Saturn motion distance during 80 Saturn days = Neptune motion
distance during 100 Neptune days.
- How (Planet 8 Days Cycle) Can Prove The Unified Motion?
- Because
- Many planets motions be done to produce One Result
- This result be the different distance 141496 km (Jupiter Diameter) (1%)
- If we deal with planets independent motions this different distance can't be created
regularly and the Cycle can't be defined.
- Because (Planet 8 days Cycle) be defined, that means, the different distance
141496 km (Jupiter Diameter) be defined regularly which can be done only if
we deal with a team motion and NOT Planets independent Motions.
- Why does Uranus depend on Pluto Day Period?(additional question)
- Pluto day period is so long (153.3 h) in comparison with the outer planets days
periods. We suppose that, Uranus Motion effect on Pluto motion causes Pluto day
extension. We know Uranus did this effect because Pluto orbital inclination = 17.2
deg but Uranus day period =17.2 hours, even if we can't catch the mechanism of
this process yet, but the data shows that's Uranus effect on Pluto motion.
A Conclusion
Planet 8 Days Cycle disproves The Planet Independent Motion Concept,
On The Contrary, The Planets Move As A Team. (A Unified General
Motion)

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8- The Sun Age Historical Proves
8-1 The Sun Circles The Earth
8-2 The Rate (1.0725)
8-3 The Sun Diameter Analysis
8-4 The Sun Rays Analysis

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8-1 The Sun Circle The Earth
- The sun gives one face always to the Earth during the motion. although Earth
revolves around the sun a complete revolution, we always see the same face of the
sun, that tells the sun rotates with us giving The Earth The Same face always
- Can this motion refer to the Sun dependency on the period (365.25 days = Earth
orbital period)?
- Can that lead us to conclude, the sun be created after the Earth and because of that,
The Sun motion takes into consideration the period 365.25 days?
- Also
- The Sun Behavior is similar perfectly to the moon behavior, both give the same
face to the Earth during the motion.
- Why we accept that the moon be created after The Earth Creation but The Sun be
Created Before The Earth Creation? if the 2 behaviors are similar it may prove
both (the sun and the moon) be created after The Earth Creation.

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8-2 The Rate 1.0725
I- Data
(A)
(1)
(2)
Earth Neptune Distance = Mercury Saturn Circumference (error 0.5%)
(3)
Jupiter Venus Distance = Venus Orbital Circumference (error 1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (error 1.2%)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (error 0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (error 0.8%)

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(B)
More Data
10. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
11. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
12. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
13. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
14. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (Error 0.6%)
15. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (Error 0.8%)
16. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (Error 0.7%)
17. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (Error 0.7%)
18. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4846
= (Error 0.5 %)
(10)
Notice
4846 mkm =the inner planets orbital circumferences total. it will be discussed with Jupiter
orbital circumference analysis
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

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II – Discussion
- We have discussed The previous data before
- 50% of all distances in the solar system be equal one another, and
- 40% of all distances in the solar system be rated one another with the same rate
(1.0725)
- We answered why 50% of distances are equal and 40% are rated
- Let's ask
-
- How this data can prove that (the sun I created after all planets creation and
motion)? let's answer in following
- Let's consider this question in following:
o The usual case is the equal distances as the data in Point no. (A), That
means, these rated distances were equal distances one another before and all
of them be effected by the rate 1.0725,
o Means,
o We have some historical order here, which is
o The distances were equal, but
o Some change caused these equal distances to be rated with (1.0725)
o And, 1.0725 = (100/100) + (7.25/100)
o We can conclude that,
o The distances were equal till The Sun Creation, and after the sun creation,
the sun obliquity on ecliptic (7.25 deg) caused the equal distances to be
rated with one another by the same one rate (1.0725).
o Means, the rate 1.0725 proves that, The Sun Is Created After All Solar
Planets Creation.
o One more feature supports this powerful argument, that

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o The same rate (1.0725) is used for different distances. Because the sun
effects on all solar planets. That shows the rate (1.0725) must be related to
the sun because it's the same one rate. If this rate be belonged to any planet,
it will not effect on different distances equally because the sun only effects
on the planets equally. That shows the rate 1.0725 is belonged to the sun effect.

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8-3 The Sun Diameter Analysis
I-Data
1- Diameters
- (6792 km Mars Diameter) / (4879 km Mercury Diameter) =1.392
- (4879 km Mercury Diameter) / (3475 km the moon Diameter) =1.392 (1%)
- (71492 km Jupiter Radius) / (51118 km Uranus Diameter) =1.392
2- Distances
- (57.9 mkm Mercury orbital distance) /(41.4 mkm Venus Earth distance)=1.392
- (108.2 mkm Venus orbital distance) /(78.3 mkm Earth Mars distance)=1.392
- (149.6 mkm Earth orbital distance) /(108.2 mkm Venus orbital distance)=1.392
- (4267 mkm Mars Neptune distance) /(3061 mkm Saturn Neptune distance)=1.392
- (778.6 mkm Jupiter orbital distance) /(550.7 mkm Mars Jupiter distance)=1.392
- (550.7 mkm Mars Jupiter distance x 2) /(778.6 mkm Jupiter orbital distance) =1.392
- (5127 mkm Pluto Jupiter distance) /(3716 mkm Jupiter Neptune distance) =1.392
(error 1%)
- (5906 mkm Pluto orbital distance) /(4267 mkm Mars Neptune distance) =1.392
- (2872.5 mkm Uranus orbital distance) /(2094 mkm Jupiter Uranus distance) =1.392
3- Periods
- (243 days Venus Rotation Period) /(175.94 days Mercury day period) =1.392
- 2.5 deg (Saturn orbital inclination) /1.8 deg (Neptune orbital inclination) =1.392
- 1.8 deg (Neptune orbital inclination) / 1.3 deg (Jupiter orbital inclination) =1.392
- 5.1 deg (The moon orbital inclination) / 3.66 =1.392
4- The Sun Diameter
- 1.392 mkm (The Sun Diameter) = Venus Diameter 12104 km x 115.2
- 1.392 mkm (The Sun Diameter) = Mars Diameter 6792 km x 205.2
- 1.392 mkm (The Sun Diameter) = Neptune Diameter 49528 km x 28.3

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II- Discussion
- The data groups no. (1,2 and 3) show that,
- The rate 1.392 is used frequently among planets different data.
- Why? because
- The solar system is similar to one machine of gears and each planet is one gear in
it. because the machine moves one unified motion, the gears data should be in
proportionality with one another.
- Similar to that, the planets creation and motion data be created in proportionality
with one another. we will prove that in details in thus paper discussion.
- Based on this vision the solar planets data be created depends on one geometrical
design.
- The Sun Diameter Be defined geometrically before the sun creation.
- Because
- The machine depends on one geometrical design and each part should be created
in harmony with this geometrical design otherwise the machine can't work.
- We should notice that, the rate 1.392 is defined with the solar system creation
geometrically, Before the sun creation and also before the planets migration.
- This definition is done geometrically, so, the value (1.392) in the data is very
ancient. But the sun diameter (1.392 mkm) be created later.
- Let's discuss the data no. (4)

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Group Data No. (4)
- The sun diameter 1.392 mkm = Venus diameter 12104 km x 115.2 = Mars
diameter 6792 km x 205.2, But
- 25.2 + 90 = 115.2 and 115.2 +90 =205.2 where (25.2 degrees = Mars Axial Tilt)
And
- The sun diameter 1.392 mkm = Neptune diameter 49528 km x 28.3
- And 28.3 degrees = Neptune Axial Tilt
Discussion
- The data shows that, Venus, Mars and Neptune diameters are rated with the sun
diameter based on Mars axial tilt and Neptune axial tilt, Means,
- We have 3 planets diameters are rated with 2 planets axial tilts based on the sun
diameter.
- More logical that, the sun diameter is created after the 3 planets diameters,
depending on their values, causing planets axial tilts be created in proportionality.
- The opposite choice can't be working. Because if the sun be created at first and all
planets later, that creates a rule for the sun effect on the planets axial tilts, by that
we have to find (not) 3 planets only their axial tilts be rated based on the sun
diameter proportionality with their diameters, but found all planets behave by this
same rule because the sun has equal effect on all of them.
- The first option is more logical, that the sun is created after the planets creation and
the sun depended on 3 planets (specifically) for its diameter creation. by that the
data be more understandable
- Shortly
- The sun diameter analysis gives support for the claim (the sun is created after all
planets creation and motion).
Notice (1)
- In kepler third law in which (D3
= P2
x constant), the constant used for the solar
planets orbital distance and periods should be = 25

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8-4 The Sun Rays Analysis
- Let's Summarize The Sun Rays Creation Idea In Following:
(Point No. A)
o The Solar Planets Move One Unified General Motion. This Idea we discuss
and prove clearly in this paper.
o The planets unified general motion causes to unify the planets velocities in
one unifies velocity = the planets velocities total.
o The 9 planets velocities total =176 km/sec,
o But
o The Earth moon is a player and its velocity should be added, where the
moon velocity be considered = Earth velocity because they aren't separated
from one another through their motions.
o The 10 planets velocities total be 176 +29.8 =205.8 km/s
o Based on this velocity (205.8 km/s) the sun rays be created
(Point No. B)
o 10921 km (The Moon Circumference) x 86400 Seconds = 940 mkm
o This Equation tells. If Earth revolves around the sun a complete revolution
(=940 mkm = Earth Orbital Circumference) in one solar day only (86400
sec), in this case, the moon circumference (10921 km) will be equal the
Earth motion distance during 1 second.
o But, because
o Earth revolves around the sun in one year (=365.25 days), so this equation
may refer to a rate of time is used by The Sun herself.
o Let's use this idea as a hypothesis. where
o 1 Day Of The Sun Motion = 365.25 Days Of Earth Motion
o We suppose that's The Sun Rate Of Time

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(Point No. C)
o The velocity (205.8 km/s) passes during a solar day (86400 s) a distance
=17.75 mkm
o 1 day of Earth motion = 237 seconds of the sun motion.
(This value depends on our hypothesis)
o Now
o The distance =17.75 mkm and the time =237 seconds what's the velocity?
o 75000 km/s = (a quarter Light Velocity) = 0.25 C
o But
o Earth has a cycle of 4 years (1461 days = 365 +365 +365 +366)
o Based on one year, the velocity be (0.25 C)
o And
o Based on the cycle (4 years), the velocity (0.25C) became (C light velocity)
o Based on this explanation the sun rays be created.
(Point No. D)
o There's one more method to create the sun rays
o 0.25 C x 4 C = C2
o Where
o The velocity 0.25 C is produced already in the previous discussion
o We need only the value (4C) to perform this equation
o Jupiter motion enjoys by light motion supposed velocity (1.16 mkm/s) this
value is very near to our required value (4C)
o Light supposed velocity travels for 1 second and passed 1.16 mkm
o And
o Earth rotates around its axis one per solar day and move a distance = its
circumference =40080 km in a solar day
o If 1 solar day of Earth motion be = 1 second of light supposed velocity
motion.

