Planet Motion Analytical Description (Revised).pdf

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawdrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
Planet Motion Analytical Description (Revised)
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 – 29th
January 2022
Abstract
Paper hypothesis
- The Solar System Be Created Out Of One Light Beam Its Velocity = 1.16
million km per second
The hypothesis explanation
- The hypothesis tells, The solar planets matters and their distances be created out of
one light beam its velocity = 1.16 mkm/s, means, the planets matters and their
distances be created out of the same one source of energy
- This hypothesis should be proved before to analyze it because it supposes many
points can create serious effects on the solar system vision
The hypothesis Proves
- If the solar system be created out of one light beam, what logical result can be
produced? The Planets Data Should Be Created Depending On One Another.
this sentence means, If we know that (Mercury orbital distance =57.9 mkm) we
should be able to predict all other planets orbital distances- NOT ONLY-
- Instead
- We should be able to predict all planets orbital distances, periods, and velocities as
well as all planets diameters, masses, orbital inclinations and axial tilts
- Shortly

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- We need a data system – its input one data only (Mercury orbital distance =57.9
mkm) and its output be the planets fact sheet data completely be predicted based
on this only one data.
- Why should this system be found?
- Because the solar planets and their distances be created out of one light beam by
that the solar system be similar to one building be consisted of stories and each
story depend on the others where the building all data can be concluded from one
another.
- Let's provide in following 7 parts of proves to prove this fact:
- PROOF NO. 1
- (A) Light supposed velocity (1.16 mkm/s) travels a distance = The Planet Orbital
Circumference in a period (T seconds)
- (B) Light known velocity (0.3 mkm/s) travels during this period (T) a distance = D
- (C) Planet moves a distance d in a period (=Ph) where D = 2π d
- And
- Ph = The Planet Orbital Period (in hours units and not solar days)
- A Conclusion
- If Planet Orbital Distance Be Known, Planet Orbital Period Can Be Predicted
- Notice
- (1.16/0.3) x 2π = 24 (error 1.3%)
- Example
- Light supposed velocity (1.16 mkm/s) needs 310 s to pass 360 mkm = Mercury
Orbital Circumference
- Light known velocity (0.3 mkm/s) passes during 310s a distance = 93 mkm = 2π x
14.8 mkm (Mercury passes 14.8 mkm in 88 hours) (error 1.3%)
- (Mercury orbital period =88 solar days)

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- PROOF NO. 2
- We know the period of time (T), and the planet orbital period, so Planet velocity
per a solar day can be defined by this equation
(T/P) = (V/1.16)
- T= The Period Of Time Be Required By Light Supposed Velocity (1.16 mkm/s)
To Pass A Distance = Planet Orbital Circumference
- P = Planet Orbital Period
- V = Planet velocity per a solar day
- 1.16 mkm = Light Supposed Velocity (1.16 mkm/s) Motion In One Second
- The Periods use the same units (seconds)
- A Conclusion
- If Planet Orbital Distance Be known, Planet Orbital Period And Its Velocity
Can Be Predicted.
- Example
- Light supposed velocity (1.16 mkm/s) needs (T= 310s) to pass 360 mkm= Mercury
Orbital Circumference
- and
- (P = Mercury Orbital Period = 88 days)
- (T/P) = (V/1.16) that means (310/88) = (V/1.16)
- 4.095 mkm per solar day = Mercury Velocity

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- PROOF NO. 3
- Now we know that, the period (T) is the backbone of the planets data definition,
the next equation defines the period (T) among the planets
- T2
= 4T0 (T-T0)
- Where
- T = The period of time Light supposed velocity (1.16 mkm/s) needs to pass a
distance = Planet Orbital Circumference
- T0 = the period of time Light supposed velocity (1.16 mkm/s) needs to pass a
distance = The Previous Planet Orbital Circumference
- The equation tells simply each planet period of time (T) depends on its previous
direct neighbor planet period of time (T0) –
- The equation has 3 exceptions which are
- Earth depends on Mercury and doesn't on Venus
- Mars depends on Venus and doesn't on Earth
- Pluto depends on Uranus and doesn't on Neptune
- We note that, The data considers the planet Ceres between Mars and Jupiter
- A Conclusion
- If One Planet Orbital Distance Be known, All Planets orbital distances, periods
and velocities can be defined simply
- Example
- T2
= 4T0 (T-T0) based on that (586 s)2
= 4x 310s (586 s – 310s)
- Where
- T = 586 s = The period Light supposed velocity (1.16 mkm/s) needs to pass 680
mkm = Venus Orbital Circumference
- T0 = 310 s = The period Light supposed velocity (1.16 mkm/s) needs to pass 360
mkm = Mercury Orbital Circumference
- Another Example
- T2
= 4T0 (T-T0) based on that (7765 s)2
= 4x 4222.6s (7765 s – 4222.6s)

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- Where
- T = 7765 s = The period Light supposed velocity (1.16 mkm/s) needs to pass 9007
mkm = Saturn Orbital Circumference
- T0 = 4222.6 s = The period Light supposed velocity (1.16 mkm/s) needs to pass
4900 mkm = Jupiter Orbital Circumference
- PROOF NO. 4
- Planets velocities be complementary one another
Example.
- Mercury (47.4 km/s) moves during 6.8 hours a distance = 1.16 mkm and
- Uranus (6.8 km/s) moves during 47.4 hours a distance = 1.16 mkm and
Example.
- Earth (29.8 km/s) moves during 2 x 5.4 hours a distance = 1.16 mkm and
- Neptune (5.4 km/s) moves during 2 x 29.8 hours a distance = 1.16 mkm and
- Means, the planets velocities are defined as velocities for one system members
and by that the solar group be seen as one machine of gears.
- PROOF NO. 5
A RULE
(Planet (A) orbital circumference / Planet (B) orbital circumference) = (VB/VA)2
- This rule be used sufficiently for all planets distances and velocities
Example.
- (940 mkm/360 mkm) = (47.4/29.8)2
(error 1.6%)
- Where
- 940 mkm = Earth Orbital Circumference
- 360 mkm = Mercury Orbital Circumference
- 47.4 km/s = Mercury Velocity
- 29.8 km/s = Earth Velocity

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- PROOF NO. 6
- (v1/v2) = (s/r)
- v1 = planet velocity in second
- v2 = neighbor planet velocity in second
- r = Planet Diameter
- s = The Planet Rotation Periods Number In Its Orbital Period
- A Conclusion
- If One Planet Orbital Distance Be known, All Planets orbital distances, periods
and velocities as well as their diameters and rotation periods can be defined.
- Example
- (v1/v2) = (s/r)
- v1 = 9.7 km/s = Saturn Velocity
- v2 = 5.4 km/s = Neptune Velocity
- r = 49528 km = Neptune Diameter
- s = 89143 Neptune rotation periods be found in Neptune orbital period
- And (59800 days = Neptune orbital period and 16.1 h = Neptune rotation period)
- Another Example
- v1 = 24.1 km/s = Mars Velocity
- v2 = 9.7 km/s = Saturn Velocity
- r = 120536 km = Saturn Diameter
- s = 24106 Saturn rotation periods be found in Saturn orbital period
- And (10747 days = Saturn orbital period and 10.7 h = Saturn rotation period)
- Notice/ the equation uses (2) by that (2 x 24106/ 120536) = (9.7/24.1)
- Notice/ The equation needs some modifications with each planet because the
velocities rate produces planets orbital inclinations –for example – 9.7/5.4= 1.8
(Neptune orbital inclination =1.8 degrees) also – 24.1/9.7 =2.5 (Saturn orbital
inclination =2.5 deg) - The planets orbital inclinations be used as rates between the
planets masses.

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- Notice/
- The equation works from Mars to Pluto because the moon rotation period = its
orbital period –that tells this data depends on a geometrical design takes the moon
motion into consideration – that may explain the question why
- The moon apogee radius =406000 km = The Solar Planets Diameters Total
- The moon perigee radius = 363000 km =The Outer Planets Diameters Total (-1%)
- PROOF NO. 7
- The solar system is one building each planet be created depending on the other
planets data
- The planets orbital circumferences total be = 100733 mkm= 2π x16033 mkm
- The planets orbital Periods total be = 2 x 197392 days = 24 x16450 mkm
- Light known velocity (0.3 mkm/s) needs 16033 seconds to pass 4810 mkm
(=Jupiter Orbital Circumference 4900 mkm error 2%)
- But
- Light supposed velocity (1.16 mkm/s) needs 2 x 16033 seconds to pass 37100
mkm (=Pluto Orbital Circumference) - that tells we deal with one geometrical
design and Jupiter Position is the central position of this system for that reason
- (a)
- Light known velocity (0.3mkm/s) needs (1200 s) to pass 360 mkm = Mercury
orbital circumference but Mercury moves during 1200 solar days = 4900 mkm
- (b)
- Light known velocity (0.3mkm/s) needs (16333 s) to pass 4900 mkm = Jupiter
Orbital Circumference but Jupiter moves during 16333 solar days = 18486 mkm
- (c)
- Light known velocity (0.3mkm/s) needs (60160 s) to pass 18048 mkm = Uranus
Orbital Circumference but Uranus moves during 60160 solar days = 35345 mkm
- (37100 mkm = Pluto Orbital Circumference with 35345mkm error 5 %)

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- The data shows the transportation process starts from Mercury to Jupiter to Uranus
to Pluto in one system proves that the solar system is one building each planet be a
part of it.
- Based on this data –the solar group vision be changed
- The Solar Group be similar to a machine of gears each planet is a gear in it, or
- The Solar Group is similar to one building and each planet is a part of this same
building, or
- The Solar Group is similar a canal water moves through the canal and causes the
rotation of 9 waterwheels. The Motion Direction of The waterwheels is different
from the water motion direction but no waterwheel can rotate without the water
motion, or
- The solar planets are similar to train carriages, they move together one unified
general motion, or
- The solar planets be similar to knots on one robe or similar to points on the same
one trajectory of energy –
- The solar planets motions be similar to chess board pieces motions, each motion
depend on the other planets motions and be done by geometrical calculations.
- The solar system distances be created in a network form and based on one
geometrical design as a chess board.
- The data shows a continuum be found in the solar system data which can be
explained only as a result of light motion features
- Notice
- The solar system creation and motion theory provides the theoretical basics for the
data explanation (The theory be provided in Point no.5-5)
Paper objective
- The paper provides the lists of data proves its hypothesis and tries to discover
the one geometrical design based on which the solar group be created.
(Please scan the figure (ORCID)

