The Moon Orbital Motion Geometry (II)

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
The Moon Orbital Motion Geometry (II)
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 –24th
February 2021
Abstract
Paper Hypothesis.
The moon velocity (=29.53 km/s) and is less than Earth velocity (29.8 km/s) with 1%
The Hypothesis Explanation.
- The apogee point (r=0.406 mkm) is the most far point the moon can reach from
Earth, the moon apogee orbital circumference =2.550973 mkm. But this distance is
shorter than the moon displacements total during (29.53 days) =2.598693 mkm
- This interesting data pushes to ask Why the moon apogee orbital circumference
doesn't = its displacements total (during the moon day period 29.53 days) ?
- These 2 distances analysis shows, The moon orbital inclination causes the moon
apogee orbital circumference to be shorter than its displacements total…. more
analysis shows the following:
o The moon velocity is less 1% than Earth velocity, this could cause
inevitably to separate the moon from the Earth through their motions.
o Because the solar system works as a machine of gears, Uranus motion
effects on the moon motion to save its accompanying with Earth.
o Uranus effect on the moon motion is seen in Metonic Cycle, where, Metonic
Cycle is 19 years because Uranus orbital distance =19 Earth orbital distance
o Uranus effect on the moon motion uses Pluto and Earth motions.
o Based on that, the moon orbital inclination (5.1 deg) is created by an effect
of Pluto orbital inclination (17.2 deg).

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o The moon orbit regression is created by Uranus & Pluto Axial Tilts Effect.
o Uranus effect on the moon motion to create the daily displacement =88000
km, by that the total displacements during (29.53 days) be greater than Earth
motion distance during 1 solar day with 1% which compensates the
difference 1% found between Earth and its moon velocities.
- But
- The fact that, the moon total displacements during (29.53 days) is greater than the
moon apogee orbit with 1%, causes the question, How the moon can reach to the
perigee point? where the displacements total 2.5986 mkm = 2πx 413600 km, that
means, if the moon daily displacement be 88000 km, the moon would to revolve
around Earth through its apogee orbit only and can't revolve through any more
near orbit (Notice, apogee orbit radius =406000 km, but this calculation tells that,
the moon would revolve around Earth through an orbit its radius =413600 km,
means the moon would revolve through even more far point than its apogee)!
o The intelligent moon uses an angle (θ) between its displacement motion
direction and its orbit horizontal level, by that the real displacement through
the moon orbit will be (L=88000 km Cos (θ)) and be shorter than 88000 km
enables the moon to revolve around Earth Through more near orbits
o means, the moon uses Pythagorean triangle technique in its orbital motion
o The Pythagorean triangle technique using by the moon motion provides 2
new tools are useful for the moon motion study which are the moon orbital
triangle and the moon orbital motion equation – The paper discusses how to
use them.
- The moon velocity reduction (1%) enables the moon to create an interaction with
Mars Motion, based on this interaction the following conclusion is reached.
Paper conclusion
The moon motion creates a harmony between the inner and outer planets motions.

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Contents
Subject Page N
1- Introduction 5
2- The Moon Orbital Motion Description 7
2-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
2-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
2-3 The Moon Orbital Motion Analysis
2-4 The Moon Orbital Motion Equation
3- The moon velocity Discussion 23
3-1 Pluto motion effect on the moon motion
3-2 Uranus motion effect on the moon motion
3-3 Earth Motion Distance During A Solar Day
3-4 How to know that, the moon velocity =29.53 km/sec?
3-5 Why the moon velocity less than Earth with 1%?
3-6 Pluto Velocity Analysis
3-7 Pluto Day Period
3-8 The Time And Distance Values Equivalence
4-Planets Unified Motions 49
5-The Moon Orbital Triangle Description 50
5-1 Preface
5-2 The Moon Orbital Triangle Description
5-3 The Moon Orbital Triangle Data Analysis
5-4 The Moon Orbital Triangle Major Points
6-The Paper Hypotheses Proves Discussion 99
6-1 The Paper Hypotheses Revision
6-2 Why the moon apogee orbital circumference doesn't = 2.598693 mkm?
6-3 Why does the moon orbit regress?
6-4 The Earth Cycle 8 years
6-5 Why the moon daily displacement =88000 km?
6-6 Why the moon day period =29.53 solar days?
6-7 The Triangle (M1RB) Data Analysis
6-8 The angle 1.1 deg effect on the moon orbit geometry
6-9 Four Planets Motions Interaction
6-10 How can the far planets effect on the moon orbital motion?

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7-The Moon Orbit Geometrical Design 140
7-1 Preface
7-2 The Triangle Geometrical Design
7-3 The moon motion angle (12.195 deg) Analysis
7-4 The Perpendicular Line BC (=86000 km)
7-5 Jupiter Motion effect on the moon orbital motion
8- The Moon orbital triangle 2nd
modification 152
8-1 Preface
8-2 The Moon Orbital Triangle 2nd
Modification
9- The Moon Orbital Triangle Geometrical Benefits 178
9-1 Preface
9-2 The Moon orbital triangle shows that (2nd
force effect on the moon motion)
9-3 The Moon orbital triangle shows that (There's 2nd Orbit for the moon motion)
9-4 The Moon orbital triangle shows that Uranus effects on the moon motion
10- The Moon Orbital Inclination Creation 183
10-1 The Moon orbital inclination creation geometrical process
10-2 Planets motions effect on the moon orbital inclination creation
10-3 The Moon Orbit Regression
10-4 Planets motions cause The Moon Orbit Regression
10-5 The Moon Orbit Regression Effect on The Earth Motion
11- Metonic Cycle Is A Proof of Uranus Effect On The Moon Motion 196
11-1 Preface
11-2 Uranus Effect On The Moon Orbital Motion
11-3 The Angle 71.9 Degrees Analysis
11-4 The Moon Orbital Triangle Angles Discussions
12- Uranus Motion Analysis 213
12-1 Uranus Motion During 1440 Of Its Days Period
12-2 Uranus Motion During 8 Pluto Days period
12-3 Uranus 144 days Cycle
12-4 The Moon Diameter Creation.
Appendix No.1 224

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1- Introduction
- The hypothesis, the moon velocity is decreased by (1%) than Earth velocity,
provides a question has an effect extending beyond the moon velocity itself.
- It supposes Planets motions depend on one another, and by that, the shortage in
the moon velocity could be compensated by other planets motions effect.
- If the solar planets are rigid bodies revolve around the sun in separated trajectories,
this description can't support the supposed cooperation…
- The hypothesis suggests, the solar system is a machine of gears, or members of
one creature body. These examples can provide the required cooperation.
- Now we have 2 different descriptions for the same machine, what's the real one?
- This simple analysis will face in the next point directly a serious challenge which
is ….. The gravitation force effect limitation defined by the gravitation equation
where the far planets can't effect on the moon motion by their gravity forces
because of the huge distances.
- One the other side, Planets data shows this cooperation as a fact can't be refuted.
- The unified motion can be a support for the paper hypothesis where there are
motions done by many planets which can be explained only as one unified motion.
(For example the Earth and its moon revolve around the sun, we expect that, the
Earth and its moon move equal distances during their revolutions, by that we
conclude this motion is one motion done by 2 planets- similar to that- there are
motions are done by many planets and the only explanation for such motions is the
cooperation between these planets motions)
- The planet is created as a geometrical building, and its data is created based on
geometrical rules and interactions which shows other planets effect in this data
creation. By that, it's simple to conclude that, the moon orbital inclination is
created by an effect of Pluto orbital inclination (and not by Mars "more near
planet), and Metonic Cycle is created by Uranus effect (and Not by Venus). The
planet data tells what's happening and they don't care for the huge distances!

