Can Light Velocity Effect On The Moon Orbital Motion? (Revised)

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
Can Light Velocity Effect On The Moon Orbital Motion? (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 –18th
March 2021
Abstract
Paper Question
- Can Light Velocity Effect On a Planet Motion?
The Question Explanation
- We observed frequently that, a Planet motion shows light motion features
- Indeed We have observed features as (different rates of time – Lorentz length
Contraction phenomenon – distance values are used as periods of time) which are
light motion features
- But,
- We couldn't catch the light motion (or any high velocity motion) can cause these
features.
- because of that, I have to provide the detailed data in best order and description to
be a source of knowledge to whom may concern to help.
- This paper provides a sufficient proof for the claim (Light Motion Features Are
Discovered In Planets Motions). Where the paper tries to provide all detailed
data and all possible support to whom tries to answer the question (How Can
These Features Be Belonged To A Planet Motion?)

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1- Introduction
- Why do we claim that (Light Velocity effects on a Planet Motion)? Because
- (1) Relativistic Effects are discovered in Planets Motions Data
- (2) A proportionality is found between Matter and Space Dimensions
- (3) Distances Equality and proportion is found in the solar system.
- The total solar eclipse phenomenon is the first observation we have discussed,
Again let's ask (Why do we see the sun disc = the moon disc?) because (the sun
diameter / the moon diameter) = (the Earth Sun distance/ the Earth Moon
distance).. So Why The Distances Rate = The Diameters Rate?!
But
- Let's suppose, (Light Velocity effects on Planet motion!), How can we prove that?
- If light & a planet motions data be in harmony can that be a proof? Let's see an
example
- The moon (27.78 km/sec) moves during 10921 seconds a distance = 303475 km,
where this distance is greater than light (0.3 mkm/sec) motion during 1 second
with (=1%) (= 3475 km= the moon diameter) where 10921 km = the moon
circumference, Can That Be A Proof?
- Shortly, This is not a proof but also this is interesting coincidences of data!
- More data may help us to see much better
More Data
- Venus (35 km/s) moves during 10921s a distance = Saturn Circumference (1%)
- Earth (29.8 km/s) moves during 10921s a distance =2 Uranus Circumference (1%)
- Jupiter (13.1 km/s) moves during 10921s a distance = Jupiter Diameter
- Uranus (6.8 km/s) moves during 10921s a distance =π2
x Pluto Circumference
- Neptune (5.4 km/s) moves during 10921s a distance = Saturn Radius (2%)
- Logically these coincidences of data can't be created by any pure coincidences, we
have to search for some geometrical reason to create this harmony of data

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- We should note that, the using of planet diameter or circumference as a period of
time is a wide using in the solar system, for example
- Uranus (6.8 km/s) moves during 7510s a distance =51118 km Uranus Diameter
- Pluto (4.7 km/s) moves during 10921 s a distance =51118 km Uranus Diameter
- Pluto (4.7 km/s) moves during 51118 s a distance =2 Saturn Diameters
- (7510 km = Pluto Circumference) - so we have to ask
- How can a planet Circumference be used as a period of time?
The Claim Difficulty
- Light motion features are discovered in planets motions, but, these features don't
form a geometrical general rule, instead, it works as pulses and because of that, its
explanation almost attributes to (pure coincidences of numbers) which covers its
real existence…
Paper Objective
- The paper analyzes the planets data showing the light motion effects on them as
clear as possible – because of that- the paper concentrates on Jupiter Data as the
most clear planet whose motion data shows light motion features clearly
Paper Contents
Subject P. N.
Point No. 2 (Paper Methodology) 4
Point No. 3 (Examples Of Light And Planet Motions Accompany) 8
Point No. 4 (Can a light beam be accompanying with the moon motion?) 32
Point No. 5 (Can Different Rates Of Time Are Found In The Solar System?) 36
Point No. 6 (A Difference 1% Is Found Between Rated Distances) 43
Point No. 7 (Light Velocity Effect On A Moving Particle (A Moving Planet) 45
Appendix No. 1 53
Appendix No. 2 55

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2- Methodology
- I use Planets Data analysis to discover the solar planets motion and origin
- The planets data analysis method aims to explain the solar planets motions based
on their own data.
- For that, this method requires to arrange the planets data in its correct order
- This method is similar to a living creature genes analysis to discover this creature
life features
- Let's use an example to explain how this method works
Example for Explanation
- I- Data
Group (No.1)
A. 1.16 mkm = 13.1 km/s (Jupiter Velocity) x 24.6 x 3600 s
B. 1.16 mkm = 29.8 km/s (Earth Velocity) x 10.7 x 3600 s (1%)
C. 1.16 mkm = 4.7 km/s (Pluto Velocity) x 17.2 x 4 x 3600 s
D. 1.16 mkm x 2 = 5.4 km/s (Neptune Velocity) x 17.2 x 7 x 3600 s (1%)
Group (No.2)
E. 1.16 mkm x 2 = 205 km/s x π x 3600 s
F. 1.16 mkm/s = 47.4 km/s (Mercury velocity) x 6.8 x 3600 s
Group (No.3)
- Jupiter Orbital Distance 778.6 mkm = 1.16 mkm/s x 671 s
- Jupiter Mercury Distance 720.7 mkm = 1.16 mkm/s x 629 s
- Jupiter Venus Distance 671 mkm = 1.16 mkm/s x 580 s
- Jupiter Uranus Distance 2088 mkm x 2 = 1.16 mkm/s x 3600 s
- Jupiter Pluto Distance 5128 mkm = 1.16 mkm/s x 4437s

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II- Discussion
- The Data Group No. (1) tells us that
o Jupiter moves during Mars rotation period a distance =1.16 mkm
o Earth moves during Saturn Day period a distance =1.16 mkm (-1%)
o Pluto moves during 4 Uranus Days periods a distance =1.16 mkm
o Neptune moves during 7 Uranus Days periods a distance =1.16 mkm (-1%)
- This data deep analysis gives more interesting results, for example Jupiter orbital
period (4331 days) =2π x Mars orbital period (687 days), which shows the distance
1.16 mkm/s passed by Jupiter during Mars rotation period has a wide effect on
both planets motions data.
- What Conclusion can we reach based on this data?
- The distance 1.16 mkm is widespread used value in the solar planets motions,
Why?
- The Planets data analysis method arranges planets data in its correct order
- What does mean (Planets Data Arrangement In Its Correct Order)?
- What's the planet? A rigid body or a light beam?
- The theory tells, Planets are rigid bodies revolves around the sun,
- But
- The planets data refer to relativistic effects and many other light motion features
discovered in Planets motions…
- So, We need only to present the data proves these features…. Showing that the
description is incorrect…
- The method tries to create a focus on planets data meanings - for example – (1)
Jupiter is Not the nearest planet to the sun –spite of its greatest Mass (2) (Mercury
– Venus – Earth) order tells that- greater mass needs greater orbital distances! (3)
the diameters rate = the distances rate! These features can't be explained by Reigid
Body Mechnics.
- Let's discuss our example data…

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- The example data.
- Many Data shows a distance (=1.16 mkm) is passed by different planets during
defined periods of time… Why? Please note, The data tells the 4 planets (Jupiter –
Earth – Neptune – Pluto) pass this distance (1.16 mkm) in defined periods (Mars
rotation period – Saturn day period- 4 Uranus days period – 7 Uranus days period),
the defined periods tell that this distance (1.16 mkm) is mentioned by the
geometrical design – means-the distance 1.16 mkm is found in the solar system
geometrical design why?
- Groups no. (2) and (3) supposes that, there's a velocity =1.16 mkm/s , The data
shows this supposed velocity (1.16 mkm/s) is used frequently – If this supposed
velocity is not found at all, from where this value (1.16 mkm) is created to be used
frequently with defined distances and periods?
Equation No. (E)
1.16 mkm x 2 = 205 km/s x π x 3600 s
- 205 km/s = the solar planets velocities total (including the moon velocity)
- Equation no. (E) tells, the supposed velocity 1.16 mkm/s passes in 2 seconds a
distance = the circumference of a distance passed by the planets velocities total
during 3600 seconds.
Equation No. (F)
1.16 mkm/s = 47.4 km/s (Mercury velocity) x 6.8 x 3600 s
- 6.8 km/s = Uranus Velocity
- Equation No. (F) tells a similar meaning, Mercury passes during 3600 seconds a
distance = 1.16 mkm based on the rate 6.8 …. Or
- Uranus passes during 3600 seconds a distance = 1.16 mkm based on the rate 47.4
- The rate (1 is equivalent to 3600) is used here also
- Where the rate (1 is equivalent to 3600) is used frequently that means, the
geometrical design takes into consideration this rate and designed the solar system
depends on it and because of that we find it frequently in the data.