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Paper Contents
Subject Page No.
1- Introduction 12
2- Methodology 17
3- The Paper Hypothesis Proves 20
3-1 Preface 21
3-2 Data No. 1 (The Main Data) 24
3-3 Data No. 2 (Planet Velocity Definition Data) 29
3-4 Data No. 3 (Planet Velocity Analysis Data) 34
3-5 Data No. 4 (The Planets Periods Equation) (The Basic Data) 38
3-6 Data No.5 (Planet Velocity And Its Orbital Circumference Relationship) 49
3-7 Data No.6 (Planet Velocity And Its Diameter Relationship) 54
3-8 Data No. 7 (Part I) (One Geometrical Design Data) 59
3-9 Data No. 7 (Part II) (One Geometrical Design Data) 61
3-10 Data No.8 (Part A) (The Distances Be In Network Form) 67
3-11 Data No.8 (Part B) (Jupiter Velocity Analysis) 70
3-12 Data No.9 (Planet Motion Data Depends On Light Motion) 75
3-13 Data No.10 (Planets Motions for 30 mkm) 79
3-14 Data No.11 (Planets Masses Rates) 84
3-15 The Data General Comment 97
4- The Data Analysis And Discussion 99
4-1 Preface 100
4-2 The Comparison between the Solar Day period and Mars rotation period 102
4-3 Jupiter Velocity More Analysis 103
4-4 Uranus Velocity Analysis 106
4-5 Can The Solar System Depend On One Geometrical Design? 112
4-6 The Moon Orbit Design Analysis 119
4-7 Uranus Effect On The Moon Motion 153
4-8 The Planets Unified General Motion 159
5- Jupiter Motion Effect On Earth And Venus Motions 171
5-1 Preface 172
5-2 Jupiter Motion Effect On Earth And Venus Motions 176
5-3 Jupiter Orbital Circumference Analysis 182
5-4 Saturn Velocity Analysis 186
5-5 The Solar System Creation and Motion Theory 207
6- The Solar System Creation (Questions and Discussions) 214
7- The Solar Planets Motions Use Different Rates Of Time 225
7-1 Preface 226
7-2 The Planets Motions Rates Of Time 228
8- The Solar Planets Rates Of Time Analysis 232
8-1 Preface 233
8-2 Venus Motion Rate of time 234

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8-3 Earth Motion Rate of time 236
8-4 Mars Motion Rate of time 238
8-5 Jupiter Motion Rate of time 240
8-6 Saturn Motion Rate of time 242
8-7 Uranus Motion Rate of time 244
8-8 Neptune Motion Rate of time 246
8-9 Pluto Motion Rate of time 248
8-10 The Planets Orbital Distances Test 250
8-11 One Law Controls The Planets Orbital Periods And Distances 255
8-12 The General Discussion 256
9- The Planets Motions Rates Of Time Effect Analysis 259
9-1 Preface 260
9-2 Planets Motions Rates Of Time And Distances Data 261
9-3 The Data Equal Distances 265
9-4 The Data and the planets velocities. 273
9-5 The Data Distances And Rates Of Time Interaction 277
9-6 The Data General Discussion 287
9-7 Mars, Jupiter and Saturn Motions Analysis 289
9-8 Why Saturn And The Moon Use Equal Rates Of Time? 294
9-9 Why Mercury Use A Double Of Its Orbital Distance? 299
9-10 The Rate (4.61) be used between Pluto and the moon motion 300
9-11 The Moon Orbital Motion Equation 304
10- Mars Migration Theory 320
10-1 Mars Migration Theory 321
10-2 Pluto Migration Theory 324
10-3 Planets Migration Theories Proves 326
10-4 Is There An Absent Planet In The Solar Group? 329
11-The Solar System Distances Be Created In A Network Form 332
11-1 Preface 333
11-2 The Continuum effect Through the Solar System Distances 335
11-3 The Solar System Distances Distribution 340
11-4 The Solar System Distances Dependency On One Another 344
12- The Continuum Effect Proof 346
12-1 The Continuum Effect Proof 347
12-2 Saturn Motion Analysis 353
12-3 Planet Diameter Analysis 379
12-4 Why do the planets revolve around the sun if there's no sun gravity? 381
13- Planet Mass Effect On its Motion 382
13-1 Preface 383
13-2 Planet Mass effect on its Motion 385
13-3 Saturn and Earth Motions Interaction 396
13-4 Planets Velocities Proportionality 405

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14- Saturn Motion Analysis 413
14-1 Preface 414
14-2 Saturn Diameter Analysis 416
14-3 Neptune Circumference Analysis 426
14-4 Neptune Day Period Analysis 435
14-5 Mercury Motion effect on Jupiter and Neptune Motions 444
14-6 Earth Motion Distance Daily Analysis 451
14-7 Uranus Day Period Analysis 455
14-8 The Inner Planets Motions Analysis 461
15- The Sun Age Description 469
15-1 Preface 470
15-2 The Sun Circles The Earth 471
15-3The Rate (1.0725) 472
15-4 The Sun Diameter Analysis 475
15-5 The Sun And Earth Motions Rate Of Time (1 day =365.25 days) 478
15-6 The Sun Rays Creation 486
16- Mercury Jupiter Distance Analysis 492
16-1 Mercury Jupiter Distance Analysis (720.7 mkm) 493
16-2 (Jupiter And Mercury Motions Analysis) 506
16-3 Jupiter Distances Analysis 510
Appendix No.1 The Solar System Equal Distances List 511
Claims Be Sent Against The Decision Of Nobel Prize For Physics 2021 513
References and Biography 518

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1- Introduction
- Let's summarize this paper main idea in following:
- The paper hypothesis tells
- The solar system be created out of one light beam. Means, the planets matters and
their distances be created out of one light beam.
- The logical result should be that (The Solar System Is One Building)
- What does mean "one building"? Or "one body"?
- The horse liver is a big one but the mouse liver is a small one.
- "One building" means, the parts dimensions are created in proportionality with one
another – so it means – We Can Conclude All Parts Data From One Data
i.e.
- If we know that (Mercury orbital distance =57.9 mkm), based on this data only we
should conclude all other planets orbital distances, periods and velocities as well as
their diameters, rotation periods, orbital inclinations and masses – we don't need to
observe the planets to know this data, Because we simply can conclude all of them
theoretically from the data we know (Mercury orbital distance =57.9 mkm)
- Here we catch the useful result of this paper – and we see it provides a new vision
for the solar system motion –
Shortly –
- We know that (Mercury orbital distance =57.9 mkm), let's summarize how to
predict all planets fact sheet data based on this data
- Let's summarize that in short steps
- (1)
- We know that (Mercury Orbital Distance be =57.9 mkm) – So
- (2)
- By the equation (T2
= 4T0 (T- T0))

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- T = The period of time Light supposed velocity (1.16 mkm/s) needs to pass a
distance = Planet Orbital Circumference
- T0 = the period of time Light supposed velocity (1.16 mkm/s) needs to pass a
distance = The Previous Planet Orbital Circumference
- We Can Conclude All Planets Orbital Distances And Periods Values
- (3)
- By the equation ((T/P) = (V/1.16))
- T= The Period Of Time Be Required By Light Supposed Velocity (1.16 mkm/s) To
Pass A Distance = Planet Orbital Circumference
- P = Planet Orbital Period
- V = Planet velocity per a solar day
- 1.16 mkm = Light Supposed Velocity (1.16 mkm/s) Motion In One Second
- The Periods use the same units (seconds)
- We Can Conclude All Planets Velocities In A Solar Day
- (4)
- By the equation (v1/v2) = (s/r)
- We Can Conclude All Planets Diameters And Rotation Periods
- (5) By the equation
- We can see one geometrical design be found behind the solar system

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- (5) By the rate (v1/v2)
- We Can Conclude All Planets Orbital Inclinations + Axial Tilts
- (6) By the rate (m1/m2)
- m1 = planet mass
- m2 = another planet mass
- The rate (m1/ m2) = planet orbital inclinations
- We Can Conclude All Planets Masses
- Shortly
- The paper provides proves for this new vision (the solar system is one building)
- And what kind of motion the solar planets move based on this new vision?
- Based on this vision, the solar system motion must depend on one geometrical
design, means
- The solar planets must move one unified motion as train carriages move together.
Or similar to ships sail over a sea move by this sea waves
- Also the solar group can be similar to 9 waterwheels be built on one canal and this
canal water causes to rotate the 9 waterwheels. Although the motion direction of
the waterwheels rotation be different from the water motion direction, but no
waterwheel can rotate without the water motion.
- The solar system also must be similar to a chess board each planet moves a
calculating motion in comparison with the other planets motions
- In fact the solar group motion should be similar to a great clock each planet be
similar to one gear in it and all planets motions be unified in one general motion
causes the clock motion.

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- Here we have 3 basic concepts be proved in this paper
- (1st
Concept) The Solar System Be Created Out Of One Light Beam
- (2nd
Concept) The Solar System Is One Building
- And
- (3rd
Concept) A light beam its velocity 1.16 million km per second be the solar
system parent
- This 3rd
concept tells, the solar system be created out of one light beam its velocity
be =1.16 mkm/s
- Because this light beam velocity be registered in the planets creation and motion
data we can discover its existence –
- We know this light beam with supposed velocity (1.16 mkm/s) because we can
conclude the planets fact sheet (all) data by dependency on the 2 light beams
velocities (the known velocity 0.3 mkm/s) and (the supposed velocity 1.16 mkm/s)
and by using any one data in the sheet – for example – Mercury orbital distance
=57.9 mkm
- Only based on these 3 numbers (1.16 mkm/s, 0.3 mkm/s and 57.9 mkm) we can
conclude All Data In The Solar Planets Fact Sheet Theoretically
- Let's Take A Look On The Paper Contents
- The Paper Hypothesis Proves Data And Discussion Be Found In 4 Points
- (These Points Number Are No.3, No.4, No.5 and No.6) the fours points be used as
one group
- Point No. (3) Provides the Proves Data Only (11 Data)
- Points No. (4 and 5) Provide The Data Discussion And Analysis
- Point No. (6) Provides Answers for the left questions in the points (3,4 and5)
- The Solar Planets Motions Use Different Rates Of Time
(Be discussed in points no. 7-8-9 of this paper) and

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(The moon orbital motion equation be discussed in point no.9-11)
- Mars Migration Theory
(Be discussed in point no. 10 of this paper)
- The Solar System Distances Be Created In A Network Form
(Be discussed in point no. 11 and 12 of this paper)
- Planet Mass Effect On its Motion
- Saturn Motion Analysis
- The Sun Age Description
- Mercury Jupiter distance analysis