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- There's one more method by which the far planets can effect on the moon orbital
motion in addition to the gravity force, by this method these planets can create
their motions required cooperation by which many defects can be solved (as the
moon velocity reduction 1% than Earth velocity)
- Let's refer to this paper contents in following
- Point no. (2) provides the moon orbital motion description
- Point no. (3) discusses The moon velocity value
- Point no. (4) provides 3 examples of planets unified motions
- Point no. (5) discusses the moon orbital triangle structure and its major points
- Point no. (6) discusses the paper hypothesis proves
- Point no. (7) discusses the moon orbital triangle geometrical design
- Point no. (8) discusses a modification for the moon orbital triangle
- Point no. (9) discusses the benefits of the moon orbital triangle
- Point no. (10) discusses the moon orbital inclination creation
- Point no. (11) discusses Uranus Effect on the moon motion (Metonic Cycle)
- Point no. (12) analyze Uranus Motion.

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2- The Moon Orbital Motion Description
2-3 The Moon Orbital Motion Analysis

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- 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….

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- 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.

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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

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2-3 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)

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- 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…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1)

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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.

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2-4-1 The Equation Concept
2-4-2 The Equation Test and Accuracy
2-4-1 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….

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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

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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

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2-4-2 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
[

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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

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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
[

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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

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2-4-3 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 deeply (Metonic Cycle
Discussion point no 11)

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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)…

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3- The moon velocity Discussion
- The moon daily displacement =88000 km but the moon apogee orbit (r=406000
km) permits for a daily displacement = 86400 km only, the difference =1600 km
- Pluto motion causes to create this difference (1600 km), and based on this creation
the angle (10.96 degrees) is created
- Then based on this angle (10.96 degrees) the perigee orbit radius is created
(r=363000 km)
- The difference (1600 km) causes 1% difference in distance between the moon
displacements total during (29.53 days) and the Earth motion distance per solar
day and also cause 1% difference between the Earth motion distance per solar day
and the moon apogee orbital circumference, that creates 3 difference distances
each differ from the other with 1%.
- The data shows clearly how this process is done , so let's discuss the data in
following with some details
I-Data
(3-1)
88000 km – 86400 km = 1600 km
(3-2)
7510 (Pluto Circumference) = 4.7 km/sec (Pluto velocity) x 1598 seconds (1600)

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(3-3)
86400 km = Cos (10.96 degrees) x 88000 km
(3-4)
Sin (10.96 degrees) x 406000 km = 77237 km
(3-5)
413600 km = 4.7 km/sec x 88000 seconds
(3-6)
17.4 degrees +10.96 degrees =28.36 degrees
(3-7)
(28.3 deg / 26.7 deg) = (26.7 deg / 25.2 deg) (25.2 deg / 23.4 deg)
II-Discussion
- The distances values are used as periods of time frequently in the solar system
geometry (in point No 3-8 we discuss this feature)
- The data shows what happens, Pluto by its circumference (7510 km) and velocity
(4.7 km/sec) causes to create the distance 1600 km (which is used as a period of
time 1600 seconds) and this distance is used as a difference between the moon
daily displacement 88000 km and the maximum displacement (86400 km)
available by its apogee orbit (r=406000 km) (where 86400 km x 29.53 days = 2π x
406000 km).
- Where …the moon daily displacements total during 29.53 solar days = 2598693
km =2π x 413600 km
- But the moon apogee radius =406000 km and NOT 413600 km because of this
difference (1600 km) between the 2 displacements (88000 km -86400 km), where
the value (1600 km) is defined by Pluto motion we conclude that, Pluto motion
effect causes to decrease the moon apogee radius to be (406000 km) in place of
(413600 km)

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- Let's ask
- Does Pluto cause to create this value 1600 km? based on this equation (7510 km
Pluto circumference =4.7 km/sec Pluto velocity x 1600 seconds).
- The answer is
o Jupiter (13.1 km/sec) moves during 10921 seconds a distance =142984 km =
Jupiter diameter (where 10921 km = the moon circumference)
o Uranus (6.8 km/sec) moves during 75101 seconds a distance =51118 km =
Uranus diameter (where 7510 km = Pluto circumference)
- I try to show that, it's a usual using in the solar system geometry, and if the moon
and Pluto circumferences are created by the previous equations through Jupiter and
Uranus motions effect, So, the distance 1600 km is created by Pluto motion effect
on the moon motion (this feature is discussed widely in point no. 3-8)
Equation No. (3-3)
86400 km = Cos (10.96 degrees) x 88000 km
- After the creation of the distance 86400 km from the original one 88000 km, the
angle (10.96 degrees) is created automatically, This angle is discussed also with
specific Pythagorean triangle (1,2 and 51/2
) in more details.
Equation No. (3-4)
- By the angle 10.96 degrees and the apogee orbital radius (406000 km) which are
defined already in the previous process, the value 77237 km is created
- During 29.53 days (the moon day period), if the moon daily displacement be
77237 km, so the total displacements will be =2.28 mkm = The moon perigee
orbital circumference.
- That means, the moon perigee radius (363000 km) is created depending on the
apogee radius (406000 km) and the angle (10.96 degrees)

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Equation No. (3-5)
- This equation shows how deep Pluto effect on the moon motion data… Pluto
moves during a solar day (86400 second) a distance =406000 km (apogee radius)
- And moves during (88000 sec) a distance =413600 km (the moon displacements
total during 29.53 days /2π)
- And moves during (77237 sec) a distance =363000 km (perigee radius)
- Again the distances values are used as periods of time.
- The equation no (3-5) tries to prove that, Pluto motion effect on the moon motion
data is created by a geometrical mechanism and can't be coincidences of numbers
because All moon motion data is created based on Pluto data
Equation No. (3-6)
17.4 degrees +10.96 degrees =28.36 degrees
- Equation no. (3-6) shows, the angle 10.96 degrees is created by an effect of Pluto,
where 17.2 deg= Pluto orbital inclination and 28.3 deg= Neptune axial tilt
- So, the angle 10.96 deg is created by an effect of Pluto and Neptune data.. but
Neptune data here is a common value, where Neptune effect already on all planets
regardless Pluto effect … the following equation may support this meaning
Equation No. (3-7)
(28.3 deg / 26.7 deg) = (26.7 deg / 25.2 deg) (25.2 deg / 23.4 deg)
28.3 deg = Neptune Axial Tilt 26.7 deg = Saturn Axial Tilt
25.2 deg = Mars Axial Tilt 23.4 deg = Earth Axial Tilt
- Equation no. (3-7) shows a continues effect of Neptune axial tilt on different
planets data by that Pluto effect of the moon Pluto has to interact with Neptune
effect which is found already and to create the angle 10.96 degrees.

