The Solar System Motion Analysis

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Solar System Motion Analysis
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 –4th
June 2021
Abstract
Paper hypothesis
- The Motion Is Transported Among The Solar Planets.
The hypothesis Explanation
- The paper suggests that, the solar group is similar to one machine of gears and
each planet is a gear in it. or it's similar to one building and each planet is a part of
this building.
- The paper vision tells that, the solar planets are created based on One Geometrical
Design and because of that the planets data is created complementary with each
other.
- Based on this vision, The Motion Must Be Transported Among The Solar Planets.
- The motion transportation doesn't force us to attribute mechanical features for the
space, because the motion transportation isn't done mechanically.
- As a result for the motion transportation through the solar system, the Planets
motions use different rates of time.
- As example of the different rates of time
- 1 hour of Mercury Motion = 9.18 hours of Pluto Motion. But
- 1 hour of Earth Motion = 6.387 hours of Pluto Motion
- The solar planets motions different rates of time be unified and created one general
rate of time.

I do this research
2nd
2
Paper Objective
- The paper discusses and proves The Motion Transportation Concept among the
solar planets
- And
- The paper discusses and proves the solar planets motions using of different rates of
time.
Please scan the figure (ORCID)

I do this research
2nd
3
Paper Contents
Subject Page No.
1- Introduction 4
2- Methodology 5
3- The Paper Theory Summary 7
3-1 The Paper Theory (Part No. I) 7
3-2 The Paper Theory (Part No. II) 9
4- The Moon Orbital Motion Description 10
4-1 Why Does The Moon Use Pythagorean Triangle In Its Motion? 11
4-2 How Does The Moon Use Pythagorean Triangle In Its Motion? 12
4-3 The Moon Orbital Motion Analysis 14
4-4 The Moon Orbital Motion Equation 18
5 Pluto Day Period Creation 27
5-1 Preface 27
5-2 Pluto Day Period Creation 29
5-3 Planet Motion (6 Days Cycle) 34
5-4 The Earth Moon Orbit Creation 37
6- Can The Solar System Be A Clock? 41
6-1 The Point Objective 41
6-2 Pluto and Mercury Rate Of Time 43
6-3 Pluto and The Moon Rate Of Time 46
6-4 Pluto and Earth Rate Of Time 48
7- Jupiter Motion Analysis 49
7-2 Mars Motion Analysis 50
7-3 Mars And Mercury Motions Analysis 53
7-4 Jupiter Motion Analysis 58
8- Can A Light Beam Its Velocity 1.16 Mkm/S Be Found? 67
8-2 Jupiter Data depends on a velocity (1.16 mkm/s) 68
Appendix No.1 75
References and Biography 77

I do this research
2nd
4
Introduction
- The Earth Moon Daily Displacement =88000 km
- During 29.53 solar days (the moon day period), the displacements total be =
2598693 km.
- This distance should be the moon apogee orbital circumference. and in this case
the moon apogee orbital radius should be 413600 km. where in fact the moon
apogee radius =406000 km.
- The decreasing in the moon apogee radius (2%) is done as a result of the motion
transportation among the solar planets.
- There are 2 basic results are done together,
- (1st
result) the moon apogee orbital radius is decreased an be 406000 km.
- (2nd
result) Pluto day period be extending and became 153.3 hours.
- The paper proves the motion transportation process and uses these 2 results as
applications for the concept.
- Let's refer to the paper contents.
- Point No. (2) discuses the paper methodology
- Point No. (3) provides the paper concepts and ideas in a summary
- Point No. (4) provides The Moon Orbital Motion Analysis to discuss The Moon
Apogee Radius Creation.
- Point No. (5) discusses Pluto Day Creation as a result of the motion transportation
among the solar planets.
- Point No. (6) discusses the planets motions using of different rates of time.
- Point No. (7) provides Jupiter Motion Analysis, this is the basic point of proof
for The Motion Transportation Concept. All other discussions depend on this
point proves and arguments.
- Point No. (8) provides a hypothesis tells (A light beam with velocity 1.16 mkm/s
be found in the solar system). This hypothesis proves be discussed in details also
in this point.

I do this research
2nd
5
2-Methodology
(I)
- I use the solar planets data analysis to discover the planets origin and motions.
- This method is so useful because it creates a comparison between the physics
theories and the planets data to show if the planets data follow the theories really.
- For example, we notice that, Jupiter is the 5th
planet in order to the sun, while the
gravitation equation tells (More Mass Needs Shorter Distance). Jupiter position
doesn't disprove the equation but refers to a contradiction needs explanation.
- That's the useful result of planets data analysis using.
(II)
- I suppose the solar group is a machine of gears and each planet is a gear in it. or it's
one building and each planet is a part of it. this hypothesis leads us to conclude
that, the solar planets motions must create one unified motion. based on that the
solar planets motions can be similar to the train carriages move tougher one
unified motion we call it (Train Carriages Motion)
(III)
- Because the solar planets are parts in the same one building, their data be created
complementary with each and by that their motions be also complementary
creating together one unified motion.
- The best example to explain the (complementary data) is the double production
experiment. Where an electron be created with a positron from gamma ray, that
shows the 2 particles charges are created depending on each other because of the
charge reservation law. But the 2 particles move depend on their charges, means,
the 2 particles motions are still controlled by (Gamma ray features) and they move
a complementary motion.
(IV)
- Based on this hypothesis, the solar planets data should be created complementary
with each other. means, There's One Equation Controls All Solar Planets Data

I do this research
2nd
6
- Let's discuss an example for better explanation.
- The Diameter =12430
- There's no a planet in the solar group its diameter =12430km, this number I have
invented it. let's test it to know if it has any effect on the solar system motion.
o The (supposed) diameter 12430 km in the one middle value between Venus
diameter (12104 km) and earth diameter (12756 km).
o (Earth diameter / its moon diameter) = (Jupiter diameter) / (12430 x π)
o Mars (24.1 km/s) moves during (12430 seconds) a distance = 300000 km =
light (0.3 mkm/s) motion distance for 1 second.
o Saturn (13.1 km/s) moves during (12430 seconds) a distance = 120536 km
= Saturn Diameter
- We compare between 2 visions.
- The classical vision tells that, the solar planets are rigid separated bodies revolve
around the sun independent in their creation and motion data from one another.
- We provide a new vision…
- The solar system is created based on one geometrical design, the planets are parts
in one building and their data (must) be complementary one another, Because they
are created based on geometrical rules. As in a triangle we know 2 angles (60 deg
and 80 deg) what's the third one?
- The diameter 12430 km proves this vision. It behaves perfectly as similar as any
planet (real) diameter. let's remember
o Jupiter (13.1 km/s) moves during 10921 sec a distance = 142984 km (=
Jupiter diameter) where (10921 km = the Earth moon circumference)
o Uranus (6.8 km/s) moves during 7510 sec a distance = 51118 km (=Uranus
diameter) where (7510 km = Pluto circumference)
- Based on that, the (supposed) diameter 12430 km is designed based on the solar
system geometrical design but be not created.
- Which supports the paper vision about the solar system creation and motion.

I do this research
2nd
7
3- The Paper Theory Summary
3-1 The Paper Theory (Part No. I)
- Paper Concept
- The paper claims (The Motion is Transported Among The Solar Planets)
- In Jupiter Motion Analysis (Point No.7) the paper discusses this hypothesis and
discusses the proves.
- (The Motion Transportation Among The Solar Planets) is the Concept based on
which the paper discussion is built.
- The paper doesn't support the classical description that the solar planets are
separated rigid bodies independent from each other in their motions. On the
contrary, the paper supposes the planets motions create one unified motion which
can be similar to a train carriages motions.
- Based on the paper vision, the solar system is built based on one geometrical
design and because of that any planet motion effects on other planets motions.
- Paper Argument
- Pluto Day Period (153.3 hour) is so long in comparison with the outer planets days
periods.
- Why does Pluto day period so long?
- The paper suggests that, Pluto Day Period is created depending on The Earth
Moon Day period (708.7 h).
- Pluto Day Period Creation should be used as an application for the motion
transportation concept among the solar planets.
- Pluto day period is extending by an effect of Venus motion on Pluto and the Earth
moon motions. So, both planets days periods be created depends on one another.
- Uranus motion causes to create the environment in which Venus motion can effect
on Pluto motion.
- That means, there are 3 steps to create Pluto day period which are

I do this research
2nd
8
o Pluto day period (153.3 h) is created directly depends on the Earth moon
day period (708.7 h). the 2 planets motions interaction create the 2 days
periods.
o Venus motion effect on Pluto and the Earth moon motions to create this
interaction by which both planets days periods are created
o Uranus motion effect on Venus and Pluto motion to create the environment
which contains Venus and Pluto and by that gives Venus the chance to
effect on Pluto motion.
- This process is done by a cooperation of Jupiter and Saturn motions.
- The Process Result
- Based on Pluto day period creation many results are done. The most noticeable
result is done in the Earth moon orbital circumference.
- The moon total displacements during 29.53 days should be = 2.598693 mkm, this
distance should be the moon orbital circumference. If So, The moon apogee radius
should be =413600 km but the moon apogee radius really = 406000 km.
- The difference in the moon apogee radius is done by Pluto day period creation
effect on the moon motion.
- The Point Conclusions
- 2 basic results are done by one process which are (1st
) Pluto Day Period be 153.3
hours and (2nd
) The moon apogee radius be 406000 km.