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o Based on that, the distances can be added
o 1.16 mkm + 40080 km = 1.2 million km = 4C
o By that the equation works
o 0.25 C x 4 C = C2
o That explains why, Jupiter and The Moon data are the 2 basic players in the
sun data definition. Because the sun creation depends on Earth and Jupiter
motion. in addition to Uranus motion which is seen in the Earth moon
motion.
o The data explains how the sun rays be created
(Point No. E)
o The rate of time (1 solar day of planet motion = 1 second of light motion) is
the original rate of time in the solar system. We should refer to that in the
planets unified general motion discussion.
o The idea is that, the light motion for 1 second provides the required energy
for a planet to move one solar day, because the light motion depends on the
solar day as we have discussed. By that the planets divides this great rate of
time (1 sec = 86400 sec) on their motions to work as a great clock and
produce this rate by their Unified General Motion.
o The basic part (1 sec =365.25 sec) be between the Earth and the Sun motion.
(Point No. F)
o Because the solar system be created out of a light beam its velocity 1.16
mkm/sec
o And the sun light velocity 0.3 mkm/sec
o That proves the light (0.3 mkm/sec)be created as a product from the original
one 1.16 mkm/sec
o And that proves
o The sun is created after all solar planets creation and motion.

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Extensive Discussion
- Why Does The Previous Investigation Prove The Claim?
- The claim tells that,
- The Sun Is The Solar System Last Created Piece. And that means, the sun is
created after All Planets Creation And Motion.
- If this fact can be proved, it will disprove Newton Theory of The Sun Mass
Gravity decisively.
- How to prove that the sun is created after all planets creation and motion?
- In addition to the historical proves there's one basic proof which is
- The sun rays velocity be = 300000 km
- While
- The solar planets matters and distances be created out of light beam its velocity
=1.16 mkm/sec.
- As a result, I have analyzed the planets data to prove that, this data be created
based on a light beam its velocity =1.16 mkm/sec
- After this analysis, we have to describe how the sun rays be created and the
previous data provides this description.
- Now
- Why Does This Analysis Provide A Sufficient Proof?
- The proof sufficiency depends on the arguments and discussions logic
- The discussions used hundreds of the planets data and put them in some form to
prove some idea. I can create the idea but can't create the data. that means if the
data be in harmony with the idea that supports the proof sufficiency.
- We have used hundreds of data in different discussion, if the discussion logic be
straight and the data be in harmony with it that proves the claim clearly
- The point is that, we searched for the rules based on which this data be created. so
as much as the used data be in harmony with these rules that prove these rules
sufficiency.

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- For example
- I claim (1 day of the sun motion = 1 year of Earth motion). this idea we have
accepted as a hypothesis and used it based on that
- But
- This idea is created because of the moon circumference equation (10921 km x
86400 seconds =940 mkm). Means, we have some data refer to this hypothesis.
- So we have used this idea as a hypothesis.
- We have found that, the data be in harmony and produce a velocity = (0.25 C)
- This velocity be created as a result for the hypothesis we suggested because of the
moon circumference
- But
- Earth Cycle (4 years) can change this velocity (0.25C) into ( 300000 km/s)
- I try to show that, w don't create the data
- We Rearrange The Data In New Positions Relative To One Another
- This is the basic method of the proof.
- Then we find that
- Jupiter (again) provides us a chance to create the value (C2
) (which can be a
source of the sun rays energy)
- So we used the velocity 1.16 mkm/s which we have learnt from Jupiter motion
data and added it to earth circumference to produce the required value (4C)
Notice
The planets unified general motion supports the proof with powerful arguments which
makes the proof more understandable and trustee.

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9- Mercury Jupiter Distance Analysis
9-1 The Main Idea
9-2 Venus Jupiter Distance Analysis
9-3 Jupiter and Pluto Motions For The Distance 720.7 mkm
9-4 Planets Motions Rates of Time
9-5 Planets Motions Interaction
9-6 Mars Motion For The Distance 720.7 mkm

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9-1 The Main Idea
(I)
- The paper hypothesis tells that the solar system be created by energy provided by a
light beam its velocity (1.16 mkm/sec). where this light beam energy created the
solar planets and their distances.
- The solar planets be similar to puppets on a puppets theater because their motions
depend on the motion of the light beam its velocity supposed to be (1.16 mkm/s)
- Of course light known velocity (0.3 mkm/sec) be found in the solar system (We
may consider the light known velocity 0.3 mkm/s be produced from light supposed
velocity 1.16 mkm/s)
- The solar system be created and moving based on the energy provided by light
supposed velocity (1.16 mkm/s) and the light known velocity 0.3 mkm/s.
(II)
- The motion of Light supposed velocity (1.16 mkm/s) effects on Mercury motion.
because Mercury is the point from which light supposed velocity (1.16 mkm/s)
started its motion.
- Light supposed velocity (1.16 mkm/s) effects on Mercury motion and forces it to
move a distance = 720.7 mkm during Mercury day period (4222.6 hours)
- Light supposed velocity (1.16 mkm/s) does that by the following method:
- During (4224 seconds) light supposed velocity (1.16 mkm/s) moves a distance =
4900 mkm = Jupiter Orbital Circumference.
(III)
- Let's remember the matter creation theory in following:
- The theory tells,
- The matter is created out of light but the born matter doesn't separate its light
parent, On the contrary they move together one unified motion but by using
different rates of time between matter and light motions.

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- The basic rate of time is that (1 Second Of Light Motion Be Equivalent To 1
Solar Day Of Matter Motion) by that the light motion considers (the second and
solar day) as the basic periods of motion.
- The basic rate is divided into small rates of time among the solar planets. And by
that the solar planets motions work as a great clock gears, Each planet works as a
gear in a mechanical clock to create the one general rate of time
- As a result
- (1 Second Of Light Motion Be Equivalent To 1 Hour Of Mercury Motion)
- This is the rate of time between light supposed velocity (1.16 mkm/s) motion and
Mercury motion (and this same rate be used between light known velocity
0.3mkm/s motion and Mercury motion)
(IV)
- Light supposed velocity (1.16 mkm/) travels during 4224 seconds a distance =4900
mkm = Jupiter orbital circumference, but
- 1 second of light motion be equivalent to 1hour of Mercury motion
- By this rule
- 4224 seconds of light motion = 4224 hours of Mercury motion
- Mercury day period =4222.6 hours
- Based on that, we conclude,
- The Motion Of Light supposed velocity (1.16 mkm/s) caused Mercury Day Period
To Be (4222.6 hours)!
- Light moves during (4224 seconds) a distance =4900 mkm
- And
- Mercury moves during (4222.6 hours) a distance = 720.7 mkm
- We need to compare between light motion and planet motion distances
- The rate (2π) distinguish usually between light motion and planet motion. by that
where light moves (4900 mkm), Mercury should move (778.6 mkm)
- But

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- Mercury in fact moves only (720.7 mkm)
- Because
- 778.6 mkm (Jupiter orbital distance) =1.0725 x 720.7 mkm (Jupiter Mercury distance)
- The rate (1.0725) be used among 40% of all distances in the solar system. this rate
using reason is still absent, but its using for (40%) of all distances makes it a fact
regardless any explanation. But
- The using of the rate (1.0725) in our equation effects negatively on the rule of the
rate (2π) which is used in other data between light and planet motions correctly.
- Shortly
- Light moves (4900 mkm) in (4224s) and forces Mercury to move (720.7mkm) in
the same period of time (4222.6 h) based on the rate (1 second =1 hour)
- A Question
- Why light supposed velocity (1.16 mkm/s) caused Mercury day period to be
(4222.6 hours) and not (4224 hours)?
- Because
- The difference between the 2 periods (4224 h - 4222.6 h = 1.4 h =5040 seconds)
- Light supposed velocity (1.16 mkm/s) travels during 5040 seconds a distance
=5848 mkm (where 5848 mkm = Mercury Pluto distance)
- Means
- The period (5040 seconds) is created in Mercury day period to create a connection
between Mercury and Pluto motions.

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- Based on that,
- Light supposed velocity (1.16 mkm/s) by one motion during 4224 seconds created
a connection between Mercury and Jupiter on one side and between Mercury
and Pluto on the other sides
- Shortly
- Light supposed velocity (1.16 mkm/sec) travels passing through 3 points in
sequence which are (Mercury – Jupiter - Pluto) effecting on the 3 planets motions
and by equal effect.
- How?
(V)
- Mercury moves 720.7 mkm during Mercury day period (4222.6hours)
- Light motion forces the other planets to move equal motions
- Means
- Jupiter and Pluto, each planet should move 720.7 mkm in a period (4222.6 hours)
- How to do that if the 2 planets velocities are different from Mercury velocity?
- The light motion caused the planets motions to create different rates of time in
proportionality with the defined task.
- The defined task is that
- Jupiter and Pluto, each planet should move (720.7 mkm) in equal periods (4222.6
hours) according to Mercury rate of time which is (1 hour of Mercury motion = 1
second of light motion)
- Shortly
- The 2 planets should move (720.7 mkm) and create configuration for their motions
rate of time according to Mercury Rate Of Time
- This task needs to create an effect through the solar planets motions. this effect
needs to create planets data in proportionality with the defined task

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- That's why the analysis of the distance (720.7 mkm=Mercury Jupiter Distance) is
an important analysis because through this analysis we can prove the paper
hypothesis that the solar planets motions depend on a light beam motion where
through the analysis we will see different planets create configurations for their
motions data to be in proportionality with light motion data.
(Mars should be added to the group of Jupiter and Pluto)
(VI)
- Let's summarize what we will do in this point of discussion in following:
- We have 4 planets (Mercury – Jupiter – Pluto - Mars)
- We discuss these planets motions in this order
- Each planet should move 720.7 mkm in a period (4222.6 hours) based on this
planet rate of time with Mercury rate of time
- For example
- Suppose 1 hour of mercury motion = 2 hours of Jupiter motion, based on that,
Jupiter should move (720.7 mkm) in a period = (4222.6 hours x 2)
- So, in our discussion we analyze the planets motions data to see if the data be in
harmony with the required rates for the motion distance (720.7mkm).
- That makes the discussion is useful to support the paper hypothesis proof.
- Let's start our discussion immediately in the next point.