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2- Methodology
- I use the planets data analysis to discover the solar system creation and motion
facts – The method is so useful because it protects from the imaginary ideas
- The method simply put the planets data in comparison with the theory and tries to
know if there's a consistency between both – let's use 3 examples to explain how
this method works
- Example No. 1 (Mars Migration Theory)
- Giant-Impact hypothesis tells that, a planet in Mars Size had collided with the
Earth and caused the moon creation.
- Can Mars Itself do that? NO
- The theory tells no hope Mars itself did this collision – So what does the data tell?
- If the planets be found in order –Mars position should be after Mercury –because
the 4 planets be in order for their diameters, masses and orbital distances –this data
tells that Mars position was the second planet between Mercury and Venus and
Mars may be migrant from its original orbital distance!
- Let's move with this hypothesis for a while
- Suppose Mars was the second planet after Mercury and had migrated what would
happen? Mars had collided with Venus and then with Earth –
- Giant-Impact hypothesis asks why Venus has no its own moon spite it had suffered
of a collision similar to the Earth? the answer is a simple one
- Imagine Mars was the second planet after Mercury (84 mkm) and had migrated to
its current position (227.9 mkm), in its displacement, Mars was pushed by force
and had collided with Venus and pushed all debris with it in its motion direction -
Venus found no debris around – for that Venus couldn't create its own moon-
- Another question asks about (the origin of the lunar magma ocean!) Venus, The
Lunar Magma Ocean is came from Venus, it's a part of Venus found by the
collision between Mars and Venus but Mars pushed all debris with it in its motion
direction and left Venus without debris

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- Earth gravity is greater than Venus and the debris lost some of their momentum
and by that the Earth could create its own moon where the moon rocks are
consisted of Venus, Earth and Mars debris
- The fact Mars has 2 moons is one more proof for this idea, because Mars with
small mass could attract 2 moons and Venus couldn't.
- The rest debris be attracted by Jupiter and consisted the asteroid belt
- The planets data analysis put the fact beside one another and by that the vision be
more clear
- Example No. 2 (The Moon Orbit Design)
- The moon daily displacement is 88000 km, during 29.53 days (the moon day
period) the total displacements be = 2598693 km- If this distance be the moon
apogee circumference – the moon apogee radius should be =413600 km!!
- Also
- Because the moon displacement daily is so long, the distance 2598693 km be so
long and the moon will be prisoner in the apogee orbit with radius 413600 km and
can't revolve around the Earth through any more near orbit…
- The data creates real astonishment –
- Because
- The moon apogee radius is 406000 km only and
- The moon revolves around Earth through many near orbits and even the moon can
reach to the perigee point (363000 km).
- How to explain this data??
- The intelligent moon creates an angle (θ) between its motion direction and its
orbital horizontal level – by that the real displacement through the orbit be not
equal =88000 km instead it be = 88000 km cos (θ)
- As a result the total displacements during (29.53 days) be less than 2598693km
and by this technique the moon could revolve around Earth through near orbits

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- The apogee orbital radius 406000 km still needs more explanation which is done in
the paper discussion – but the data analysis is the only method by which we can
discover the moon creation of the angle (θ) in its motion. based on this angle the
moon orbital motion equation be formed which be discussed in Point no. (9-11)
- Example No. 3 (The Solar Distances are A Network)
- The solar system distances are created based on one geometrical design and in a
network form.
- The solar system distances be created as a chess board – the distances is one group
found based on One Geometrical Design - this fact be concluded from the data
The Distances
2 x 149.6 mkm (Earth orbital distance) = 119.7 mkm (Venus Mars distance) x 2.48
227.9 mkm (Mars orbital distance) =91.7 mkm (Mercury Earth Distance) x 2.48
2 x 680 mkm (Venus orbital Circumference) =550.7 mkm (Jupiter Mars distance) x 2.48
778.6 mkm (Jupiter orbital distance) = 2π x 550.7 mkm (Venus Mercury distance) x 2.48
720.7mkm (Mercury Jupiter distance)= π x 91.7 mkm (Mercury Earth Distance) x 2.48 (1%)
1375 mkm (Mercury Saturn distance) = 550.7 mkm (Venus Mercury distance) x 2.48
1325 mkm (Venus Saturn distance) = π x 170 mkm (Mercury Mars Distance) x 2.48
2815 mkm (Mercury Uranus distance) = π x 360 mkm (Mercury orb. Circumference) x 2.48
2764 mkm (Venus Uranus distance) = 2 x 550.7 (Jupiter Mars distance) x 2.48(1%)
4267 mkm (Mars Neptune distance) = π x 550.7 (Jupiter Mars distance) x 2.48
3030 mkm (Uranus Pluto distance) = 1205 (Mars Saturn distance) x 2.48 (1.4%)
5127 mkm (Jupiter Pluto distance) = π x 655 (Jupiter Saturn distance) x 2.48
5678 mkm (Mars Pluto distance) = π x 720.7 (Mercury Jupiter distance) x 2.48 (1%)
654.9 mkm (Jupiter Saturn distance) = 84 mkm (Mars Original Point) x 2.48 (1%)
Shortly, the solar system orbital and internal distances total be = 55 distances from
which these 28 distances (50%) be rated to one another by the same one rate (2.48),
that can be done only if these distances be created together based on one Geometrical
design and in a network form (the distances Network be discussed in Point no.11)

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3- The Paper Hypothesis Proves
Paper hypothesis
The Solar Planets Matters And Their Distances Be Created Out Of One Light
Beam –This Light Beam Velocity Be 1.16 million km per second –
The Point Provides 11 Data Proves The Hypothesis
3-1 Preface
3-2 Data No. (1) (The Main Data)
3-3 Data No. (2) (Planet Velocity Definition Data)
3-4 Data No. (3) (Planet Velocity Analysis Data)
3-5 Data No. (4) (The Planets Periods Equation) (The Basic Data)
3-6 Data No. (5) (Planet Velocity And Its Orbital Circumference Relationship)
3-7 Data No. (6) (Planet Velocity And Its Diameter Relationship)
3-8 Data No. (7) (Part I) (One Geometrical Design Data)
3-9 Data No. (7) (Part II) (One Geometrical Design Data)
3-10 Data No. (8) (Part A) (The Distances Be In Network Form)
3-11 Data No. (8) (Part B) (Jupiter Velocity Analysis) .
3-12 Data No. (9) (Planet Motion Data Depends On Light Motion)
3-13 Data No. (10) (Planets Motions for 30 mkm)
3-14 Data No. (11) (Planets Masses Rates)
3-15 The Data General Comment

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3-1 Preface
- The Paper hypothesis tells
- (The Solar System Be Created Of One Light Beam Its Velocity 1.16 mkm/s)
- If this idea is a fact – one necessary result should be produced
- What's it??
- The planets data should be created based on one another. because all of them be
created out of one source of energy (one light beam)!
- i.e.
- We should conclude all planets data based on one data only,
- Shortly (if we know this data)
- Mercury Orbital Distance =57.9 mkm
- Based on this distance (57.9 mkm) only and by help of the 2 light velocities
(0.3mkm/s and 1.16 mkm /s) we should conclude the planets fact sheet data
completely
- Now we have a clear task
- We know one data tells (Mercury Orbital Distance =57.9 mkm) and we use 2
velocities of light (0.3mkm/s and 1.16mkm/s) based on these 3 numbers we have
to conclude all planets orbital distances, periods, and their diameters, masses, axial
tilts and orbital inclinations – briefly – The Planets Fact Sheet
- Can we do that? if we do that, we will have a reason to pay attention for this
hypothesis and analyze it in the paper next points.
- This point no. (3) provides the required data (11 data) by which we can conclude
the planets fact sheet with all data based on one data only which is (Mercury
Orbital Distance =57.9 mkm)
- This Point no.(3) provides no discussion or analysis for the data, just short
comments for necessaries.
- The data discussion and analysis be provided in Points No. (4 and 5)
- Let's summarize how these (11 data) can help us to perform the required task.

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- Data no. (1,2 and 3) proves that, Planets orbital distances and periods in addition
to planets velocities per a solar day, these data be defined directly based on light
motion.
- Data no (4) Is The Basic Data, because it provides one equation which defines
each period of time be required by light motion to pass each planet orbital distance
– by that – if we know (Mercury orbital distance be =57.9 mkm), based on this
data and the equation we can conclude each period of time required by light
motion to pass a distance = its orbital circumference
- So, based on this data only (Mercury orbital distance be =57.9 mkm) we can
conclude each planet orbital distance, period and its velocity per a solar day
- Data no.(5) provides a direct relationship between planet velocity and its orbital
circumference
- Data no.(6) provides a direct relationship between planet orbital period, planet
velocity and planet diameter
- (Planet diameter means planet own diameter as the Earth diameter 12756 km)
- Data no. (7) provides the data proves that, the solar system is one machine and
one building – means –the planets be as 9 stories in one building or be as 9 gears in
one machine of gears) as a result the planets move together one unified motion –
as carriages in one train or as gears in a great clock
- Data no. (8 ) provides the data proves that, the solar system distances be created
based on one geometrical design and in a network form – where no one distance
can be created independently from the other planets distances. Instead all distances
be created as a chess board based on one geometrical design and calculations
- Data no.(9) provides the data proves that, Planets orbital distance be rated with its
cycle period by the light motion velocity (both 2 velocities) – means the light two

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velocities be used as a rate between planet orbital circumference and its cycle
period – proves light motion effect on planet motion data
- Data no.(10) provides the data show, planet motion distance for 30 mkm
- Data no.(11) provides the data shows, planets masses rate can be concluded based
on planet orbital inclinations on one side and Planets velocities rate can be used as
planets orbital inclinations which shows that the masses rate can be concluded
based on the velocities rate.
- Shortly
- If we know only (Mercury orbital distance be =57.9 mkm) we can conclude all
data of The Planets Fact Sheet completely based on this one data, by using the 2
light beam velocities which are (the known velocity 0.3 mkm /s and the supposed
velocity 1.16 mkm/s)
- If this is a fact that lead us to the following conclusions
- (1)
- The Solar System (Planets Matters And Their Distances) Be Created Out Of
One Light Beam Its Velocity 1.16 mkm/s
- (2)
- There's A Light Beam Its Velocity = 1.16 mkm/sec
- (3)
- The Solar Group Be Created As One Building Each Planet Be A Story Or A
Part Of This Same Building
- Notice
- Data no. (4) provides a short discussion to answer (Why does Mercury orbital
distance =57.9 mkm?) - And
- Through the points no.(3, 4 and 5) many questions be left to save the data and
discussions ideas without confusion, these questions be written in red color.
- The left questions in points no. (3, 4 and 5) be answered in point no.(6)
- Let's try to prove these fact be the following 11 Data