I do this research
2nd
9
3-2 The Paper Theory (Part No. II)
- Paper hypothesis
- Different rates of time are used in the solar planets motions
- Because
- The Motion is Transported Among The Solar Planets
- Based on that,
- 1 hour of Mercury Motion = 9.18 hours of Pluto Motion.
- This rate of time is created because of the motion transportation
- And that means
- This rate of time is created as the (3rd
) Result of the process by which, Pluto Day
period be = 153.3 hours and the moon orbital apogee radius be = 406000 km.
- Paper hypothesis proves
- The motion transportation concept is proved in Jupiter Motion analysis (Point No.
7)
- Pluto day period creation and its effect on the moon orbit, and The rate of time
(9.18) between Pluto and Mercury Motions will be discussed in (Can the solar
system be a clock). (Point No. 6)

I do this research
2nd
10
4- The Moon Orbital Motion Description
4-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
4-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
4-3 The Moon Orbital Motion Analysis
4-4 The Moon Orbital Motion Equation

I do this research
2nd
11
4-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
- Let's summarize this question answer in following:
o The moon uses Pythagorean triangle basically to decrease its displacement
daily through its orbit
o The moon daily displacement = 88000 km and the moon has to move this
distance every day without any decreasing (later we will know why!)
o But
o If the moon moves by this displacement as its orbital displacement the moon
would revolve around Earth through its apogee orbit only (r=0.406 mkm)
o For that reason
o The moon creates an angle between its motion direction and its orbit
horizontal level to create a displacement through its orbit less than (88000
km)
o As a result of this technique, the moon can revolve around Earth through
more near orbits than apogee orbit (r=0.406 mkm)
o Simply, because the moon uses this technique the moon can revolve around
Earth through perigee orbit (r=0.363 mkm)
o Let's explain this intelligent technique with some details to show the useful
result of using Pythagorean triangle by the moon orbital motion….

I do this research
2nd
12
4-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
- The moon moves daily (88000 km) on the right triangle hypotenuse (AC), but the
moon creates an angle (θ) between its motion direction and its orbit horizontal
level, by that the real displacement through the moon orbit will be (L= 88000 km
cos (θ)), and by that, spite the moon moves 88000 km, but the real orbital
horizontal displacement be less than (88000 km) and this is the objective for which
the moon uses Pythagorean triangle –
As an example,
- If (θ) =28.63 degrees, the real displacement (L== 88000 km cos (θ)) = 77237 km,
So, if the moon real displacement daily be (77237 km), during 29.53 days the
moon will pass a distance = 2.28 million km and this will be the moon orbital
circumference, where 2.28 mkm = 2π x (0.363 mkm)
- The Moon Orbital Perigee Radius =0.363 mkm
- That means, the moon by a real displacement =77237 km can move around Earth
through the perigee orbit (radius =0.363 mkm), this is the useful result the moon
performs by using Pythagorean triangle,
- Now let's suppose the moon doesn't use Pythagorean triangle, what would happen?
- The moon daily displacement = 88000 km, during 29.53 days the moon moves a
distance = 2.598 mkm where 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving around Earth through its apogee
orbit (or even far from apogee orbit) because the total distance can't be passed
through any more near orbit around Earth…
- The data shows how Pythagorean triangle is so useful for the moon orbital motion.

I do this research
2nd
13
The Angle θ
- The angle (θ) should get our attention for its specific effect…let's summarize the
idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and by that the moon can be near or at
Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon can be near or at apogee orbit
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees),
but before we need to analyze the moon orbital motion
Notice
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with dimensions (363000 km, 373000km, 86000 km),
what's the angle (θ)? The angle (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
i.e.,
The angle (1.1 deg.) controls the moon motion from perigee to apogee, we will need
this notice later in our discussion

I do this research
2nd
14
4-3 The Moon Orbital Motion
- The moon moves per a solar day a motion typical to the Earth motion to avoid the
separation from Earth through their motions, based on this rule, the moon moves
per a solar day 2.573 million km with an angle declines on the horizontal level
0.98562 degrees as typical to Earth motion
- If there's no Lorentz Length Contraction Phenomenon effect on the moon motion,
the moon motion trajectory would to be a parallel line to Earth Motion Trajectory,
But Lorentz Length Contraction effects on the moon motion daily distance (2.573
mkm) with a rate 1.0725 and causes this distance to be contracted (2.399 mkm)
- The moon difficulties are started here, because the difference between both
distances (0.17 mkm) will cause the moon to be separated from Earth motion
inevitably
- We should notice that, these motions are done far from our observation, means, we
see nothing of this motion distance, because the moon moves on the Earth orbital
circumference revolving around the sun, but, even if we can't observe this motion
distance the motion is still fact and proved by its power, because the Earth moves
per a solar day 2.573 mkm and if the moon doesn't move this same distance every
solar day that necessities the moon to be separated from the Earth through their
motions course – based on that- the facts prove this motion regardless our
observation ability for it.
- Now the moon has an additional distance to be passed (0.17 mkm) and the moon
has to pass this distance on the same solar day to avoid the separation from the
Earth during their motions.
- Because of that, the moon moves its daily displacement (88000 km) depends on
Earth gravity force (by which we see the moon in the Earth sky), but the different
distance (0.17 mkm) to be covered still needs the moon to move one more
displacement (= 88000 km)

I do this research
2nd
15
- The previous explanation tells that, the moon has to move 2 displacements each =
88000 km, while we see one displacement only because it's done through the
moon orbital motion around Earth but the other displacement should be done also
because this total distance (0.17 mkm) is required to cover the different distance
and create the total (2.573 mkm) which saves the moon and Earth motions
accompanying.
- Now we have 2 basic information about the moon orbital motion
o (1st
information) the moon uses Pythagorean triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagorean triangle
in its motion, because the moon can't decrease its daily displacement (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.399 mkm) and Earth motion distance (2.573 mkm),
So the moon needs to move this displacement perfectly, but if it's used as a
displacement through the moon orbit, the moon would be always a prisoner in the
apogee orbit (r=0.406 mkm) as we have discussed before, because of that, the
moon creates Pythagorean triangle technique by which the moon moves actually
88000 km daily but the real displacement through the moon orbit became less (L =
88000 Cos θ) and by that the moon can achieve 2 objectives, First to pass the
required distance (88000 km) and Second to move in near orbits to Earth, that
shows the intelligent moon motion technique…

I do this research
2nd
16
Notice
- The moon motion distance daily (2.574 mkm) be decreased with the rate (1.0725)
to be 2.399 mkm. I suppose this rate be found by Lorentz length contraction
phenomenon. But whether this hypothesis is real or imaginary idea it has no effect
on the result. Because the rate (1.0725) effects on around 40% of all solar planets
distances and based on that (40%) of the distances be contracted with this rate
(1.0725). This data is discussed deeply with Jupiter motion analysis (point no. 7)
and in The Appendix No. 1 (at end of this paper).
- The idea is that, this rate (1.0725) effects on the distances and causes to contract
them. And the moon motion distance is one of these distances are effected by this
rate and then the moon has to suffer to repair this rate (1.0725) effect on the moon
daily motion distance.

I do this research
2nd
17
The Moon Orbital Motion Needs One More Orbit
- The previous explanation tells that, the moon moves 2 displacements each =88000
km, we see one of these 2 displacements but where's the other displacement?!
- We know that, the moon original motion (2.573 mkm) which is contracted to be
(2.399 mkm) isn't seen by us because the moon moves this distance revolving with
Earth around the sun along the Earth Orbital Circumference
- We may accept that, the 2nd
displacement the moon does on this same trajectory
and isn't seen by us.
- So,
- There must be one more orbit for the moon to move through this 2nd
displacement.
means,
- There's 2nd
Orbit For The Moon Motion
- But
- How can we discover this second orbit if we can't observe the 2nd
displacement
motion?
- We can discover this 2nd
orbit by the moon orbit data analysis. So we should
depend on the moon orbital triangle data analysis to define this 2nd
orbit position.
- For that we have to discuss the moon 2nd
orbit in our deep analysis of The Moon
Orbital Triangle Geometrical Structure.

I do this research
2nd
18
4-4 The Moon Orbital Motion Equation
4-4-1 The Equation Concept
4-4-2 The Equation Test and Accuracy
4-4-1 The Equation Concept
The Moon Orbital Motion Equation
(θ1= θ0 + 1.7 degrees)
- The moon orbital motion equation is created depending on the concept we have
discussed, which is (the moon uses Pythagorean triangle in its orbital motion)
- The moon uses Pythagorean triangle and by this intelligent technique the moon be
under control of the angle (θ) change
- The angle (θ) defines almost all the moon motion features.…
- The moon uses this technique, aiming to create a real displacement shorter than its
actual displacement (88000 km) based on the equation (L =88000 cos (θ)) and by
that while the moon moves a displacement =88000 km but the real displacement
(L) through its orbit be shorter than 88000 km and by that the moon can revolve
around Earth through more near orbits than its apogee orbit (r=0.406 mkm).
- The moon orbital motion equation depends on this concept and, the equation
uses (the constant) 1.7 degrees as the moon daily motion degrees, and the equation
uses the previous day angle (θ0) to produce the today angle (θ1)
- We have 3 questions in this equation study which are:
o How does this equation work?
o Is this equation trustee and correct?
o Why does the equation use the angle 1.7 degrees for the moon daily motion?
Let's try to answer….