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- Data Group No. (3) tells that, Jupiter distances to the inner planets are used as
periods of time based on this supposed velocity (1.16 mkm/s) to produce each
other.
- Again
- What does Planets data analysis Method try to do?
- The method tries to explain the planets motions based on their own data, the theory
tells, a Planet Velocity is independent from other planets velocities, but Planets
data shows that, Planets velocities total is used as one value relative to the
supposed light velocity (1.16 mkm/s) which is used frequently in the solar system
- Shortly, the method arranges planets data in its correct order to face the current
theories and by the comparison between both we may have a chance to perform a
progress in the solar system motion understanding.

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3- Examples of Light & Planet Motions Accompany
3-1 Preface
3-2 The Examples Explanation
3-3 Light & Planet Velocities Proportionality
3-4 Jupiter Distances Analysis
3-5 The Supposed Velocity
3-6 Jupiter Mars Motions Interaction
3-1 Preface
- Let's summarize the examples idea in following
o We suppose a light beam travels with Mercury, so we have 2 motions (the
light beam motion and Mercury motion)
o We use the same number of time for both motions (but the number in
seconds for light motion and in hours for Mercury motion)
o Because we have 2 velocities only, so the passed distances by both should
be rated
Notice (1)
o We suppose that, there's a light beam its velocity =1.16 mkm/sec
Notice (2)
o 1.16 mkm/s = 47.4 km/s (Mercury velocity) x 6.8 x 3600 s
o Because we use the period number in seconds (for light) and in hours (for
Mercury) so the value (3600) is taken into consideration
o 6.8 km/s = Uranus Velocity
o The value 6.8 = 2π x 1.0725
o Where 1.0725 is a rate used frequently between the solar planets distances.
I have attributed it to Lorentz Length Contraction Effect.
Notice (3)
o How to test our experiment?
o The produced distances by these 2 motions will lead us to the conclusions

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3-2 Examples For Explanation
Example No. (I)
Data
(1)
Light supposed velocity (1.16 mkm/s) travels during 4222.6 S a distance = 4900 mkm
(2)
Mercury (47.4km/s) moves during 4222.6 hours a distance = 720.7 mkm
Where
- 4900 mkm =Jupiter Orbital Circumference
- 778.6 mkm = Jupiter Orbital Distance
- 720.7 mkm = Mercury Jupiter Distance
- 4222.6 hours = Mercury Day Period
Data Analysis
- In these 2 equations we have a light motion and Mercury motion and we need to
find connections between the 2 periods of time and the 2 distances
- The Time Connection Is That:
- A second period of light motion is equivalent to An hour period of Planet
Motion (Mercury Motion)
- The Distance Connection Is That:
- 4900 mkm = 2π x 778.6 mkm
- 778.6 mkm = 1.0725 x 720.7 mkm
- (4900 mkm =720.7 mkm x 6.8)

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Example No. (II)
Data
(1)
Light supposed velocity (1.16 mkm/s) travels during 670 S a distance = 778.6 mkm
(2)
Mercury (47.4km/s) moves during 670 hours a distance = 114.3 mkm
Where
- 778.6 mkm = Jupiter Orbital Distance
- 114.3 mkm = Mercury Orbital Distance x 2 (=115.8 mkm) (error 1.2%)
- 670 mkm = Venus Jupiter Distance
Data Analysis
Motion.
- 778.6 mkm = 2π x 123.9 mkm
- 123.9 mkm = 1.0725 x 114.3 mkm

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Example No. (III)
Data
(1)
(2)
Mercury (47.4km/s) moves during 629 hours a distance = 107.3 mkm
Where
- 720.7 mkm = Jupiter Mercury Distance
- 107.3 mkm = Venus orbital Distance (= 108.2 mkm) (0.8%)
- 629 mkm = Earth Jupiter Distance
Data Analysis
Motion.
- 720.7 mkm = 2π x 114.3 mkm
- 114.3 mkm = 1.0725 x 107.3 mkm

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Example No. (IV)
Data
(1)
Light supposed velocity (1.16 mkm/s) travels during 500 S a distance = 580 mkm
(2)
Mercury (47.4km/s) moves during 500 hours x 2 a distance = 170.6 mkm
Where
- 170 mkm = Mercury Mars Distance
- 149.6 mkm = Earth Orbital Distance
Data Analysis
Motion.
- 580 mkm = 2π x 92.3 mkm
- (91.7 mkm = Mercury Earth Distance – error 0.6%)
Notice (1)
Mars (24.1 km/s) moves during 580 hours a distance = 50.3 mkm
Mars (24.1 km/s) moves during 580 days a distance = 1205 mkm
Where
50.3 mkm Mercury Venus Distance
1205 mkm Mars Saturn Distance
Notice (2)
670.4 mkm = Venus Jupiter Distance
629 mkm = Earth Jupiter Distance

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Discussion
- Let's summarize what we did in the previous 4 examples:
1. We have 2 players,
2. A light beam its supposed velocity =1.16 mkm/sec
3. Mercury motion its velocity = 47.4 km/ sec
4. We have 4 periods of time (4222.6, 670, 629 and 500) we use the same period
in seconds units for light motion and in hours units for Mercury motion
The Result Distances
- We have 8 resulted distances which are:
4900 mkm = Jupiter Orbital Circumference
778.6 mkm = Jupiter Orbital Distance
720.7 mkm = Jupiter Mercury Distance
114.3 mkm = Mercury orbital distance 57.9 mkm x 2 (1%)
107.3 mkm = Venus orbital distance 108.2 mkm (1%)
170 mkm = Mercury Mars Distance
91.7 mkm = Mercury Earth Distance
- These 8 distances are created (4 distances by light motions and 4 distances by
Mercury motions)
- All Distances Are The Planets Distances – Why?
- We use 2 players in 2 motions,
- Based on the current theories, Mercury Velocity is independent from light velocity
based on that, if we use the 2 velocities in motions the results must be independent
from each other – means- the produced distances should be random distances!
- We have No random distances, we have distances from the inner planets to Jupiter
or to the sun or internal distances between the inner planets,
- Means, The produced distances are complementary each other.
- Can that be a proof for a proportionality between Mercury and light velocities?

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- Let's analyze the principle equation as possible in following:
o 47.4 km/sec (Mercury velocity) x 6.8 x 3600 = 1.16 mkm/sec (1)
o 6.8 km/sec (Uranus velocity) x 47.4 x 3600 = 1.16 mkm/sec (2)
o The experiment uses 1 second of light motion (1.16 mkm/sec) as equivalent
to 1 hour of Mercury Motion (47.4 km/sec) where the constant is 6.8 = 2π x
1.0725
o But we can also use the experiment in another form as following …
o To use 1 second of light motion (1.16 mkm/sec) as equivalent to 1 hour of
Uranus Motion (6.8 km/sec) where the constant is 47.4. let's avoid that for
now
o Questions
Can we consider the supposed velocity 1.16 mkm/sec to be as a
fact based on the previous data?
Can a Planet Velocity be defined as a function in this supposed
light velocity (1.16 mkm/sec)? If yes why? and How?
What relationship can be found between the light supposed
velocity 1.16 mkm/s and a planet velocity, Why planet velocity
should depend on it?
More analysis may help our discussion
Let's see the following data…

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3-3 Light Planet Velocities Proportionality
I-Data
(1) 1.16 mkm = 47.4 km/s (Mercury Velocity) x 6.8 x 3600
(2) 1.16 mkm = 35 km/s (Venus Velocity) x 9.2 x 3600
(3) 1.16 mkm = 29.8 km/s (Earth Velocity) x 10.8 x 3600
(4) 1.16 mkm = 27.78 km/s (The Moon Velocity) x 11.6 x 3600
(5) 1.16 mkm = 24.1 km/s (Mars Velocity) x 13.4 x 3600
(6) 1.16 mkm = 13.1 km/s (Jupiter Velocity) x 24.6 x 3600
(7) 1.16 mkm = 9.7 km/s (Saturn Velocity) x 33.4 x 3600
(8) 1.16 mkm = 6.8 km/s (Uranus Velocity) x 47.4 x 3600
(9) 1.16 mkm = 5.4 km/s (Neptune Velocity) x 59.7 x 3600
(10) 1.16 mkm = 4.7 km/s (Pluto Velocity) x 68.6 x 3600
Data Analysis
The Blue Rates
- 6.8 km/sec = Uranus velocity, and 47.4 km/sec = Mercury velocity, where
- 1.16 mkm = 3600 x 6.8 x 47.4
- That means, Mercury velocity uses Uranus velocity as a rate to perform the light
supposed velocity 1.16 mkm/s (1 second of light is equivalent to 1 hour of planet)..
- Uranus velocity – on the other side – does the same – and uses Mercury velocity as
a rate to perform the supposed velocity
- It must be very happy coincidence to create this supposed velocity 1.16 mkm/sec
based on the 2 planets velocities…. Otherwise a geometrical reason can be found
behind, but What's This Geometrical Reason?