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9-2 Venus Jupiter Distance Analysis
I- Data
(i)
Light known velocity (0.3 mkm/s) travels during 671 sec a distance =201.1 mkm
(ii)
Light supposed velocity (1.16 mkm/s) travels during 671 sec a distance =778.6 mkm
(iii)
4222.6 mkm = 2π x 671 mkm
(iv)
687 days x 24 hours = 671 x 24.6 hours (Mars Rotation Period)
(v)
4331 days = 2π x 687 days
(vi)
929 mkm x 108.2 mkm= 149.6 mkm x 671 mkm
(vii)
(929 mkm/629 mkm) = (940 mkm/ 637 mkm)
(viii)
1.1318 mkm/day x 637 days = 720.7 mkm
(ix)
1.16 mkm= 13.1 km/s x 24.6 hours x 3600 s
But
4900 mkm= 1.16 mkm x 4222.6
(x)
0.3 mkm /s x 5906 seconds =1775 mkm
But
0.406 mkm/day x 1775 days =720.7 mkm

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Where
- 671 mkm = Venus Jupiter Distance
- 778.6 mkm = Jupiter Orbital Distance
- 4222.6 hours = Mercury day period (if 1 hour =1 mkm be 4222.6 mkm)
- 4331solar days = Jupiter Orbital Period
- 629 mkm = Jupiter Earth Distance
- 929 mkm = Jupiter Earth Distance (the 2 planets on the 2 sun Sides)
- 1.1318 mkm = Jupiter Motion Distance During A Solar Day
- 13.1 km/s = Jupiter velocity
- 5906 mkm =Pluto Orbital Distance
- 0.406 mkm =Pluto Motion Distance During A Solar Day
- Why Do We Need This Data?
- Venus Jupiter Distance provides incredible data very rich for the discussion
- But
- We analyze this data for one reason which is
- Light known velocity (0.3 mkm/s) during 671 seconds travels 201 mkm (while
Jupiter moves during 4222.6 hours a distance = 199.2 mkm error 1%)
- And
- Light known velocity (0.3 mkm/s) during 5906 seconds travels 1775 mkm (while
Pluto during 1775 solar days moves 720.7 mkm)
- Where
- 671 mkm = Venus Jupiter distance (used as 671 seconds for light motion)
- 5906 mkm = Pluto orbital distance (used as 5906 seconds for light motion)
- (1 mkm = 1 second for light motion and 1 mkm= 1 day for planet motion)

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II- Discussion
Equation no. (i)
Light known velocity (0.3 mkm/s) travels during 671 sec a distance =201.1 mkm
Equation no. (ii)
Light supposed velocity (1.16 mkm/s) travels during 671 sec a distance =778.6 mkm
-
- Equations no. (i) tells that,
- Light known velocity (0.3 mkm/s) uses a period (671 seconds) to pass distance
201.1 mkm. (where Jupiter moves during Mercury day period 4222.6 h a distance
=199.2 mkm has a different 1% with the distance 201.1 mkm)!
- Means,
- light known velocity (0.3 mkm/s)moves during (671s) a distance = Jupiter motion
distance during Mercury day period 4222.6 h (why?)
- Equations no. (ii) tells,
- Light supposed velocity (1.16 mkm/s) uses equal period (671 seconds) to pass a
distance = 778.6 mkm = Jupiter Orbital Distance
- That tells
- Light motion be using the period (671 seconds) mentioned Jupiter data in both
motions (201 mkm and 778.6 mkm)
- Why?
- 671 mkm = Venus Jupiter distance and be used as 671 seconds for light motion.
- Because
- (1 second of light motion be = 1 day of planet motion)
- By that
- (1 mkm= 1 second for light motion and 1 mkm = 1 day for a plant motion)
- This rate enable to use the distances values as periods of time

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Equation no. (iii)
4222.6 mkm = 2π x 671 mkm
- Mercury day period =4222.6 hours
- And
- 1 second of light motion =1 hour of mercury motion
- Based on that
- 1 hour of mercury motion be = 1 mkm
- By that
- 4222.6 h be = 4222.6 mkm = 2π x 671 mkm
- Why Venus Jupiter distance (671 mkm) is seen in every where here? why be used
time after time in different data?!
- How to understand this data
- Please remember
- The difference = 1.3%
- Why Venus Jupiter Distance be almost = Venus Orbital Circumference?!
- Notice
- The basic difficulty is this equation is the rate (2π)
- This rate is used between light motion and planet motion. by that the distance
4222.6 mkm should be considered as light motion but 671 mkm should be
considered as a planet motion. Is this consideration a correct one? I have no
answer yet…

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Equation no. (iv)
687 days x 24 hours = 671 x 24.6 hours (Mars Rotation Period)
Equation no. (v)
4331 days = 2π x 687 days = 24.6 hours x 4222.6
- Mars orbital period (687 solar days) =(671 Mars rotation periods)
- And
- 4331 days (Jupiter orbital period) = (2π) x Mars orbital period (687 days)
- Is 4331 days produced by a light motion and 687 days be produced by planet
motion?! I don't know yet…
- Why the rate (2π) be used so frequently? What geometrical meaning it tells?
- If
- 1 mkm = 24.6 hours based on that
- 671 Mars rotation period = 671 mkm Venus Jupiter distance!
- Why these numbers are the same numbers? How to create a geometrical meaning
between 671mkm and 671Mars rotation periods?
-
- Notice
- 680 mkm = Venus Orbital Circumference (error 1% with 687 mkm)
- If
- 1 mkm = 1 day, so
- 687 mkm = Mars Orbital Period
- I want to say
- Both periods (687 solar days and 671 mars rotation periods) are created based on
Venus distances (680 mkm and 671 mkm)
- Means
- Mars cycle periods be produced depending on Venus distances values
- But why?!

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- Notice (1)
- The previous 5 equations are found to explain the using of the value 671,
- Although the data is so rich and interesting we can't catch the real using of the
value or its geometrical effect
- That's happened because of one reason … let's try to explain it in following:
- Jupiter (13.1 km/s) moves in (24.6 h = Mars rotation period) a distance =1.16 mkm
- This information is the basic on behind all data
- Because
- Jupiter orbital circumference (4900 mkm) = 1.16 mkm x 4222.6
- That means, Jupiter orbital circumference be designed based on 1.16 mkm.
- Now,
- Light supposed velocity (1.16 mkm/s) motion for 1second passes a distance be
passed by Jupiter motion during 24.6 hours (Mars rotation period) and by that the
data which be explained based on 1 second period of time should be explained
again based on a period =24.6 hours –
- Notice (2)
- Jupiter orbital period (4331 days) be the other picture on mirror because
- 4331 days =103944 hours = 4222.6 Mars rotation periods = 680 Pluto days periods
= 2 x 5200 Jupiter days periods (error 1%) = 1775 x 58.65 hours
- Where 58.65 days = Mercury Rotation Period
- And
- During 1775 days Pluto moves a distance 720.7 mkm
- Notice (3)
- During 5200 hours Pluto moves 88 mkm (Mercury Orbital Period 88 days if 1
mkm =1 day)

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Equation no. (vi)
929 mkm x 108.2 mkm= 149.6 mkm x 671 mkm =100733 mkm
Equation no. (vii)
(929 mkm/629 mkm) = (940 mkm/ 637 mkm)
Equation no. (viii)
1.1318 mkm/day x 637 days = 720.7 mkm
- Equation no. (vi) we have discussed before
- The basic equation is the number (viii) because
-
- Jupiter (1.1318 mkm/day) moves during 637 days a distance =720.7 mkm. This is
the required distance. Jupiter moves it in a period 637 days
- But the rule tells
- Jupiter should move 720.7 mkm in a period (4222.6 hours = Mercury day period)
- How the period 637 days can be equal 4222.6 hours?
-
- By using a different rate of time
- we have 2 jobs here
- (1st
job) to know how Jupiter motion passes the required distance
- (2nd
job) to know the rate of time by which 637 solar days be = 4222.6 hours
- Notice
- 1.16 mkm/s x 346.6 seconds = 2 x 671 seconds x 0.3 mkm/s

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- (1st
Job) Jupiter Motion Discussion
- Jupiter moves during (4222.6 hours = Mercury day period) a distance =199.2 mkm
- Light known velocity (0.3 mkm/s) travels during 671s a distance =201.1 mkm
(error 1%)
- Light supposed velocity (1.16 mkm/s) travels during 671s a distance =778.6 mkm
= Jupiter orbital distance
- That tells
- Jupiter motion for the distance (199.2 mkm) during (4222.6 h) is accompanying
motion with light beam motion.
- The light uses 199.2 mkm (or 201.1 mkm) to pass the required distance (720.7
mkm) this process we should discuss in details in the next point, means
- The first job should be performed in the point
- (5-3 Jupiter and Pluto Motions For The Distance 720.7 mkm)
- And
- The second job should be performed in the point
- (5-4 Planets Motions Rates of Time)

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9-3 Jupiter and Pluto Motions For The Distance 720.7 mkm
For Jupiter
Equation no. (vii)
(929 mkm/629 mkm) = (940 mkm/ 637 mkm)
Equation no. (viii)
1.1318 mkm/day x 637 days = 720.7 mkm
- Jupiter moves during 637 days the required distance 720.7 mkm
- But
- The rule tells
- The motion should be in 4222.6 hours (Mercury day period)
- How to do that?
- Equation no. (vii) tells a part of the story
Equation no. (vii)
(929 mkm/629 mkm) = (940 mkm/ 637 mkm)
- Where
- 629 mkm = Earth Jupiter distance
- 929 mkm = Earth Jupiter distance (when the 2 planets on the sun 2 sides)
- 940 mkm = Earth orbital circumference
- Shortly
- Jupiter depends on Earth distances values to create the value (637mkm) and then
use it as 637 days.
- Why? because
- 629 mkm = π x 199.2 mkm
- Means
- The distance Jupiter moves during 4222.6 hours be a radius of a circle whose
circumference = Earth Jupiter distance. By that Jupiter depends on its distance to
Earth in different geometrical forms to create the required value (637 mkm)

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- Let's see how Jupiter did that in details in following:
Jupiter Motion Details
- Jupiter moves during 4222.6 hours 199.2 mkm
- And
- Jupiter (1.1318mkm/day) moves during 201 days a distance = 227.9 mkm
- Jupiter (1.1318mkm/day) moves during 227.9 days a distance = 257.9 mkm
- Jupiter (1.1318mkm/day) moves during 257.9 days a distance = 292 mkm
- Jupiter (1.1318mkm/day) moves during 330.4 days a distance = 374 mkm
- Jupiter (1.1318mkm/day) moves during 423.2 days a distance =479 mkm
- Jupiter (1.1318mkm/day) moves during 479 days a distance = 542 mkm
- Light supposed velocity (1.16 mkm/s) travels during 542 sec a distance =629 mkm
- Data Analysis
- The data show that, Jupiter uses light motion strategy, in which
- Jupiter moves a distance and uses it as a period of time to pass another distance
and by that the motion creates a network of distances based on which Jupiter
moves the required distance (720.7 mkm) depends on the original distance (199.2
mkm) which Jupiter passes in (4222.6 hours = Mercury day period)
- Notice
- 5848 mkm (Mercury Pluto Distance) = 2π x 929 mkm
- Where 5848 mkm = light motion during 5040 seconds (light supposed velocity
1.16 mkm/sec during 5040 sec moves 5848 mkm)
- 929 mkm expresses planet motion.
- That connects Jupiter motion for the distance (720.7 mkm) with Pluto motion.
- This conclusion will be supported by different data found in the following
discussions.