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- The previous process shows the direct geometrical effect of Pluto motion data to
create the moon apogee and perigee radiuses, but this process covers the main
force behind which pushes Pluto to do this role
- Where, Pluto does this effect based on a force provided to it by other planets
Shortly
- There's an interaction of motion between 5 planets which are (Uranus, Saturn,
Earth, the moon and Pluto), this interaction pushed Pluto to do this effect on the
moon orbital motion – the following data can help our analysis
I-Data
(3-8)
6.8 km/sec x 1600 seconds = 10921 km (the moon circumference)
(3-9)
Uranus (6.8 km/sec) moves during (378675 seconds) a distance = 2574990 km
(3-10)
Cos (19.367 deg) x 1600 km = 1507 km
(where cos (19.367 deg) = tan (43.33 deg))
II-Discussion
Equation no. (3-8)
6.8 km/sec x 1600 seconds = 10921 km (the Earth moon circumference)
- Uranus moves during 1600 seconds a distance = the moon circumference, we
remember that, Pluto (4.7 km/sec) moves during 1600 seconds a distance = 7510
km = Pluto Circumference, means, the same value (1600 s) is used by Pluto and
Uranus motions to create Pluto and the moon circumferences, that tells the value
(1600 s) is defined by Uranus and used by Pluto
- That tells us, the value (1600 seconds or km) is defined by some force greater than
Pluto and the moon motions
- Equation no. (3-9) tells that Uranus moves during (378675 seconds) =2574990 km

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- Earth motion distance per solar day = 2573483 km
- The difference = 2574990 km - 2573483 km = 1507 km
- This difference between the 2 motions causes the difference 1600 which is used by
Pluto and Uranus to create Pluto and the moon circumferences, the next equation
tells how that happens…
Equation no. (3-10)
Cos (19.367 deg) x 1600 km = 1507 km
(where cos (19.367 deg) = tan (43.33 deg))
- In the moon orbital triangle (is discussed in the next points), the angle BCS =
19.367 degrees and the angle RCK = 43.33 degrees
- In the moon orbital triangle we discuss these angles and their effect on the moon
orbital radiuses creation
- Equation no (3-10) tells that, these angles are created between Earth and Uranus
motions to use the difference 1507 km to create the difference 1600 km which is
used by Pluto and Uranus motions
- Means, the difference 1600 km is created based on the difference of Earth and
Uranus motions, and based on that, the difference 1600 km is found as a result for
the difference 1507 km and Pluto effect on the moon motion should be considered
as a result of the interaction of Uranus and Earth motions
- This data needs more analysis because Uranus motion distance (2574990 km)
depends on the period (378675 sec) (where 378675 km = Saturn Circumference),
So why Uranus depends on this value?! We should discuss that in the next points
in more details
Notice
The difference between 2598693 km (the moon displacements total during 29.53
days) and 2550973 km (the moon apogee orbital circumference) = 47720 km but
Uranus motion distance 2574990 km be in the middle for this 2 values and by that
Uranus distance is different from both value with 23860 km

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I - Data
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
(Uranus motion distance – Saturn motion distance = π x 10921 km)
II-Discussion
- Why These Distances Are Equal Approximately?
- Let's imagine that, we have a motion done by many planets… for example, Earth
revolves around the sun, and the moon revolves around Earth, So we conclude
simply that, the Earth and moon motions distances must be equal because they
move together and not as separated planets from each other
- The equal distances tell that, these distances are created by one unified motion….
that tells, all planets move together one unified motion (regardless our observation)
but the equal distances shows it.
- The next question will be how that can be occurred? And we need to review the
solar system motion description… where the solar planets are considered rigid
bodies revolves around the sun in separated trajectories of motions from each
other, by this description we should have serious difficulties to explain the
previous data, but let's provide another suggested description
o Let's imagine that, the solar planets is a theater of puppets, all puppets are
connected with each other, and their data is created to be complementary
with one another ….
o The double production experiment is a suitable example to explain that

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o From Gamma rays an electron and a positron is produced, that explains the
meaning of (complementary with one anther), where the products are equal
in masses and opposite in charges and without observation, I expect to find a
produced positron with the electron because of the charge reservation law.
o The original source (Gamma rays) forces the products data to be created
complementary to each other
o By a similar method the planets data is created complementary to each,
which tells us there's one source for these planets data and by that the
unified motion can be produced logically
o And as in the puppets theater there's one line connects all puppets together,
we may imagine one line of energy creates the planets and their distances
data,
BUT
o The simple vision of the double production experiment can't tell us all
details found behind the planets data creation, where the planets data is
created based on geometrical interactions and complexities…. For example
Any animal body be examined shows clearly the deep complexities based on
which the body is created – the body members details shows the deep
interaction between each other – and very similar to that- the planets data is
created complementary to one other but with complex geometrical process
o Based on that
o The distances which are equal (approximately) aren't equal (perfectly)
because of geometrical interactions and requirements
o As we have seen, Uranus motion distance (2574990 km) divides the distance
47720 km into 2 equal parts tells that, the moon displacements total
(2598693 km) and its apogee orbital circumference (2550973 km) both are
created by Uranus effect (through Pluto) on the moon motion

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o Also the difference between Uranus and Saturn motions distances (2550973
km) - (2540779 km) = π x 10921 km (the moon circumference)
o Also
o The Earth moon distance at The Total Solar Eclipse = 378675 km =Saturn
Circumference
o This data shows that, a complex geometrical process is found behind the
solar system creation. This paper tries to prove only that, the moon velocity
(=29.53 km/s) and is less that Earth velocity (29.8 km/s) with 1%
Notice
o In this paper there are 3 examples of the planets unified motions (which are
discussed in Point no. 4 of it)
o (1st
) The Earth motion distance during a solar day which we discussed
o (2nd
) The planets motions to pass a distance = Uranus orbital circumference
o (3rd
) The Pluto 8 Days Cycle

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- Let's summarize the idea in following:
o The moon velocity is (29.53 km/sec) and less than Earth velocity (29.8
km/sec) with 1%
o The planets motions is similar to machine of gears, because of that, The
planets motions compensate the difference 1% between Earth and the moon
velocities to save the moon & Earth motions accompanying
o This compensation process is seen in the moon motion as Metonic Cycle –
that means- Metonic Cycle is found to compensate the difference in
velocities between Earth and the moon
o By the decreasing of the moon velocity than Earth velocity (1%), the moon
creates an interaction with Mars motion and this motions interaction works
to connect the inner planets motions with the outer planets motions – in
more specific words – this motions interaction creates a connection between
Mercury Motion and Jupiter motion… this connection depends on the
interaction in motions between the moon and mars where this interaction
depends on the Moon Metonic Cycle.
o Our process became clear– we need to prove that- Metonic Cycle is found as
a result of the moon velocity reduction (1%) than Earth velocity where
Metonic cycle is found to compensate this difference in velocities and to
save the moon and Earth motions accompanying.
o Then we have to prove that, Mercury (the head of the inner planets)
connects with Jupiter (the head of the outer planets) through Metonic Cycle.
o These are 2 different jobs – the 2nd
job will be discussed in point (3-5), and
let's do the 1st
job in following, trying to prove that- Metonic Cycle is found
as a result for the moon velocity (29.53 km/sec) which is less than Earth
velocity (29.8 km/sec) with 1%.