I do this research
2nd
19
How to use this equation?
- Perigee Radius =0.363 mkm, so Its Orbital Circumference =2.28 mkm
- Suppose the moon will revolve around Earth through perigee orbit only during
29.53 days, so
- (2.28 mkm /29.53 days) = 77237 km
- This is (the real displacement = L = 88000 km Cos θ = 77237 km),
- What's the angle θ value? the angle θ = 28.63 degrees
- Suppose the moon stand on this point yesterday with the angle (θ) =28.63 degrees,
where the moon will move today?
- From Perigee (the most near point to Earth) the moon will move in Ascending
motion because it moves from perigee (0.363 mkm) to apogee (0.406 mkm)
- In Ascending motion we use (-1.7 degrees) because the angle (θ) is decreased
where the real displacement (L) is increased, So let's do that in following
o (θ1= θ0 - 1.7 degrees)
o (θ1= 28.63 degrees - 1.7 degrees) = 26.93 degrees
o L = 88000 Cos (26.93 degrees) = 78454 km
o During 29.53 days so (78454 km x 29.53 days = 2.316 mkm)
o 2.316 mkm = 2π x 368722 km
That means
o The moon was (before motion) on Perigee radius (r=0.363 mkm) and starts
its motion displacement 88000 km. For day motion the equation uses 1.7
degrees, that means, the moon on perigee uses Pythagorean triangle with
angle (28.63 degrees) and during one solar day the moon uses - 1.7 degrees
and by that the angle will be (26.93 degrees)…... The angle 1.7 degrees
expresses The Moon Daily Motion
o By using Pythagorean triangle its angle (θ) = 26.93 deg, the displacement
(88000 km) will create a real displacement through the moon orbit = 78454
km and the moon will finish its motion today at a distance 368722 km

I do this research
2nd
20
means the moon is far from perigee radius with (368722 km-363000 km
=5722 km )
o So, the moon after 1 day motion will be at the point 368722 km and will
have the Pythagorean triangle its angle 26.93 degrees.
The Descending Motion
o When the moon moves from apogee (0.406 mkm) to perigee (0.363 mkm),
so the angle (1.7 degrees) will be positive (+1.7 degrees) because the angle
(θ) is increased and the real displacement (L = 88000 Cos (θ)) be shorter.
So
o If the moon in apogee radius (r=0.406 mkm), what's the angle (θ)?
o The apogee orbital circumference = 0.406 mkm x2π =2.55 mkm = 29.53
days x 86400 km, the angle (θ) = 10.96 degrees (=11 deg approx.)
o The moon moves from apogee to perigee (descending motion)
o (θ1= θ0 + 1.7 degrees) means (θ1= 11 degrees + 1.7 degrees) = 12.7 deg.
o L = 88000 Cos (12.7 degrees) = 85847 km
o During 29.53 days so (85847 km x 29.53 days = 2.535 mkm)
o 2.535 mkm = 2π x 403467 km
So
o After one day the moon will be on 403467 km far from apogee (406000 km)
with 2540 km
Now let's see this equation test and efficiency in following

I do this research
2nd
21
4-4-2 The Equation Test and Accuracy
- I have tested the Equation with real data for 2 months June 2020 and October 2020
- The results are very good and I provide the results here for better vision
concerning the equation efficiency
1st
Test June 2020
Day Registered Data The Results (1.7) Difference
6-6-2020 369418 km
7-6-2020 373729 km 374772.5 - 1044
8-6-2020 378917 km 378821.5 96
9-6-2020 384534 km 383667.7 867
10-6-2020 390096 km 388890 1206
11-6-2020 395156 km 394000 1156
12-6-2020 399345 km 398604.2 741
13-6-2020 402395 km 402361.3 34
14-6-2020 404153 km 405052.8 -900
15-6-2020 404574 km ---- ---
16-6-2020 403718 km 401848.5 1870
17-6-2020 401733 km 400876.1 857
18-6-2020 398840 km 398640.7 200
19-6-2020 395303 km 395417.4 115
20-6-2020 391409 km 391521.2 -113
21-6-2020 387432 km 387273.4 159
22-6-2020 383607 km 382968.4 639
23-6-2020 380110 km 378852 1258
24-6-2020 377044 km 375107 1937
25-6-2020 374451 km 371836.5 2615
26-6-2020 372338 km 369077 3262
27-6-2020 370703 km 366855.6 3847
[

I do this research
2nd
22
The 1st
Test Results Analysis:
- The Total Results Are 20 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 2 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 3 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 20) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
Category of results defines the moon position in error range from (1300
km to 2000 km) = error (4.5%), that means (2 values of 20) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (3 values of 20) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

I do this research
2nd
23
2nd
Test October 2020
Day Registered Data Results (1.7) Difference
5-10-2020 405,690 km --- ---
6-10-2020 404,171 km 403125.3 km 1046 km
7-10-2020 401,649 km 401390 km 259 km
8-10-2020 398,073 km 398545.6 Km - 473 km
9-10-2020 393,464 km 394568.8 km -1105 km
10-10-2020 387,944 km 389510 km -1567 km
11-10-2020 381,763 km 383520 km -1758 km
12-10-2020 375,302 km 376875.3km -1574 km
13-10-2020 369,063 km 369981km -919 km
14-10-2020 363,617 km 363363.4km 254 km
15-10-2020 359,530 km 357612 km 1918 km
16-10-2020 357,269 km 353307 km 3962 km
17-10-2020 357,105 km ---- --
18-10-2020 359,048 km --- --
19-10-2020 362,851 km 364979.7 km - 2129 km
20-10-2020 368,058 km 368579.3 km -522 km
21-10-2020 374,101 km 373492.4 km 609 km
22-10-2020 380,412 km 379168.3 Km 1244 Km
23-10-2020 386,497 km 385059.3Km 1438 km
24-10-2020 391,989 km 390694.3 km 1295 km
25-10-2020 396,659 km 395729.5 km 930 km
26-10-2020 400,395 km 399958.7 km 437 km
27-10-2020 403,181 km 403299 km 112 km
28-10-2020 405,059 km 405738.5 km -680 km
29-10-2020 406,104 km 407359.4 km -1256 km
[

I do this research
2nd
24
The Test Results Analysis:
- The Total Results Are 22 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 5 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 2 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 22) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
km to 2000 km) = error (4.5%), that means (5 values of 22) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (2 values of 22) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

I do this research
2nd
25
4-4-3 The Value 1.7 degrees
- The 3rd
question was, why the equation uses 1.7 degrees?
Because
1.7 degrees = 0.98562 degrees + 0.712 degrees
Where
- 0.98562 degrees = Earth motion daily degrees, and it equals the moon daily
motion degrees because the moon has to move an equal distance to Earth motion
daily distance to save their motions accompanying
- This question and the angle 0.712 degrees is discussed in the previous paper
(The Moon Orbital Motion Geometry (II))
Please see the references

I do this research
2nd
26
The Moon Motion Difficulties
- There are 2 basic difficulties are observed in the moon orbital motions, let's refer
to them in following:
o (1st
Difficulty) The moon moves per day different distances from perigee to
apogee…..
o We know the moon moves from perigee to apogee (go and back) during
Anomalistic month (27.55 solar days)
o (43000 km x 2) / 27.55 days = 3122 km
o The moon doesn't use this rate (3122 km) in its motion, instead the moon
can move (6000 km) on one day only and on another day may move only
2500 km (or even less)!
o The moon orbital equation tries to solve this difficulty by using the rate 1.7
degrees in the equation (θ1 = θ0 + 1.7 degrees), the value 1.7 degrees is a
great number and enables the moon to move around (5000 km) per solar day
and by that if the moon moves per solar day 4000 km the different distance
will be 1000 km and if the moon moves 6000 km the different will be
– 1000 km, it’s the same difference, and by that, the error be minimized as
possible enabling the equation to be more efficient..
o (2nd
Difficulty) The moon stays in perigee and apogee points long time….
o That means, while the moon be on perigee or apogee, the moon doesn't use
the equation and doesn't change its distance to perigee or apogee for long
days…we may notice that in the equation tests, when the moon reach to
perigee or apogee the equation stops its work and stays 2 or 3 days to return
to its work… because the moon consumes long time to leave the points
(perigee and apogee)…

I do this research
2nd
27
5- Pluto Day Period Creation
5-1 The Point Objective
5-2 Pluto Day Period Creation
5-3 Planet Motion (6 Days Cycle)
5-4 The Earth Moon Orbit Creation
5-1 Preface
- Earth (29.8 km/s) moves during a solar day (24 h) a distance = 2574720 km
- Pluto (4.7 km/s) moves during its day (153.3) a distance = 2593836 km
- The moon displacements total during (29.53 days) = 2598693 km
- Pluto and the moon motion distances be equal but they different from Earth motion
distance with (1%).
- What geometrical result can be produced by these equal distances?
- We can't conclude this geometrical effect. But
- This geometrical effect is a real one because it's seen in other data, for example
(I)
- (Earth velocity / Pluto velocity) = (Pluto Day period/ earth day period)
- Pluto orbital distance 5906 mkm = Earth orbital circumference 940 mkm x 2π
- Pluto orbital period 90560 days = Earth Cycle (1461 days) x 2π3
(II)
- (406000 km /88000 km) = (708.7 h/153.3 h)
- Where
- 406000 km = Pluto motion distance during a solar day
- 88000 km = The Moon daily displacement
- 708.7 hours = the moon day period
- 153.3 hours = Pluto Day Period

I do this research
2nd
28
- The equal distances is a geometrical method by which the solar system geometrical
design causes the motion transportation. Through the motion transportation the
data be transported in proportionality with each other.
- The motion transportation created unified motion from planets motions. And this
unified motion depends on data complementary with each other.
- That explains how the previous data is created.
- I want to say that,
- The motions distances equality is a basic method depends on which the solar
system general design. For that reason Planets motions different cycles depend on
this same concept. As we have discussed Planet (8 days cycle) and (planet 6 days
cycle) (the 6 days cycle we will review in this paper). These all cycles depend on
the same concept that, the equal distances are the method to create a bridge among
planets motions.
- Pluto day Period Creation is a good example to see the effect of the planets
motions distances equality.
- Let's discuss it in following…

I do this research
2nd
29
5-2 Pluto Day Period Creation
I- Data
(1)
88 days /58.66 days = 4222.6 h /2802 h = 1.5 =5.1 deg /3.4 deg
(2)
(5848 mkm / 57.9 mkm ) =101
(3)
153.3 hours = 1.5 x 101 hours (error 1%)
(4)
101 h x 3600s x 47.4km/s =17.2 mkm
(5)
57.9 mkm = 17.2 mkm x 3.366
(6)
708.7 hours = 101.2 h x 7 (The Discussion Basic Data)
II –Discussion
Equation no. (1)
88 days /58.66 days = 4222.6 h /2802 h = 1.5 =5.1 deg /3.4 deg
- Where
- 88 days = Mercury Orbital Period
- 58.66 days = Mercury Rotation Period
- 4222.6 hours = Mercury Day Period
- 2802 hours = Venus Day Period
- 3.4 degrees = Venus orbital inclination
- 5.1 degrees = The moon orbital inclination
- Frequently we have asked why Mercury day period= 3 Mercury rotation period?
And Why Venus Day Period = 2 Mercury rotation period?