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For Pluto
Equation no. (x)
0.3 mkm /s x 5906 seconds =1775 mkm
But
0.406 mkm/day x 1775 days =720.7 mkm
- Where
- 5906 mkm =Pluto Orbital Distance
- 0.406 mkm = Pluto Motion Distance Per A Solar Day
- Notice
- 1.16 mkm/sec x 5127 seconds = 5906 mkm (Pluto orbital distance) (error 0.7%)
- 0.3 mkm/sec x 5127 seconds = 2x 778.6 mkm (Jupiter orbital distance) (error 1%)
- That tells
- Pluto and Jupiter orbital distances depend on equal periods of time by using
different light beam velocities (1.16 mkm and 0.3 mkm respectively).
-
- The data has a great depth, because the light motion be used in both planets
motions (Jupiter and Pluto) to force the 2 planets to move 720.7 mkm or a distance
related to it
- But
- The geometrical machine by which the light motion forces the 2 planets for their
motions is still unclear.
- Notice
- (1775 Days /346.6 Days) = 5.12 = (Mars Velocity / Pluto Velocity)

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9-4 Planets Motions Rates Of Time
1st
Point (Pluto Rate Of Time)
- 1 Hour Of Mercury Motion =9.18 Hours Of Pluto Motion
- This rate of time be concluded from the data (1407.6 hours =153.3 hours x 9.18)
- Where
- But
- The rate of time is created because the planets velocities are different.
- means,
- the task requires Pluto to move 720.7 mkm in a period (=4222.6 h = Mercury day
period) as Mercury does
- but
- Pluto velocity is different from Mercury velocity where Mercury velocity 47.4
km/s = 10.08 x Pluto velocity 4.7 km/s
- The rate of time is created to cause equality between the planets motions periods of
time by change the rates of time
- Based on that, the rate should be 10.08 means (1hour of Mercury motion should be
= 10.08 hours of Pluto motion) but it's not the fact, on the contrary the fact is that
1hour of mercury motion be =9.18 hours of Pluto motion
- How this rate perform the Job?
- Let's see the following data

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I- Data
(A)
(9.18 /10.08) = (654.9 mkm/720.7 mkm)
(B)
654.9 mkm= 86400 x 7510 km (error 1%)
(C)
0.406 mkm/day x 1622.7 days = 654.9 mkm (error 0.6%)
II-Discussion
Equation no. (A)
(9.18 /10.08) = (654.9 mkm/720.7 mkm)
- Where
- 654.9 mkm = Jupiter Saturn Distance
- 720.7 mkm = Jupiter Mercury Distance (The required Distance)
- Equation no. (A) tells that rate of time (9.18) is rated with the velocities rate
(10.08) based on the distance 654.9 mkm ration with the required distance 720.7
mkm
- Why? what's importance of this distance 654.9 mkm for Pluto motion?
Equation no. (B)
654.9 mkm= 86400 x 7510 km (error 1%)
- Where
- 7510 km =Pluto Circumference
- Equation no. (B) tells that,
- If the distance 654.9 mkm be passed in one solar day only, Pluto circumference
will be equal a distance of this motion for a period 1second
- This method of calculations I have learn from the moon because it has a similar
equation let's see it in following

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- 10921 km x 86400 seconds =940 mkm
- Where
-
- The equation tells
- If the earth revolves around the sun a complete revolution (940 mkm) in one solar
day only (86400 s) in this case the moon circumference be equal a distance of
Earth motion for 1 second period of time
Equation no. (C)
0.406 mkm/day x 1622.7 days = 654.9 mkm (error 0.6%)
- Where
- 1622.7 mkm =Uranus Neptune Distance
- 0.406 mkm/day = Pluto motion distance per a solar day
- Notice
- Because the rate of time be confirmed by the equation no.(B) and this method is
confirmed by the moon circumference data I have guessed that may this is the
method to define the rate of time and for that I have used this method to define
Jupiter rate of time
- Let's do that in the next point

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1st
Point (Jupiter Rate Of Time)
- 1 Hour Of Mercury Motion =17.2 Hours Of Jupiter Motion
- I have used the same method to define this rate of time,
- 142984 km (Jupiter diameter) x 86400 seconds = 720.7 mkm x 17.2
- I have used Jupiter diameter in place of its circumference
- Notice (1)
- 1407.6 hours x 17.2 =24211 hours = 2x 12063 hours
- Where
- But
- 24211 hours =9.9 h (Jupiter day period) x 2437
- (where 2437 km = Mercury Radius)
- Notice (2)
- We remember that 671 mkm is a basic value in Jupiter motion analysis where
(671 mkm = Venus Jupiter Distance)
- Because of that
- 17.2 h = 3.4 x 5.12 (error 1%)
- Where
- 3.4 degrees = Venus orbital inclination, but
- 3.4 hours = 12104 seconds where (12104 km= Venus diameter
- Also
- The rate 5.12 controls great part of data in the solar system motion

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- Notice (3)
- (Venus velocity / Uranus velocity) = 5.12
- (Mars velocity / Pluto velocity) = 5.12
- (the moon velocity / Neptune velocity) = 5.12
- (3717 mkm Jupiter Neptune Distance / 720.7 mkm) = 5.12
- (3413.5 mkm/ 671 mkm) = 5.12
- (where 3413.5 mkm =the 4 inner planets orbital circumferences total)
- (671 mkm = Venus Jupiter distance)
- Jupiter orbital distance / Earth orbital distance = 5.12 (error1%)
- But
- Light known velocity (0.3 mkm/s) travels during 3717 sec a distance = 2 x 550.7
mkm (Mars Jupiter distance) (error1%)
- Light supposed velocity (1.16 mkm/s) travels during 3717 sec a distance =4331
mkm (where 4331 days = Jupiter orbital period)

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9-5 Planets Motions Interaction
I- Data
(1)
8 x 9.9 x 3600 s = 285120 s = π x 90756.2 seconds
(90756.2 seconds – 90560 seconds) = 196.5 seconds
(2)
196.5 sec x 1.16 mkm/sc = 227.9 mkm (Mars Orbital Distance)
(3)
(90560 mkm x 2) = 227.9 mkm x 198.7
(4)
170 sec x 1.16 mkm/sec = 197.2 mkm
(5)
655.7 sec x 0.3 mkm/sec = 196.7 mkm
(6)
196.5 sec x 27.78 km/sec = 10921 km/2
(7)
194 sec x 35 km/sec = 6792 km
(8)
193 sec x 0.3 mkm/sec =57.9 mkm (Mercury Orbital Distance)
(9)
193.7 sec x 1.16 mkm/sec =224.7 mkm
224.7 sec x 1.16 mkm/sec =521.3 mkm
(10)
30589 days = 58.65 days =521.3
(11)
606.3 mkm = 193 mkm x π = 1.16 mkm/sec x 522 sec
(12)
643.3 sec x 0.3 mkm/sec = 193 mkm

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(13)
3735072 km = 196.5 x 19000
720.7 mkm =3.735 mkm x 193
And
(5687.1 mkm = 19000 sec x 0.3 mkm/sec)

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II –Discussion
- Let's summarize the analysis basic idea in following
- The theory tells, the light supposed velocity (1.16 mkm/s) started from Mercury
passing Jupiter, Pluto and Mars
- Light supposed velocity (1.16 mkm/s) motion effects on the 4 planets caused the 4
planets to move equal distances (720.7 mkm) in equal periods of time (4222.6h)
- Because of this motion
- 3 Planets motions depend on the period (196.5 seconds)
- The data discussion will explain the meaning of planet motion dependency on the
period of time 196.5 seconds.
- Because the 3 planets motions depend on equal periods of time (196.5 s) their
motions be integrated and unified into one general unified motion
- As A Result
- Mercury, Mars and Jupiter moves One Unified General Motion.
- This Unified Motion Be Proved in the 3 planets motions data.
- This point of discussion shows that, the 3 planets motions depend on the period
(196.5 seconds).

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Equation no. (1)
8 x 9.9 x 3600 s = 285120 s = π x 90756.2 seconds
(90756.2 seconds – 90560 seconds) = 196.5 seconds
- Where
- 9.9 hours = Jupiter day period
- 90560 days =Pluto orbital period
- (8 days of Jupiter motion creates Jupiter 8 days cycle)
- The equation uses Pluto orbital period (90560 days) in seconds unit (90560
seconds) why?
- We know this answer, and keep it by heart
- (Light Motion For 1 Second Causes A Planet Motion For 1 Solar Day)
- By that, all periods in solar day units for planets motions be considered in second
units for light motion.
- What we need from this equation?
- The period 196.5 seconds
- This period is the connection reason among the 4 planets motions (Mercury – Mars
– Jupiter and Pluto)
- In fact our discussion cornerstone is the period 196.5 seconds
- Why is this period 196.5 s important for the planets motions?
- Because it's created by interaction between Jupiter and Pluto motions. and the
paper main equation which we have discussed deeply before tells
Equation no. (I) (The Paper Main Equation)
1.16 mkm per second x 2 x 86400 seconds =2 x 100733 mkm = (37100 mkm –
4900 mkm) x 2π = 28255 mkm + 2 x 86400 mkm
- We have discussed this equation before