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3-2 Data No. 1 (The Main Data)
The following Data Proves,
Planet Motion Data Be Defined By Light Motion
- (A) Light supposed velocity (1.16 mkm/s) travels a distance = The Planet Orbital
Circumference in a period (T seconds)
- (B) Light known velocity (0.3 mkm/s) travels during this period (T) a distance = D
- (C) Planet moves a distance d in a period (=Ph) where D = 2π d
- And
- Ph = The Planet Orbital Period (in hours units and not solar days)
- Let's Test That In Following:
- (1) Mercury Motion
- During 310.4s light supposed velocity (1.16 mkm/) travels 360 = 2π x 57.9 mkm
- During 310.4 s light known velocity (0.3 mkm/) travels 93.1 = 2π x 14.8 mkm
- Where
- 57.9 mkm = Mercury Orbital Distance
- 88 days = Mercury Orbital Period
- Mercury moves during 88 hours a distance = 14.8 mkm (error1.4 %)
- (2) Venus Motion
- During 586s light supposed velocity (1.16 mkm/) travels 680 = 2π x 108.2 mkm
- During 586s light known velocity (0.3 mkm/) travels 175.8 = 2π x 28 mkm
- Where
- 680 mkm = Venus Orbital Circumference
- 224.7 days = Venus Orbital Period
- Venus moves during 224.7 hours a distance =28 mkm (error1%)

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- (3) Earth Motion
- During 810s light known velocity (0.3 mkm/) travels 243 = 2π x 38.7 mkm
- Where
- 940 mkm =Earth Orbital Circumference
- 149.6 mkm= Earth Orbital Distance
- 365.25 days = Earth Orbital Period
- Earth moves during 365.25 hours a distance = 38.7 mkm (error 1.2%)
- (4) Mars Motion
- During 1235s light known velocity (0.3 mkm/) travels 370.6 = 2π x 59 mkm
- Where
- 1433 mkm = Mars Orbital Circumference
- 227.9 mkm= Mars Orbital Distance
- 687 days = Mars Orbital Period
- Mars moves during 687 hours a distance = 59 mkm (error 1%)
- (5) Jupiter Motion
- During 4222.6s light supposed velocity (1.16 mkm/) travels 4900=2π x 778.6 mkm
- During 4222.6s light known velocity (0.3 mkm/) travels 1267 = 2π x 201.7 mkm
- Where
- 4900 mkm =Jupiter Orbital Circumference
- 778.6 mkm = Jupiter Orbital Distance
- 4331 days = Jupiter Orbital Period
- 4222.6 h = Mercury Day Period
- Jupiter moves during 4331 hours a distance = 201.7 mkm (error 1.3%)

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- (6) Saturn Motion
- During 7765s light supposed velocity (1.16 mkm/) travels 9007 = 2π x 1433 mkm
- During 7765s light known velocity (0.3 mkm/) travels 2330 = 2π x 371 mkm
- Where
- 9007 mkm = Saturn Orbital Circumference
- 1433 mkm = Saturn Orbital Distance
- 10747 days = Saturn Orbital Period
- Saturn moves during 10747 hours a distance = 371 mkm (error 1%)
- (7) Uranus Motion
- During 15559s light supposed velocity (1.16 mkm/) travels 18048=2π x 2872 mkm
- Where
- 18048 mkm = Uranus Orbital Circumference
- 2872 mkm = Uranus Orbital Distance
- 30589 days = Uranus Orbital Period
- Uranus moves during 30589 hours a distance = 742 mkm (error 1%)
- (8) Neptune Motion
- During 24348s light supposed velocity (1.16 mkm/) travels 28244 =2π x4495 mkm
- Where
- 28244 mkm = Neptune Orbital Circumference
- 4495.5 mkm = Neptune Orbital Distance
- 59800 days = Neptune Orbital Period
- Neptune moves during 59800 hours a distance = 1163 mkm

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- (9) Pluto Motion
- During 31983s light supposed velocity (1.16 mkm/) travels 37100= 2π x 5906mkm
- Where
- 37100 mkm = Pluto Orbital Circumference
- 5906 mkm = Pluto Orbital Distance
- 90560 days = Pluto Orbital Period
- Pluto moves during 90560 hours a distance = 1527 mkm
- (1527 = 2π x 243)
- (10) The Earth Moon Motion
- During 639s light supposed velocity (1.16 mkm/) travels 741 mkm
- During 639s light known velocity (0.3 mkm/) travels 191.6 = 2π x 30.5 mkm
- But
- 30.5 mkm = 88000 km x 346.6 days
- The moon displacements total during 346.6 (solar days) = 30.5 mkm
- Where
- 346.6 solar days= The nodal year
- 88000 km = The moon displacement for a solar day
- 742 mkm = The distance be passed by Uranus in 30589 hours (error 1%)
- This data tells, the moon orbit regression be done by effect of Uranus motion.
- Notice
- All defined periods (the red colored periods) are the planets orbital periods in
hours units. (Data Max Error is 1.4%)
- Notice
- The moon motion be done in a period 346.6 solar days and not 346.6 hours (that
creates a special case for the moon motion) – because we need to know why the
moon uses the period (346.6) in days units and not in hours as the others.

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- Notice
- The inner planets distances in blue color ( 93.1, 175.8, 243, 370.6) can be used as
cycles periods because
- 175.8 solar days =Mercury day period
- 243 solar days = Venus rotation period
- 370.6 solar days are near to 365.25 days (error 1.5%),
- (93.1 days x 2π= 584 days =the period of periodical meeting of Earth and Venus)
- This equality depends on the rate 1mkm = 1 day
- That tells each planet creates its previous neighbor cycle period
- And the outer planets distances in blue color (1267, 2330 , 4664, 7305, 9595)
- 2330 = 2 x 1165
- 4664 = 4 x 1165
- 7305 = 2π x 1165
- This data we need later for analysis -
- Please note/
- Neptune motion is the most accurate data because it has Zero Error!
- Comments On Data No. (1)
- The data shows all planets use the same system sufficiently -

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3-3 Data No. 2 (Planet Velocity Definition Data)
The following equation defines The Planet Velocity,
- Where
- T = The period of time be required by light supposed velocity (1.16 mkm/s) to pass
a distance = a planet orbital circumference (Please review Data no.1)
- P = This Planet Orbital Period (in seconds units)
- Vd= This Planet Velocity For A Solar Day
- The periods use the same units (seconds)
- Let's Test This Equation In Following
- (a) Mercury Velocity
- T = 310.4 seconds (Please review Data no.1)
- P = 88 seconds
- (310.4 s /88s) = Vd /1.16 mkm
- V (Mercury velocity per a solar day) = 4.095 mkm daily
- Mercury orbital period = 88 solar days
- (b) Venus Velocity
- T = 586 seconds (Please review Data no.1)
- P = 224.7 seconds
- (586 s /224.7s) = Vd /1.16 mkm
- V (Venus velocity per a solar day) = 3.024 mkm daily
- Venus orbital period = 224.7 solar days
mkm
1.16
P
T d
V
=

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- (c) Earth Velocity
- T = 810 seconds
- P = 365.25 seconds
- (810 s /365.25s) = Vd /1.16 mkm
- V (Earth velocity per a solar day) = 2.574 mkm daily
- Earth orbital period = 365.25 solar days
- (d) Mars Velocity
- T = 1235 seconds
- P = 687 seconds
- (1235 s /687s) = Vd /1.16 mkm
- V (Mars velocity per a solar day) = 2.082 mkm daily
- Mars orbital period = 687 solar days
- (e) Jupiter Velocity
- T = 4222.6 seconds
- P = 4331 seconds
- (4222.6 s /4331 s) = Vd /1.16 mkm
- V (Jupiter velocity per a solar day) = 1.13184 mkm daily
- Jupiter orbital period = 4331 solar days
- Notice
- (Jupiter velocity per a solar day 1.13184 mkm is different from 1.16 mkm by 2.5%
only, this feature we should analyze in point no. 4)
- (f) Saturn Velocity
- T = 7756 seconds
- P = 10747 seconds
- (7765 s /10747 s) = Vd /1.16 mkm
- V (Saturn velocity per a solar day) = 0.838 mkm daily
- Saturn orbital period = 10747 solar days

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- (g) Uranus Velocity
- T = 15559 seconds
- P = 30589 seconds
- (15559 s /30589 s) = Vd /1.16 mkm
- V (Uranus velocity per a solar day) = 0.5875 mkm daily
- Uranus orbital period = 30589 solar days
- (h) Neptune Velocity
- T = 24348 seconds
- P = 59800 seconds
- (24348 s /59800 s) = Vd /1.16 mkm
- V (Neptune velocity per a solar day) = 0.466884 mkm daily
- Neptune orbital period = 59800 solar days
- (i) Pluto Motion
- T = 31983 seconds
- P = 90560 seconds
- (31983 s /90560 s) = Vd /1.16 mkm
- V (Pluto velocity per a solar day) = 0.406 mkm daily
- Pluto orbital period = 90560 solar days
- Notice
- Data errors are less than 1%
- Notice (The Moon Motion)
- T = 639 seconds and P= 346.6 seconds
- (639.6s /346.6 s) = 1.16 mkm/ V
- V =2.14 mkm (a velocity has error 2.8% with Mars daily velocity 2.082 mkm)!
- The moon velocity is 2.4 mkm per a solar day as we should discuss later

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- Comments On Data No. (2)
- Notice
- Planet orbital period (P) be used in seconds units because (One Day Of Planet
Motion = One Second Of light Motion) this concept be provided by The Solar
System Creation And Motion Theory, based on that, all planets orbital periods can
be used in seconds units for light motion.
- Because the 2 periods (T and P) are useful and enable use to define all planets
velocities per a solar day accurately , we will put these 2 periods rate in one table
to get more clear vision about the data development
Planet T/ P The values rate
Mercury 88/310.4 = 0.283 28.3% (----)
Venus 224.7/ 586 = 0.383 38.3 % (---)
Earth 365.25 /810 = 0.45 45% (--)
Mars 687/1235 = 0.55 55.6% (-)
Jupiter 4331 /4222.6 = 0.97 97.5%
Saturn 10747/7765 =1.384 72.2% (+)
Uranus 30589/15559 = 1.966 50.8% (++)
Neptune 59800 /24348 = 2.456 40.7% (+++)
Pluto 90560/31983 =2.8315 35.3% (++++)
- The values rate measures the rate between the 2 periods of time (T and P), and
because of that (for example), the period T = 15559 while and P (Uranus orbital
period) =30589, and for that the period 15559 be 50.8% of Uranus orbital period.
- The data shows Jupiter is the nearest point for the equality of the 2 periods (T and
P) (The difference is only 2.5% - because Jupiter velocity 1.1318 is less than light
supposed velocity 1.16 with 2.5%)
- The inner planets data shows the period (T) is less than planet orbital period (P)