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The Brown Rates
Equation No. (3)
- 1.16 mkm = 29.8 km/s (Earth Velocity) x 10.8 x 3600
- 10.8 = 10.7 hours (Saturn Day period = Saturn rotation period) (error 1%)
- Equation no. (3) tells that,
- Earth (29.8 km/s) during Saturn day period (or rotation period) moves a distance
= 1.16 mkm = light supposed velocity for 1 second. (-1%)
Notice (1)
10747 days (Saturn orbital period) = 365.25 d ( Earth orbital period) x 29.53
(29.53 days = the moon day period)
Notice (2)
Earth (29.8 km/s) moves during Saturn Day period (10.7 hours) a distance = 1147896
km which needs 12104 km to be 1160000 km (accurately)
But 12104 km = Venus Diameter
Equation No. (6)
- 1.16 mkm = 13.1 km/s (Jupiter Velocity) x 24.6 x 3600
- 24.6 hours = Mars Rotation period
- Equation no. (6) tells that,
- Jupiter (13.1 km/s) during Mars rotation period moves a distance = 1.16 mkm =
light supposed velocity for 1 second.
- 4331 days (Jupiter orbital period) = 687 d (Mars orbital period) x 2π
- Jupiter moves 1.16 mkm during Mars rotation period (the only planet 0 error)
The Red Rates
Equation No. (5)
1.16 mkm = 24.1 km/s (Mars Velocity) x 13.4 x 3600
- Equation no. (5) tells
- Mars during 13.4 hours moves a distance = 1.16 mkm = light supposed motion for
1 second. (13.2 = 25.2 deg (Mars Axial Tilt) x 0.52 deg (Mars motion daily))

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The Discussion
- The analysis of the value 1.16 mkm shows that more harmony is found between
this supposed velocity 1.16 mkm/sec and the solar planets velocities, based on this
harmony the planets motions and light motion (during comparing periods) can pass
complementary distances as we have seen between Mercury and the light supposed
velocity motions analysis…
- Can we consider that, this velocity 1.16 mkm/sec is a fact?
- Can we consider that, some proportionality is found between this supposed light
velocity (1.16 mkm/sec) and the planets velocities (For Geometrical Necessities).
- Let's remember (1.16 mkm =3600 x 47.4 x 6.8) and we tested Mercury Velocity,
let's here use Uranus velocity as the other player in this same equation
(A)
- Mercury (47.4km/s) moves during 4222.6 hours a distance = 720.7 mkm
- Uranus (6.8 km/s) moves during 30589 hours a distance = 748.8 mkm
(Error 3.7%)
(B)
- Mercury (47.4km/s) moves during 670 hours a distance = 114.3 mkm
- Uranus (6.8 km/s) moves during 4900 hours a distance = 120 mkm
(Error 2%) (120 mkm = Venus Mars Distance)
(C)
- Mercury (47.4km/s) moves during 629 hours a distance = 107.3 mkm
- Uranus (6.8 km/s) moves during 720 hours x2π a distance = 110.8 mkm
(Error 3.2 %)
- The data shows that, Uranus also moves proportional distances (inner planets
distances) during the same periods, and also Uranus uses its own data (30589 days
= Uranus orbital period) in the required form to perform this task.
- That may explain why (30589 days =347.6 x 88 days Mercury orbital period)
(and 346.6 days = the nodal year).

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The Discussion (Continued)
- Many questions are left behind in the previous discussion, let's refer to some of
them in following:
(1) Why Mercury Velocity (Specifically) is in harmony with this light motion? (If
this supposed velocity 1.16 mkm/s is found really!)
(2) How to manage the distances between Jupiter and the other planets? Why their
values are so soft in calculations (Mars is still exceptional)?
(3) Suppose, there's a light velocity =1.16 mkm/s, Can planets velocities be
created as functions of this velocity? If this a fact how why that's happened?
(1st
Question) Why Mercury Velocity specifically be in so harmony with light
supposed velocity (1.16 mkm/sec)?
Equation No. (I)
- 205 km/s x π x 3600 = 1.16 mkm x 2
o 205 km/s = the solar planets velocities total (including the moon velocity)
o Equation no. (I) tells that, The Circumference of the distance passed by the
planets velocities total during 1 hour = light supposed velocity motion for 2
second (1.16 mkm/s) …Let's remember our equation
- 47.4 km/s x 6.8 x 3600 = 1.16 mkm,
Equation No. (II)
- 205 km/s x π = 47.4 km/s x 6.8 km/s x 2
o Mercury and Uranus Velocities can define the solar planets velocities total,
that shows clearly why both (Mercury and Uranus velocities) are so specific
velocities in the solar system
o The moon velocity massive effect is found almost by Uranus effect.
o 202.6 km/s = 29.8 km/s (Earth Velocity) x 6.8 km/s, This data tells that,
Earth with Uranus can perform a similar behavior as Mercury does
(2nd
Question) Why does Jupiter position depend on the velocity 1.16 mkm/s?

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3-4 Jupiter Distances Analysis
I-Data
Data Group No. (1)
o 4222.6 sec x 1.16 mkm/s = 4900 mkm (Jupiter orbital Circumference)
o 670.4 sec x 1.16 mkm/s = 778.6 mkm (Jupiter Orbital Distance)
o 629 sec x 1.16 mkm/s = 720.7 mkm (Jupiter Mercury Distance) (1.2%)
o 580 sec x 1.16 mkm/s = 670.4 mkm (Jupiter Venus Distance)
o 534 sec x 1.16 mkm/s = 629 mkm (Jupiter Earth Distance) (1.5%)
o 470 sec x 1.16 mkm/s = 550.7 mkm (Jupiter Mars Distance) (1%)
o 3600 seconds x 1.16 mkm/s = 2x 2088 mkm (Jupiter Uranus Distance)
(470 mkm = π x 149.6 mkm Earth Orbital Distance)
(534 mkm = π x 170 mkm Mercury Mars Distance)
- More Data
- 778.6 mkm = 1.0725 x 720.7 mkm
- 720.7 mkm = 1.0725 x 670.4 mkm
- 670.4 mkm = 1.0725 x 629 mkm
- 629 mkm = 1.0725 x 580 mkm Notice 1.16 = (1.0725)2
Data Group No. (2)
o 2094 seconds x 0.3 mkm/s = 629 mkm (Jupiter Earth Distance)
o 4331 seconds x 0.3 mkm/s = 2 x 654.6 mkm (Jupiter Saturn Distance) (1%)
o 1835 seconds x 0.3 mkm/s = 550.7 mkm (Jupiter Mars Distance)
o 12400 seconds x 0.3 mkm/s = 3717 mkm (Jupiter Neptune Distance)
(12390 seconds x 24.1 km/s = 300000 km) and
(12390 seconds x 9.7 km/s = 120536 km)
Notice
654.9 mkm (Jupiter Saturn distance) x π = 2057.5 mkm (error 1.5 % with 2088 mkm)
2094 mkm = Jupiter Uranus Distance

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Notice
Jupiter Pluto Distance
(I)
Data
(a)
Light supposed velocity (1.16 mkm/s) travels during 86400 S x 2 a distance = 200448
mkm
(b)
(Pluto Orbital Circumference 37110 mkm) – (Jupiter orbital Circumference 4900
mkm) x 2π = 202372 mkm
The difference between No. (1) and No. (2) = 1%
(II)
(Neptune orbital circumference 28255 mkm–Venus orbital circumference 680 mkm)
=27575 mkm But
27575 mkm x 2π = 2 x 86400 mkm
Where
2 x 86400 mkm + 28255 mkm (Neptune orbital circumference) = 201055 mkm
The data shows, the light energy = the orbital circumferences between Neptune and
(Mercury and Earth) where Venus is used as a middle point for both.
Jupiter Orbital Circumference =4900 mkm
4900 mkm = 360 mkm +680 mkm + 940 mkm +1433 mkm +1433 mkm
Where
360 mkm = Mercury Orbital Circumference
680 mkm = Venus Orbital Circumference
940 mkm = Earth Orbital Circumference
1433 mkm = Mars Orbital Circumference
1433 mkm = Saturn Orbital Distance