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- This equation shows, the solar system basic energy is sent from Jupiter to Pluto
and then reflected from Pluto to Neptune, means, the interaction between Jupiter
and Pluto deal with the solar system basic energy…
- This conclusion shows a great significance for the interaction between Jupiter and
Pluto motions. As a result, the value 196.5 seconds should be considered as a very
important period in the solar system motion.
- But
- 8 Jupiter days create (planet 8 days cycle for Jupiter). Can this period 196.5
seconds have any relationship with (Jupiter 8 days cycle?) we should in equation
no. (13) of this discussion.
- Notice
- The period 196.5 sec is defined by an interaction between Jupiter and Pluto
motions, and
- Light known velocity (0.3mkm/sec) needs 193 sec to pass Mercury orbital distance
(57.9 mkm).
- That means, Mercury orbital distance be defined by an effect of Jupiter and Pluto
motions interaction.
- That tells, Mercury Jupiter distance is defined at first and then the light beam built
Pluto orbital circumference and after that Mercury orbital distance be defined. (or
the sun position be defined)

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Equation no. (2)
Equation no. (8)
- The period 193 seconds is different from the period 196.5 seconds with (2%)
- We neglect this difference (2%) for now and consider the 2 periods are equal
- Now
- By equal periods of time (196.5 seconds), the 2 light beams defined the 2 planets
orbital distances…
- Light supposed velocity (1.16 mkm/sec) during (196.5 sec) travels a distance =
227.9 mkm = Mars orbital distance
- Light known velocity (0.3 mkm/sec) during (193 sec) travels a distance = 57.9
mkm = Mercury Orbital Distance
- By equal periods of time the 2 light different velocities defined the 2 planets
orbital distances.
- What Effect Can Be Produced By The Equal Periods Of Time?
- We deal with light motion…. let's use one example for help
- In light motion, if there's a phase between the light waves the experiment will
produce an interference but if there's no phase we will see light waves without
interference
- The phase creates the differences in motions
- But without phase
- The motions will be unified in one, means
- Mercury, Mars And Jupiter Move One Unified General Motion Because The
3 planets Use Equal Periods Of Time…

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- Let's summarize what we have…
- The 3 planets use equal periods of time (196.5 seconds) and by that their motions
will be unified in one unified general motion.
- The period 196.5 seconds is born by interaction between Jupiter and Pluto
motions. where this interaction express the solar system basic energy
- The unification of motions between Mercury, Mars and Jupiter can be produced by
Jupiter dependency on Mercury and Mars motions because they are the first and
last planets in the inner group and by that Jupiter creates a connection with the
inner planets through its connection with Mercury and Mars

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Equation no. (2) (Continued)
Equation no. (8)
- We know that mercury, Venus and Earth define their orbital circumferences to be
equal their distances to Jupiter.
- means, Mercury position is defined in the space relative to Jupiter position.
- Mars defined its orbital circumference to be equal Saturn orbital distance…
- That tells, Mercury and Mars positions are defined relative to Jupiter and Saturn
positions.
- And
- Their distances to the sun be defined by using equal periods of time (196.5 sec)
- Let's ask, if the 2 planets positions are defined, what result of their using of equal
periods of time?
- They almost define the sun position in the solar system
- The theory tells, the sun is created after all solar planets creation and motion. the
equal periods be used almost to define the sun position in the solar system relative
to Mercury and Mars positions relative to Jupiter position.
- Mercury (4.095 mkm/day) moves during its rotation period (58.65 days) a distance
= 243mkm (error 1%)
- Mars (2.082 mkm/day) moves during Venus day period (116.75 days) a distance =
243 mkm
- Where, 243 mkm x 0.8 =196.5 mkm
- This motion is almost related to the sun creation process
Notice
- 243 mkm x 2 =40080 km (Earth Circumference) x 12104 km (Venus diameter)
- 29.8 km/sec x 12104 seconds = 3 x 120536 km

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Equation no. (3)
(90560 mkm x 2) = 227.9 mkm x 198.7
Equation no. (4)
170 sec x 1.16 mkm/sec = 197.2 mkm
- Where
- 90560 days = Pluto Orbital Period
- 227.9 mkm = Mars Orbital Distance
- 170 mkm = The Distance Between Mercury And Mars
- Equation no.(4) we have discussed in the distances series discussion (point no. 3)
of this current paper.
- Equation no. (3) tries to show that Pluto orbital period be in proportionality with
the period (196.5s)
- Notice
- The error (2%) between 193 seconds and 197 seconds we don't discuss because,
the machine is so complex
- We usually find the data in rates (100%, 99%, and 98%) these rates are repeated
frequently, and these rates may be created because of one geometrical necessity,
- For example Earth moves during its day period a distance =2.574 mkm (100%) but
the moon displacement total during 29.53 days =2.598 mkm (101%) and the moon
apogee orbital circumference =2.550 mkm (99%)
- The rates (98%, 99% and 100%) are used frequently, I can't write all data which
use these rates and can't explain the reason of it.

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Equation no. (5)
655.7 sec x 0.3 mkm/sec = 196.7 mkm
Equation no. (6)
196.5 sec x 27.78 km/sec = 10921 km/2
Equation no. (7)
194 sec x 35 km/sec = 6792 km
- Where
- 27.78 km/sec = The Moon Velocity
- 6792 km = Mars Diameter
- What do we need from these equations?
- Shortly
- We need equation no. (6) because it shows that, the moon circumference depends
on the period (196.5 seconds)
- We remember that,
- The 3 Planets (Mercury – Mars and Jupiter) depend on equal periods (196.5 sec) in
their motions. and these equal periods (196.5 sec) causes the 3 planets motions to
be unified in one general unified motion.
- Means, the period (196.5 seconds) is the cornerstone of these 3 planets motions.
- Now
- The moon circumference (10921 km) depends on the period (196.5 sec) also
- So, we have more players be under the same one unified general motion. because
the moon uses the same period, so no phase will be between the moon motion and
Jupiter motion, on the contrary, they move together one unified motion…
- That may explains the important following equations:
(NO. A)
- 4900 mkm (Jupiter orbital circumference) = 10921 km x 449197 km

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- Where
- 449197 km = Jupiter Circumference
(NO. B)
- Jupiter (13.1 km/sec) moves during 10921 sec a distance = 142984 km = Jupiter
diameter.
(NO. C)
- The moon (27.78 km/sec) moves during 5127 sec a distance = 142984 km =
Jupiter diameter.
- (where 5127 mkm = Jupiter Pluto distance)
- Notice
- Light supposed velocity (1.16mkm/sec) travels during 4900 sec a distance =
5678.1 mkm (Mars Pluto Distance)
- Notice
- The moon orbital apogee radius 406000km = 2jupiter diameters + 1 Saturn
Diameter.
Notice
- Why the moon velocity be =27.78 km/s? because the moon moves daily a
distance = Earth motion distance daily because they aren't separated from one
another. means the moon daily motion distance =2.574 mkm but this distance be
effected by the rate 1.0725. (we have referred to that frequently before). We don't
know why the distance 2.574 mkm be effected by the rate (1.0725) but 40% of all
distances in the solar system be effected by this same rate 1.0725 and appendix
no.1provides a list for these distances. By that the moon distance 2.574 mkm be
=2.4 mkm daily which =27.78 km/sec.

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Equation no. (8)
Equation no. (9)
193.7 sec x 1.16 mkm/sec =224.7 mkm
224.7 sec x 1.16 mkm/sec =521.3 mkm
Equation no. (10)
30589 days / 58.65 days =521.3
- Where
- 224.7 days = Venus Orbital Period
- 30589 days = Uranus Orbital Period
- 58.65 days = Mercury Rotation Period
- Equations (8), (9) and (10) show that, the rate between Uranus orbital period and
Mercury rotation period depends on the period (193 seconds) by light motion.
- The data shows that, the series creates a connection between the 2 planets
- But
- Mercury rotation period is related to Uranus motion by some complex geometrical
interaction because
- Mercury Uranus distance = 2814.6 mkm
- But
- 2815.2 hours = 2x 1407.6 h (Mercury rotation period)
- If 1 mkm = 1 hour that makes the distance 2815.2 mkm= 2 Mercury Rotation
Period, And
- Mercury moves during its rotation period a distance = 243mkm (error 1%)
- Where 243 days = Venus rotation period
- If 243 mkm be equivalent to 243 days that tells 1 mkm = 1 day
- By that Mercury creates with Uranus the rate of time (1 hour of mercury = 24
hours of Uranus) we discuss that in Planets motions use different rates of time
point no.(5) of this paper.

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Equation no. (11)
606.3 mkm = 193 mkm x π = 1.16 mkm/sec x 522 sec
- This equation gives 3 branches of data let's summarize each one in following
- (1st
Branch)
- 606.3 seconds x 1.16 mkm/sec = 703.3 mkm
- 703.3 seconds x 1.16 mkm/sec = 816 mkm
- 816 seconds x 1.16 mkm/sec = 946.6 mkm
- Where
- We remember that
- 5848 mkm (Mercury Pluto distance) – 4900 mkm (Jupiter orbital circumference) =
948 mkm (Earth Orbital Circumference) (error 1%)
- That means,
- The value (606.3 mkm) is connected with light supposed velocity (1.16 mkm/sec)
motion trajectory from Mercury to Jupiter and then to Pluto. This trajectory is the
1st
trajectory we have studied…
- (2nd
Branch)
- 193 seconds x 0.3 mkm/sec = 57.9 mkm (Mercury Orbital Distance)
- 193.7 seconds x 1.16 mkm/sec = 224.7 mkm (Venus orbital period 224.7 days)
- 224.7 seconds x 2 x 1.16 mkm/sec = 521 mkm
- (3rd
Branch)
- 30589 days (Uranus orbital period) = 58.65 days (Mercury rotation period) x 521
- Equation no. (11) connects many basic data together.

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Equation no. (12)
643.3 sec x 0.3 mkm/sec = 193 mkm
- Equation no. (12) is important because
- Jupiter moves during 636 days a distance = 720.7 mkm (where 636 days is
different from 643.3 days with 1%)
- The period (643.3 days) be used as (643.3 sec ) for light motion
- Equation no. (12) tells that, light known velocity (0.3 mkm/s) travels during (643.3
s) a distance = 193 mkm
- That causes the period 193 second be found related to the distance 720.7 mkm
- The basic difficulty is that the light creates geometrical interactions aren't clear for
our analysis and by that many geometrical effects be created but aren't known for
us because we don't realize perfectly how the machine works.