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- But
- This fact is reversed in the outer planets data where the planet orbital period (P) be
greater than the period (T). That shows Jupiter is the balancing point between all
planets data.
- I try to put the data in a clear form before to add any idea of explanation.
- Because the data provides a very good proof for the following basic concepts and
conclusions which are:
(1) Planet Motion Data Be Defined Based On Light Motion.
(2) There's A Light Beam Its Velocity Be 1.16 Mkm/Sec
(3) One Second Of Light Motion = One Solar Day Of Planet Motion
(4) The Planets Distances Be Created In A Network Form
(5) There's A Continuum In The Solar System
- These conclusions are so effective on the solar system vision, for example, based
on these conclusions we can consider any planet orbital period to be used in units
of seconds and not solar days – that creates a general change for the planets data
reading –for that reason I try to provide different forms of the light motion effect
on planets motions data – I try to do that without to add any idea – I try to provide
the data alone without any negative effect on them – then after the data be
completed we should discuss how this data be created and what kind of motion the
solar system moves based on this data – so we have to provide a wide discussion
after the data providing completion.
- Notice/ Why does Mercury Day Period =4222.6 hours?
- To use the original period of time (T) as a cycle period is happened also with
Uranus only – because the original period of time (T) in Uranus Data be = 15559s
which = 15559 seconds x 4 = Uranus Day Period (17.2 hours) –
- This question is answered in point no.(4-6) (vii)

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3-4 Data No. 3 (Planet Velocity Analysis Data)
Planets Velocities Are Complementary One Another
(a)
1160000 seconds = 3600 seconds x 322.2 hours
(b)
322.2 = 47.4 km/s (Mercury velocity) x 6.8 km/s (Mercury velocity)
322.2 = 35 km/s (Venus velocity) x 2 x 4.7 km/s (Pluto velocity) (error 2%)
322.2 = 29.8 km/s (Earth velocity) x 2 x 5.4 km/s (Neptune velocity)
322.2 = 24.1 km/s (Mars velocity) x 13.1 km/s (Jupiter velocity) (error 2%)
322.2 =35 km/s (Venus velocity) x 9.7 km/s (Pluto velocity) (error 5%)
322.2 =27.78 km/s (The moon velocity) x (24.1/2) (error 4%)
(c)
300000 seconds x 4 = 3600 seconds x 333.3
333.3 = 35 km/s (Venus velocity) x 9.7 km/s (Pluto velocity) (error 1.8%)
333.3 = 24.1 km/s (Mars velocity) x 2 x 6.8 km/s (Uranus velocity) (error 1.7%)
(d)
(1.16 /0.3) = (24.1/2π)
(e)
83.33 = 2π x 13.26
(f)
322.2 =24.6 x 13.1

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Comments On Data No. (3)
- We have noticed that,
- The solar planet velocity per a solar day be created as a function in light supposed
velocity (1.16 mkm/sec),
- Because of that
- The planets velocities be created complementary one another where the light
supposed velocity (1.16 mkm/s) be used as a rate between them.
- The rate 322.2 = 1160000 /3600, means, Light supposed velocity (1.16 mkm/s)
travels for 1 second, it passes a distance =1160000 km which be seen for us as
1160000 seconds = 322.2 hours
- Data no. (b) shows, the planets velocities are created rated to one another based on
this rate (322.2) –
- The data shows the planets velocities be complementary, and that shows specific
relationship between these planets, simply, the rate 322.2 create a classification for
the planets into couples with one another.
- Notice
- The simple direct meaning of the data is clear – for example – 322.2 = 6.8 x 47.4,
this data tells – Mercury during 6.8 hours moves 1.16 mkm- and also Uranus
during 47.4 hours moves 1.16 mkm
- But
- The proportionality of planets velocities don't tell all facts. Many facts are still
hidden and need to be discovered. For example the equation (322.2 = 35 x 4.7 x 2)
(error 2%) shows that some proportionality be found between Venus velocity (35
km/s) and Pluto velocity (4.7 km/s) but more facts aren't mentioned here – because
Venus during 12104 s moves a distance =the distance Pluto moves in 90560 s =
Uranus motion distance in its day period (where 12104 km= Venus diameter and
90560 days = Pluto orbital period). I want to say, this proportionality of velocities
of Venus and Pluto causes great effect on their motions but we can't discover it

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because other data should be analyzed and refer to its proportionality with planets
velocities.
- But, why the planets velocities be created rated to one another as seen in the data
clearly?
- Because the velocity is used as a period of time – this is the simple direct meaning
we have referred -
- Simply
- The planet velocity can be used as a period of time (as a planet day period)
Let's discuss the point equations in following:
Equation no. (d)
(1.16 /0.3) = (24.1/2π)
- 1.16 mkm/s = Light Supposed Velocity
- 0.3 mkm/s = Light Known Velocity
- 24.1 km/s = Mars Velocity
Equation no. (e)
83.33 = 2π x 13.26
- 83.33 = 300000 /3600s
- 13.1 km/s = Jupiter Velocity (error 1.2% with 13.26)
Equation no. (e)
322.2 =24.6 x 13.1
- 24.6 hours = Mars Rotation Period
- 13.1 km/s = Jupiter velocity
- These 3 equations shows that, Jupiter and Mars have specific velocities, we have
to explain that in more details in the data discussion. But Mars Data shows specific
feature we need to refer in the following notice

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- Notice
- We accept that planet velocity can be used as a period of time. Based on that some
very interesting addition be found in Mars Data which is
- (24.6 h +24.6 h = 25.2 h +24.1 h)
- 24.6 hours = Mars Rotation Period
- 24.7 hours = Mars Day Period
- 25.2 degrees = Mars Axial Tilt (be used as a period of time also)
- 24.1 km/s = Mars velocity
- The accurate addition tells that these 4 values are used for one process and that
tells they are used as periods of time.
- Shortly
- Jupiter moves during Mars rotation period (24.6h) a distance = 1.16 mkm where
Jupiter orbital circumference be designed based on the rate (1.16) and that makes
Mars rotation period (24.6 h) is so effective period of time on Jupiter motion data,
and on all other planets motions.
- We discuss Mars rotation period in point no.(4) of this paper.

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3-5 Data No. (4) (The Planets Periods Equation) (The Basic Data)
T2
=4 T0 (T - T0)
- Where
- T = The Period Of Time Be Required By Light Supposed Velocity (1.16 mkm/s)
To Pass A Distance = The Planet Orbital Distance
- T0 = The Period Of Time Be Required By Light Supposed Velocity (1.16 mkm/s)
To Pass A Distance = The Previous Planet Orbital Distance
- Let's test this Equation in following
(1) Venus Motion
- T= 586.2 seconds
- T0= 310 seconds
- (586.2)2
= 310 x 4 (586.2 - 310) (No Error)
- 310 s Light supposed velocity (1.16 mkm/s) needs to pass a distance =360 mkm =
Mercury Orbital Circumference
- 586.2 s Light supposed velocity (1.16 mkm/s) needs to pass a distance =680 mkm
= Venus Orbital Circumference
- Venus depends on Mercury
(2) Earth Motion
- T= 810 seconds
- T0=310 seconds
- (810)2
= 310 x 4 (810 - 310) (Error 2.8%)
- 310 s Light supposed velocity (1.16 mkm/s) needs to pass a distance =360 mkm =
- 810 s Light supposed velocity (1.16 mkm/s) needs to pass a distance =940 mkm=
Earth Orbital Circumference

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(3) Mars Motion
- T= 1235.4 seconds
- T0= 586.2 seconds
- (1235.4)2
= 586.2 x 4 (1235.4 - 586.2) (No Error)
- 1235.4s Light supposed velocity (1.16 mkm/s) needs to pass a distance =1433
mkm = Mars Orbital Circumference
(4) Ceres Motion
- T= 2248 seconds
- T0= 1235.4 seconds
- (2248)2
= 1235.4 x 4 (2248- 1235.4) (Error 0.5% )
- 2248 s Light supposed velocity (1.16 mkm/s) needs to pass a distance =2608
mkm= Ceres Orbital Circumference and Ceres orbital distance =415 mkm
- Ceres depends on Mars
(5) Jupiter Motion
- T= 4224 seconds
- T0= 2248 seconds
- (4224)2
= 2248 x 4 (4224- 2248) (No Error)
- 4224s Light supposed velocity (1.16 mkm/s) needs to pass a distance =4900 mkm
= Jupiter Orbital Circumference
- Jupiter depends on Ceres
(6) Saturn Motion
- T= 7765 seconds
- T0= 4224 seconds
- (7765)2
= 4224 x 4 (7765- 4224) (Error 0.4%)
- 7765s Light supposed velocity (1.16 mkm/s) needs to pass a distance =9007 mkm
= Saturn Orbital Circumference
- Saturn depends on Jupiter

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(7) Uranus Motion
- T= 15559 seconds
- T0= 7765 seconds
- (15559)2
= 7765 x 4 (15559-7765) (No Error)
- 15559s Light supposed velocity (1.16 mkm/s) needs to pass a distance =18048
mkm = Uranus Orbital Circumference
- Uranus depends on Saturn
(8) Neptune Motion
- T= 24348 seconds
- T0= 15559 seconds
- (24348)2
= 15559 x 4 (24348- 15559) (Error 4%)
mkm = Neptune Orbital Circumference
- Neptune depends on Uranus
(9) Pluto Motion
- T= 31983 seconds
- T0= 15559 seconds
- (31983)2
= 15559 x 4 (31983-15559) (No Error)
mkm = Pluto Orbital Circumference
- Pluto depends on Uranus
- Notice
- The Equation has 3 exceptions
- Pluto depends on Uranus and doesn't on Earth
- The data analysis may explain these 3 exceptions reason

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- Notice
- The equation works sufficiently but has great error only with Earth (2.8%) and
Neptune (4%) – the other planets errors are less than 1%
- (1st
Point)
- Data No. (4) is the basic data because it creates a relationship between all planets
periods of time – So, If we know that- Light supposed velocity (1.16mkm/s) needs
310 sec to pass a distance = 360 mkm =Mercury Orbital Circumference, based on
this equation we can define all other periods of time be required by light supposed
velocity (1.16mkm/s) to pass the other planets orbital circumferences – based on
these periods we can define these planets orbital periods and velocities
- Means,
- We here in very important moment. Because the paper hypothesis can be proved
based on this equation –the solar planets be similar to one building consisted of 9
stories – Mercury is the first story – based Mercury we can define Venus period
and Earth period –based on Venus we can define Mars period ….etc
- Here we need one data (Mercury orbital distance = 57.9 mkm) and all planets
orbital distances can be defined, based on that, all planets orbital periods and
velocities can be defined – one data can produce all the planets 3 types of data –the
next step is –If we know the planet velocity can we define its diameter? in next
data we answer this question -