I do this research
2nd
30
- As the data shows, we deal with some mathematical system of data… the data be
created based on mathematical and geometrical calculations, so, the questions are
simply valid questions and have geometrical answers…
- The rate 1.5 is mentioned to be produced geometrically
- Even Mercury moves during its day period 720.7 mkm but Venus moves during its
day period 353 mkm and 720.7 mkm = 353 mkm x 2.04
- Mercury day period = Mercury orbital period x 2
- Venus day period = Mercury rotation period x 2
- It's shown clearly that some geometrical system is found behind this data.
- Let's suppose that, the geometrical mechanism required the rate (1.5) to be used
- Notice, in our discussion we follow the number (101) closely, it's the paper main
data of discussion.
Equation no. (2)
(5848 mkm / 57.9 mkm ) =101
- Where
- 5848 mkm = Mercury Pluto Distance
- 57.9 mkm = Mercury Orbital Distance
- Mercury Pluto Distance (5848 mkm) is very important distance in our discussion,
that because, the solar system geometrical design refers to it strongly. Let's shows
that in following
- Mercury day period =4222.6 hours needs 84 minutes to be 4224 hours =176 days
- 84 minutes =5040 seconds
- If there's a light beam its velocity (=1.16 mkm/s) it can travel during 5040 seconds
a distance = 5848 mkm (Mercury Pluto Distance). We discuss the supposed light
velocity with Jupiter motion analysis.
- But

I do this research
2nd
31
- We need this data here because the rate (101) which controls Mercury and Pluto
distances controls also their cycles periods of time. That shows we deal with a
mathematical and geometrical design of data.
Equation no. (3)
153.3 hours = 1.5 x 101 hours (error 1%)
- Where
- 1.5 = the rate between Mercury orbital period and rotation period
- Equation no. (3) tells that, Pluto Day period is crated based on 2 players, the period
(101 hours) and the rate (1.5)
- W know that rate (1.5), and that shows the interaction between Mercury and Pluto
motions data to define Pluto day period.
- But
- What's the period (101 hours)? Why we need it in this discussion?
- The next equation tells us the answer
Equation no. (4)
101 h x 3600s x 47.4 km/s =17.2 mkm
- Mercury (47.4 km/s) moves during (101 hours) a distance =17.2 mkm but we
know that Pluto orbital inclination = 17.2 degrees. Where we know that, Mercury
motion creates a rate 1 mkm = 1 degree because Mercury orbital circumference
=360 mkm =360 degrees.
- Equation no. (4) tells that, Mercury motion during (101 h) is related to Pluto
orbital inclination (17.2 deg). this is the information which we need because the
distance (17.2 mkm) will be the motor of the whole process.

I do this research
2nd
32
Equation No. (5)
57.9 mkm = 17.2 mkm x 3.366
- Where
- 57.9 mkm = Mercury Orbital Period
- 3.366 = 3.4 deg (Venus orbital inclination) (error 1%)
- Equation no. (5) tells that, Mercury orbital distance (57.9 mkm) depends on the
distance 17.2 mkm by a rate = Venus orbital inclination.
- Let's write Venus orbital inclination analysis in following:
o Venus diameter = 12104 km
o Venus diameter is used frequently as 12104 seconds
o 12104 seconds = 3.366 hours
o Where Venus orbital inclination =3.4 deg
o But
o 17.4 degrees = 3.4 deg x 5.1 deg (the moon orbital inclination)
o And we know that
o 17.4 deg = The inner Planets orbital inclinations total
o 17.2 deg = Pluto orbital inclination = 17.4 deg x 0.99
o 23.6 deg = The outer Planets orbital inclinations total
o 23.4 deg = Earth Axial Tilt = 23.6 deg x 0.99
o 23.6 deg + 17.2 deg = 23.4 deg +17.4 deg =41 deg
o The data is created based on mathematical and geometrical system

I do this research
2nd
33
Equation No. (6)
708.7 hours = 101 h x 7 (The Discussion Basic Data)
- Where
- 7 degrees = Mercury Orbital Inclination
- This is the basic discussion equation because based on it Pluto day period (153.3
hours /1.5) (=101 hours) be created as a function in the moon day period (708.7
hours).
- 17.2 deg (Pluto orbital inclination) = 7 deg + 2 x 5.1 deg (The moon orbital
inclination).
- Mercury motion distance during (101 hours) which is (17.2 mkm) is used here as
17.2 degrees.
Notice
- This process will be more clear through the moon orbit creation discussion. But
before we need to define (6 days cycle) by which Uranus creates the environment
in which Venus and Pluto can be found together and gives the chance for Venus
motion to effect on Pluto motion.
- So, in next point we should discuss (Planet Motion 6 Days Cycle)

I do this research
2nd
34
5-3 Planet Motion (6 Days Cycle)
The Cycle Description
- Uranus (6.8 km/s) moves during its day period (17.2 h) a distance = 421056 km
- Venus (35 km/s) moves during 12104 seconds a distance = 423640 km
- Pluto (4.7 km/s) moves during 90560 seconds a distance = 425632 km
- Where
- 12104 km = Venus Diameter
- 90560 days = Pluto orbital Period
- The previous data gives the base on which the (6 time units Cycle) is created – and
based on that, the provided data uses (6 Uranus days) and (6 x 12104 s) and (6 x
90560 s) to pass the mentioned distances and creates the defined differences
- Based on that,
(1)
Venus (35 km/s) moves during 12104 seconds (x 6) a distance = 2541840 km
Uranus moves during 6 Uranus days period (103.2 h) a distance = 2526336 km
2541840 km – 2526336 km = 15504 km (Mercury Circumference error 1%)
(2)
Pluto (4.7 km/s) moves during 90560 s x 6 a distance = 2553792 km
Venus (35 km/s) moves during 12104 seconds (x 6) a distance = 2541840 km
2553792 km – 2541840 km = 11952 km (Venus Diameter 12104 km error 1.2%)
(3)
Pluto (4.7 km/s) moves during 90560 s x 6 a distance = 2553792 km
Uranus moves during 6 Uranus days period (103.2 h) a distance = 2526336 km
And
2553792 km – 2526336 km = 27456 km (= 86400 km /π)
(4)
Jupiter (13.1 km/s) moves during (6) of Jupiter days (59.4 h) a distance = 2.8 mkm =
2π2
x 142984 km (Jupiter Diameter)

I do this research
2nd
35
(5)
Neptune (5.4 km/s) moves during (6) of Neptune days (96.6 h) a distance = 1.87 mkm
= Jupiter (13.1 km/s) motion distance during 142984 seconds.
Basic Questions
- The cycle of (6 units) doesn’t depend on planet day period Why?
- Because
o The planet diameter is used as a period of time, and also planet orbital
period is a period of time
o These all data consist together the planet motion required periods of time
but in categories- as we have seen in Saturn analysis – let's remember it
o Saturn day period =10.7 hours but Saturn diameter =120536 km and can be
used as 120536 seconds where 10.7 h x 3600 s x π = 120536 s
o That means, Saturn diameter in fact = Saturn day period but be seen in
another form – Saturn orbital period should be a third form for the same
period of time
o But we have no equation can express these 3 values in each planet..! that's
the difficulty because we don't know the geometrical rule based on which
the planet diameter is created (again we need to ask, how the matter is
created?)
Notice
- Pluto moves during (6 x 90560 s) a distance = 2553792 km
- Pluto moves during its day period (153.3 h) a distance =2593836 km
- The difference = 40080 km = Earth Circumference
- Please remember 40080 km = the 5 inner planets diameters total
- Pluto moves during 8520 seconds a distance =40080 km.

I do this research
2nd
36
The Cycle Using Examples
Example No. (1) Saturn orbital Distance 1433.5 mkm
I- Data
(A)
Jupiter moves during (512 Jupiter days x 6) a distance =1434.267648 mkm
(B)
Saturn moves during (640 Saturn days x 6) a distance = 1434.79296 mkm
(C)
Neptune moves during (12430 hours x 6) a distance = 1449.8352 mkm
(D)
Uranus moves during (64 Pluto days x 6) a distance =1441.069056 mkm
(E)
Mercury moves during (1407.6 h x 6) a distance = 1441.157184 mkm
(F)
Mars moves during (2802 h x 6) a distance = 1458.60912 mkm
(G)
Neptune moves during (778.6 x 16.1 h x 6) a distance = 1462.136055 mkm
II- Data Analysis
(E) – (D) = 88100 km
(F) – (E) = 17.4519 mkm
A Comment
- Mercury moves during its rotation period a distance = 243 mkm (error 1%)
- And Mercury moves during its day period a distance = 720.7 mkm
- The distance 720.7 mkm between Mercury and Jupiter is the orbital radius and the
diameter needs 2 x 720.7 mkm which need 2 x 4222.6 hours (= 6 x 1407.6 hours
Mercury rotation period)
- And during (6 Mercury rotation periods) Mercury movies a distance = 1433 mkm,
that creates a connection between (243 mkm) and (1433 mkm).