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Equation no. (13)
3735072 km = 196.5 x 19000
720.7 mkm =3.735 mkm x 193
And
(5687.1 mkm = 19000 sec x 0.3 mkm/sec)
- Where
- 3735072 km=the distance be passed by Jupiter during 8 Jupiter days (please
review Jupiter 8 days cycle)
- 19000 seconds the light known velocity (0.3 mkm/sec) needs to pass the distance
from Mars to Pluto (5678.1 mkm)
- We remember that
- 4900 sec x 1.16mkm/sec = 5687.1 mkm = (Mars Pluto Distance)
- 720.7 mkm (Mercury Jupiter distance) = 3.735 mkm x 193
- That tells, the period (193 seconds) is defined by a geometrical mechanism takes
into consideration Mars Pluto Distance.
- Because, in fact
- Mars connection with Mercury and Jupiter be created through Pluto. where the
light supposed velocity (1.16 mkm/s) travels from Mercury to Jupiter directly and
doesn't touch Mars in the way.

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9-6 Mars Motion For The Distance 720.7 mkm
I-Data
(a)
47.7 km/s (Mercury velocity) x 4222.6 h x 3600 s = 720.7 mkm
2.082 mkm/day (Mars velocity daily) x 346.6 days = 720.7 mkm
1.1318 mkm/ day (Jupiter velocity daily) x 636 days = 720.7 mkm
0.406 mkm/ day (Pluto velocity daily) x 1775.2 days = 720.7 mkm
Notice
720.7 mkm/ 196.5 mkm = 3.66
(b)
2.082 mkm/ day x 346.6 days =720.7 mkm
And
(5848 mkm x π) = 720.7 mkm x 25.2 (error 1%)
(c)
0.406 mkm/ day x 1775.2 days =720.7 mkm
But
(90560 days x 2) /1775 days = 102
(d)
(940 mkm/720.7 mkm) = (119.7 mkm/91.7 mkm) = (720.7/550.7) = (1557.2 mkm/
1205 mkm)=(2094 mkm/ 1622.7 mkm) = (5848 mkm/ 4495.1 mkm) = 1.3

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II-Discussion
Equation no. (b)
2.082 mkm/ day x 346.6 days =720.7 mkm
- where
- 2.082 mkm / day = Mars Velocity Daily
- 346.6 days = The Nodal Year
- Equation no. (b) tells, Mars moves during 346.6 days (the nodal year) a distance
=720.7 mkm = Mercury motion distance during Mercury Day Period (4222.6h)
- Why?
- The 2 numbers are mentioned because 346.6 days= the nodal year and 720.7 mkm
= Mercury motion distance…. That means, this distance be mentioned to be passed
by Mars during this defined period of time (346.6 days) but why?
-
- My answer is (the machine is A Complex One)
- The following data may help a little
o Mercury day period needs 5040 seconds to be = (4224 hours)
o 5040 seconds = 84 minutes
o But
o Mercury moves during 84 solar days a distance = 344 mkm
o Mercury moves during 346.6 solar days a distance = 1419 mkm
o Mercury moves during 1433.5 solar days a distance = 5848 mkm (Mercury
Pluto Distance) (error 0.4%)
o This system of motion we have seen before in light motion, when the light
passes a distance and then uses this distance as a period of time to move
another distance and then uses the second one to move a third one
o We notice

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o The period 346.6 days be used in Mercury motion which concerning the
distance 5848 mkm. By that we conclude the period 346.6 days has a
relationship with the distance (5848 mkm)
o Where
o The distance 5848 mkm expresses the light motion effect on the 4 planets
motions to pass equal distances (720.7 mkm)
o By that
o Mars uses (346.6 days) to pass the require distance (720.7 mkm)
o That proves Mars motion for the distance 720.7 mkm be found by an effect
of light motion to pass the distance (5848 mkm) as similar to the other
planets.
o Of this explanation be insufficient
o The difficulty is in the period (346.6 days) which is used by Mercury and by
Mars. This using creates an interaction between the 2 planets motions
o (Notice, the point no.3-3 explains in details the planets motions interaction).
Notice (1)
(1407.6 h / 24.6 h) = (180 /π)
But
(4222.6 h /24.7 h) =170
- Where
- 4222.6 hours = Mercury Day Period
- 24.7 hours = Mars Day Period
- 170 mkm = Mercury Mars Distance
Notice (2)
(5848 mkm x π) = 720.7 mkm x 25.2 (error 1%)
- Where 25.2 degrees = Mars Axial Tilt

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Equation no. (c)
0.406 mkm/ day x 1775.2 days =720.7 mkm
But
(90560 days x 2) /1775 days = 102 = (5848 mkm/57.9 mkm) (error 1%)
- Where
- 0.406 mkm / day = Pluto Motion Distance During A Solar Day
- 90560 days =Pluto Orbital Period
- 57.9 mkm =Mercury Orbital Distance
- 5848 mkm =Mercury Pluto Distance
- Equation no.(c) tells that,
- Pluto moves during 1775.2 days a distance =720.7 mkm
- This is the distance be required by light motion effect on the 4 planets motions
- Now
- The period 1775.2 days should be related to Mercury day period (4222.6 h)
- Equation tells that, the period 1775.2 days be rated with Pluto orbital period
(90560 days) by the same rate between (5848 mkm =Mercury Pluto distance) and
57.9 mkm (Mercury orbital distance)
- The equation tells, the period 1775.2 days is defined based on Pluto basic data in
proportionality with Mercury data.
- Shortly
- Pluto motion data be created takes into consideration the period 1775.2 day, that
tells the distance 720.7 mkm effects on Pluto motion also

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Equation no. (d)
(940 mkm/720.7 mkm) = (119.7 mkm/91.7 mkm) = (720.7 mkm /550.7 mkm) =
(1557.2 mkm/ 1205 mkm)=(2094 mkm/ 1622.7 mkm)=(5848 mkm/ 4495.1 mkm) =
1.3
- Where
- 720.7mkm = Mercury Jupiter Distance (our examined distance)
- 119.7 mkm = Venus Mars Distance
- 91.7 mkm = Mercury Earth Distance
- 550.7 mkm = Mars Jupiter Distance
- 1557.2 mkm = Jupiter Orbital Distance (1557.2 mkm =2 778.6 mkm)
- 1205 mkm = Mars Saturn Distance
- 2094 mkm = Jupiter Uranus Distance
- 1622.7 mkm = Uranus Neptune Distance
- 5848 mkm = Mercury Pluto Distance
- 4495.1mkm = Neptune Orbital Distance
- 1.3 degrees = Jupiter Orbital Inclination
- Equation no. (d) shows that the planets distances be created in proportionality with
one another.
- Ion different many data I try to show that, no single distance be created
interpedently, we deal along time with one geometrical system, means the
distances be created based on one geometrical design.

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Earth Jupiter Distance Analysis
Equation no. (I)
100733 mkm = 929 mkm x 108.2 mkm = 671 mkm x 149.6 mkm
- Where
- 108.2 mkm = Venus orbital distance
- 149.6 mkm = Earth orbital distance
- 929 mkm = Earth Jupiter Distance (The 2 Planets Be On The Sun 2 Sides)
- Light supposed velocity (1.16 mkm/s) travels during a solar day (86400 s) a
distance = 100733 mkm
- Equation no. (I) tells, from one source of Energy the 4 distances be created
- Specifically
- From the distance 100733 mkm, the 4 distances are created
- These 4 distances be created from the same one distance (100733 mkm) in
comparison with one another. that means these 4 distances are created based on
one geometrical design which proves the network theory of distances.
Equation no. (II)
(940 mkm/680 mkm) = (929 mkm / 671 mkm)
- Where
- The distances are created for geometrical reasons and based on geometrical
mechanisms.

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Equation no. (III)
(778.6 mkm / 929 mkm) = (5848 mkm/4900 mkm) = 0.838
- Where
- 778.6 mkm = Jupiter Orbital Distance
- 5848 mkm =Mercury Pluto Distance
-
- I try to show, the distance 929 mkm is built based on a geometrical mechanism
which creates a system for the distances creation, or a network of distances
- That means,
- The distance 929 mkm be taken into consideration by the light motion from
Mercury to Pluto (5848 mkm) because it expresses the planet motion (929 mkm)
in comparison with the light motion (5848 mkm=2πx 929 mkm)
- In following we analyze the distance 929 mkm to see if this distance can be found
geometrically…

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Earth Jupiter Distance Analysis (More Discussion)
- The previous analysis tries to show that, the distance between Earth and Jupiter
across the sun (929 mkm = 778.6 mkm +149.6 mkm) is defined as a basic point for
Mercury Pluto Distance (5848 mkm) where
- 5848 mkm = 929 mkm x 2π
- Let's analyze this distance in following
- The figure shows the sun and the 3 planets (Mercury – Venus – Earth)
- The 3 planets be drawn on 2 sides of the sun. and Jupiter is the planet (J)
- Now let's start our analysis
- The distance between Earth and Jupiter (on one side) = 778.6 -149.6 = 629 mkm
- The distance between Earth and Jupiter (on 2 sides) = 778.6 +149.6 = 929 mkm
- The distance between Earth and Mercury (on 2 sides) =57.9 +149.6 = 207.5 mkm
- The distance between Earth and Venus (on 2 sides) =108.2 +149.6 = 257.8 mkm
- The question is
- Are The Distances Across The Sun Be Considered Geometrically For The
Planets Motions And Orbital Distances?
- Let's test that with Venus Earth Distance in the following example

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Example no. (1)
Additional Equation (i)
Sin (23.6 deg) x 257.8 mkm = 103.4 mkm
- S= The Sun
- V = Venus
- E= The Earth
- ES = Earth Orbital Distance = 149.6 mkm
- SV = Venus Orbital Distance = 108.2 mkm
- EV = 103.4 mkm
- Notice
- The distance between Venus and Earth = 41.4 mkm and NOT 103.4 mkm. So how
this triangle be built?!
- let's move step by step:
o Earth orbital distance 149.6 mkm + Venus orbital distance 108.2 mkm =
257.8 mkm (This distance be found when Earth and Venus be on 2 different
sides from the sun with an angle 180 degrees between the 3 players)
o 257.8 mkm x sin (23.6 deg) = 103.4 mkm (=EV distance In The Triangle)
o 23.45 deg = Earth axial tilt (has a difference with 23.6 deg be 1%)
o In our triangle (SEV) the angles be defined geometrically as following
o The Angle (S) = 43.6 degrees
o The Angle (V) = 90 degrees
o The Angle (E) = 46.4 degrees (= 2 x 23.2 deg = Earth Axial Tilt error 1%)
o Now let's analyze this triangle in following

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- Is this triangle (SEV) be an imaginary one? because the distance between Venus
and earth be 103.4 mkm and the real distance = 41.4 mkm?
- We see that, the distance between the Earth and Venus can be even = 257.8 mkm
when the 2 planets be on 2 different sides from the sun. So if the distance can be =
257.8 mkm, that means, it can be also 103.4 mkm. That means this triangle can't
be an imaginary one.
- Why this triangle is useful?
- Because of the angle (E) = 2 x 23.2 degrees (Earth Axial Tilt) (error 1%)
- That tells, Earth axial tilt is produced by some interaction be found among Earth,
Venus and The Sun Data
- Why Venus angle =90 degrees?
- What Useful Results We Get From This Triangle? 2 Basic Results
- (1st
Result)
- The planets distances across the sun are used geometrically in the solar system
geometrical design.
- (2nd
Result)
- The Sun Be Perpendicular On Venus Planet in this triangle, while Earth axial tilt
be defined. Can that effect geometrically on the sun position relative to Venus and
Earth positions?
Notice
41.6 mkm = 0.384 mkm x 108.2 mkm (Venus Orbital Distance)
- 41.4 mkm = Venus Earth Distance (different from 41.6 mkm with 0.4%)
- 0.384 mkm = The Earth Moon Orbital Distance
- The data tells, The Earth moon orbital distance also be defined accordingly
- The paper provides an interesting approach to explain the solar planets motions.