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- (2nd
Point)
- The equation in fact is the planets orbital distances equation – we have discovered
it from long time ago - let's remember it in following:
- The Distance Equation
- d2
=4 d0 (d - d0)
- where
- d = A Planet orbital distance
- d0= Its direct Previous neighbor planet orbital distance –
- Let's test it in following:
(1) Venus Motion
- d = 108.2 mkm (Venus Orbital Distance)
- d0 = 57.9 mkm (Mercury Orbital Distance)
- d2
=4 d0 (d - d0)
- (108.2)2
= 4 x 57.9 x (108.2-57.9) (No Error)
- Venus depends on Mercury
(2) Earth Motion
- d = 149.6 mkm (Earth Orbital Distance)
- d0 = 57.9 mkm (Mercury Orbital Distance)
- d2
=4 d0 (d - d0)
- (149.6)2
= 4 x 57.9 x (149.6-57.9) (error 2.5%)
(3) Mars Motion
- d = 227.9 mkm (Mars Orbital Distance)
- d0 = 108.2 mkm (Venus Orbital Distance)
- d2
=4 d0 (d - d0)
- (227.9)2
= 4 x 108.2 x (227.9-108.2) (Error is less than 1%)

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(4) Ceres Motion
- d = 415 mkm (Ceres Orbital Distance)
- d0 = 227.9 mkm (Mars Orbital Distance)
- d2
=4 d0 (d - d0)
- (415)2
= 4 x 227.9 x (415-227.9) (Error is less than 1%)
- Ceres depends on Mars
(5) Jupiter Motion
- d = 778.6 mkm (Jupiter Orbital Distance)
- d0 = 415 mkm (Ceres Orbital Distance)
- d2
=4 d0 (d - d0)
- (778.6)2
= 4 x 415 x (778.6- 415) (Error is less than 1%)
- Jupiter depends on Ceres
(6) Saturn Motion
- d = 1433.5 mkm (Saturn Orbital Distance)
- d0 = 778.6 mkm (Jupiter Orbital Distance)
- d2
=4 d0 (d - d0)
- (1433.5)2 = 4 x 778.6 x (1433.5- 778.6) (Error is less than 1%)
- Saturn depends on Jupiter
(7) Uranus Motion
- d = 2872.5 mkm (Uranus Orbital Distance)
- d0 = 1433.5 mkm (Saturn Orbital Distance)
- d2
=4 d0 (d - d0)
- (2872.5)2
= 4 x 1433.5 x (2872.5- 1433.5) (Error is less than 1%)
- Uranus depends on Saturn
(8) Neptune Motion
- d = 4495.1 mkm (Neptune Orbital Distance)
- d0 = 2872.5 mkm (Uranus Orbital Distance)

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- d2
=4 d0 (d - d0)
- (4495.1)2
= 4 x 2872.5 x (4495.1- 2872.5) (4% Error)
- Neptune depends on Uranus
(9) Pluto Motion
- d = 5906 mkm (Pluto Orbital Distance)
- d0 = 2872.5 mkm (Uranus Orbital Distance)
- d2
=4 d0 (d - d0)
- (5906)2
= 4 x 2872.5 x (5906- 2872.5) (Error is less than 1%)
- Pluto depends on Uranus and doesn't on Neptune
- Notice
- The equation has 3 exceptions which are:
- Earth depends on Mercury and Not on Venus
- Mars depend on Venus and not on Earth
- Pluto depend on Uranus and not on Neptune
- We hope the next short discussion explain these exceptions and answer why does
Mercury orbital distance =57.9 mkm or why its period of time =50 s (310s)?
- Notice
- As we have seen, the period of time (T) can help to define planet orbital distance,
period and velocity as well as this planet diameter, rotation period, orbital
inclination and masses rate
- That tells, the planet data be defined based on its position!
- Can this sentence mean?- If Mars be in Earth Position – The Planet Mars Data
should be changed into Earth Data – but that's not true –because we have
discovered Mars Migration because Mars data isn't suitable to its position – means
– in the planet creation its data can be defined based on its orbital distance but
later the planet may increase its mass and that may prevent the change in data if
the planet changed its position -

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- Notice
- We notice, The 2 equations are very near because–the 2 equations are rated by
(1.16) simply - Also
- The equation can be used by light known velocity (0.3mkm/s) accurately as with
light supposed velocity (1.16 mkm/s), it works simply sufficiently.
A Short Discussion
T2
=4 T0 (T - T0)
- Where
- T = The Period Of Time Be Required By Light Supposed Velocity (1.16 mkm/s)
To Pass A Distance = The Planet Orbital Distance
- T0 = The Period Of Time Be Required By Light Supposed Velocity (1.16 mkm/s)
To Pass A Distance = The Previous Planet Orbital Distance
- (Point No. 1) (The equation exceptions)
- Earth depends on Mercury and Not on Venus
- Mars depend on Venus and not on Earth
- Pluto depend on Uranus and not on Neptune
- Why?
- Earth Motion
- 810 sec – 310sec = 500 sec
- Where
- 810 sec be required by light supposed velocity (1.16 mkm/s) to pass 940 mkm=
Earth Orbital Circumference
- 500 sec be required by light known velocity (0.3 mkm/s) to pass 150 mkm = Earth
Orbital Distance (149.6 mkm)
- The data tells for a geometrical necessity Earth depends on Mercury

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- Mars Motion
- 1235 sec – 586 sec = 655 sec
- Where
Mars Orbital Circumference
Venus Orbital Circumference
- 655.7 hours = The Moon Rotation Period
- That tells Mars depends on Venus because of the interaction be found between
Mars and the moon (where the moon depends on Venus)
- Pluto Motion
- 31983 sec = 16166 sec +15559 sec
- Where
- 31983 sec be required by light supposed velocity (1.16 mkm/s) to pass 37100 mkm
= Pluto Orbital Circumference
- 16166 sec = the total periods of all planets from Mercury to Saturn
- 15559 sec be required by light supposed velocity (1.16 mkm/s) to pass 18048 mkm
= Uranus Orbital Circumference
- Means
- All planets periods total (except) Neptune = Pluto Period of time
- The data shows Pluto depends on Uranus for a geometrical necessity

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- (Point No. 2)
- (Why Mercury period =310 s or 50 s)??
(1)
- Light known velocity (0.3 mkm/s) passes a distance =100733 mkm =The Planets
Orbital Circumferences Total in a period =335777 seconds
- 335777 seconds = 3600 x 93.2 hours
- The period of time (310s) which is required by light supposed velocity
(1.16mkm/s) to pass 360 mkm = Mercury Orbital Circumference –during this
period (310s) light known velocity (0.3mkm/s) travels 93 mkm which can be seen
as 93 hours – That makes Mercury the origin point of the solar system.
(2)
- (88056 sec /310 sec) = 285 = (The Sun Diameter / Mercury Diameter)
- The Sun Diameter = 1.392 million km
- Mercury Diameter = 4879 km
- 88060 seconds = 86825 seconds + 1235 seconds
- During 86825 s light supposed velocity (1.16 mkm/s) passes a distance =100733
mkm = The Planets Orbital Circumferences Total (0.5% error with the solar day
86400s) - During 1235 s light supposed velocity (1.16 mkm/s) passes a distance
=1433 mkm (Mars orbital circumference)
- The point here is that
- 4900 mkm =360 mkm +680 mkm +940 mkm +1433 mkm +1433 mkm
- The data tells Jupiter orbital circumference (4900 mkm) = Mercury's (360) +
Venus' (680)+ Earth's (940) + Mars's (1433) + 1433
- This second value (1433 mkm) be considered as the moon orbital circumference
around the sun – we can't use (940 mkm) instead because the total will not be =
4900 mkm – we have to use (1433 mkm) for second time and in this case the total
accurately will be = (4846 mkm) =4900 mkm (error 1%)

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- In Mercury data the value (1433) be used for second time also because the total
86825 s can be 88060 s only if we add additional 1235s which be used for 1433
mkm – by this addition the rate (88060 s/310s) = (1.392mkm/4879km) accurately
which tells there's a geometrical necessity for this equality.
- Notice
- 100733 mkm (The Planets Orbital Circumferences Total) =2π x 16033 mkm
- 16033 mkm = 1.16mkm/s x 13821 seconds (=3.83 hours)
- But
- 1.16 mkm/s (light supposed velocity) /0.3 mkm/s (light known velocity) =3.866
- (the difference between 3.83 and 3.866 is 1%)
- The data tells the planets orbital distances depends on One Geometrical Design

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3-6 Data No. (5) (Planet Velocity And Its Orbital Circumference Relationship)
A RULE
- The rule is a simple and sufficient, let's test it in following…
Example No. (1)
- (940 mkm/360 mkm) = (47.4/29.8)2
(error 1.6%)
- Where
- 940 mkm = Earth Orbital Circumference
- 47.4 km/s = Mercury Velocity
- The rule works sufficiently
- Let's test one more data
Example No. (2)
- (18048 mkm/4900 mkm) = (13.1/6.8)2
- Where
- 18048 mkm = Uranus Orbital Circumference
- 4900 mkm = Jupiter Orbital Circumference
- 13.1 km/s = Jupiter Velocity
- 6.8 km/s = Uranus Velocity
- The rule works sufficiently
- Let's test one more data

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Example No. (3)
- (37100 mkm /680 mkm) = (35/4.7)2
(error 1.6%)
- Where
- 37100 mkm = Pluto Orbital Circumference
- 680 mkm = Venus Orbital Circumference
- 35 km/s = Venus Velocity
- 4.7 km/s = Pluto Velocity
- The rule works sufficiently and the error is in normal range but
- Notice
- Should Planet velocity be decreased while its orbital distance be increased? The
planet velocity equation tells something different – let's remember -
- (T/P) =(V/1.16)
- While planet orbital distance be increased the period of time (T) be required for
light supposed velocity (1.16 mkm/s) to pass its orbital circumference be increased
also –that tells – longer distance should cause speeder velocity –
- So why the velocity be decreased with longer distance? Because of planet orbital
period which consumes the distance increasing rate and the velocity (supposed)
increasing rate and forces the velocity to be decreased.
- The planets orbital periods be used for the different rates of time of the planets
motions. We discuss the planets using different rates of time in points (7,8 and 9)