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II-Discussion
(1)
- Jupiter has 9 distances to the other planets and the sun, from these
- 5 distances are defined depending on the light supposed velocity 1.16 mkm/sec
- 4 distances are defined depending on the light known velocity 0.3 mkm/sec
(2)
The 5 distances are,
- Jupiter Orbital Distance 778.6 mkm = 1.16 x 671 s
- Jupiter Mercury Distance 720.7 mkm = 1.16 x 629 s
- Jupiter Venus Distance 670.4 mkm = 1.16 x 580 s
- Jupiter Uranus Distance 2088 mkm x 2 = 1.16 x 3600 s
- Jupiter Pluto Distance 5128 mkm = 1.16 x 4437s
The 4 distances are,
- Jupiter Earth Distance 629 mkm = 0.3 x 2094 s
- Jupiter Mars Distance 550.7 mkm = 0.3 x 584π s
- Jupiter Saturn Distance 654.9 mkm x 2 = 0.3 x 4331 s (1%)
- Jupiter Neptune Distance 3717 mkm = 0.3 x 12400 s
Also
- 670.4 mkm (Venus Jupiter distance) x π = 2106 mkm (Uranus Jupiter distance) (1%)
(3)
- The 5 distances which depends on 1.16 mkm/s use other distances as periods of
time – so the distance 671 mkm (Venus Jupiter Distance) can be used as 671
seconds and during it the light supposed velocity 1.16 mkm/s travels a distance =
778.6 mkm = Jupiter Orbital Distance
- The other values follows this same rule

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(4)
- The 4 distances which depend on light known velocity (0.3 mkm/s) are
o Jupiter Uranus distance (2094mkm) is used as 2094 seconds, where light
known velocity during 2094 s travels a distance = 629 mkm (Jupiter Earth
Distance)
o 4331 days (Jupiter orbital period) is used as 4331 seconds by light known
velocity to pass a distance = 2 x 654.9 mkm (Jupiter Saturn Distance)
o 584 x π =1835 seconds, during which the light known velocity (0.3mkm/s)
travels a distance = 550.7 mkm = Mars Jupiter Distance
o Where 584 days is the meting period between Venus and Earth
o During 12400 seconds, light known velocity (0.3mkm/s) travels a distance =
3717 mkm = Jupiter Neptune Distance
Notice
- The value 12400 second is so effective in the solar system design because
o 3717 mkm (Jupiter Neptune Distance = 12400 sec x 0.3 mkm/s
o 300000 km (light for 1 s) = 12400 x 24.1 km/s (Mars velocity)
o 1.16mkm (supposed velocity) = π x 12400 x 29.8 km/s (Earth velocity)
o 120536 km (Saturn Diameter) = 12400 seconds x 9.7 km/s (Saturn velocity)
o 142984 km (Jupiter Diameter) = π x 12400 x 3.66
o (where 3.66 = Earth diameter / the moon diameter)

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More Data
(a)
- Mars orbital Circumference = Saturn orbital distance = Neptune orbital distance/π
(b)
- Mars (24.1 km/s) during 5040 sec. moves a distance = Saturn Diameter (0.7%)
- Saturn (9.7 km/s) during 5040 sec. moves a distance = Neptune Diameter (1.1%)
- Neptune (5.4 km/s) during 5040 sec. moves a distance = 86400km /π (1%)
(c)
- (Mars rotation 24.6 h/ Neptune rotation 16.1 h) = (Neptune rotation 16.1 h/ Saturn
rotation 10.7 h) (error 1.5%)
- The data tells, some deep connection is found between the 3 planets, we still
search why these 3 planets distances depend on light known velocity (0.3 mkm/s)
- We know that Mars had migrated and changed its orbital distance from 84 mkm
into 227.9 mkm
- We conclude that, because Mars has changed its orbital distance, that may cause
Mars to lose its connection with the original light beam (1.16 mkm/s), specially
because Saturn is created after Mars Migration – that tells – The light known
velocity (0.3 mkm/s) is found after the planets migration ….
- Neptune also had changed its original orbital distance, because Pluto occupied its
position and Neptune had to be more near to the sun in its new orbital distance
(4495.1 mkm) where Neptune original orbital distance was (5906 mkm= The
Current Pluto orbital distance). (for that reason Neptune used 14% of Jupiter
energy to create its orbital circumference because Neptune has no orbit).
(5)
- The next question is, Why do the other 5 distances depend on the supposed
velocity 1.16 mkm/s?

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More Discussion
- How can we make progress for this discussion? Let's summarize the question to be
in front of us?
- Planets motions Data show frequently light motion effects and features, but no
high velocity motion we can catch through the solar system which can produce
these features and effects….
- Our options are so limited,
- We simply can't just avoid the data claiming that no light motion effects or features
are found in the planets motions
- On the other side we can't find the source of these features and effects
- What can we do?
- Let's provide one idea tries to solve this question in following:
o Imagine, the solar group works as a machine of gears.
o Imagine, the planets velocities are added with each other (as gears) to create
one unified velocity (which equal All Planets Velocities Total).
o The planets velocities total will be produced relative to one point only
(which is the sun)
o Means, each planet moves by its own velocity relative to the other planets
but relative to the sun point only all planets velocities are added together
forming one unified general velocity (= The Planets Velocities total)
o The planets velocities total =176 km/sec (without The Earth moon)
o The planets velocities total =205 km/sec (with The Earth moon velocity)
o Means, the sun doesn't see any planet velocity, but feels only one velocity =
=205 km/sec.
o Light known velocity (0.3mkm/s) =205 km/sec x 1461 seconds
o If 1 second of light velocity = 1461 seconds of planets velocities total
o In this case the sun will see this total velocity (205 km/s) as light velocity

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o We have 2 points here
o (1) this total velocity is NOT Created for any planet But just for The Sun
o (2) we need different rates of time in the solar system to produce the
required velocity (because 1 second must be =1461 seconds)
o The light Mercury Motions data suggested that 1 second of light supposed
velocity motion should be equivalent to 1 hour of Mercury Motion,
o This rate we have defined by geometrical necessity because the data forced
us to provide it as a hypothesis (1 hour is equivalent to 1 second)
o What we need to do here, is to see, if it's possible to have this rate of time in
the solar system motion ….
Shortly
o If we can prove that, 1 Second Of Light Motion Is Equivalent To 1 Hour
Of Mercury Motion That Will Create Different Rates Of Time In The
Solar System.
But
o On the other side, Can different rates of time be created by geometrical rules
doesn’t use high velocity motion? means, Can the solar planets velocities
interaction produce different rates of time without using high velocity
motion?
o Can the rate (1 second =3600 seconds) be used for the supposed light
velocity (1.16 mkm/s) but the rate (1 second =1461 seconds) be used for the
light known velocity (0.3 mkm/s)? but we need to prove that this supposed
velocity (1.16 mkm/s) is a fact art first
Left Question (1)
How can 2 planets velocities be added to each other and form one velocity?

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3-5 The Supposed Velocity Analysis
I-Data
(1)
1.16 mkm x 2 = 205.8 km/s x π (old data)
(2)
1.16 mkm = 47.4 km/s x 6.8 x 3600 (old data)
(3)
47.4 km/s x 6.8 x2 = 205 km/s x π (old data)
(4)
29.8 km/s (Earth velocity) x 6.8 = 202.6 km/s (differing with 205.8 with 1.5%)
II-Discussion
- The 3 equations (1, 2 3) are old data, let's discuss Equation no. (4)
Equation no. (4)
29.8 km/s x 6.8 = 202.6 km/s
- The planets velocities total =205.8 km/s, But
- Earth Velocity 29.8 km/s x 6.8 = 202.6 km/s (error 1.5%)
- Based on that, Earth velocity behaves as Mercury velocity in Equation no. 2, if
that's true – what distances Earth pass by the same periods Mercury had used in its
motion Example? Let's see in following:
- Earth (29.8 km/s) moves during 4222.6 hours a distance = 453 mkm
- Earth (29.8 km/s) moves during 670 hours a distance = 71.8 mkm
- Earth (29.8 km/s) moves during 720 hours a distance = 78.3 mkm
- Earth (29.8 km/s) moves during 500 hours x 2 a distance = 107.3 mkm
Where
- 453 mkm = 2 x 227.9 mkm (Mars orbital distance) (error 0.7%)
- 71.8 mkm= The Moon motion distance during 29.53 days (error 1%)
- 78.3 mkm= Mars Earth distance (error 0.7%)
- 107.3 mkm = Venus orbital Distance (108.2 mkm) (error 0.8%)