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- While the solar planets motions description inherited from Newton Theory of the
sun mass gravity tells that each planet motion is independent from the other
planets motions. As a result each planet orbital distance be defined independently,
The suggested description tells that, each planet motion takes into consideration
the other planets motions.
- We here deal with a building stories found depending on one another. based on
this vision, because Venus orbital distance =108.2 mkm, that causes Earth orbital
distance to be 149.6 mkm
- And because of the 2 planets orbital distances be (108.2 mkm and 149.6 mkm) the
Earth moon orbital distance be (0.384 mkm).
- The vision is completely different from what Newton told us. while Newton
supposed the space is limitless and can contain hundreds of planets moving
without effect on each other. the fact is the 9 planets motions effect on one another
and forces each planet to define each motion in proportionality with the other
planets. The vision is different because Newton vision of (NO Effect) created
separated trajectories of motions. And based on this vision we see the planets as
cars move through track independently from one another (even the real cars can't
be independent because the track is the same one and all have to move in harmony
otherwise they will collide). The suggested description tells that, the motions be
built as a building and each motion depend on the other motions and by that they
need one geometrical design controls and manages all of them.

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Discussion
- This triangle I put here because it shows the distance between the earth and Venus
can be =257.8 mkm by put the 2 planets on 2 different sides from the sun.
- This behavior perfectly is similar to Earth Jupiter distance (929 mkm) while the 2
planets be on different 2 sides from the sun.
- I try to show that, it's part of the distances distribution system to use the planets
positions across the sun.
Notice
- The distance between Earth and Mercury across the sun be = 149.6 mkm + 57.9
mkm =207.5 mkm
- Pluto (0.406 mkm/day) moves during 511 days a distance = 207.5 mkm
- Where
- 511 degrees = The Solar Planets Axial Tilts Total
- A Summary
- The distances between Mercury and Pluto be distributed based on geometrical
design to create a proportionality between the 3 planets motions for periods (720.7
mkm). (the 3 planets are Mercury – Jupiter – Pluto)
- The distance distribution be performed by light motion effect on te planets
distances creation.

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10- The Solar Planets Motions Analysis
10-1 Preface
10- 2 The Equal Distances Method
10-3 The Inner And Outer Planets Motions Be Distinguished

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10-1 Preface
- The paper claims that,
- The solar planets move a type of motion be different from the motion type Newton
had imagined based on The Sun Mass Gravity Concept.
- The previous discussion proves that:
1. The Planets Move One Unified General Motion, Disproving Newton Concept
Of Planet Independent Motion
2. The Planets Motions Depend On Different Rates Of Time.
- These Features Contradicts Newton Theory Of The Sun Mass Gravity.
- But
- There are more interesting features which prove that, the solar planets motions be
different in nature from the motion Newton had mentioned by The Sun Mass
Gravity.
- Among these features we study 2 basic features which are
o The Equal Distances Method
o The Distinguish between The inner and outer Planets Motions.
- Let's start our discussion in following…

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10- 2 The Equal Distances Method
- The Equal distances Method is A Feature be discovered in the solar system motion
- 2 Planets move equal distances in defined periods of time. And as a result this 2
planets data be in proportionality with one another.
- This feature be analyzed as deep as possible to know why it's used and what
geometrical effect it creates. Let's explain that by using examples in following:
Example No. (1) (Earth and Pluto Data)
(A)
- Earth moves during its day period a distance = Pluto motion distance during its day
period = the Earth moon displacements total during its day period (error 1%).
(B)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ Earth day period) = 6.37 (0.8%)
(C)
- (Pluto motion distance daily406000 km/ the moon displacement daily88000 km)=4.61
- (The Moon Day Period 708.7 h / Pluto Day Period 153.3 h) = 4.61
- Tan (12.2) x 708.7 h (The Moon Day Period) = 153.3 hours (Pluto Day Period)
- Tan (13.17) x 655.7 h (The Moon rotation Period) = 153.3 hours (Pluto Day Period)
- (Where 13.17 deg = the moon daily motion degrees, and 12.2 deg =13.17 deg – 0.9856
deg Earth motion daily degrees)
- We have discussed this data before.
- The proportionality of data may be created by the equal distances of motions or by
the rate of time. Or may both be created based on each other.
- The fact is that,
- The 2 planets move equal distances in defined periods of time and their data be
created in proportionality with one another.

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- Let's analyze the previous example as possible:
o Equal Distances Be Passed By Earth And Pluto Motions,
o We should pay attention for the equal distances nature, because the distances
are equal. But these distances are not random distances. On the contrary,
these are mentioned distances because the planets pass these distances
during their days periods.
o It's necessary to put this data in front of our eyes (always). Because the
equal distances aren't any random equal distances but distances be passed
during defined periods of time and by that these distances be mentioned to
be equal.
o Defined period of time means a period be defined by a planet motion. as a
planet orbital period or day period or rotation period …etc
o Shortly a period depends on a planet motion
Example No. (2) (Mercury and Mars Data)
- Mercury moves during its rotation period a distance =243 mkm (error 1%)
- Mars moves during Venus day period a distance =243 mkm
- We have 2 equal distances! Where's the data proportionality?
- (Mercury Day Period)/ (Venus Day Period) = (Mercury Orbital Period / Mercury
Rotation Period)
- That tells, Venus day period is created as a function in Mercury cycles periods.
Can Mars motion for the distance 243 mkm be the reason of this proportionality?
Notice
243 solar days = Venus Rotation Period (if 1 mkm = 1 solar day), so this distance
(243 mkm) may refer to Venus Rotation Period
On the other side
Venus moves during (365.25 days) a distance = 1106 mkm =2π x 175.94 mkm
175.94 solar days = Mercury Day period (means, 1 mkm = 1 solar day is a rate
known by Venus and Mercury).

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The Equal Distances Using Range
- Why Do We Study The Equal Distances Method?
- Because
- This Method is almost the planets motions method. That means the solar planets
don't move by the sun mass gravity but move by the equal distances method (or at
least the outer planets move by the equal distances method).
- That means,
- The solar system motion depends on the equal distances method greatly.
- Why Do We Need to Know That? Even If Be A Fact
- Because
- This explains the heavy using of the equal distances methods in the solar system
motion. and by that we explain the following facts:
o 50% Of All Distances In The Solar System Be Equal One Another
o And
o 40% of all distances in the solar system be rated one another with the same
one rate (1.0725)
o This data be in appendix no. 1 of this paper
- Notice
- We have discussed this data before but the data of (50% of all distances are equal)
and (40% of all distances are rated) be put in the appendix no. 1 to be used as
reference

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Let's summarize the equal distances using range in following:
(1)
The data in appendix no. (1) shows that
50% of all distances in the solar system are equal one another
and
40% all distances in the solar system are rated with the same rate (1.0725)
(2)
Planet 8 days Cycle depends on the equal distances using
We have discussed Planet 8 days Cycle in The planets Unified General Motion
(3)
Different examples we have discussed as Earth and Pluto equal distances
And also Mercury and Mars equal distances
(4)
Planet diameter or circumference using as a period of time depends on equal distances
let's provide some examples in following:
o Jupiter (13.1 km/s) moves during 10921 second a distance =142984 km =
Jupiter diameter.
o Uranus (6.8 km/s) moves during 7510 second a distance =51118 km =
Uranus diameter.
o Pluto (4.7 km/s) moves during 10921 second a distance =51118 km =
Uranus diameter.
o Pluto (4.7 km/s) moves during 51118 seconds a distance =2 x 120536 km =
Saturn Diameter.
o Pluto (4.7 km/s) moves during 2 x 120536 seconds a distance =1.1318 mkm
= Jupiter motion distance during a soar day.
o Where
o 10921 km = The Earth Moon Circumference
o 7510 km = Pluto Circumference

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The Equal Distances Using Reason
- I provide here an explanation why the equal distances method be used
- I want to say,
- This explanation (whether be correct or incorrect) can't effect on the fact that (the
Equal distances is the reason of the solar planets motions) because this fact
depends on The Equal Distances Method Wide Range Of Using.
- So, let's ask
- Why Be The Equal Distances Used As The Method Of The Planets Motions?
- Because
- This is the method be used between light motion and planet motion. and between 2
planets motions.
- That means,
- If a light beam energy be used to cause a planet motion, this energy be used
through the equal distances method
- And
- If a planet motion energy be used to cause another planet motion, this energy be
used through the equal distances method
- It's the same one method by which the motion be transported from a light beam to
a planet or from a planet to a planet. One method be used for 2 jobs
- As a result, we find distances be passed by light motion and be passed (again) by
planets motions.
- That means, because The Matter Is Created Out Of Light
- The Solar Planets motions depend on the equal distances method

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Example
- Light known velocity (0.3 mkm/s) travels during a solar day (86400 seconds) a
distance= 25920 mkm
- But
- All planets motions distances per a solar day (=17.75 mkm) x 1461 days = 25920
mkm
- If the period of time is not the cycle (1461 days), this equation would have no any
value. But because the period of time is a cycle (1461 days =365 +365 +365 +366)
defined by Earth motion. that tells the distance 25920 mkm be passed by 2 players.
(A light beam and all planets motions total).
- To prove this is a light motion. we see the value 25920 mkm as a period of time
which is (the precession cycle 25920 years)
- Also
- The moon (2.4 mkm/ daily) moves during 10747 days a distance =25920 mkm
- The data tries to show that, the equal distances be use by light and planet motions
or by planet and planet motions.
- That makes this method so useful and suitable for the solar planets motions

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10-3 The Inner And Outer Planets Motions Be Distinguished
10-3-1 The Inner Planets Motion
I- Data
(1)
Earth Mass = The Masses total of (Mercury + Venus + Mars)
(2)
Earth Circumference = The 5 inner planets diameters total
(Means the diameters total of (Mercury + Venus + The Moon + Earth + Mars))
(3)
2 x Earth Circumferences = 2 Mars Circumferences + Venus Circumference
(4)
Earth Circumference = The moon freedom space of motion
II- Discussion
- The double production experiment tells us
- From Gamma ray one electron and one positron can be created, 2 particles equal in
mass and opposite in charge.
- Equation no. (1) tells, Earth Mass = The 3 Inner Planets Masses Total (error 1%)
- Can we imagine that, a great gamma ray be used to create the inner planets.
- Earth is (The Great Electron) (means, Earth atom has a positive nucleus and
negative charges revolve around)
- The 3 Inner Planets Consist Together (The Great Positron) (means, the inner
planets Mercury – Venus – Mars , Their atoms have negative nucleus and
positive charges move around)
- If these hypotheses be working, the inner planets can create One Electric Field to
cause their motions around the sun.
- The error (1%) refers to a part of mass be consumed to create the required energy
for these planets motions through their orbits.