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- 4.095 mkm (Mercury daily velocity) x 57.9 mkm (Mercury orbital distance) = 237
- 3.024 mkm (Venus daily velocity) x 108.2 mkm (Venus orbital distance) = 327
- 2.4 mkm (The moon daily velocity) x 149.6 mkm (Earth orbital distance) = 359
- 2.574 mkm (Earth daily velocity) x 149.6 mkm (Earth orbital distance) = 385
- 2.082 mkm (Mars daily velocity) x 227.9 mkm (Mars orbital distance) = 475
- 1.13184 mkm (Jupiter daily velocity) x 778.6 mkm (Jupiter orbital distance) = 881
- 0.838 mkm (Saturn daily velocity) x 1433 mkm (Saturn orbital distance) = 1200
- 0.5875 mkm (Uranus daily velocity) x 2872 mkm (Uranus orbital distance) = 1688
- 0.46688mkm (Neptune daily velocity) x 4495 mkm (Neptune orbital distance) = 2099
- 0.406 mkm (Pluto daily velocity) x 5906 mkm (Pluto orbital distance) = 2398
- More Details
Planet Velocity (v) Distance (d) Vd Difference
Mercury 4.095 mkm 57.9 mkm 237 237+237 =474
Venus 3.024 mkm 108.2 mkm 327 327-237=90
The moon 2.4 mkm 149.6 mkm 359 359-327 =32
Earth 2.574 mkm 149.6 mkm 385 385-327=58
Mars 2.082 mkm 227.9 mkm 474.5 474-385=90
Jupiter 1.1318 mkm 778.6 mkm 881.2 881-474=406
Saturn 0.838 mkm 1433 mkm 1200 1200-881=319
Uranus 0.857 mkm 2872 mkm 1687.3 1687.3-1200= 487.3
Neptune 0.4668 mkm 4495 mkm 2099 2099-1687=412
Pluto 0.406 mkm 5906 mkm 2398 2398-2099=299
Where ( (v) = Planet velocity / a solar day and (d) = Planet Orbital Distance)- the data
uses the planets orbital distances as periods of time (1mkm =1 solar day)
- The data shows, there's a general geometrical distribution for this value (vd)
because planet orbital distance be increased by a rate greater than the velocity
decreasing by that the value (vd) shows a continuous increase.

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- (Point No. A)
- The Inner Planets
- Mercury is the master point in the inner planets and Mercury defines the data, let's
see how that's done
- Mercury value is 237 mkm
- The inner planets depend on (237 mkm x 2) = 474 mkm (=Mars Value)
- Also
- (237 mkm x 1.5) = 359 mkm (=The Moon Value)
- We know that
- Mercury day period (175.94 solar days) =2 x 88 days (Mercury orbital period)
- 88 days (Mercury orbital period) =1.5 x 58.65 days (Mercury rotation period)
- That means, Mercury defines the positions and velocities of Mars and the moon
based on Mercury own cycles periods rates
- Then
- From the 2 terminal points (Mercury and Mars), the design left (90) and put the
planet by that Venus be far from Mercury with (90) and Earth be far from Mars
with (90)
- By that the five planets data be defined based on Mercury cycles periods rate.
- Notice / 237+327+385+474.5 =1424= 4 x 356 (the moon value =359 error 0.8%)
- The Outer Planets
- 406.7 +319+487+412=1624.7 = 4x 406.2
- That shows the 4 planets (Jupiter, Saturn, Uranus and Neptune) be distributed
geometrically
- Notice
- The moon value shows that Venus effects on the moon motion and defines its
velocity. (The moon velocity/ solar day 2.4 mkm we prove that in point) (no. 9-11)

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- (Point No. B)
- We see that Mars value =474 = 237 x 2
- This connects Mars with Mercury. In Fact Mercury and Mars be connected
because
- Light known velocity (0.3 mkm/s) needs 193 seconds to pass 57.9 mkm (Mercury
orbital distance)
- And
- Light supposed velocity (1.16 mkm/s) needs 196.5 seconds to pass 227.9 mkm
(Mars Orbital Distance) (between 193s and 196.5s the error 2%)
- This using of data be repeated before
- The period 16166 s = The periods total of all planets from Mercury to Saturn
- Now
- Light known velocity (0.3 mkm/s) travels during 16166 s a distance = 4850 mkm
(= 4900 mkm Jupiter Orbital Circumference error 1%)
- Light supposed velocity (1.16 mkm/s) travels during 16166 s x 2 a distance =
37500 mkm (= 37100 mkm Pluto Orbital Circumference error 1%)
- I want to say that
- This is a geometrical rule or a data system be using by light motion to create a
connection between the planets data
- As example
- Mercury Day Period needs 5040 seconds to be = 4224 hours why?
- Because
- Light supposed velocity (1.16 mkm/s) travels during 5040 s a distance =5848 mkm
= Mercury Pluto Distance

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3-7 Data No. (6) (Planet Velocity And Its Diameter Relationship)
- (v1/v2) = (s/r)
Also
- The rate (v1/ v2) defines Planet Orbital Inclination –
- Let's test this rule for the solar planets data in following
(a)
(89143 /49528) = (9.7 /5.4)
(b)
(24106 x2 /120536) = (9.7 /24.1)
(c)
(42683 /51118) = (5.4/6.8) (error 5%)
(d)
(14178 /2390) = (27.78/4.7)
(e)
(10500 / 142984) = (13.1/29.8 x8)
(f)
671/6792 = (4.7 /24.1 x2) (error 1.3%)

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II- Discussion
- (v1/v2) = (s/r)
- v1 = Planet Velocity In Second
- v2 = Neighbor Planet Velocity In Second
Equation no. (a)
(89143 /49528) = (9.7 /5.4)
- Neptune Orbital Period (59800 solar days) has 89143 Neptune rotation periods
(16.1 h)
- 49528 km = Neptune Diameter
- 9.7 km/s = Saturn Velocity
- 5.4 km/s = Neptune Velocity
- The data follows the rule perfectly – error less than 1%
- Also
- 9.7 /5.4 = 1.8 where 1.8 degrees = Neptune Orbital Inclination
Equation no. (b)
(24106 x2 /120536) = (9.7 /24.1)
- Saturn Orbital Period (10747 solar days) has 24106 Saturn rotation periods (10.7h)
- 120536 km = Saturn Diameter
- 9.7 km/s = Saturn Velocity
- Also
- 24.1/9.7 = 2.5 where 2.5 degrees = Saturn Orbital Inclination

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Equation no. (c)
(42683 /51118) = (5.4/6.8) (error 5%)
- Uranus Orbital Period (30589 solar days) has 42683 Uranus rotation periods
(17.2h)
- 51118 km = Uranus Diameter
- 5.8 km/s = Neptune Velocity
- 6.8 km/s = Uranus Velocity
- The data has an error 5%
- Also
- 5.4/6.8 = 0.8 where 0.8 degrees = Uranus Orbital Inclination
Equation no. (d)
(14178 /2390) = (27.78/4.7)
- Pluto Orbital Period (90560 solar days) has 14178 Pluto rotation periods (153.3 h)
- 2390 km = Pluto Diameter
- 27.78 km/s = The moon Velocity
- The moon orbital inclination (5.1 degrees) is defined as a rate between its velocity
and Neptune velocity (27.78/5.4) = 5.1
Equation no. (e)
(10500 / 142984) = (13.1/29.8 x 6)
- Jupiter Orbital Period (4331 solar days) has 10500 Jupiter rotation periods (9.9 h)
- 142984 km = Jupiter Diameter
- 13.1 km/s = Jupiter Velocity
- Also

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- (29.8/13.1) x 2 = 4.5 where (Jupiter orbital inclination 1.3 deg + Jupiter axial tilt
3.1 deg) = 4.4 deg
Equation no. (f)
671/6792 = (4.7 /24.1 x2)
- Mars Orbital Period (687 solar days) has 671 Mars rotation periods (24.6 h)
- 6792 km = Mars Diameter
- The data has error 1.3%
- Also
- Mars orbital inclination 1.9 deg be defined as a rate between Mercury and Mars
velocities
- The rule doesn't work with the Earth almost because the moon orbital period = the
moon rotation period – and there are difficulties to us the rule with Mercury and
Venus because the rotation period is so great in comparison with the orbital period
- The previous data is important because the planets velocities rates be in function
with their diameters, rotation periods and orbital periods,
- Because, the time equation (T2
= 4T0 (T-T0), this equation defines all planets
periods of time that connects all planets orbital distances, periods and velocities in
one equation – our current Data (No. 6) add one more connection between planets
velocities and their diameters, rotations periods and orbital periods by that all
planet data be connected by one equation- clearly – all planets data be controlled
by one equation – and can be defined based on one data only for example
(Mercury orbital distance be =57.9 mkm) This data only enough to rewrite the
planets fact sheet data concluding by geometrical rules - That proves the solar
system is one machine created out of one light beam its velocity 1.16mkm/s

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Notice
(1)
- (9.7 /5.4) = 4π / 7
- (6.8 /5.4) = 4/ π (1%)
- (9.7 /6.8) = π2
/ 7 (1%) And
(9.7/5.4) + (13.1/9.7) = π
(2)
(27.78 x π2
/ 47.4 x2) = 2.86 (error 1%)
(35 x2 /24.1) = 2.86 (error 1.5%)
(4 x29.8 / 13.1π) = 2.86 (error 1.3%)
(27.78 /9.7) = 2.86 (Zero error)
(24.1 x 2 / 5.4 π) = 2.86 (error less 1%)
(2π x 13.1)/(5.4)2
= 2.86 (error 1.3%)
(9.7 x 2/6.8) = 2.86 (error less 1%)
(6.8 x 2/ 4.7) = 2.86 (error 1%)
(1.16/0.406) = 2.86 (Zero error)
- 47.4 km/s = Mercury velocity 35 km/s = Venus velocity
- 29.8 km/s = Earth velocity 27.78 km/s = The Earth Moon velocity
- 24.1 km/s = Mars velocity 13.1 km/s = Jupiter velocity
- 9.7 km/s = Saturn velocity 6.8 km/s = Uranus velocity
- 5.4 km/s = Neptune velocity 4.7 km/s = Pluto velocity
- 1.16 mkm = Jupiter motion distance during mars rotation period
- 0.406 mkm = Pluto motion distance during a solar day
- This data shows great importance for Pluto velocity per a solar day because its rate
with light velocity (2.86) controls all planets velocities rates.