I do this research
2nd
37
5-4 The Earth Moon Orbit Creation
I- Data
(A)
2598693 km = 112.2 km/s x 23160 seconds
(B)
406000 km = 35 km/s x 23610 seconds.
(C)
(406000 km/88000 km) = (708.7 h/153.3h) = (417800 km/90560 km)
(D)
2 x 153.3 h = 4.61 x 66.8
But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
(E)
3.024 mkm = 88000 km x 2 x 17.2
(F)
243 mkm = 595 mkm x 0.406 mkm (error 1%).
II- Discussion
- Let's summarize the whole story in following
o The interaction between the moon and Pluto motions data is done under
Venus motion effect. But this process needed a complex geometrical design.
o The 3 planets (Mercury, Venus and Earth) unified their velocities together
and created one unified velocity =112.2 km/sec
o This process is one by an effect of Jupiter motion, where the data which
shows the interaction between Jupiter and the inner planets motions supports
this claim strongly (please review Jupiter data analysis Point no. 7)
o The 3 planets velocities total (112.2 km/sec) effects on the moon
displacement total during (29.53 days) (= 2598693 km) which creates the
period of time 23160 seconds.

I do this research
2nd
38
Equation no. (A)
2598693 km = 112.2 km/s x 23160 seconds
- Equation no. (A) shows this process, the period 23160 seconds is created by an
effect of the 3 planets velocities total on the moon displacements total
Equation no. (B)
406000 km = 35 km/s x 23610 seconds.
- Equation no. (B) shows that, Venus during 23610 seconds moves 406000 km = the
moon orbital apogee radius = Pluto motion distance during a solar day
- That means, Venus motion effect on the moon creation process but this effect is
done based on the 3 planets velocities total using
- Why the 3 planets velocities total be used?
- The 3 planets velocities total (112.2 km/s) works as one gear (1st
Gear) and Pluto
velocity (4.7 km/s) works as another gear (2nd
Gear).
- The motion of the 2 Gears creates the rate (112.2 = 4.7 x 23.9) which means, 1
hour of the (3 planets velocities total motion) be equivalent to (1 earth rotation
period 23.9 h) of Pluto motion.
- By that, 1 hour of a planet motion be = 23.9 hours of another planet motion. the
process creates a different rate of time
- The different rate of time is discussed deeply in the next point. But here we refer
only to the concept because we interest for the technical process based on which
the moon orbit be created.
- Shortly
- Venus motion creates the distance (406000 km) and effect on the moon and Pluto
motions, and by that the 2 planets adopt this distance as their basic distance of
motion (406000 km = the moon orbital apogee radius = Pluto motion distance
during a solar day).

I do this research
2nd
39
Equation no. (C)
(406000 km/88000 km) = (708.7 h/153.3h) = (417800 km/90560 km) =4.61
- Where
- 90560 days = Pluto Orbital Period
- 417800 mkm = different 1% with 413600 km
- Equation no (C) tells clearly what's happening…
- Pluto orbital period (90560 days) which is used in (planet 6 days cycle) is used
here again as a distance
- The moon displacements total 2598693 km = 2π x 413600 km
- The distance (417800 km) is different from 413600 km by (1%).
- Equation no. (C) shows that the geometrical mechanism of the used distances
effects on the 2 planets days periods. And by that, the proportionality of the planets
days periods be created by the proportionality of the distances.
Equation no. (D)
2 x 153.3 h = 4.61 x 66.8
But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
- Where
- 708.7 hours = The moon Day period
- 655.7 hours = The Moon Rotation Period
- 10.7 hours = Saturn Day Period
- 9.9 hours = Jupiter Day Period
- The rate (4.61) we get from equation no. (C). where these rate is found between
the distances and days periods of the moon and Pluto.
- The equation produces the rate 66.8 which is the rate between the moon cycles
periods on one side and Saturn and Jupiter cycles periods on the other side.

I do this research
2nd
40
Equation no. (E)
3.024 mkm = 88000 km x 2 x 17.2
- Where
- 3.024 mkm = Venus motion distance per a solar day
- 88000km = The moon displacement per a solar day
- 17.2 deg = Pluto orbital inclination
- Equation no. (E) tries to support the claim of Venus motion effect on the moon and
Pluto motions.
Equation no. (F)
243 mkm = 595 mkm x 0.406 mkm (error 1%).
- Where
- 595 mkm =1980 seconds x 0.3 mkm/sec (light known velocity)
- And
- 1980 mkm = (360 mkm +680 mkm +940 mkm)
- The distance 1980 mkm be used as 1980 second by light motion.
- 0.406 mkm = Pluto motion distance during a solar day
- This equation also tries to support the same claim (the equation discussion be more
clear after reading Jupiter data analysis).

I do this research
2nd
41
6- Can The Solar System Be A Clock?
6-2 Pluto and Mercury Rate Of Time
6-3 Pluto and The Moon Rate Of Time
6-4 Pluto and Earth Rate Of Time
- Does the second period of time be equal for all solar planets?
- Earth Day Period =24 hours, but Earth rotation period =23.9 hours, the difference
between these 2 periods force us to divide the solar day into 24 hours and each
hour into 60 minutes and each minute into 60 seconds.
- I try to show, The second period of time is defined based Earth motion features.
- But
- The solar planets motions features are different from one another. because of that,
the period of time (a second) should be different from one planet to another. that
means, different rates of time are found in the solar system motion.
- The paper basic hypothesis is that (The Motion Be Transported Among The
Solar Planets)
- If the planets rates of time are different from one another, how the motion can be
transported among the planets?
- Based on this description we conclude, one unified general rate of time must be
created by the solar planets motions.
- If this one unified rate of time is found in the solar system, that means, the solar
system must work as a clock. In which 1 second of one planet be = 1 minute of
another planet, and equal to 1hour of a third planet. By that, the planets motions
work as the clock gears.

I do this research
2nd
42
- The planets rates of time should depend on their velocities. By that, Mercury
velocity (47.4 km/s) = 10.08 x Pluto velocity (4.7 km/s) and the rate of time should
be 1 hour of Mercury motion be =10.08 hours of Pluto motion.
- But, In fact
- 1 hour of Mercury motion be = 9.18 hours of Pluto motion.
- This change in the rate of time is done because of the planets motions interaction.
The Discussion Point Objective
- The point proves that the rate of time between Mercury and Pluto is 9.18 and not
10.08.
- Also the point discusses the time rate of Uranus Pluto Motions and explain how
these rates be defined.

I do this research
2nd
43
6-2 Pluto and Mercury Rate Of Time
I- Data
(1)
112.2 km/s = 4.7km/s x 23.9
(2)
1407.6 hours =153.3 hours x 9.18
(3)
90560 days = 4222.6 hours x 9.18 x 2.33 (error 2%)
(4)
(2 x 153.3 h x 3600 s) /5040 s =23.9 x 9.18
(5)
((4.6 x 24.7)/24) =9.18 /1.9379
But
(47.4/24.1) =(Mars mass /Mercury mass) = (6792 km /4879km) =1.9379
(6)
4222.6 h =153.3 h x 27.5
II- Discussion
Equation (1)
112.2 km/s = 4.7 km/s x 23.9
- Where
- 112.2 km/s = The 3 Planets Velocities Total
- 4.7 km/s = Pluto Velocity
- We know this equation. We suppose that, the 3 planets velocities be unified in one
velocity =112.2 km/s, So this velocity works as one gear and Pluto velocity works
as another gear, move relative to one another.
- By this supposed motion of 4 planets, the rate (1 hour = 23.9 hours) be created and
that cause 1 hours of a motion be = 23.9 hours of another motion in the solar
system.

I do this research
2nd
44
- This process must be supposed process but it shows its effect greatly on the solar
planets motions rates of time because it adds one more velocity which isn't found
among the planets. This process also is very important because if the planets
motions rates of time depend only on the planets velocities that makes the max rate
is (10.08) means, 1 hour of a planet motion be = 10.08 hours of another planet
motion. and this very narrow range of the rates of time. The 3 planets velocities
total creates an extension for this process and enable the system to create the rate
(1h = 23.9 h).
Equation (2)
1407.6 hours =153.3 hours x 9.18
- Where
- 1407.6 hours = Mercury Rotation Period
- 153.3 days = Pluto Day Period.
- (9.18) = The rate of time between Mercury and Pluto motions.
- We conclude that mercury and Pluto motions rates of time is (9.18) based on this
equation.
- By this same method we conclude the rate of time between Pluto and the moon
motions. It was the rate between the moon day period and Pluto day period. We
should remember this rate of time in the next sub-point.
Equation (3)
90560 days = 4222.6 hours x 9.18 x 2.33 (error 2%)
- Where
- 90560 days = Pluto Orbital Period
- 4222.6 hours = Mercury day period
- 2.33 = ??
- Mercury velocity (47.4 km/s) = 10.08 x Pluto velocity (4.7km/s). if the planets
rates of time depend on their velocities. That means 1 hour of Mercury motion =

I do this research
2nd
45
10.08 hours of Pluto motion. and in this case 24 hours of Pluto motion = 2.37
hours of Mercury Motion.
- That because the equation uses (Pluto orbital period 90560 solar days) in
comparison with 4222.6 hours (Mercury day period). The equation proves that the
rate (9.18) is the used one between the 2 planets.
Equation (4)
(2 x 153.3 h x 3600 s) /5040 s =23.9 x 9.18
- Where
- 5040 seconds are required for Mercury day period to be 176 solar days
- 153.3 h = Pluto Day Period.
- 23.9 = the between the 3 planets velocities total (112.2 km/s) and Pluto velocity
(4.7 km/s).
- This equation also proves that the rat of time between Mercury and Pluto is (9.18)
Equation (5)
((4.6 x 24.7)/24) =9.18 /1.9379 But
(47.4/24.1) =(Mars mass /Mercury mass) = (6792 km /4879km) =1.9379
- Where
- 24.7 h = Mars day period
- 24 h = Earth day period
- 47.4 km/s = Mercury veloicity 24.1 km/s = Mars veloicity
- 6792 km = Mars Diameter 4879 km = Mercuy Diamter
- Equation no. (5) tells that, the rate (9.18) between Mercury and Pluto motions
effect on Mars motion data and creates its rate (4.6) (where 1 hour of Mars motion
=4.6 hours of Pluto motion).
Equation (6)
4222.6 h =153.3 h x 27.5
- 4222.6 h = Mercury Day Period 153.3 h = Pluto Day Period
- 27.5 days = The moon perigee month

I do this research
2nd
46
6-3 Pluto and The Moon Rate Of Time
I- Data
(7)
(406000 km/88000 km) = (708.7 h /153.3 h) = 4.61
(8)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
II- Discussion
Equation (7)
(406000 km/88000 km) = (708.7 h /153.3 h) = 4.61
- Where
- 406000 km = Pluto motion distance during a solar day
- 88000 km = the moon displacement per a solar day
- 708.7 hours = The Moon Day Period
- Based on this same rule we have defined the rate (9.18) between Mercury and
Pluto motions. But here we use the moon day period and with Mercury we have
used Mercury rotation period.
- The basic difference is that, the moon an Pluto days periods are rated with their
motions distances per a solar day but in Mercury case we use the rate between
Mercury rotation period and Pluto day period (9.18) avoiding their motions
distances per a solar day or velocities (10.08)
- This is done because, the 3 planets velocities total causes this change and the
velocities total effect on some planets and not on all planets.