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- By that the error (1%) is found for a physical necessity and not be found as an
error in measurements.
- The previous description provides a suitable base to create an eclectic field
- The electric field is better for the planets motions than the sun gravity force.
- Newton has no one proof for the sun mass gravity force but the electric field
theory depends on the masses equality fact.
- The next discussion can support the electric field theory.
Equation no. (2)
Earth Circumference = The 5 inner planets diameters total
(Means the diameters total of (Mercury + Venus + The Moon + Earth + Mars))
Equation no. (3)
2 x Earth Circumferences = 2 Mars Circumferences + Venus Circumference
- For what we search? For Electric Charges
- If we can prove that, some electric charge be found in the inner planets. That can
support this idea greatly
- Why these 2 equations refer to charges in the inner planets data?
- Because of Gauss Law!
- Gauss law tells, The Charges Be On The Outer Surface, that means, the outer
surface can be effected Geometrically by electric charges. Because more wide
outer surface receives more charges. That tells The Charge Electric Field Effect on
the matter dimensions must be A Geometrical Effect
- I want to say
- If some planet be charged its outer surface formation and dimensions should be
effected Geometrically because of the electric charge.
- Means, the geometrical effects should be used as a proof for charges existence.

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- The previous 2 equations no. (2 and 3) can support this idea because the inner
planets circumferences and diameters be created in proportionality which can refer
to geometrical effects be produced by electric charges on their creation.
Equation no. (4)
Earth Circumference = The moon freedom space of motion
- The moon orbital Perigee radius =363000 km
- The moon orbital apogee radius =406000 km
- The difference 43000 km
- But the space without the moon diameter =43000 km- 3475 km = 39500km= Earth
Circumference (error 1.3%)
- This data is used for the same argument, A geometrical effect be found on the
surface distances which can be used as a reference for an electric field effects on
the matter dimensions
- Notice
- We know the distance can't be charged unless under specific conditions, but I try
to show geometrical effects are found. It's not necessary for this effect to be a
direct one, but it can be found through interactions.
- Notice
- 43000 km (the distance from perigee to apogee) x 3475 km (the moon diameter)
=149.6 mkm (Earth Orbital Distance). I try to show that, geometrical features are
found behind the dimensions and distances definition in the inner planets. And
these geometrical features supports the electric field effect claim.
- Notice
- We have discovered that the planets motions can't be independent from the other
planets motions on the contrary the planets motions be similar to chess board
motions where each motion be calculated geometrically. This features (Chess

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Board Motion) we have discussed in the planets unified general motion. and I want
to say that, the electric field is the best method can provide this type of motion
(Chess Board Motion)
- That supports the claim, (The Inner Planets Motions Depend On Electric Field)
A Conclusion
- The electric field idea aims to create a source of motion for the inner planets. By
that the moving planets are the inner planets. That tells the inner planets motions
depend on the inner planets electric field. But the outer planets motions depend on
the motion transportation from the inner planets motions to the outer planets
motions.
- By that we have 2 different types of motions but both originated from the same
one source by 2 different methods and that enable the planets motions to be
integrated in one unified general motion
- If the 2 types of motions be proved that disprove Newton Theory of The sun mass
gravity.

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10-3-2 The Outer Planets Motion
The Outer Planets Motions Are Distinguish By 2 Features
- The outer planets motions be distinguished from the inner planets motions by 2
basic features which are
- (1st
Feature) Planet 8 days Cycle doesn't use any inner planet motion.
- (2nd
Feature) The outer planet moves during its day period a distance equal to, or
rated with its circumferences which is a feature doesn't be used by the inner planets
motions.
- Planet 8 days cycle we have studied before
- So,
- Let's discuss in following feature no. (2)

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(2nd
) Planet moves during its day period a distance = Its Circumference
The outer planet moves during its day period a distance equal to, or rated with its
circumferences
I-Data
(1)
Saturn moves During Its Day Period A Distance=Its Circumference (error 1.3%)
(2)
Jupiter moves During Its Day Period A Distance =Its Circumference (error 4%)
(3)
Uranus moves During Its Day Period A Distance =260% of Its Circumference
(4)
Neptune moves During Its Day Period A Distance =200% of Its Circumference
(5)
Pluto moves During Its Day Period A Distance =345.6 x Its Circumference

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II-Discussion
- We should notice that, Jupiter (4%) causes Planet 8 days Cycle
- And in Planet 8 days cycle we have seen that, a different distance = Jupiter
diameter be defined between Jupiter motion distance and its circumference which
creates planet 8 days cycle for Jupiter motion.
- I want to say that, the planet circumference is a player in this planet motion. and
because of that, the (4 great planets circumferences) have deviations from their
circumferences because of Planet 8 days cycle.
- Why does the outer planets be different from the inner planet?
- The outer planets move by observing their circumferences but the inner planets
moves observing their orbital circumferences. That may be happened because of
the electric field controls the inner planets motions and causes their motions be
controlled by their orbital circumferences. But the outer planets depend on the
motion transportation from the inner planets, and by that the outer planet almost
uses its circumference as a measurement for its motion.
Saturn Motion Analysis
- We analyze Saturn motion as an example for the 4 planets motions, in following
- Saturn (9.7 km/s) moves during its day period (10.7 hours) a distance = 373644 km
= 98.67 % of Saturn Circumference (378675 km).
- The difference between the distance (373644 km) and Saturn Circumference =
5031 km. (= 1.3%)
- Why is this interesting?
- Because if a planet moves during its day period a distance = its circumference that
will create a rate between this planet orbital period and day period
- Means
- (Planet Orbital Period/ Day Period) = (Planet Orbital Circumference / Planet
Circumference)
- For that reason, we interest in planet motion distance during its day period.

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- Saturn motion performs this rate
- But
- It shows that, the rate in Saturn data be changed because of geometrical necessities
which I claim they are produced as a result for (planet 8 days cycle). Let's
summarize the idea in following
o The inner planets move under their electric field because of that their
motions be defined by this field effect, but
o The outer planets receive the motion by transportation of motion. because of
that, the outer planets have no a guide in their motion, they use their
circumference as a guide and measurement.
o If this rule be working, the outer planet would move during its day period a
distance = its circumference - But
o Because planet 8 days cycles which causes the motion transportation among
the outer planets, that creates an interaction which changes the rule and
caused the outer planets to move during their days periods distances (NOT)
equal their Circumferences, but (rated with their circumferences)
o Jupiter (8 day cycle) shows clearly that, the planet circumference is a player
in the cycle and proves the interaction which is found among the outer
planets as a result for planet 8 days cycle.
o This explanation details disproves clearly Newton Description because the
motion (here) is transported from one planet to another clearly and with
strong proves which disproves the sun masses gravity concept
o The point is that, the interaction among the outer planets motions cause the
change in the rule and creates planet 8 days cycle. But. The rule still works
spite the deviation.
o I want to say that,

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o Saturn moves During Its Day Period A Distance=Its Circumference
(error 1.3%), But the machine feels that no error in Saturn motion.
because this error is found as a result for the configuration with Planet
8 days cycle
By this same explanation all outer planets motions be controlled. The errors in the
planets motions from the rule (Outer Planet moves during its day period a distance =
its circumference) still works sufficiently, the errors be found as a result for planet 8
days cycle. (hypothesis)

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Appendix no. (1)
The Solar System Equal Distances List
(A)
(1)
(2)
Earth Neptune Distance = Mercury Saturn Circumference (0.5%)
(3)
Jupiter Venus Distance = Venus Orbital Circumference (1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (1.2%)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (0.8%)
II- Discussion (A)
The previous distances form around 50% of all distances found in the solar system
(All orbital and internal distances)… Why These Distances Are Equal One Other?
We may notice that – the distances equality can be produced more easily by light
motion than the rigid body motion - for example – when we push a ball toward a wall
the ball after collision with the wall will return a distance (NOT) equal the original
one - because the collision causes to decrease the ball motion momentum – but the
light can be reflected at equal distances easily – means – equal distances can be
produced by light motion more easy than the Rigid Body Motion.

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(B)
19. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
20. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
21. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
22. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
23. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (0.6%)
24. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (0.8%)
25. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (0.7%)
26. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (0.7%)
27. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4864
= (0.5 %)
(10)
Discussion (B)
The same rate (1.0725) is used for all equations (around 18 distances = 40% of all
solar system distances) – why?
Suppose the equal distances are produced by light reflection and that cause these
distances to be equal – as I have supposed in the previous point (A).
Now suppose– part of these equal distances – is passed through another frame relative
to us – so this part of distances will suffer from Lorentz Length Contraction Effect
which is seen in the rate 1.0725
(Another frame can be found in the solar system because we deal with light motion) –
This explanation can answer why some distances are equal and others are rated with
the same rate (1.0725) – it's simply a feature of light motion.
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

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References and Biography
(Preprint No. 1) Mars Migration Theory
https://www.academia.edu/49051037/_Preprint_No_1_Mars_Migration_Theory
(Preprint No. 2) The Moon Orbital Motion Equation
https://www.academia.edu/49051029/_Preprint_No_2_The_Moon_Orbital_Motion_Equation
The Solar Planets Motions Description (Revised)
https://www.academia.edu/49464125/The_Solar_Planets_Motions_Description_Revised_
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944p=15209
Mr. Gerges Francis Tawdrous +201022532292
Physics Department- Physics Mathematics Faculty
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
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Jupiter and mercury motions analysis (revised) (2)

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