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Metonic Cycle Discussion (1st
Point) Pluto Velocity
- Why did I claim that, the moon velocity is less than Earth velocity with 1%?
- This conclusion I have reached based on Pluto velocity analysis because
- 29.53 km/sec = 4.7 km /sec (Pluto Velocity) x 2π
- This is an accurate calculation, where the result of (4.7 km /s x 2π) = 29.53 km/s
and NOT =29.8 km/s (Earth Velocity)….
- This data pushes me to conclude that, this velocity 29.53 km/s is almost the moon
velocity, because the moon orbit 3 radiuses are defined based on Pluto velocity – I
found that- the rate (29.53/2π) shows the value 29.53 as a velocity (29.53 km/s)
and not as a period of time (29.53 solar days =the moon day period)
- Based on that I have concluded, the moon velocity is 29.53 km/sec
Also
- The moon orbital circumference =2550973 km which decreases than the Earth
motion distance during a solar day with 1% a we have discussed in the previous
point– Where the distance of Earth motion during its day period = the moon total
displacements during 29.53 days (the moon day period) = Pluto motion distance
during Pluto day period (153.3 hours) (error 1%) . This data tells that, if the moon
revolves around Earth (means the moon moves 2550973 km) that means, the moon
motion distance during 29.53 days is less than Earth motion distance during 1 solar
day with 1%. That means, Earth velocity (in its day) is greater than the moon
velocity (in its day) by 1%, means, means, the moon velocity =29.53 km/sec
- These are 2 different ways lead us to the same conclusion that, the moon velocity
=29.53 km/sec, and spite of this velocity less than Earth with 1% the moon doesn't
separate from Earth through their motions course
- So, the previous data analysis gives a reason to provide the hypothesis that, the
moon velocity is (29.53 km/s) and less than Earth velocity (29.8 km/s) with 1%
- Now let's test if we can prove this hypothesis in following

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(2nd
Point) The Velocity Compensation
I-Data
(3-11)
(610.7 mkm /9.4 mkm) = 2609022 km /40080 km
(3-12)
18048.5 mkm = 610.7 mkm x 29.53
Equation No. (3-11)
(610.7 mkm /9.4 mkm) = 2609022 km /40080 km
- The moon displacements total during 6939.75 solar days (Metonic Cycle period) =
610.7 mkm
- The paper hypothesis tells, The moon velocity is less (1%) of the Earth velocity,
But Earth Orbital Circumference = 940 mkm, means, Earth moves during 1 year
a distance = 940 mkm, but the moon velocity is less than Earth velocity with 1%
that means, the moon moves the same distance (minus) 9.4 mkm
- 40080 km = Earth Circumference
- But
- What's this 2609022 km
- We know that,
o 5.1 degrees = The moon orbital inclination
o 2598693 km = the moon displacements total during 29.53 days
o 2598693 km = cos (5.1) x 2609022 km
Please Note
o The moon orbital inclination (5.1 deg) is created as an angle between the
total displacements 2598693 km and the dimension 232000 km of the
triangle built on the difference 47720 km (Between the total displacements
2598693 – the moon apogee orbital circumference 1550973 km).

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- The value 2609022 is this triangle hypotenuse found geometrically depends on the
difference 47720 km and this value is rated to Earth circumference with the rate
(610.7 mkm /9.4 mkm)
- This analysis shows that, the values are created based on geometrical interaction
taking into consideration the value (9.4 mkm) which expresses the moon decrease
velocity 1%.
- (This triangle discussion is in the moon orbital inclination creation Point no. 9)
Notice
- Please remember, Saturn moves during its day period a distance =its circumference
- In calculations 9.7 km/s (Saturn velocity) x 10.7 h x 3600 = 373644 km = Saturn
Circumference (error 1.3%).
- This discussion we have did before and through it we have discovered Jupiter 8
days cycle, also we know that, Neptune moves during its day period a distance = 2
Neptune circumferences & Uranus moves a distance = 2.6 Uranus circumferences.
- This discussion shows an interaction between planet motion and its circumference,
and by reference to Equation (no.3-11) that tells us the value 9.4 mkm is taken into
consideration with Earth motion.
- That shows the value 9.4 mkm has a geometrical effect on the Earth motion.
Equation No. (3-12)
18048.5 mkm (Uranus orbital circumference) = 610.7 mkm x 29.53
Where
610.7 mkm the moon total displacements during Metonic Cycle (6939.75 days)
Notice
Earth motion distance during Metonic Cycle (6939.75 days) =17859.3 mkm
- Earth motion distance during Metonic Cycle is less than Uranus orbital
circumference with 1%

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- That tells us, because of the difference 1% in Earth and its moon velocities, Uranus
effects on the moon motion and makes the moon displacements total during 29.53
solar days is greater than Earth motion distance per a solar day with 1% to
compensate the difference in velocities…
Notice
- Uranus Motion distance during 1440 of its days = 606.32 mkm where the moon
displacements total during Metonic Cylce =610.69 mkm (the difference 0.7%)
- That tells Uranus motion effect on the moon motion causes to create the moon
daily displacement to be =88000 km,
- By making the moon displacements total during 29.53 days = 2598693 km, Uranus
enable the moon motion to compensate the difference of its velocity with Earth
velocity (1%).
Notice
- In the following there's a wide discussion is provided about Pluto Velocity analysis
where an interaction is found between Uranus and Pluto motions, this interaction is
extending to the 8 days cycle where Uranus uses 8 of Pluto days in this cycle.
- That tries to show how this process is done, where the moon velocity is less 1% of
Earth velocity and causes a continuous shortage for the moon motion, Uranus has
to create a continuous compensation for the moon motion, because of that Uranus
depends on Pluto because of Pluto velocity effect on the moon motion data.

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Notice
9.4 mkm = 106.8 x 88000 km (the moon daily displacement)
366556 km = 105.5 x 3475 km (The moon diameter)
Where
366556 km = 153.3 x 2390 km (Pluto diameter)
But
Pluto rotation period = -153.3 hours
The moon rotation period = (104.4 x 2π) =655.7 hours
(The values 106.8, 105.5, and 104.4 are different with 1%, showing that the value 9.4
mkm has a geometrical effect on the moon motion data)
366556 km = The Outer Planets Diameters Total
366556 km = 363000 km (perigee radius) + 3475 km (The moon diameter)
Notice
- Let's translate how this process is done in the moon motion data ....
o Uranus effects on Pluto motion and Uranus day period (17.2 hours) is seen
in Pluto motion data as (17.2 degrees = Pluto orbital inclination)
o The moon orbital inclination (5.1 deg) is created depending on Pluto orbital
inclination where (17.2 deg = 5.1 de x 2 + 7 deg Mercury orbital inclination)
o The moon displacements total during a year = 88000 km x 365.25 days = 2π
x 5.1 million km
o If 1 deg = 1 mkm, So 5.1 mkm be = 5.1 deg (the moon orbital inclination)
o 1 mkm = 1 degree because Mercury orbital circumference =360 mkm =360
degrees.