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More Discussion
- What does the data tell?
- We know (205 km/s = the planets velocities total) x π x 3600 = 2 x 1.16 mkm
- Earth velocity 29.8 x Uranus Velocity 6.8 = 202.6
- Where 202.6 is very near to 205, means it can behave similar behavior
- (202.6 km/s = x π x 3600 = 2 x 1.16 mkm (1%))
- But
- We know that, Earth moves during 10.7 h (Saturn day period) a distance = 1.16
mkm (1%) how to understand that?
- Because 6.8 x π = 2 x 10.7
- Where 6.8 km/s = Uranus Velocity But 10.7 h = Saturn Day Period
- Again …What does the data tell us?
- The data is created to do defined task for that the data is created complementary
with each other – that's what I argue about
- While the theory considers that, the solar planets are rigid bodies independent from
one another in their motions data, the data shows that, they are created for some
specific tasks, and because of that, they are created complementary one another – I
wish I can explain the idea as possible…
- Suppose we have an electron and a positron, but we found them by random search,
so the theory tells us, the electron belongs to some atom but the positron belongs to
some positive rays –the theory gives us this description clearly –
- But
- After a while we discovered that, a gamma ray was found here before the electron
and positron were found, that will change the description, now we will claim that
the electron and positron are produced from Gamma rays (the double production
experiment), the origin will be changed based on the new description – although
neither the electron nor the positron is changed but their origin theory will be
changed

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- This historical context goes in wrong way, basically because, the electron and
positron data are complementary one another (because their charges total = Zero)
and that means they are created for one another – means- without this story – we
have to know that – these 2 particles have one origin because their data is created
complementary one another
- That's the major benefit we receive from the planets data analysis, because the data
provides us the correct direction of their origin- by that – we have a method to
avoid all unreal explanations –
- So, if the data is created based on one another, we can't suppose that these planets
are independent in their motions or origins, on the contrary, we have obligation to
follow the data direction –
A Major Conclusion
- The paper discusses different data to show that, this data are created based on each
other in different forms, by that, even if the geometrical rules based on which this
data is created aren't known, that doesn't effect negatively on this conclusion,
because the data shows clearly their unified origin
- And based on that, we can easily conclude the solar planets form together one
geometrical building – so they are not rigid bodies are separated from one another
and revolve around the sun – on the contrary- they are one building or one
machine and each planet is a part of this same building.
- So, we can consider the solar system as (One Trajectory of Energy and each
planet is a point on this same trajectory)
- Or we can suppose that, the solar system is consisted of one light beam its velocity
1.16 mkm/s and each planet is a point on it.

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3-6 Jupiter Mars Motions Interaction
I-Data
(1)
Jupiter (13.1 km/s) moves during Mars rotation period (24.6 h) a distance = 1.16 mkm
(accurately)
(2)
4331 days (Jupiter Orbital Period) = 2π x 687 days (Mars orbital period)
(3)
4331 days (Jupiter Orbital Period) = 4222.6 (Mars Rotation period)
(4)
687 days (Mars orbital period) = 671 (Mars Rotation Period)
(5)
(24.6/9.9) = (24.1/9.7) = (0.3 mkm/120536 km)
I-Discussion
- Equation No. (1,2 and 3) we have discussed before, let's start directly with (4)
Equation No. (4)
687 days (Mars orbital period) = 671 (Mars Rotation Period)
- Jupiter data tells us that, some general geometrical design is found under it but we
can't catch it yet
- Let's try to be near for this design as possible ….
- Jupiter (13.1 km/s) moves during Mars rotation period (24.6 h) a distance = 1.16
mkm = light supposed velocity (1.16 mkm/s) for 1 second, why Jupiter uses Mars
cycles periods?
- Let's summarize the idea in following
o Mars orbital period 687 days (24 h) = 671 Mars rotation period (24.6 h)
o 671 mkm = Venus Jupiter Distance
o 680 mkm = Venus orbital circumference
o 687 mkm = increased with (1% ) than Venus orbital circumference

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o That means,
o Mars Cycles Periods (687 solar days and 671 Mars rotation) are created
based on the distance between Venus and Jupiter
o We have seen this using before where
o 671 seconds x 1.16 mkm/s = 778.6 mkm (Jupiter orbital distance) , so
o 671 is used as (Mars rotation) in place of (second)
o Also this using we have seen before where
o During 4222.6 seconds light (1.16 mkm/s) travels a distance =4900 mkm
o 8But Jupiter orbital period 4331 solar days = 4222.6 (Mars rotations)
o There's a geometrical mechanism behind causes these using, but we don't
catch it yet- any way we have clear data tells us
o Mars Cycles Periods are defined based on the distance between Jupiter and
Venus, and because of that Jupiter uses Mars rotation period
Equation No. (5)
(24.6/9.9) = (24.1/9.7) = (0.3 mkm/120536 km)
- Where
- 24.6 hours = Mars Rotation Period
- 9.9 hours = Jupiter Rotation Period
- 24.1 km/s = Mars Velocity
- 9.7 km/s = Saturn Velocity
- 120536 km = Saturn Diameter
- Equation no. (5) shows the rate between Mars and Jupiter rotation periods, where
this same rate controls many different data, for example Mars Saturn velocities,
and also light motion for 1 second relative to Saturn diameter

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Notice (1)
Jupiter orbital period (4331 days) = 103944 hours
The moon orbital are from perigee to apogee =103944 mkm2
If
1 hour = 1 mkm
So these 2 values are equal
Notice (2)
(24.1 km/s Mars velocity / 9.7 km /s Saturn velocity) =
(13.1 km/s Jupiter velocity / 5.4 km /s Neptune velocity) (error 2.4%)

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4- Can a light beam be accompanying with the moon motion?
4-1 the Moon Motion data depends on Light Velocity for 1 second
4-2 Light Motion Features are found in the moon motion!
4-1 The Moon Motion data depends on Light Velocity for 1 second
I-Data
(1)
- This is the moon orbital triangle (we have studied before)
- The triangle ACE, its dimensions are (AE =449197 km, AC =121622 km and CE
= 373000 km)
- The perimeter of the triangle ACE = 943819 km = π x 300000 km
- (where light velocity = 300000 km/sec)
(2)
- The lunar eclipse umbra length =1.392 mkm
- But the distance EA = 449197km (where E is Earth)
- 1.392 mkm - 449197 km = 943819 km
II-Discussion
- The data shows that, the value 943819 km is the unit on which the moon orbital
triangle base is created as well as the lunar eclipse umbra length…
- Where this value (943819 km) is the circumference for the distance passed by
light motion during 1 second.

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The Moon Motion Data Depends On Light Motion For 1 Second (More Data)
This information we can conclude from the moon motion data, let's try to prove that:
I-Data
(1) 10921 km x 86400 seconds = 940 mkm (Earth Orbital Circumference)
(2) 10921 km x 27.3 days = 300000 km (The Moon Circumference =10921 km)
(3) 43000 km x 27.3 days x 0.99 =1.16 mkm (43000 km = Perigee Apogee Distance)
II-Discussion
Equation No. (1)
10921 km x 86400 seconds = 940 mkm (Earth Orbital Circumference)
- Equation no. (1) tells that, If Earth revolves around the sun a complete revolution
in one day only, so the moon circumference will be = a distance of Earth motion
for 1 second period of time.
- Equation No.(1) refers to some other rate of time, Because Earth revolves the sun
in 1 year. So it may tell that (1 Year on Earth = 1 Day on The Sun)
Equation No. (2)
10921 km x 27.3 days = 300000 km (light motion for 1 second)
- Equation no. (2) tells that, If the moon rotates around its axis one rotation per solar
day (as Earth does), so the moon would to move during its orbital period (27.3
days) a distance = 300000 km = light motion distance during 1 second
- Equation No.(2) refers to some other rate of time.
- So it may tell that (1Second of light motion = 27.3 Days of the moon motion)
Equation No. (3)
43000 km x 27.3 days x 0.99 =1.16 mkm
- Equation no. (3) tells, If the moon is a point without diameter, and moves daily
from perigee to apogee, so during its orbital period (27.3 days) the moon would
move a distance = 1.16 mkm = light supposed velocity (1.16 mkm/s) during 1
second.