I do this research
2nd
47
Equation (8)
2 x 153.3 h = 4.61 x 66.8 But
(708.7 h /10.7 h) = (655.7 h /9.9 h) =66.23 (error 1%).
- Where
- 708.7 hours = The moon Day period
- 655.7 hours = The Moon Rotation Period
- 10.7 hours = Saturn Day Period
- 9.9 hours = Jupiter Day Period
- The rate (4.61) we define from equation no. (6). where this rate is found between
the distances and days periods of the moon and Pluto.
- The equation uses the rate 66.8 which is found between the moon day period and
Saturn day period, and found between the moon rotation period and Jupiter
rotation period.
- That tells this process is effected by Saturn and Jupiter motions.
- Notice
- The rate (4.61) is the same in the 2 equations, no. (6) and (7) which shows that the
rate of time is a fact. where in the moon Pluto motions rate we found it (4.61) in
different data and also for mercury Pluto motions rate (9.18) we found it in
different data also. That shows, the rate of time is a defined data for the planet
motion (as the planet registered data) .

I do this research
2nd
48
6-4 Pluto and Earth Rate Of Time
Equation (9)
(29.8 /4.7) = (153.3 h /24 h)
- Where
- 29.8 km/s = Earth Velocity
- 4.7 km/s = Pluto Velocity
- 153.3 h = Pluto day period
- 24 h = Earth day period
- It's the same rule we have used with the other planets. Just with Mercury the
periods rate doesn't = the passed distances rate but with the moon and Earth the
periods rate be = the passed distances. This is the effect of the 3 planets velocities
total.
- The equation no. (9) tells that, 1 hour of earth motion = 6.387 hours of Pluto
motion.
- this meaning is so important, because
- 1 hour of Pluto motion be = 4.61 hours of the earth moon motion
- based on that,
- 1 hour of 1 earth be = 4.61 x 6.387 =29.53 hours of the moon motion.
- this meaning we get simply because the moon displacements total during 29.53
days = Earth motion distance during a solar day (error 1%)
- that's why 1 day of Earth motion be = 29.53 solar days of the moon motion.

I do this research
2nd
49
7- Jupiter Motion Analysis
7-2 Mars Motion Analysis
7-3 Mars And Mercury Motions Analysis
7-4 Jupiter Motion Analysis
- The point claims (The Motion Must Be Transported Among The Solar Planets)
- That because
- Planet Motion Data Is Used By Another Planet Motion Data
- The used data isn't an unique random data, on the contrary, A System Of Data be
used for different planets motions.
- That means (A System Of Data) be transported from one planet to another.
- We ask how this (A System Of Data) be transported from one planet to another if
the planets are rigid bodies separated and independent from one another in their
motions?
- Jupiter Motion Data Analysis proves this claim. Because,
- Jupiter Motion Data shows that, Jupiter Motion data depends on the inner planets
motions data clearly and strongly.
- It's similar to " Small Gears Move To Cause A Motion For One Greater Gear"
this description perfectly expresses Jupiter Motion data.
- Mars Motion Data can be a usual approach for this discussion. For that we discuss
Mars Motion Data (at first) and then we discuss Jupiter Motion Data.

I do this research
2nd
50
7-2 Mars Motion Analysis
I- Data
- Mars Orbital Period = 687 days
- Jupiter Orbital Period = 687 days x 2π = 4331 days (error 0.3%)
- Also 4331 days x π = 13606 days
- And 13606 days x π = 42745 days
Group No. (1)
- Mars moves during 687 d. a distance= 1433 mkm= Mars Orbital Circumference
- Mars moves during 4331d. a distance= 9010 mkm= Saturn orbital circumference
- Mars moves during 13606 d a distance= 28255 mkm= Neptune orbital circumference
Group No. (2)
- Mercury moves during 687 d. a distance= 2815 mkm= Mercury Uranus Distance
- Mercury moves during 4331d. a distance= 2815 mkm= Mercury Uranus Circum.
Group No. (3)
- Jupiter moves during 687 d. a distance= 778.6 mkm= Jupiter Orbital Distance
- Jupiter moves during 4331 d. a distance= 4900 mkm= Jupiter Orbital Circum.
- Jupiter moves during 13606 d. a distance= 15394 mkm (later be discussed).
Group No. (4)
- Saturn moves during 687 d. a distance = 576 mkm
- Saturn moves during 4331 d. a distance = 3630 mkm
- Where
- 1433.5 mkm (Saturn Orbital Distance) = 576 mkm x 2.5
- 9010 mkm (Saturn Orbital Circumference) = 3630 mkm x 2.5
- And (2.5 degrees = Saturn Orbital Inclination)

I do this research
2nd
51
II- Discussion
- How Does This Data Be Created?
- Let's consider one great river has many canals.
- The water motion through these canals depend on one another. Because the
original water amount (total energy) define each canal (water) capacity and by that
defines the motion (velocity).
- The (water) motion in this river and its canals should be considered one unified
motion (or a network of motions). because it depends on one source and the
motions range is limited by this source.
- Now, the moving water can carry some ships or without. Still the motion be done
through the river and canals.
- We try to explain Mars Motion Data
- Because of that,
- We can't see the motion source, Because we stand on one canal of the river and try
to define this canal water motion features. So we don't see the source, but we see
one canal motion only…. But
- Because the (water) motion depend on each other, for that, we discover the motion
data be in proportionality with other canals motions data. That's what we see in
Mars Previous Data.
- Let's deepen our analysis for this example,
- Imagine one canal is narrow and has rocks prevent the water motion, that will
effect on its neighbor canals and increase the water level in them.
- By measurement we discover that, the water level be lower in this canal than its
neighbors, that can be explained by analysis for the canal passage
- The paper vision for the solar system motion tells
o The solar planets aren't separated rigid bodies independent in their motions
from one another. on the contrary

I do this research
2nd
52
o The solar group is one machine of gears, or a network of motions. each
planet motion be done depending on other planets motions and because of
that (The Motion Is Transported Among The Planets). And that explains
how some planets motions us (A system of Data) for other planets motions.
Because the data is transported with the transported motion.
o Also
o Let's ask 9in our example) Why the canals motions data is in proportionality
with one another? because of the water amount in each canal. That's the
same answer for the solar planet motion. from one source of energy all
planets are created and this energy is distributed among the planets, because
of that, the planets data be complementary with each other, because their
total is the same source.

I do this research
2nd
53
7-3 Mars And Mercury Motions Analysis
(Point no. I)
- 4900 mkm =1.16 mkm x 4222.6,
- where
- 4900 mkm = Jupiter Orbital Period
- 1.16 mkm = a distance Jupiter passes in =24.6 h (= Mars rotation period)
- 4222.6 hours = Mercury Day Period
- This data tells, some interaction must be found between Mercury & Mars motions.
- i.e.
- The motions be interaction between Mars and Mercury and based on this motions
interaction Jupiter motion depends.
- What we need in this point of discussion is, to see if the 2 planets motions data be
created depends on one another and to know why, let's see the data in following:
(I)
(1)
- Mercury (47.4 km/s) moves during 5040 seconds a distance = 238896 km
- Mars (24.1 km/s) moves during 5040 seconds a distance = 121464 km
(2)
- Mercury (47.4 km/s) moves during 1407.6 hours a distance = 240.3 mkm
- Mars (24.1 km/s) moves during 2802 hours distance = 243 mkm
(3)
- Mercury (47.4 km/s) moves during 687 days a distance = 2815.2 mkm
- Mars (24.1 km/s) moves during 687 days a distance = 1433.5 mkm
(4)
- Mercury (47.4 km/s) moves during 346.6 days a distance = 1419.5 mkm
- Mars (24.1 km/s) moves during 346.6 days distance = 720.7 mkm
- Why the planets motions during (defined periods) pass (defined distances)?
- What point of view this question asks about?

I do this research
2nd
54
- Imagine the solar planets are rigid separated bodies independent in their motions
from the other planets. In this case each planet moves its orbital circumference (a
defined distance) in its orbital period ( a defined period of time).
- means, for example,
- Venus doesn’t know the period (365.25 days) and doesn't consider it in its motion.
because Venus motion trajectory is independent from Earth motion trajectory.
- The data disproves this description…
- Each planet (knows) the other planets cycles periods of time and takes these
periods into consideration in its motion. why and how?
- We don't deal with separated points of motions. On the contrary, we deal with one
chess board, each motion of any piece effects on the other pieces motions. And
because of that the motions data be seen in other planets motions.
- Do we have an obligation to attribute mechanical features for the space to use it as
a conveyor belt for the transporting motion among the planets? I'm not sure.
- Because we don't know the geometrical mechanism by which the motion is
transported. So it's not sure the transportation be done mechanically. In the light
supposed velocity (1.16 mkm/sec) discussion we discuss that more widely.
- Now let's return to the data,
- Data no (1) Mercury moves during 5040 sec a distance = 2 Saturn diameters (-1%)
- Mars moves during 5040 sec a distance = Saturn Diameter (+1%)
- (Mercury day period needs 5040 sec to be = 176 solar days)
- Data no (2) Mercury moves during its rotation period (1407.6 h) a distance =
240.3 mkm
- Mars moves during Venus day period (2802 h) a distance = 243 mkm
- (Venus rotation period =243 Solar Days)
- Let's ask (why does Mars use Venus day period 2802 h?)
- We answer this question in the next point.