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- This is done for a specific role done by the moon motion, which is ….
o The moon motion works to unify the outer planets motions with the inner
planets motions
o Specifically
o The moon motion connects between Mercury and Jupiter motions, because
both planets is considered the head of its group,
o Means, the inner planets (whose head is Mercury) is unified and connected
with the outer planets (whose head is Jupiter)
o This effect of the moon motion is done by Uranus effect on the solar system
The following data should help our discussion
I- Data
(3-13)
1507 seconds x 47.4 km/sec (Mercury Velocity) = 71492 km (Jupiter Radius)
(3-14)
10921 seconds x 13.1 km/sec (Jupiter velocity) = 142984 km (Jupiter Diameter)
II- Discussion
Equation no. (3-13)
- We remember the difference in distances between Earth and Uranus motions =
1507 km, based on this distance, the distance 1600 km is created based on which
the moon & Pluto circumferences are created
- Mercury uses this distance (1507 sec) to create Jupiter Radius
Equation no. (3-14)
- Jupiter (13.1 km/sec) moves during 10921 seconds a distance = 142984 km =
Jupiter diameter, it's the same method by which both planets motions are related to
the moon data (1507 km and 10921 km), which explains the moon motion effect
on the solar system.

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- The previous data is a summarized part of data, because the relationship between
these planets is discussed before deeply …. Fro example
o The moon circumference 10921 km x Jupiter circumference 449197 km=
Jupiter orbital circumference 4900 mkm
o Mercury moves during its day period (4222.6 hours) a distance = 720.7
mkm = Mercury Jupiter distance
Notice (1)
- Mercury moves during (8 of Mercury days period) a distance =5745 mkm= Uranus
Orbital Diameter
- Because Mercury 8 days refer to the cycle of (8 days) follows by all planets that
shows Uranus effect on Mercury motion.
Notice (2)
- This also explains Mars and the moon motions interaction, which is seen in their
data, let's remember it in following
o Mars orbital period 687 days = the moon orbital period 27.3 days x 25.2
o 25.2 deg (Mars Axial Tilt) = 1.9 deg (Mars orbital inclination) x 13.177
(The moon motion per solar day =13.177 degrees)
o The moon day period 708.7 h = Mars day period 24.7 h x (180/ 2π)
Notice No. (3)
- Jupiter orbital period 4331 days = Mars orbital period 687 days x 2π, shows The
Moon And Mars motions interaction aims to create an interaction with Jupiter
motion…
Notice No. (4)
- The suggested role for the moon motion may explain some old data which is
- The solar planets diameters total = the moon apogee radius
- The outer planets diameters total = the moon perigee radius (error 1%)
- The inner planets diameters total = The distance between perigee and apogee
(space without the moon diameter) (error 1%)

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A Question
Why Uranus uses Saturn Circumference to pass the distance 2574990 km
Where
Uranus (6.8 km/sec) x 378675 seconds = 2574990 km
But
378675 km = Saturn Circumference…
But Also
Saturn (9.7 km/sec) moves during 6.8 x (10.7 hours) = 2540779 km
- On the other side Saturn uses Uranus velocity (6.8) as a rate of its days periods to
move the equal distance (approximately)!
- How to understand this interaction of data? Let's answer this question in the next
point…

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I-Data
(1)
10921 seconds x 4.7 km/sec= 51329 km (but 7510 seconds x 6.8 km/sec =51068 km)
(3)
51118 seconds x 4.7 km/sec = 2 x 120127 km (Saturn diameter =120536 km)
But
12756 seconds x 4.7 km/sec = 59953 km (Saturn Radius = 60268 km)
Where
7510 km = Pluto Circumference
10921 km = the moon Circumference
51118 km = Uranus Diameter
12756 km = Earth Diameter
II-Discussion
- The previous data tells that, the moon circumference and Saturn diameter are
created by Pluto motion depending on Uranus Diameter 51118 km!
- The moon circumference is the period (10921 s) used by Pluto velocity (4.7
km/sec) to pass a distance = Uranus diameter
- And
- Uranus diameter is the period (51118 s) used by Pluto velocity (4.7 km/sec) to pass
a distance = 2 Saturn Diameters
- So, we have 2 values are created depending on each other by Pluto motion (Uranus
diameter, Saturn diameter and the moon circumference)
- That gives some light to answer why Uranus uses (specifically) Saturn
circumference to pass the required distance (2574990 km)
- But
- Saturn uses Uranus velocity (6.8 km/sec) as a rate for its days period to pass the
required distance 2540779 km

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- That tells,
o The Planet Diameter Is A Function Of The Planet Velocity
o Uranus and Saturn motions interaction is a deep interaction effects on all
planet data….
Notice
- This discussion will be more clear with the discussion of the 2 examples of the
unified motions
o The motion distance = Uranus orbital circumference
o The 8 days Pluto Cycle
o We should discuss these subjects in point (no. 4) of this paper.

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I- Data
Tan (12.2) x 708.7 hours =153.3 hours
Tan (13.159) x 655.7 hours =153.3 hours
- 708.7 hours = the moon day period and 655.7 hours = The moon rotation period
- 153.3 hours = Pluto day period and (-153.3 hours) = Pluto rotation period
- But
o 13.177 degrees the moon daily motion (360 deg /27.32 days)
o 12.19 deg =13.177 deg -0.985 deg (Earth daily motion 360 deg/365.25 days)
- The moon moves both motions, the displacement 88000 km which creates the
angle 13.177 degrees and the Earth motion distance per day which creates the
angle 12.19 degrees.
- The Data tells that, Pluto day and rotation periods effect on the moon day and
rotation periods based on the 2 used angles in the moon motion
- If we can imagine that, Pluto uses the 2 values (708.7 hours and 655.7 hours) to
produce one value (153.3 hours) by that, Pluto effect on the moon cycles periods
of time
Notice
- (29.8 Earth velocity /4.7 Pluto velocity) = (152.17 hours /24 hours)
- Cos (7) x 153.3 hours =152.17 hours
- 7 degrees = Mercury Orbital Inclination

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- Frequently in the solar system data analysis, it's found, planet motion uses another
planet circumference as a period of time!
- Because this using is a common and very wide using in the solar system data, I
have to accept that, this using is found based on geometrical reason
- Although I have tried so hardily, but I can't reach how to explain this data – where
the data is strong enough to be real and challenge our logical and analysis ability
- In following I provide a great part of this data I have found to be a source for
whom can solve this question – I would to provide more help as possible – for this
brave person who can solve this hard question – let's provide the data in following
I-Data
(1)
o Jupiter (13.1 km/sec) moves during 10921 seconds a distance =142984 km
= Jupiter diameter (where 10921 km = the moon circumference)
(2)
o Uranus (6.8 km/sec) moves during 7510 seconds a distance =51118 km =
Uranus diameter (where 7510 km = Pluto circumference)
(3)
o Pluto (4.7 km/sec) moves during 10921 seconds a distance =51329 km
(where 51118 km= Uranus diameter and 7510 km = Pluto circumference)
o Pluto (4.7 km/sec) moves during 51118 seconds a distance = 2 x 120127 km
(Saturn diameter =120536 km)
o Pluto (4.7 km/sec) moves during 12756 seconds a distance = 59953 km
(Earth diameter =12756 km and Saturn Radius = 60268 km)
(4)
o Venus (35 km/sec) moves during 12104 seconds a distance = 421056 km
(421056 km = the distance passed by Uranus during its day period) (12104
km= Venus diameter)