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4-2 Light Motion Feature is found in the moon motion!
I-Data
(4)
300000 km/sec x 8 seconds = 2.4 mkm
(5)
300000 km/sec x 8.66 seconds = 2.59869 mkm
(6)
(8.662/8.64) = (361/360)
(7)
86400 km = 8 x 10800 km (the moon Circumference =10921 km 1%)
II-Discussion
Equation No. (4)
300000 km/sec x 8 seconds = 2.4 mkm
- Equation no. (4) tells that, light known velocity (0.3 mkm/sec) travels during 8
seconds a distance = 2.4 mkm
- What's the distance 2.4 mkm?
- We remember that, the moon motion a distance = Earth distance per a solar day
2.576 mkm but the moon distance is suffered from a contraction (1.0725) and
caused this distance to be 2.4 mkm which needs necessary the moon displacement
daily 88000 km to be doubled (88000 km x 2) to cover the difference distance
- So, the distance 2.4 mkm is the moon contracted motion distance which needs 2
displacements (each equal 88000 km) to cover the different distance with Earth
motion distance.
- Light needs 8 seconds to pass this distance! What's this information significance?!
- The number (8) is important because there's a cycle for 8 days, the cycle is found
for Jupiter and for Uranus… but how does this cycle effect on this 8 second?!

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Equation No. (5)
300000 km/sec x 8.66 seconds = 2.59869 mkm
- The moon displacements total during 29.53 days = 2.59869 mkm
- Light travels this distance is 8.66 seconds
- This equation is important because of its next one
Equation No. (6)
(8.662/8.64) = (361/360)
- 86400 seconds = the solar day = 8.64 x 10000
- 10000 = (light Velocity / Earth Velocity)
- The other value (8.662 s) is the period needs by light to pass a distance = the moon
displacements total during 29.53 days (2.59869 mkm)
- 360 degrees = a cycle degrees
- 361 degrees = Metonic Cycle degree, because the moon orbit regresses 19 degrees
per year and Metonic Cycle continues for 19 years, means, the total degrees be 361
degrees.
- Equation no. (6) tells that, the solar day production process needs a contribution
from the moon motion. It means, the solar day is produced by an interaction
between Earth and its moon motions, but both motion depends on light motion!
Equation No. (7)
86400 sec = 8 x 10800 sec (the moon Circumference =10921 km 1%)
- We used frequently the moon circumference as a period of time (10921 seconds)
in different equations, the noticeable data is that, this value (10921 sec) = 1/8 of
the solar day (with error 1%) and we have seen that the light moves during 8
seconds a distance = the moon motion distance during a solar day –
- This equation connects again the moon motion data with the number 8 (or the
cycle of 8 days), spite we still don't know how this connection is found.

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5- Can Different Rates Of Time Are Found In The Solar System?
5-1 Can 2 or 3 Planets velocities be added together?
5-2 Can 1 hour be equivalent to 1 solar day?
5-1 Can 2 or 3 Planets velocities be added together?
I- Data
(1)
29.5 km/s = 2π x 4.7 km/s
And
(a) 43000 km = 29.5 km/s x 1461 seconds
(b) 7510 km = 29.5 km /s x 254 seconds
(2)
Mercury Velocity (47.4 km/s) + Venus Velocity (35 km/s) + Earth Velocity (29.8
km/s) = 112.2 km/s = 24 x 4.7 km/s (Pluto Velocity)
(3)
3600 seconds x 1.16 mkm/sec (light supposed velocity) =2 x 2088 mkm
(4)
3600 seconds x π x 205 km/s = 2 x 1.16 mkm
(5) (Old Equation)
3600 seconds x 47.4 km/s x 6.8 = 1.16 mkm /sec (more Discussion)
(6)
8 Mercury Days =1407.6 solar days
II- Discussion
Equation No. (1)
29.5 km/s = 2π x 4.7 km/s
- I have supposed that, the moon velocity =29.5 km/s which is less with 1% than
Earth velocity (29.8 km/s), this idea we have discussed in the previous paper. The
basic proof depended on Pluto Velocity (4.7 km/s) because, 2π x 4.7 km/s = 29.5

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km/s and Not 29.8 km/s (Earth Velocity), that told us that, a value =29.5 km/s
must be found, and so I have concluded that this is the moon velocity
- That paper told us, the moon real velocity =29.5 km/s but because the solar group
works as a machine of gears, the other planets could compensate this difference in
velocity between the Earth and its moon which help to save the moon and Earth
motions accompany and prevented their separation from each other which was the
logical result for this difference 1%.
- Now let's ask, Can this idea be a true one? how to prove that the moon velocity
=29.5 km/s and be less than 1% of Earth velocity (29.8 km/s)? why Pluto velocity
can be a proof for this claim?
- The answer we can conclude from the light and Mercury motions comparison
which we did in point no. (2) of this paper… we have found that.. light motion
produces a distance = 2π x Mercury Passed Distance, this feature we have found
in all 4 examples we have used… let's suppose it’s a general behavior done
between 2 velocities one great velocity and the other small velocity, the small
velocity is Pluto (4.7 km/s) and so the great velocity will be =2π x 4.7 = 29.5 km/s
but, we don't know any velocity =29.5 km/s…! So we may conclude that the
velocity 29.5 km/s is the moon velocity.
- Let's try to test our conclusion in following…
Equation No. (a)
43000 km = 29.5 km/s x 1461 seconds
- 1461 days =Earth cycle = 365 +365 +365 +366
- Let's use this value in seconds unit
- 43000 km = The Distance Between Perigee And Apogee
- Equation No. (a) tells that, the moon moves during 1461 seconds a distance =
Perigee Apogee Distance – can that be a proof for the moon velocity =29.5 km/s?

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Equation No. (b)
7510 km = 29.5 km /s x 254 seconds
- 7510 km = Pluto Circumference
- 254 months = Metonic Cycle (27.3 days x 254 months)
- The value 254 is used as 254 seconds,
- Equation No. (b) tells that, the moon moves during 254 seconds a distance = Pluto
Circumference
Notice (1)
- The periods of time are used by different units, this using is a supposed using and
not real one, what we try to do, is that, we try to attribute the data to one another –
so if the moon moves during 1461 seconds – there's no cycle by this value 1461 s
but there's a cycle 1461 days. These 2 values may be different from each other –
but – as we have seen before – one hour of Mercury motion is considered
equivalent to 1 second of light motion – we still try to prove this idea- but –
supposing the proves are found – we use this same rule in our analysis – so – light
may use the period 1461 days as 1461 seconds and by that may the moon velocity
can do a similar behavior
- We don't confirm that this using is a fact but it's a method to attribute the data to its
origins, and by that, the value 1461 second should be attributed to Earth and its
moon motions….
Notice (2)
- Why do we need to review the moon velocity in this subject?
- Because
- The velocity 29.5 km/s = Mars velocity 24.1 km/s + Neptune Velocity 5.4 km/s
- Let's remember the question …
- Can 2 or 3 planets velocities be added to each other? Yes
- The moon velocity 29.5 km/s is produced as Mars and Neptune Velocities Total

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Equation No. (2)
Mercury Velocity (47.4 km/s) + Venus Velocity (35 km/s) + Earth Velocity (29.8
km/s) = 112.2 km/s = 24 x 4.7 km/s (Pluto Velocity)
- Equation no. (2) shows that, 3 Planets Velocities Total = 24 x Pluto Velocity
- Why do we need this information?
- Because
- The 3 planets motions for 1 hour pass a distance passed by Pluto in 24 hours
- Means,
- 1 Day Of Pluto Motion Is Equivalent To 1 Hour Of 3 Planets Motions
- The last sentence provides an example of (Different Rates Of Time)
- Here we need to answer 2 questions which are
o (1st
Question) Do these 4 planets motions relative to each other create 2
different rates of time where (1 hour be equivalent to 24 hours)?
o (2nd
Question) Do these different rates of time be created by some
geometrical rules without using high velocity motion?
These 2 questions will be answered in separated points(3-2 )
Now let's complete our discussion

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Equation No. (3)
3600 seconds x 1.16 mkm/sec (light supposed velocity) =2 x 2088 mkm
- In this equation we uses light supposed velocity (1.16 mkm/s) to move 3600
seconds (1 hour) … why?
- The comparison of motions between light and Mercury which is done in point (no.
2) of this paper, suggested that, Mercury motion for 1 hour is equivalent to light
motion for 1 second … so here we use light motion for 1 hour to see how this
value can be effective in the solar system motion….
- Let's see what's in this equation ….
- During 3600 seconds, light supposed velocity (1.16 mk/s) travels a distance = 2
x 2088 mkm (Jupiter Uranus Distance) Why? let's see more data
o 4.095 mkm/s (Mercury velocity daily) x 511 days = 2092 mkm
o 3.02 mkm/s (Venus velocity daily) x 687 days = 2075 mkm
o 2.573 mkm/s (Earth velocity daily) x 1623 days = 2088 mkm x 2
o 2.41 mkm/s (The moon velocity daily) x 1738 days = 2094 mkm x 2
o 2.082 mkm/s (Mars velocity daily) x 1000 days = 2075 mkm
o 1.1318 mkm/s (Jupiter velocity daily) x 3717 days = 2104 mkm x 2
o 0.838 mkm/s (Saturn velocity daily) x 5040 days = 2112 mkm x 2
o 0.5875 mkm/s (Uranus velocity daily) x 3600 days = 2097 mkm
o 0.466 mkm/s (Neptune velocity daily) x 4495 days = 2097 mkm
o 0.406 mkm/s (Pluto velocity daily) x 5128 days = 2082 mkm
- Notice
- 511 degrees = The Planets Axial Tilts Total
- 687 days = Mars orbtial period
- 1623 mkm = the distance Between Uranus and Neptune
- 1738 km = The Moon Radius