I do this research
2nd
55
- Data no (3)
- During Mars orbital period (687 solar days) Mars moves its orbital circumference
1433 mkm and Mercury moves its distance to Uranus 2815 mkm. The data support
the concept clearly. If we deal with separated moving points, Mercury should
move during (687 mkm) any random distance, but it moves a distance = Mercury
Uranus Distance (2815 mkm) because the planets motions depend on one
geometrical design.
- Data no (4)
- During the nodal year (346.6 days) Mercury moves a distance =1419 mkm (= 99%
of 1433 mkm = Mars Orbital Circumference) (a distance related to Mars)
- And Mars during the same period (346.6 days) a distance = 720.7 mkm = Mercury
Jupiter Distance (a distance related to Mercury)
- Now let's answer the question (Why Does Mars Use Venus Day Period?)

I do this research
2nd
56
(Point no. II) (The Cycles Periods Interaction)
(A)
- Mercury Day Period 4222.6 hours = 3 x 1407.6 hours (Mercury rotation period)
- Mercury Day Period 4222.6 hours = 2 x 2112 hours (Mercury orbital period)
- Venus Day Period 2802 hours = 2 x 1407.6 hours (Mercury rotation period)
- (Mercury rotation period 1407.6 h)/ (Mars rotation period 24.6 h) = (180/π)
(B)
- Mercury (47.4 km/s) moves during 1407.6 hours a distance = 240.3 mkm
- Mars (24.1 km/s) moves during 2802 hours distance = 243 mkm
- But
- Venus (35 km/s) moves during 365.25 solar days a distance =1105 mkm
(1105 mkm = 2π x 175.94 mkm)
(175.94 solar days = Mercury Day Period)
(243 solar days = Venus Rotation Period)
(C)
- Venus Rotation period 243 solar days x 3 = 729 days.
- (Mercury moves during its day period 4222.6 h a distance =720.7 mkm), (error
1%)
- That means, Mercury moves during its rotation period a distance (240 mkm) can
be used as (243 days = Venus rotation period). Mars did the same result by using
Venus day period (116.75 days).
(D)
- 687 days (Mars orbital period) x 24 h = 671 x 24.6 h (Mars rotation period),
- Where
- 671 mkm = Venus Jupiter Distance
- Also
- 680 mkm = Venus orbital circumference (as a period be 687 days different from
Mars orbital period 687 days with 1%)

I do this research
2nd
57
- And
- Mars (24.1 km/s) moves during 329.8 solar days a distance = 687 mkm,
- Where
- 329.8 days = the moon sidereal year (327.6 days) but this year doesn't end with
sidereal month (27.3 days) but end with a synodic month (29.53 days).
- The distance 687 mkm, if 1mkm = 1 day will be 687 days (Mars Orbital Period)
Also
- Venus (35 km/s) moves during 108.2 solar days a distance = 327.6 mkm
(327.6 days = the Lunar Sidereal Year)
- I try to show that, it’s usual using in the solar planets data to use the distances as
periods of time.
Conclusions
- The data shows we deal with a system of data.
- Mercury and Mars motions cycles are interacted with each other. This interaction
makes (the unified motion) of the 2 planets more clear than the others in data. That
help us to see this fact.
- Venus day period (and other cycles) are defined in proportionality with Mercury
cycles periods. For that reason Mars uses Venus day period.
- Also,
- Because Jupiter orbital circumference depends on the 2 planets data that tells they
are the (representatives) of the inner planets motions, based on them Jupiter
motion depends.

I do this research
2nd
58
7-4 Jupiter Motion Analysis
Preface
I-Data
(I)
Jupiter (13.1 km/s) moves during (24.6 hours = Mars rotation period) a distance =
1.16 mkm
(II)
4900 mkm (Jupiter Orbital Circumference) = 1.16 mkm x 4222.6 (Mercury Day
Period = 4222.6 hours)
II-Discussion
- This point discussion aims to prove that (The Motion Be Transported Among
The Solar Planets),
- Let's suppose it's a fact that, the motion be transported among the solar planets,
What result can prove that?
- Jupiter motion data provides a good proof as a result, because
- Jupiter orbital circumference is designed based on the inner planets motions data.
- The previous 2 equations give us a clear approach for this meaning…. Jupiter
Orbital Circumference (4900 mkm) be designed geometrically based on the inner
planets orbital circumferences. This fact should be proved by many data we
discuss in following. The Idea Is That, Jupiter be as a great gear its motion
depends on other small gears.
- Because of that, Jupiter orbital circumference be = the inner orbital circumferences
total (will be discussed). Because of this fact Jupiter orbital circumference is
dividing on the distance (1.16 mkm) and (Mercury Day Period 4222.6 h)
- The following data will show how Jupiter orbital circumference be designed as a
function in Mars and Mercury motions data, we should discuss the theoretical
explanation with the data analysis because the data gives more clear vision how
this dependency is created.

I do this research
2nd
59
Notice
- In the next page we discuss The Theory Proves In details
- But
- The data can have another reading, let's see it in following
4900 mkm (Jupiter Orbital Circumference) = 1.16 mkm/sec x 4222.6 seconds
- This data tells
- If there's a light beam its velocity (=1.16 mkm/sec), So this light beam can pass
Jupiter orbital circumference (4900 mkm) in a period = 4222.6 seconds.
- The idea is that,
- The distances between Jupiter and the most other planets depend on the rate (1.16),
- If we try to explain all Jupiter distances data based on the fact that (Jupiter moves
during 24.6 hours a distance = 1.16 mkm) these explanations will be so complex.
- And
- These explanations will be so simple and clear if we suppose that a light beam its
velocity (=1.16 mkm/sec) be found.
- I want to say that,
- This (supposed) velocity (1.16 mkm/sec) should be considered the basic column
on which Jupiter Motion Data depends.
- Because Of That
- I Put the supposed light velocity (1.16 mkm/sec) discussion in a separated point
after Jupiter motion analysis discussion, and by that we provide 2 readings for the
same data. And we aim to show that the motion is transported among the solar
planets which is a fact proved by Jupiter motion analysis. But why the motion is
transported among the solar planets? Because the matter is created out of light and
by that the light motion feature (continuum motion) be effected on Planet motion
features and causes the motion transportation among the solar planets.

I do this research
2nd
60
Jupiter Motion Analysis
(1st
Point)
- 4846 mkm = 360 mkm +680 mkm+940 mkm +1433 mkm +1433 mkm
- 4900 mkm = Jupiter Orbital Circumference
- 360 mkm = Mercury Orbital Circumference
- 680 mkm = Venus Orbital Circumference
- 940 mkm = Earth Orbital Circumference
- 1433 mkm = Mars Orbital Circumference - But
- The difference between 4900 mkm and 4846 mkm is (1%)
Notice
- The total (4846 mkm) uses 2 values of the distance 1433 mkm (Mars Orbital
Circumference) because the geometrical design defines the Earth moon orbital
around the sun be = 1433 mkm = Mars orbital circumference. This definition is
done based on some geometrical necessity we still search for it. but some data may
prove this idea because (1433 mkm = 550.7 days x 2.6 mkm)
- We know that, Earth motion during its day period (2.574 mkm) = Pluto motion
during its day period (2.593836 mkm) = the moon displacements total during
29.53 days (= 2.598693 mkm). Where the moon and Pluto motion are greater than
Earth motion with (1%) and the moon distance is greater even than Pluto distance.
The data uses the distance (2.6 mkm) which is greater than others and be very near
to the moon distance. That means, Mars orbital circumference 1443.5 mkm is
defined based on Mars Jupiter Distance 550.7 mkm (uses as 550.7 days) by the
moon distance 2.598 mkm. This data can be used as a proof for the claim (the
geometrical design uses 1433 mkm as the moon orbital circumference in place of
940 mkm).

I do this research
2nd
61
More Data
(A)
- Mercury moves during its day period a distance = 720.7 mkm= Mercury Jupiter
distance
(B)
- Venus moves during its orbital period a distance = 680 mkm (Venus Orbital
Circumference) - But
- Venus Jupiter distance =671 mkm (the difference = 1.3%)
(C)
- Earth moves during its orbital period a distance = 940 mkm (Earth Orbital
Circumference) - But
- Earth Jupiter distance = 928 mkm (the difference = 1.3%)
- (Note, in this case the Earth and Jupiter should be on 2 different sides from the sun
and by that the distance 928 mkm = 778.6 mkm +149.6 mkm)
(D)
- Mars moves during its orbital period a distance = 1433 mkm (Mars Orbital
Circumference) (please remember 1433 mkm = 550.7 mkm x 2.6) (where 550.7
mkm = Mars Jupiter Distance)
- The data tries to show that,
o The inner planets define their orbital circumferences based on their distances
to Jupiter.
o And
o Jupiter defines its orbital circumference based on the inner planets orbital
circumferences total.
- i.e.
- Some mutual effect is found between Jupiter and the inner planets motions. This
effect is seen clearly in both data.