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(5)
o Venus (35 km/sec) moves during 10921 seconds a distance = 378675 km
(Saturn Circumference) (error 1%)
(6)
5040 sec. x 6.8 km /s (Uranus velocity) = 34309 km (= 10921 km x π)
34309 sec x 13.1 km/s (Jupiter velocity) = 449197 km = Jupiter Circumference
34309 sec x 4.7 km/s (Pluto velocity) = 160592 km = Uranus Circumference
34309 sec x 27.78 km/s (the moon velocity) = 943817 km
(943817 km = the perimeter of the moon orbital triangle ACE) (error 1%)
(7)
o Mercury (47.4 km/sec) moves during 1507s = 71492 km (Jupiter Radius)
(8)
o Saturn (9.7 km/sec) moves during 142984 seconds a distance = 1392000 km
(1392000 km The Sun Diameter But 142984 km = Jupiter diameter)
o Saturn (9.7 km/sec) moves during 49528 seconds a distance = 4 x120536
km (120536 km Saturn Diameter But 49528 km = Saturn diameter)
(9)
o Uranus (6.8 km/sec) moves during 12756 seconds a distance = 86400 km
(12756 km Earth Diameter But 86400 seconds = The Solar Day)
(10)
o Neptune (5.4 km/sec) moves during 2390 seconds a distance = 12756 km
(Earth Diameter) but (2390 km = Pluto diameter)
o Neptune (5.4 km/sec) moves during 120536 seconds a distance = 4 x
160592 km (Uranus circumference) but (120536 km = Saturn diameter)
(11)
o Mars (24.1 km/sec) moves during 7510 seconds a distance = 181000 km
(Please review the specific Pythagorean triangle 1,2 and 51/2
in the moon
orbital triangle major points discussion)

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Discussion
- The data leads to the following conclusions
- Jupiter and Uranus are the basic 2 planets in the solar system because they pass
distances =their own diameters but the other planets depend on others'
- The using of planet diameter or circumference as a period of time is a general
using in the solar system motion.
- The previous data is the most near and clear data… because the solar system
depends on geometrical mechanism many other data is used but unknown to us
because they aren't planets data but found inside the geometrical interactions…
- For example Mars moves a distance =181000 km (Data No. 11), the value 181000
km is very important value in the moon orbit geometrical structure because it’s
found based on the Pythagorean triangle (1,2 and 51/2
) – i.e. The value 181000 km
is defined directly based on the perigee (363000 km) and apogee (406000 km)
radiuses definition… that makes this value 181000 km effective geometrically on
the moon orbital motion but it's not known data for us– many similar data can be
added but I have removed because they are unknown…
The left questions are
(1) Why does the solar system geometry use the planet diameter (or
circumference) as a period of time?
(2) By what geometrical mechanism this is done?

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Notice
- I have tried to explain the previous data, I have to suppose that the matter is
created out of light…
- In the solar system motion there are many features of light motion, these features I
have discussed deeply before with a great amount of data that proves these features
and gives no room for any doubt…
- Planets motions indeed have features of light motion
- But
- On the other side, the geometrical interactions which are used usually for the rigid
body motion are still found as planets motions features – by that- both motions
features (light motion and rigid body motion) are found in the planets motions
- Based on that, the hypothesis "The matter (planet) is created out of light" is so
limited hypothesis, because the light passes through many geometrical interactions
to produce the matter…
- That means, the matter indeed is created out of light but the giving birth process is
done in some geometrical workshop, where thousands of geometrical rules are
used to perform the job.
- And because these rules are still unknown for use, we are still astonished before
the imaginary hypothesis (the matter is created out of light)
- Among these rules, our golden one, by which The Geometrical Design Uses
Planet Diameter Or Circumference As A Period Of Time – which we still try to
know why and how the higher intelligent geometry does that?

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4-Planets Unified Motions
The paper provides 3 examples concerns the meaning of planets unified motions, in
following these examples data is mentioned but their discussions are found in their
defined points because of the discussions logical structure…
Examples 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
(Example no. 1 is discussed in point no. 3-3 Earth Motion Distance During A
Solar Day)
Examples no. 2
(1)
Earth moves during (6939.75 solar days) a distance = 17859.325 mkm
(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)
This example is discussed in point no. 11 (Metonic Cycle Discussion)

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Examples no. 3
1. Uranus (6.8 km/s) moves during 8 days of Pluto days period (1226.4 hours) a
distance = (4415040 seconds x 6.8 km/s) = 30.022272 million km
2. Jupiter (13.1 km/s) moves during 64 days of Jupiter days period (64 x9.9 h =
633.6 hours) a distance = (2280960 seconds x 13.1 km/s) = 29.880576 million km
3. Saturn (9.7 km/s) moves during 80 days of Saturn days period (80 x10.7 h = 856
hours) a distance = (3081600 seconds x 9.7 km/s) = 29.891520 million km
4. Neptune (5.4 km/s) moves during 100 days of Neptune days period (100 x16.1 h =
1610 hours) a distance = (5796000 seconds x 5.4 km/s) = 31.298400 million km
This example is discussed in point no. (12) (Uranus Motion Analysis)
Notice
The following data can support the discussion of example no. 1
- Mercury day period needs 5040 seconds to be 176 solar days
o Mercury moves during 5040 s a distance = 2 Saturn diameters (1%)
o Mars moves during 5040 s a distance = 1 Saturn diameter
o Saturn moves during 5040 s a distance = Neptune diameter(1.3%)
o Venus moves during 5040 s a distance = 2 x 88000 km
o Uranus moves during 5040 s a distance =10921 km x π

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5-The Moon Orbital Triangle Description
5-1 Preface

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5-1 Preface
- In this point we discuss how to create this moon orbital triangle and to define its
distances & angles, also we discusses the major points of this triangle geometrical
design, where these major points are required in the paper hypothesis proves
discussion.
- This triangle is created by creation a vertical line BC perpendicular on the triangle
base. This vertical line should be used 2 times in the triangle, one time when the
moon be in perigee and the second time when the moon be in apogee. For that
reason the triangle creates one form for each case and then created also one
combined form for the 2 cases.
- The moon using of Pythagorean triangle is discovered by analyze the moon motion
basic points which are
o Perigee point (r=363000 km), the nearest point the moon can reach to Earth
o Pongee point (r=406000 km), the far point the moon can reach from Earth
o T.S. Eclipse (r=373000 km), the moon creates total solar eclipse at it
o The distance (r=384000 km) which is registered as the moon orbital distance
The following data proves their using of Pythagorean rule.
These 4 points are defined based on each other by Pythagorean rule:
o (363000 km)2
+ (86000 km)2
= (373000 km)2
o (373000 km)2
+ (86000 km)2
= (384000 km)2
o (384000 km)2
+ (86000 km)2
= (393000 km)2
o (393000 km)2
+ (86000 km)2
= (406000 km)2
(Error 1%)
- By this data it's discovered the moon using of Pythagorean triangle in its motion
Notice
- The perpendicular Line (BC) which we use to create the moon orbital triangle its
length =86000 km.
Let's know how to create the moon orbital triangle

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- When we use the vertical line BC to be perpendicular on the moon in the perigee
point, the triangle form be as following..
- When we use the vertical line BC to be perpendicular on the moon in the apogee
point, the triangle form be as following..
- The combined form be as following..