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- 3717 mkm = The Distance between Neptune and Jupiter
- 5040 secinds = a period of time required for Mercury day to be 176 days
- 4495 mkm = Neptune orbtial distance
- 5128 mkm = The distance from Jupiter to Pluto
Equation No. (4)
3600 seconds x π x 205 km/s = 2 x 1.16 mkm
- 205 km/s = The Solar Planets Velocities Total
- Equation no. (4) tells that, the circumference of the distance passed by planets
velocities total during 1 hour (3600 seconds), equal to the distance passed by light
supposed velocity (1.16 mkm/s) during 2 seconds.
- Why 2 seconds? Why not 1 second?
- Notice No. (1)
- Please note the harmony of data, Planets velocities total is in proportionality with
light supposed velocity based on the rate (1 second of light motion is equivalent
to 1 hour of planets motions), the harmony of data is so clear and strong, that
may help our proving process even for the light supposed velocity (1.16 mkm/s)
existence….
- Notice No. (2)
- The value 205 km/sec =the solar planets velocities total with the moon velocity,
but the 9 planets velocities total = 176 km/s, if we use this velocity in place of
205km/s, the result can't be 1.16 mkm/s x 2 but less (2 x 0.99 mkm), means, the
solar system design takes into consideration the moon velocity to perform the total
velocities 205 km/s! but why? for what job this velocities total is required?
Equation No. (6)
8 Mercury Days =1407.6 solar days Let's summarize the idea in this equation
- 1407.6 hours = Mercury Rotation Period
- (8 days) should refer to a cycle, because, Mercury during 8 of its days moves a
distance = Uranus orbital distance x 2 = Uranus orbital diameter
- The idea is that, the produced 1407.6 days is used as 1407.6 hours by Mercury
itself and Mercury uses this rate (1 hour = 1 day) for its motion…

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5-2 Can 1 hour be equivalent to 1 solar day?
I-Data
4222.6 h = 153.3 h x 27.55
Where
- 4222.6 h = Mercury Day period
- 153.3 h = Pluto Day period
- 27.55 days = Anomalistic month (27.55 solar days)
II- Discussion
- We know that, (Mercury + Venus + Earth) Velocities total = 24 Pluto Velocity,
- And we know also
- (light motion for 1 second causes planet motion for 1 solar day)
- But For Mercury
- (light motion for 1 second causes Mercury motion for 1 hour)
- Based on that, the rate 1 hour is equivalent to 1 day is produced by interaction
between Pluto on one side and the three planets (together) on the other side
Notice
- The paper provides a huge number of data depends on this rate (1 second is
equivalent to 1 hour) starting from its methodology and through different
discussion points the data uses this rate frequently – the idea is that – if no truth in
this rate this data can't be created based on it – but the frequent using tells that –
this rate is a basic one in the solar system motion and because of that – this rate
must be real one
- The left question is,
- Can planets motions interaction create this rate (1 second is equivalent to 1 hour)
without using of high velocity motion?

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6- A Difference 1% Is Found Between Rated Distances
Data no. 1
a- The Moon Orbital Circumference at apogee radius = 2550973 km (100%)
b- Earth Daily Motion Distance = 2573483 km (101%)
c- Pluto moves during 153.3 hours =2593836 (102%)
d- The displacements 88000 km total during (29.53 days) = 2598693 km (102%)
e- Uranus (63.8 km/s) moves (during 378675 seconds) = 2574990 km (101%)
f- Saturn (9.7 km/s) moves during 6.8 x (10.7 hours) = 2540779 km
Data no. 2
(1)
Earth moves during (6939.75 solar days) a distance = 17859.325 mkm (1%)
(2)
Pluto moves during (6939.75) x (153.3 hours) a distance = 18000.57 mkm
(3)
The moon displacements total during (6939.75) x (29.53 days) = 18034.278 mkm
(4)
Uranus Orbital Circumference = 18048.449 mkm
(5)
Mercury moves during (4331 days) a distance = 17737 mkm
(6)
Mars moves during (2x 4331 days) a distance = 18036.3 mkm
(7)
Venus moves during (5906 days) a distance = 17860 mkm
(notice Venus distance = Earth distance accurately)

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II- Discussion
- Can the difference 1% be a proof of light motion effect on the moon motion?
- Fizu experiment shows that, the light velocity in liquids is less than its velocity in
vacuum by around 1%
- That means, if we have 2 light beams one passes through a basin of some liquid
and the other passes in vacuum (through equal distances), the first light beam will
pass the distance in time shorter than the other with (1%).
- That shows the equal distances which have difference 1%
- I want to say that, this difference 1% can be used as a proof for light motion effect
on the moon orbital motion – but – Not Now
- Before we have to know why these distances are different by 1%, means, we need
to know, the geometrical effect of this difference on the solar system geometry..
- Let's try to explain that,
- Fizu experiment measures the light velocity through liquids, and this difference
1% is observed as a feature of light motion through a liquid
- But
- This feature (for us) is seen as an effect of 2 components (light velocity and a
liquid) where in the solar system geometry, this feature is one of the general
geometrical design. Means, this feature is one of many other features all of them
form the geometrical design and we need to discover this design and by that we
will know the complete effect of this difference 1% and we will be able to answer
the hard question – how the matter is created? -
- Because, as we can conclude easily, matter is created almost out of light – but this
hypothesis is so limited to answer the question (how the matter is created) because
there are many geometrical rules the light has to pass through to give a birth to a
piece of matter…!

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7- Light velocity effect on a moving particle ( a moving planet)
7-1 Preface
7-2 Light velocity effects on the moon motion (proves discussion)
7-3 Earth Velocity Analysis
7-1 Preface
- How does light velocity effect on a moving particle? We know that from Lorentz
Transformations Discussion - let's summarize it in following
o Lorentz Length contraction is done for distances and particles lengths (as if
both are equivalent one another!). where the same equation controls both
o Our old question, Why The Particle Length Is Contracted By Its High
Velocity Motion?
o We can't answer because no theory tells us
o But, Lorentz transformations show effects on particles properties, even we
can't explain why, but the experiments and transformations equations prove
this fact
o We have discussed this idea deeply before – why we need it here?
o We don't know (how the matter is created?) but on the other side, we
consider that, the rigid body and the light beam are 2 different things
(absolutely). Although we don't know how the matter is created but we
know the matter can't be a light beam….
o In this paper I provide planets data to show that, the matter and light are so
near to each other – we don't know how the matter is created –but it's almost
created out of light (by some geometrical methods).
o The words can't help the proving process because the light beam passes
through many geometrical rules before to be born as matter – we neither
know these rules nor how the light beam became matter –but the data shows
that what's now matter was a light beam – I provide the data to show that
there's some fact under this claim

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7-2 Light velocity effects on the moon motion (proves discussion)
I-Data
(a)
(3.02 mkm/0.3 mkm) = (47.4 /4.7)
(b)
300000 km = 88000 km x 3.41
(3.41/3.4) = (361/360)
(c)
17.2 sec x 0.3 mkm /sec =5.16 mkm
(d)
1.16 mkm = 88000 x 13.18
(e)
2.408 mkm =2.32 mkm + 0.088 mkm
(f)
1.16 mkm x 27.7 sec = 88000 km x 365.25 days = 0.3 mkm/sec x 107.2 sec