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3-8 Data No. (7) (Part I) (One Geometrical Design Data)
Can The Solar Planets Motions Depend On One Geometrical Design?
(A)
- Light known velocity (0.3 mkm/s) needs 1200 seconds to pass a distance = 360
mkm = Mercury Orbital Circumference – But
- Mercury (4.095 mkm /day) moves during 1200 solar day a distance = 4914 mkm
(where 4900 mkm = Jupiter Orbital Circumference)
(B)
- Light known velocity (0.3 mkm/s) needs 16333 s to pass a distance = 4900 mkm =
Jupiter Orbital Circumference – But
- Jupiter (1.1318 mkm /day) moves during 16333 solar days a distance = 18486.4
mkm (where 18048 mkm = Uranus Orbital Circumference – error 2.4%)
(C)
- Light known velocity (0.3 mkm/s) needs 60161 s to pass a distance = 18048 mkm
= Uranus Orbital Circumference – But
- Uranus (0.5875 mkm /day) moves during 60161 solar days a distance = 35346
mkm (where 37100 mkm = Pluto Orbital Circumference – error 4.8%)
(D)
- Light known velocity (0.3 mkm/s) needs 123667 s to pass a distance = 37100 mkm
= Pluto Orbital Circumference –
- Notice
- Light known velocity (0.3 mkm/s) needs 3134 s to pass a distance = 940 mkm =
Earth Orbital Circumference – But
- During 2 x 3134 solar days, the moon displacements total = 550.7mkm =Mars
Jupiter Distance.

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Comments On Data No. (4) (Part I)
- This data idea tells that, there's one thread connects Mercury, Jupiter, Uranus and
Pluto together. That can be noticed because the 4 planets motions depends on the
value (4900).
- Jupiter Orbital Circumference = 4900 mkm
- Uranus needs 4900 solar days to move a distance = 2872.5 mkm (Uranus Orbital
Distance)
- Pluto needs 3 x 4900 solar days to move a distance = 5906 mkm (Pluto Orbital
Distance) (error 1%)
- Notice
- Neptune needs 2 x 4900 solar days to move a distance = 4495.1 mkm (Neptune
Orbital Distance) (error 1.7%)
- The data tells some geometrical design must be found behind which we need to
discover
- For that reason we examine all planets motions to discover the effect of this value
(4900) on them
- Notice
- (4890 days/687days) =7.1= (14.14 days/ 2 days)

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3-9 Data No. (7) (Part II) (One Geometrical Design Data)
(A)
mkm = Mercury Orbital Circumference – But
- Mercury (4.095 mkm /day) moves during 1200 solar day a distance = 4914 mkm
(Where 4900 mkm = Jupiter Orbital Circumference)
- Notice
- Mercury orbital distance 57.9 mkm and Mercury needs 14.14 days (=340 h) to
pass this distance (but light known velocity (0.3mkm/s) moves in (340s) a distance
102 mkm). (Note Please, 4914 mkm =102 mkm x 48.2)
(B)
- Light known velocity (0.3 mkm/s) needs 2266.7 seconds to pass a distance = 680
mkm = Venus Orbital Circumference – But
- Venus (3.024 mkm /day) moves during 2266.7 solar day a distance = 6855 mkm
- Notice
- Venus orbital distance 108.2 mkm and Venus needs 35.8 days (=859.2 h) to pass
this distance (but light known velocity (0.3mkm/s) moves in (859.2s) a distance
257.8 mkm). (Note Please, 6855 mkm =257.8 mkm x 26.6)
(C)
mkm = Earth Orbital Circumference – But
- Earth (2.574 mkm /day) moves during 3134 solar day a distance = 8067.5 mkm
- Notice
- Earth orbital distance 149.6 mkm and Earth needs 58.1 days (=1394.5 h) to pass
418.35 mkm). (Note Please, 8067.5 mkm =418.35 mkm x 19.28)

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(D)
mkm = Mars Orbital Circumference – But
- Mars (2.082 mkm /day) moves during 4776 solar day a distance = 9946.2 mkm =
2 x 4973 mkm
- Notice
- Mars orbital distance 227.9 mkm and Mars needs 109.5 days (=2626.8 h) to pass
788 mkm). (Note Please, 9946.2 mkm =788 mkm x 12.62)
- Also
- (Mars Axial Tilt =25.2 deg = 2 x 12.6) and (778.6 mkm=Jupiter orbital distance
and with 788 mkm the error 1.2%) and (9946.2 mkm = 2 x 4973 mkm where 4900
mkm is Jupiter Orbital Circumference with 4973 mkm be 1.5%)
(E)
mkm = Jupiter Orbital Circumference – But
- Jupiter (1.1318 mkm /day) moves during 16333 solar day a distance = 18487 mkm
(18048 mkm = Uranus Orbital Circumference) (error 2.4%)
- Notice
- Jupiter orbital distance 778.6 mkm and Jupiter needs 688 days (=16510 h) to pass
this distance (but light known velocity (0.3mkm/s) moves in (16510 s) a distance
4952 mkm). (Note Please, 18487 mkm =4952 mkm x 3.73)
(4900 =Jupiter Orbital Circumference with (4952) the error 1%)
(F)
mkm = Saturn Orbital Circumference – But
- Saturn (0.838 mkm /day) moves during 30023 solar day a distance = 25162 mkm

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- Notice
- Saturn orbital distance 1433.5 mkm and Saturn needs 1710 days (=41037 h) to
pass this distance (but light known velocity (0.3mkm/s) moves in (41037 s) a
distance 12311 mkm(Note Please, 25162 mkm =12311 mkm x 2.04)
(G)
- Light known velocity (0.3 mkm/s) needs 60160 seconds to pass a distance =18048
mkm = Uranus Orbital Circumference – But
- Uranus (0.5875 mkm /day) moves during 60160 solar day a distance = 35345
mkm
- Notice
- Uranus orbital distance 2872.5 mkm and Uranus needs 4889 days (=117341 h) to
pass this distance (but light known velocity (0.3mkm/s) moves in (117341 s) a
distance 35202 mkm (Note Please, 35345 mkm =35202 mkm, error 0.4%)
(H)
mkm = Neptune Orbital Circumference – But
- Neptune (0.4668 mkm /day) moves during 94147 solar day a distance = 43925 mkm
- Notice
- Neptune orbital distance 4495.1 mkm and Neptune needs 9635 days (=231229 h)
to pass this distance (but light known velocity (0.3mkm/s) moves in (231229 s) a
distance 69369 mkm (Note Please, 43925 mkm = 69369 mkm x 0.6333)
(I)
- Light known velocity (0.3 mkm/s) needs 123667 seconds to pass a distance =
37100 mkm = Pluto Orbital Circumference – But
- Pluto (0.406 mkm /day) moves during 123667 solar day a distance = 50219 mkm

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- Notice
- Pluto orbital distance 5906 mkm and Pluto needs 14544 days (=349054 h) to pass
this distance (but light known velocity (0.3mkm/s) moves in (349054 s) a distance
104716 mkm (Note Please, 50219 mkm = 104716 mkm x 0.47957)
- Notice
- Jupiter moves during (24.6 h =Mars Rotation Period) a distance =1.16 mkm (light
supposed velocity passes in 1 second a distance = 1.16 mkm) - And
- Uranus moves during (24.6 h =Mars Rotation Period) a distance =0.6 mkm
- (light known velocity passes in 2 seconds a distance = 0.6 mkm)
- Also
- Uranus moves in 2 solar days a distance =1.16 mkm (error 1.3%)
- Let's see the following Data
- That shows One Geometrical Design be found for the solar system motion

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- Because the 2 distances are equal in Uranus Data (35345 mkm =35202 mkm,
error 0.4%), We conclude that a geometrical design be found behind the planets
velocities distribution.
- Before Uranus, Planet motion distance be greater than light motion distance in all
planets but the difference be decreasing gradually.
- After Uranus, Planet motion distance be shorter than light motion distance in
(Neptune and Pluto) where the difference be increasing gradually.
- For the 3 planets (Uranus, Saturn and Neptune) the rates be function in their orbital
distance – (if Uranus be 1), So Saturn be (1/2) and Neptune be (π/2) because
Uranus orbital distance 2872.5 mkm = 2 Saturn orbital distance 1433 mkm =2/ π
Neptune orbital distance 4495.1 mkm.
- For another group of 3 planets (Jupiter, Saturn and Uranus), their velocities per a
solar day almost be in order (2, 3, and 4), where Uranus moves in a solar day
0.5875 mkm (Light 0.3mkm/s moves during 2 sec a distance 0.6 mkm error 2%)
and Saturn moves 0.838 mkm (light 0.3 mkm/s in 3 sec 0.9 mkm error 7.25%) and
Jupiter moves 1.1318 mkm (light 0.3 mkm/s in 4 sec 1.2 mkm error 6%)
- We should discuss these great errors later, because the error (7.25%) depends on
the rate (1.0725) which be used as a rate between 40% of all distances in the solar
system –for that reason –the error explanation needs more area for extending
discussion.

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- Notice
- (0.9 mkm- 0.838 mkm = 61920 km where accurately 61920 s=17.2 h = Uranus
Day Period
- That tells, Uranus Day Period (17.2h) be created by Saturn motion effect, that
explains why Saturn moves in (17.2 h) a distance= 600000 km
- Many questions still be left behind - one basic question is that – How to explain
the using of the distance 4900 mkm? in different data we have seen this distance
and also Uranus needs around 4900 days to pass its orbital distance –How to
explain that

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3-10 Data No. (8) (Part A) (The Distances Be In Network Form)
The Solar System Distances Creation In A Network Form
- The solar system distances be created in a network form. The following data can
prove this fact clearly…
(A)
- 0.3 mkm /s (light known velocity) passes in 2094 seconds a distance =629 mkm
- 629 mkm = Earth Jupiter Distance
- 2094 mkm= Jupiter Uranus Distance
- The data shows that, a light beam started from the Earth to Jupiter and passed from
Jupiter to Uranus
- The data shows that the 2 distances (629 mkm and 2094mkm) are created together
based on a geometrical mechanism
(B)
- 0.3 mkm /s (light known velocity) passes in 2 x 2723 sec a distance = 1634 mkm
- 2723 mkm = Earth Uranus Distance
- 1622.7 mkm = Uranus Neptune Distance (with 1634 mkm error0.7%)
- The data shows that, a light beam started from the Earth to Uranus and passed
from Uranus to Neptune -
- The data uses the rate (2) because the distance 2723 mkm x 2 = Earth Uranus
Orbital Diameter through the revolution around the sun – the data shows (two)
distances be used in equivalence to (one) distance- which is a known feature of the
solar system motion- as –Uranus orbital distance = 2 Saturn orbital distances
- The data shows that the 2 distances (2723 mkm and 1622.7 mkm) are created
together in a network form

Planet Motion Analytical Description (Revised).pdf

Planet Motion Analytical Description (Revised).pdf

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Planet Motion Analytical Description (Revised).pdf