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The Triangle Data Summary
- The moon moves on its orbital plane (the Red Line) from Perigee to apogee
- This distance is defined by M1 and M2 distance (=43000 km) and the distance BD
=42800 km be a very similar to it
- The line BC is perpendicular on a point parallel to the perigee point
- So the triangle CBD expresses the moon motion from perigee to apogee
- This triangle data is
o The angle BCD = 26.46 degrees
o The line BC = 86000 km
o The hypotenuse CB = 96062 km
Notice
- This figure I have brought from internet to use in the Explanation -
- We have supposed, the inner circle is the Perigee orbit and the outer circle is the
apogee orbit, And we have calculated the tangent DB = 181843 km
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, The triangle (ODB) angles are 26.564 deg. and 63.435 deg.
- But the triangle (BCD) in our triangle is a similar to this triangle (ODB), their
dimensions are rated and their angles are equal, both are created as a specific
Pythagorean triangle (1, 2 and 51/2
).
Why is this specific Pythagorean triangle (1,2 and 51/2
) is a necessary tool for the
moon orbital motion? The paper answers this question.

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The Moon Orbital Triangle Building
(1st
Point) The Earth Position (Point E)
- The Point (T) refers to The Earth Center
- The Point (M1) refers to The Moon Center (The moon in Perigee Point).
- The Points (T, Q and Y) are on The Earth Ecliptic Line
- The Red Line (TM) is the moon orbit plane with an inclination 5.1 degrees on the
Earth ecliptic line.
- The Green Line (BE) is the moon triangle base, the distance BE = 363000 km, I
choose it and accordingly I have to define the point (E) position.
- The line BC is a perpendicular on the triangle base (BE), its length =86000 km
- The line BC is perpendicular on the triangle base (BE) on the point (B), parallel to
the moon perigee point. (The 1st
Case).
- The angle CBE =90 degrees but the angle CYT = 89.557 degrees.
- The points (Q and P) are the intersection points of CE with the ecliptic and the
moon orbit plane respectively.
- The line TX is a perpendicular from the Earth Center on the base BE
- K is the intersection point between the triangle base (BE) & the moon orbit plane.
- The angle is Zero between the points ( A, B , K , X and E).
- The line EC connects between the points C & E where BC =86000 km and BE =
363000 km (As The Triangle Creation Requirements).

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(2nd
Point) The Moon Motion (From Perigee To Apogee)
- The moon moves on its orbit planet (MT) with an inclination 5.1 degrees on the
ecliptic, from Perigee (M1) (r=363000 km) to Apogee (M2) (r=406000 km).
- The distance M1 M2 = 43000 km (=The Perigee Apogee Distance)
- The line M1B is perpendicular on the triangle Base (EA) on The perigee point.
Notice
- M1B and M2D are perpendicular on the moon orbital triangle base (EA) (the
Green Line) …… BUT
- M1B and M2D are perpendicular on the triangle Base EA on (x-y plain) but the
line BC is perpendicular on the base (EA) on the (z-axis)
- Based on that
- The distance BD is parallel to M1R, and the moon motion from perigee to apogee
(M1M21) can be expressed on the triangle base by the distance (BD) where the
distance (M1M2) =43000 km and the distance BD =42800 km (error 0.4%)
- The blue line is the moon equator line, where the triangle Base (EA) has 1.1
degrees above the moon equator and has 0.443 degrees under the ecliptic.

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- Let's define the Earth Point in following:
(1) In the Triangle ATK
o The angle ATK = 5.1 degrees (the moon orbital inclination)
o The angle TAK =0.443deg (an angle between the base and ecliptic)
o The angle AKT = 174.457 degrees
o The angle BKM1 = 5.543 degrees
(2) In the Triangle M1BK
o The angle M1KB = 5.543 degrees
o The angle KM1B = 84.457 degrees
o The angle RM1M2 = 5.543 degrees
o The distance M1B = 31604 km
o The distance M1K = 327188 km
o The distance BK = 325658 km
o The distance KT = 35812 km
o The distance BX = 361300 km
(3) In the Triangle RM1M2
o The angle M2M1R = 5.543 degrees
o The angle RM2M1 = 84.457 degrees
o The angle M1M2N = 6.643 degrees
o The distance M2R = 4153 km
o The distance M1R = 42800 km
(4) In the Triangle KTX
o The angle XKT = 5.543 degrees
o The distance TX = 3460 km
o The distance KX = 35644 km

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(5) In the Triangle TM1Y
o The angle TM1Y = 84.457 degrees
o The angle TYM1 = 90.443 degrees
o The angle M1TY =5.1 degrees
o The distance TM1 = 363000 km
o The distance YT = 361313 km
o The distance M1Y = 32269.5 km
o The distance YB = 665 km
o The distance M1B = 31604 km
(6) In the Triangle KTE
o The angle E = 63.87 degrees
o The angle ETK = 110.6 degrees
o The angle ETQ = 115.7 degrees
o The distance TX = 3460 km
o The distance TE = 3854 km
o The distance XE = 1700 km (to make the distance BE =363000 km)
o The distance KE = 37344 km (= 35644+1700)
(7) In the Triangle EPK
o The angle EPK = 161.1 degrees
o The angle EKP = 5.543 degrees
o The angle PEK = 13.328 degrees
o The distance PK = 26604 km
o The distance PE = 11147 km
(8) In the Triangle EPT
o The angle TEP = 50.54 degrees
o The angle ETP = 110.57 degrees (84.457+26.12)

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o The angle EPT = 18.89 degrees
o The distance TP = 9190 km
(9) In the Triangle QTP
o The angle TPQ = 161.1 degrees
o The angle T = 115.72 degrees
o The angle PTQ = 5.1 degrees
o The angle TQP = 13.78 degrees
o The distance TQ = 12491 km
o The distance QP = 2529 km
o The distance EQ = 13673 km = 11144 + 2529
Data Analysis
(1)
o The Triangle TXE
o The distance TX = 3460 km The distance XE =1700 km
o The moon diameter =3475 km and the moon radius =1737.5 km, both are
equal the triangle 2 dimensions (error around 2%). That shows geometrical
interaction in this distances definition.
(2)
o The Point (E) is found inside the Earth but far from its center with 3854 km
with an angle 63.8 degrees where its level is far from the Earth center with a
perpendicular distance =1700 km.
(3)
o The line M1B has an angle 90 degrees (M1BK) but the angle M1YT
=90.443 degrees.

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(3rd
Point) The Point (A)
- The Point (A) is a point on the Ecliptic Line I have choose and caused to create it
with an angle =0.443 degrees under the ecliptic line. By that the triangle base (AB)
be found under the Ecliptic with 0.443 degrees and above the moon equator line
(the blue line) with 1.1 degrees.
- That means, the triangle base (AB) depends on the Earth ecliptic line.
- The triangle ABC is a closed triangle where the point (A) is the intersection point
between the ecliptic line, the triangle base AB and the triangle dimension AC
- I choose the distance AB =86000 km.
- The line BC is a perpendicular on the point B, (which is parallel to the perigee
point M1 with a radius r=363000 km). (1st
Case)
- The line BC length =86000 km (I choose it).
Notice
- The moon equator line (the blue line) doesn't intersect neither with the ecliptic nor
the moon orbital triangle AB on the point (A),
- The moon equator line (the blue line) will intersect the ecliptic line beyond the
point (A) with a long distance

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- Let's define this intersection point position in following:
o The moon orbit plane declines on the Ecliptic line with 5.1 degrees, means,
far distance be found between the Earth and moon will cause longer
perpendicular distance between the moon center and the ecliptic line
o For that, we use the moon distance on a apogee because it's the most far
point the moon can reach from Earth
o ON APOGEE …
o Earth moon distance on apogee point = 406000 km
o The perpendicular distance from the moon center to the ecliptic line = 36091
km, because of the moon orbital inclination (5.1 degrees)
o But
o The angle between the ecliptic line and the moon equator line =1.543 deg
o So these 2 lines will be intersected each other at a distance =1340318 km
o i.e.
o The ecliptic line will intersect with the moon equator line after the apogee
point with a distance =1340318 km
o but the distance from perigee to apogee =43000 km
o i.e. The ecliptic line will intersect with the moon equator line after the
perigee point with a distance =1383318 km
o Notice, the lunar eclipse umbra length =1392000 km (error 0.6%)
The Useful Result :
The triangle base (AE) has an angle = 1.1 degrees with the moon equator line.