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II-Discussion
Equation No. (a)
(3.02 mkm/0.3 mkm) = (47.4 /4.7)
Where
o 3.02 mkm = Venus motion distance per a solar day
o 0.3 mkm = Light known Velocity
o 47.4 km/sec = Mercury Velocity
o 4.7 km/sec = Pluto Velocity
- Equation no. (a) is important because Mercury velocity is the solar system greatest
velocity and Pluto is the lowest one, means, the rate (47.4/4.7) defines the solar
planets velocities range.
- This range is defined by the rate of Venus velocity (per solar day) rated with light
velocity for 1 second!
- We should remember that, the value (1 second) is related to the moon motion- and
we still analyze to know if other planets depends on this same period (1 second) or
it's related to the moon only
- Equation no. (a) can't help us, because Venus motion effects greatly on the moon
motion, means, it’s possible the moon relation to this period (1 second) be
inherited from Venus.
Equation No. (b)
300000 km = 88000 km x 3.41
(3.41/3.4) = (361/360)
- 3.4 degrees = Venus Orbital Inclination
- 88000 km = The moon displacement daily
- 300000 km = light known velocity (0.3 mkm/sec) motion per second
- Equation no. (b) tells that, Venus orbital inclination has an effect on the moon
daily displacement (in comparison with light motion for 1 second)

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- It tells, light (0.3mkm/s) motion per 1 second 300000 km is equivalent to the moon
displacement per a solar day (88000 km) because of Venus effect – means- Venus
connects the moon daily displacement with light motion velocity through its
orbital inclination-
- Equation no (b) can support the meaning of Equation no. (a) and by that we can
conclude one meaning that (Venus is the connection point between the moon
displacement and the light velocity!)
Equation No. (c)
17.2 sec x 0.3 mkm /sec =5.16 mkm
- Equation no. (c) tells that, light known velocity (0.3 mkm/sec) moves during 17.2
seconds a distance = 5.16 mkm
- But We know that
- 1 mkm = 1 degree so 5.16 mkm =5.16 deg (the moon orbital inclination)
- Means, the moon orbital inclination is created depends on 17.2
- This information we knew before because we concluded that, the moon orbital
inclination (5.1 deg) is created depending on Pluto orbital inclination (17.2 deg)
- But here the value 17.2 degrees is used as 17.2 seconds Why?
- In fact the original value is 17.2 hours (Uranus day period) and this value effects
on Pluto motion and be seen as 17.2 deg (Pluto orbital inclination) and effects on
the moon motion and be seen as 17.2 seconds
- It's one value (17.2 hours) (Uranus Day Period) but this value is found as a point
on a light beam and because of that, this light beam travels from Uranus to Pluto
and also travels from Uranus to the moon – and by that – the data is transported in
different units according to its effect nature on the received planet (Pluto or the
moon) – even if this explanation is imaginary – the data still needs to be explained
geometrically.

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Equation No. (d)
1.16 mkm = 88000 x 13.18
- 1.16 mkm/sec = light suppose velocity
- 88000 km = the moon daily displacement
- 13.18 degrees = the moon daily motion degrees (360 deg /27.3)
- This equation is so important but I can't reach its meaning yet- why this value
13.18 is used here? it tells in 13.18 days the moon displacement total be = 1.16
mkm = light supposes velocity motion for 1 second
- So, This equation also tells the moon motion data is defined in 1 second
- But why 13.18 is found here?!
Equation No. (e)
2.408 mkm =2.32 mkm + 0.088 mkm = 2 x 1.16 mkm + 0.088 mkm
- We know that, the distance 2.4 mkm is passed by light known velocity (0.3
mkm/s) in 8 seconds
- The same distance is defined by 2 seconds of 1.16 mkm and 1 day displacement
(88000 km).
- Some deep interaction must be found here
Equation No. (e)
1.16 mkm x 27.7 sec = 88000 km x 365.25 days = 0.3 mkm/sec x 107.2 sec
- This equation is consisted of 2 equations,
- It tells, The moon displacements total during a year (365.25 days) = light known
velocity (0.3 mkm/sc) motion during 107.2 seconds
- What's this 107.2 seconds?
- Suppose it's 107.2 degrees, So 107.2 deg = 90 deg +27.2 degrees

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- It's tells the moon orbital period (27.3 days) depends on this equation!
- The Equation second part tells that
- The moon displacements total during a year (365.25 d) = light supposed velocity
(1.16 mkm/sc) motion during 27.7 seconds
- What's this 27.7 seconds? (27.7 x 0.99 = 27.4 seconds)
- We have seen that 1 second of light motion may cause a planet (the moon) to move
one solar day, so this value 27.7 seconds is the moon orbital period (27.3 days) but
in seconds units and with error 1%
- that tells the moon orbital period is a value acceptable by the 2 light velocities
which makes it’s a specific value in the solar system.

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7-3 Earth Velocity Analysis
- let's ask, can any planet moves with light velocity?
- Imagine Earth moves by light velocity, and the moon moves by 99% of light
velocity, what do we will see? We see only the difference 1%, it will be so low
velocity and no one will imagine that we move by light velocity – but – the data
round us will show the high velocity motion effect on the other planets and
distances
- I want to say,
- Let's suppose the solar planets move with high velocity relative to the sun point,
that means, we see them move in low velocities which are the differences between
our motion velocity and them, but for the sun which we may consider as a slow
motion point relative to them – that can create light velocity as total for the planets
velocities relative to the sun

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A General Comment
- This paper is prepared to be a source of ordered data to whom concerns to help the
question answer –
- For that reason, the provided data explanations are just tries to form a general
frame for it– let's explain that in following…
- Sometimes the data be so strange and has no any explanation, in this case I provide
the data and write any simple or even imaginary idea to explain. I do that because
I want to provide the data to the readers – so the explanation idea has no value –
- Sometimes, the explanation idea be a good one but not complete yet, in this case
also I provide the data and provide the available part of explanation
- I want to say…
- The data shows light motion features in planets motions, and this paper is prepared
to be a source of data to whom may concern to explain how light motion features
can be found in planets motions data.. Because of that the paper should provide all
available data regardless its explanation because it's the material by which the
paper claim is proved …
- That's why many data is provided with (not strong) explanations
- Also
- There are data still needs more analysis – for example – Jupiter distances with the
other planets need more analysis, because it has more treasure even than what we
have discovered …
- In appendix no. 1 I provide the equal and proportional distances in the solar system
to explain that this equality and proportionality should be produced as light motion
features
- In appendix no. 2 I provide more analysis for the proportionality between light
velocity and the solar planets velocities

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Appendix No.1
Is There Lorentz Length Contraction Effect In The Solar System?
i.e.
(Are There Relativistic Effects In The Solar System?)
Lorentz Length Contraction Effect is the near possible answer to explain the planets
data, in following I provide one example of such planets data to prove that, this
conclusion is the most near one to explain it.
I- Data (A)
Why These Distances Are Equal?
(1)
Saturn Orbital Distance = Saturn Uranus Distance
= Mars Orbital Circumference
= Pluto Neptune Distance
= Pluto eccentricity Distance
= 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
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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I-Data (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)
(4894
nce
Circumfere
Orbital
Jupiter
= (1.5 %)
(10)
I-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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Appendix No. 2
- Light Known Velocity (0.3 mkm/s)
(I)
- Light velocity (0.3 mkm/s) travels during 2061 minutes a distance = 37100 mkm
- Light velocity (0.3 mkm/s) travels during 79.6 minutes a distance = 1433 mkm
(II)
o Light velocity (0.3 mkm/s) travels during 328 minutes = 5906 mkm
o Light velocity (0.3 mkm/s) travels during 250 minutes= 4495.1 mkm
o Light velocity (0.3 mkm/s) travels during 160 minutes= 2872.5 mkm
o Light velocity (0.3 mkm/s) travels during 4778 second = 1433.5 mkm
- The data no (I) shows the required periods by light known velocity (0.3mkm/s) to
pass the planets orbital circumferences (the planets are ordered from Pluto to
Mercury)
- The data no (II) shows the required periods by light known velocity (0.3mkm/s) to
pass the planets orbital distances (the planets are ordered from Pluto to Mercury)

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- Light Supposed Velocity (1.16 mkm/s)
(I)
- Light velocity (1.16 mkm/s) travels during 254 minutes a distance=2π x 2815 mkm
- Light velocity (1.16 mkm/s) travels during 7770 sec a distance = 9000 mkm
(II)
o Light velocity (1.16 mkm/s) travels during 5092 sec = 5906 mkm
o Light velocity (1.16 mkm/s) travels during 3875 sec = 4495.1 mkm
o Light velocity (1.16 mkm/s) travels during 196.5 sec = 227.9 mkm
o Light velocity (1.16 mkm/s) travels during 93.27 sec = 108.2 mkm
- The data no (I) shows the required periods by light supposed velocity (1.16mkm/s)
to pass the planets orbital circumferences (the planets are ordered from Pluto to
Mercury)
- The data no(II) shows the required periods by light supposed velocity (1.16mkm/s)
to pass the planets orbital distances (the planets are ordered from Pluto to
Mercury)

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References
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
or
https://www.slideshare.net/Gergesfrancis/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
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
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
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Facebook https://www.facebook.com
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJhl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

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