I do this research
2nd
62
(2nd
Point)
- Jupiter (13.1 km/s) moves during (24.6 hours = Mars Rotation Period), a distance
=1.16 mkm, means
- Jupiter during 4222.6 ( x 24.6 hours) moves a distance = 4900 mkm
- This data has 2 readings (as we have seen). In this point we try to show the effect
of the number (4222.6) on the inner planets motions. Where this effect can help to
conclude the suitable reading for the data.
(I)
- 4222.6 hours = Mercury Day Period - Where
- (Mercury rotation period 1407.6 h / Mars rotation period 24.6 h) = (180/π)
- Some geometrical mechanism is found between Mercury, Mars and Jupiter.
(II)
4222.6 mkm = 2π x 671 mkm (Venus Jupiter Distance)
(III)
(4222.6s / 5040s) = (778.6 mkm/928 mkm).
Where
778.6 mkm = Jupiter orbital distance
928 mkm = Earth Jupiter Distance (be on 2 different sides from the sun)
And Mercury Day period needs 5040 seconds to be = 176 solar days.
The data shows that we don't deal with separated independent motions, but we deal
with one unified motion is done by the inner planets and Jupiter motions. By that the
data proves the hypothesis, that, the motion is transported among the solar planets.
Notice
680 mkm + 940 mkm =1620 mkm
1.1318 mkm (Jupiter velocity per solar day) x 1433 days = 1620 mkm
The data shows that, More Interactions of Motions must be found between the inner
planets and Jupiter motions.

I do this research
2nd
63
(3rd
Point)
Data Group No. (I)
(a)
778.6 mkm = 671 sec x 1.16 mkm/s (where 778.6 mkm = Jupiter orbital distance)
(b)
672.8 mkm = 580 sec x 1.16 mkm/s
(c)
720.7 mkm = 629 sec x 1.16 mkm/s (error 1%)
(d)
629 mkm = 542 sec x 1.16 mkm/s
Data Group No. (II)
(e)
778.6 mkm = 1.0725 x 720.7 mkm (Mercury Jupiter Distance)
(f)
720.7 mkm = 1.0725 x 671 mkm (Venus Jupiter Distance)
(g)
671 mkm = 1.0725 x 629 mkm (Earth Jupiter Distance)
(h)
629 mkm = 1.0725 x 580 mkm (error 1%)
(i)
580 mkm = 1.0725 x 542 mkm
- The data group no. (I) uses the light supposed velocity reading which we discuss in
details later.
- the data shows that, the distances between Jupiter and the inner planets are defined
geometrically, where the rates (1.16 and 1.0725) control these distances.
- I try to show that, the data is created based on geometrical reasons by using these 2
rates. We here don't need to explain why these 2 rates are used but only we try to
prove that the distances are planned geometrically and defined based One
General Geometrical Design. The planets motions as separated rigid bodies
independent from each other is completely disproved by the data. We deal with a
network of motions define each planet motion distance based on the same one
geometrical design. The motion transportation must be fact based on that.

I do this research
2nd
64
Notice
- The rate (1.0725) effects on around 40% of all distances in the solar system.
- That makes Jupiter distances be as a part only of many similar effected distances.
- Appendix no. (1) of this paper provides these effected distances and provides an
explanation why these distances are effected by this rate (1.0725)

I do this research
2nd
65
(4th
Point) (The distance 720.7 mkm)
- The distance 720.7 mkm = Mercury Jupiter Distance is important distance in our
discussion.
- But
- It depends on (the light supposed velocity discussion) which will done in the next
point. We have to put this distance discussion here,
- So, We discuss the distance 720.7 mkm in following depends on the light
supposed velocity discussion.
- Let's remember the data
- 4900 mkm = 1.16 mkm/sec x 4222.6 seconds
- The data tells, light supposed velocity (1.16 mkm/sec) travels during 4222.6
second a distance = 4900 mkm = Jupiter orbital circumference
- The light motion provides the energy required for Mercury motion. for that reason,
mercury day period =4222.6 hours, because 1 second of light motion be equivalent
to 1 hour of Mercury motion.
- The energy is transported by light supposed velocity to Mercury and from Mercury
to the other inner planets. That's why the value (4222.6) effects on the 4 inner
planets data as we have discussed before.
- The value (4222.6) effects by different forms on the inner planets because it's a
light motion value.
- The idea is that, the distance (720.7 mkm) gives a comparison between light
motion and planet motion during the same (supposed) period. Because light moves
during (4222.6 sec) and Mercury moves during (4222.6 h) where (1 second = 1
hour)
- Light supposed motion be (4900 mkm = 2π x 778.6 mkm) but planet motion be
(720.7 mkm) where (778.6 mkm = 720.7 mkm x 1.0725)
- That may tell light move along (the circle circumference) but the planet moves
with its radius (still the rate 1.0725 needs explanation)

I do this research
2nd
66
- The point is that, Jupiter (planet) itself should move (720.7 mkm) during this same
period (4222.6 hours) because by that, the light motion will be related to Jupiter
motion. and the 2 motions be unified in one source
- But
- Jupiter (13.1 km/s) moves during (4222.6hours) a distance = 201 mkm
- But
- 629 mkm (Earth Jupiter distance) = 201 mkm x π
- Where
- 720.7 mkm = 1.0725 x 671 mkm (Venus Jupiter distance)
- And
- 671 mkm = 1.0725 x 629 mkm (Earth Jupiter distance)
- That tells us, the system of data on which Jupiter distance are defined is a
necessary tool to create the unification of motion between Jupiter and Mercury
motion ( or the unification of motion between light motion and a planet motion)
- Regardless any explanation is provide here, the distance system of data is required
geometrically and proves the planets unified motion concept. (Train Carriages
Motion).
Notice
720.7 mkm = 1.1318 mkm x 636.8 days (where 636.8 = 1.16 x 550.7)
And
Light known velocity (0.3 mkm/s) travels during 671 seconds a distance = 201 mkm

I do this research
2nd
67
8- Can A Light Beam Its Velocity 1.16 Mkm/S Be Found
8-2 Jupiter Data depends on a velocity (1.16 mkm/s)
- This point of discussion tries to prove the following hypothesis…
- A Light Beam With Velocity 1.16 Million Km Per Second Is Found In The
Solar System.
- Jupiter data is the basic data used for this hypothesis proof. That because
o All Jupiter distances are defined clearly on the rate (1.16)
o Jupiter distances are designed geometrically in clear form than any other
planets
o Jupiter motion data shows light motion features clearly more than many
other planets.
- The following discussion aims to explain the data and we find that the best
explanation should use the hypothesis (a light beam its velocity 1.16 mkm/s be
found).

I do this research
2nd
68
6-2 Jupiter Data depends on a velocity (1.16 mkm/s)
I- Data
Group No. (1)
(a)
4900 mkm =1.16 mkm/sec x 4222.6 seconds.
(b)
778.6 mkm = 1.16 mkm/sec x 671 seconds.
(c)
720.7 mkm = 1.16 mkm/sec x 621 seconds.
(d)
671 mkm = 1.16 mkm/sec x 580 seconds. (but 580 mkm =1.16 x 500s)
(d)
5906 mkm = 1.16 mkm/sec x 5127 seconds (and 5127 mkm =1.16 mkm/s x 4437s)
(e)
636 mkm =1.16 mkm x 550.7 seconds (during 636 days Jupiter moves 720 mkm)
(f)
2 x 2094 mkm = 1.16 mkm/sec x 3600 seconds (but 2094 sec x0.3 mkm/s =629 mkm)
(g)
564 mkm = 1.16 mkm/sec x 2 x 243 seconds (but 655 mkm = 1.16 x 564s)
Group No. (2)
(h)
201467 mkm = 2 x The Planets Orbital Circumferences Total 100733.5 mkm
(i)
201467 mkm = 2 x 86400 seconds x 1.16 mkm/sec
(j)
201467 mkm = (Pluto Orbital Circumference – Jupiter orbital circumference) x 2π
(k)
201467 mkm = 28255 mkm + 2 x 86400 mkm
(l)
201467 mkm = (Jupiter Circumference 449378.3 km)2
Group No. (3)
(i)
5040 seconds x 1.16 mkm/s = 5848 mkm
(ii)
25920 mkm =1.16 mkm/s x 6.2 x 3600
But
4846 mkm = 6.2 x 778.6 mkm And 1419 mkm = 6.2 x 227.9 mkm

I do this research
2nd
69
II-Discussion
- Data Group no. (1) are Jupiter distances we have discussed before, let's provide a
complete list about this distances in following:
o 778.6 mkm (Jupiter orbital distance) = 1.16 mkm/s x 671s
o 671 mkm (Jupiter Venus distance) = 1.16 mkm/s x 580s
o 580 mkm = 1.16 mkm/s x 500s
o 720.7 mkm (Jupiter Mercury distance) = 1.16 mkm/s x 629 s (error 1%)
o 629 mkm (Jupiter Earth distance) = 1.16 mkm/s x 542 s
o 636 mkm = 1.16 mkm/s x 550.7 s (Mars Jupiter distance =550.7 mkm)
o 720.7 mkm = 636 days x 1.1318 mkm (Jupiter velocity per solar day)
o 5906 mkm =1.16 mkm/s x 5127 s (Jupiter Pluto distance =5127 mkm)
o 4331 mkm =1.16 mkm x 3716 s (Jupiter Neptune distance= 5127 mkm)
o Where
o 4331 days = Jupiter Orbital Period
o 5906 mkm = Pluto orbital distance
Also
o 142984 mkm =1.16 mkm x 3716 s (Jupiter Neptune distance=3716 mkm)
o Where, (142984 km = Jupiter diameter)
- The data is created almost depend on the rate (1.16) or light supposed velocity
(1.16 mkm/sec).
- I try to show that, the light supposed velocity (1.16 mkm/s) is found really in the
solar system motion. and because of that, the light motion effect is seen in different
data.. we here don't depend on some numbers found by a chance, but we depend on
the light motion nature…
- Light use time and distance values exchangeable. Because of that the light motion
is not only seen in the light velocity effect but also in the light motion nature
effect.

The Solar System Motion Analysis

The Solar System Motion Analysis

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to The Solar System Motion Analysis

Similar to The Solar System Motion Analysis (20)

More from Gerges francis

More from Gerges francis (20)

Recently uploaded

Recently uploaded (20)

The Solar System Motion Analysis