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(4th
Point) The Line BC
- The line BC is perpendicular on the triangle base on the point (B), so, the angle
ABC =90 degrees. The blue line is the moon equator line and the red line is the
moon orbit plane – the green line is the triangle Base (BA).
- Based on that,
o The angle BYA =89.557 degrees
o The angle CYA =90.443 degrees
o The angle M1NV =91.1 degrees
o The angle M2NM1 =88.9 degrees
o The angle M1NM2 =6.643 degrees
o The angle between the blue line (the moon equator) and the green line
(the triangle Base BA) = 1.1 degrees
o The distance BC = 86000 km (I have choose it)
o The distance AB = 86000 km (I have choose it)
o The distance AY = 86009 km
o The distance YB = 665 km
o The distance MB = 31604 km

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(1st
Question)
- The moon orbital triangle geometrical structure depends on 3 points (E, C and A),
- The Point (E) (found inside Earth)
- The point (C) (found on z-axis)
- But
- What's the point (A)? how this point can be created and effect on the moon orbital
motion and triangle?! Because this point is far from apogee radius with 43000 km
and the moon can't move beyond the apogee radius, means, this point (A) is found
in space and should have no effect on the moon orbital motion! so to find this point
(A) in the moon orbital triangle geometrical structure that creates a question needs
to be solved!
- Geometrically the point (A) is one pillar of the moon orbital triangle pillars,
means, the geometrical structure forces us to accept the massive importance of the
point (A).
- The paper claims that (Another force effects on the moon orbital motion in
addition to Earth gravity force and this point (A) refers to this 2nd
force)
- Our investigation in this study tries to discover if this claim can be proved based
on the moon orbital triangle geometrical design analysis.
(2nd
Question)
- The moon daily displacement 88000 km during 29.53 days creates a total distance
= 2598693 km
- But The moon orbital circumference at apogee orbit =2550973 km
- Where The apogee point is the most far point the moon can reach from Earth, that
means, the moon orbital circumference is shorter than the moon displacements
total during the moon day period (29.53 solar days) with a distance = 47720 km
- Why the moon orbital circumference at apogee doesn't =2598693 km?

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The following major points are selected from the moon orbital geometrical design
discussion because we need them to prove the paper hypotheses – let's refer to these
points in following:
5-4-1 The Necessity of Pythagorean Triangle (1, 2, 51/2
)
5-4-2 The Triangle Data (The Combination Form)
5-4-3 The Value 1290 degrees
5-4-4 The Trapezoid CDM2M1
5-4-5 The Triangle CDM2
5-4-6 The angle 17.4 degrees
5-4-7 The moon orbital triangle modification

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5-4-1 The Necessity of Pythagorean Triangle (1, 2, 51/2
)
(1st
Point) The Moon Motion Limits Definition
- In this moon orbital triangle I have added the line CA2 to create a total angle =137
degrees – based on that
(A)
- The angle ECA2 =137 degrees
- The distance BA2 = 150628 km
- The distance A2A = 64628 km
- The hypotenuse C A2 = 173450 km
- The perimeter of the triangle BCA2 = 173450 +150628 +86000 = 410080 km
- The triangle perimeter (BCA2) =410080 km= the apogee radius (406000 km)
(error 1%)
(B)
- The perimeter of the triangle (A CA2) =121622 + 173450 +64628 = 359700 km
- Perigee radius = 363000 km (error 1%)
A Conclusion
- The triangle BCA2 defines the moon motion limits from perigee to apogee by a
geometrical mechanism depends on The angle 137 degrees……. Why & How?

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(2nd
Point) The Rate 0.08
Why Pythagorean Triangle (1,2, 51/2
) Is Required?
This figure is discussed before.
- The inner circle refers to the perigee orbit
- The outer circle refers to the apogee orbit
- OB = 406000 km = Apogee Radius
- OR = 363000 km = Perigee Radius
- DB = 181843 km
- Perigee Orbital Circumference = 2.28 mkm
- Apogee Orbital Circumference = 2.55 mkm
I - Data
(1)
(DB / Perigee Orbital Circumference) = (181843 km/2.28 mkm) = 0.08
(2)
10.96 = 137 (The basic Angle) x 0.08
(3)
(4)
Cos (10.96 degrees) 88000 km = 86400 km
II – Discussion
- Why is the Pythagorean triangle (1,2,51/2
) required for the moon orbital motion?
- Because, the rate (0.08) is required to create interaction with the angle (137 deg),
and based on this interaction, the valuable angle (10.96 degrees) will be created,
and based on this angle (10.96 degrees) most of the moon orbital motion data will
be created.
- That answers the question why the rates (1,2,51/2
) were required necessary for the
moon orbital motion? because based on these rates the rate (0.08) will be produced
which will be used to produce the angle (10.96 degrees)…… So

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- Based on the angle (CA2E =137 degrees), the moon orbital motion receives 3
basic data which are
o The apogee point radius (r=0.406 mkm) which is defined by the triangle BC
A2) Perimeter
o The Perigee point radius (r=0.363 mkm) which is defined by the triangle AC
A2) Perimeter
o And the rate (0.08) which is defined between the tangent DB (181843 km)
and the perigee orbital circumference (2.28 mkm)…….. then
o 10.96 = 137 x 0.08
o The valuable angle (10.96 degrees) is created.
Equation No. (3)
- This equation tells the story in more clear way….
- The value 77237 km is very important…. If the moon moves daily a displacement
= 77237 km, during 29.53 days, the total distance will be = 2.28 mkm = the moon
orbital circumference at perigee orbit (r= 363000 km)
- Means, the perigee orbital circumference = 29.53 displacements each =77237 km,
that tells the value (77237 km) is defined by perigee radius (r=0.363 mkm) and the
moon day period (29.53 solar days), whatsoever the moon apogee radius be ….
Now the angle (10.96 deg) is defined before (10.96 = 137 x 0.08), and by that the
apogee radius is defined….
- I try to show that, we deal here with few players are created depending on each
other , all of them has one origin which is the angle 137 degrees, and has one
result which is the angle (10.96 deg)… what I try to do here is to show how the
data is arranged in a clear direction, by that, I may prove this is Directed Data.
Equation No. (4)
Cos (10.96 degrees) 88000 km = 86400 km
- The analysis is still complex and we need to consider it deeply in following…..

The Moon Orbital Motion Geometry (II)

The Moon Orbital Motion Geometry (II)

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The Moon Orbital Motion Geometry (II)