The Moon Orbital Motion Description (Revised)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Moon Orbital Motion Description (Revised)
The Author Authorized To Be Used By
Mr. Gerges Francis Tawdrous
A Student–Physics Department- Physics
& Mathematics Faculty –
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Dr. Budochkina, Svetlana Aleksandrovna
Associate Professor (Mathematical Analysis
and Theory of Functions Department)
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Phone +201022532292
E-Mail: mrwaheid@gmail.com
Curriculum Vitae http://vixra.org/abs/1902.0044
Phone +7 (495) 952-35-83
E-Mail: budochkina-sa@rudn.ru, sbudotchkina@yandex.ru
Website
http://web-local.rudn.ru/web-local/prep/rj/index.php?id=2944&p=19024
The Assumption Of S. Virgin Mary -Written in Cairo –Egypt – 26th
January 2021
Abstract
Paper hypothesis No. (1)
There's 2nd
force effects on The Earth Moon Orbital Motion
Paper hypothesis No. (2)
Uranus Motion effects on the Earth moon orbital motion and creates Metonic Cycle
Paper Argument
- The moon displacement daily = 88000 km
- During 29.5 days (The Moon Day Period), the total displacements will =2.59 mkm
this distance should be = the moon orbital circumference, and this distance = the
moon orbital circumference at apogee orbit radius (r=0.406 mkm) (Error 1%)
- That means, if the moon uses this displacement (88000 km) as a real displacement
through its orbit, the moon would revolve around Earth only through its apogee
orbit (r=0.406 mkm), and be always on the most far point from Earth and can't
revolve around Earth through any near orbits.
- To solve this dilemma, the moon uses Pythagoras triangle technique
- By this technique, the moon creates an angle (θ) between its displacement (88000
km) and its orbit horizontal level, So, the real displacement be (L =88000 cos(θ))
by this displacement (L) the moon passes through its orbit although the moon
actual displacement = 88000 km daily.
- By Pythagoras technique using the moon can revolve around Earth through more
near orbits as perigee orbit (r=0.363 mkm)

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- The moon using of Pythagoras triangle technique causes the moon orbit to be
created in a triangle form, which we analyze in this paper
- The moon orbital triangle data analysis lead to the following hypothesis (A 2nd
Force must effect on the Moon Orbital Motion in addition to Earth Gravity
Force)
- Also the 2nd
hypothesis is created based on this first one, (Uranus effects on the
moon orbital motion, causing to create Metonic Cycle).
Paper Main Idea
- The paper claims, the moon orbital triangle is created based on logical geometrical
structure, can't be created by the Sun and Earth gravities effects only.
- Planets (As Jupiter and Uranus) can effect by their gravities on the moon orbital
motion and these effects are seen in the moon orbital motion data and features.
- The Proving process depends on that, The triangle geometrical structure is created
by many players effects. This fact is proved by the triangle geometrical analysis,
and this fact shows different planets gravities effects on the moon orbital motion.
- These planets different effects are defined under one title (2nd
force effects on the
moon orbital motion)
- This conclusion supports the hypothesis (Uranus Motion causes to create Metonic
Cycle in the Moon Orbital Motion)

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Contents
Subject Page N
1- Introduction 4
2- The Moon Orbital Triangle Description
2-1 The Moon Orbital Triangle Description
2-2 The Moon Orbital Triangle Data Analysis
5
3- The Moon Orbital Motion Analysis
3-1 Why Does The Moon Use Pythagoras Triangle In Its Motion?
3-2 How Does The Moon Use Pythagoras Triangle In Its Motion?
3-3 The Moon Orbital Motion Analysis
3-4 The Moon Orbital Motion Equation
14
4-The Moon Orbit Geometrical Design
4-1 Preface
4-2 The Necessity Of Pythagoras Triangle (1, 2, 51/2
)
4-3 Why The Moon Displacement Daily =88000 km?
4-4 The Moon Motion Angle (12.195 degrees) Analysis
4-5 Why The Moon Day Period =29.53 solar days?
30
5- The Moon Orbital Triangle Benefits
5-1 Preface
5-2 The Moon orbital triangle shows that (2nd
force effect on the moon motion)
5-3 The Moon orbital triangle shows that (There's 2nd
Orbit for the moon motion)
5-4 The Moon orbital triangle shows that Uranus effects on the moon motion
48
6- Metonic Cycle Is A Proof of Uranus Effect On The Moon Motion
6-1 Preface
6-2 Uranus Effect On The Moon Orbital Motion
6-3 Uranus, The Moon And Pluto Motions Interaction
6-4 The Moon Orbital Triangle Angles Discussions
6-5 Moon Day Period Analysis (29.53 Solar Days)
53
7- Uranus Motion Analysis
7-1 Uranus Motion During 1440 Of Its Days Period
7-2 Uranus Motion During 8 Pluto Days period
7-3 Uranus 144 Days Cycle
74
8- Appendix No.1 95

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1- Introduction
- The moon uses Pythagoras triangle as one of the moon orbital motion techniques
- The moon motion 4 basic points were the method by which I have discovered that,
the moon motion 4 basic points are:
o Perigee radius (r=0.363 mkm), the most near point the moon can reach to
Earth.
o Apogee radius (r=0.406 mkm), the most far point the moon can reach from
Earth
o Total Solar Eclipse radius (r= 0.373 mkm), the moon creates A total solar
eclipse when the moon be at this distance from Earth or Shorter.
o The Moon Orbital distance (r=0.384 mkm), this value is the registered one
in the moon data sheet as the moon orbital distance.
- These 4 points are defined based on each other by Pythagoras 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%)
- Based on this data, the concept is discovered that, The Moon Uses Pythagoras
Triangle As One Of The Moon Motion Techniques
- But 2 questions are raised with this concept (1st
question) why does the moon use
Pythagoras triangle (2nd
question) what's this dimension (86000 km) which is used
frequently in the previous data?
- By answering these question, the following results are produced
o The moon motion data analysis refers to The Moon Orbital Triangle
o The moon motion data analysis refers to The Moon Motion Equation
- Let's discuss these tools in following in addition to discover how the moon uses
Pythagoras triangle and what benefits the moon receives for this using…

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2- The Moon Orbital Triangle Description
2-1 The Moon Orbital Triangle Description

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2-1- The Moon Orbital Triangle Description
- This is the suggested moon orbital triangle, let's explain it:
- Let's start with the triangle above the Ecliptic
(1st
) The Triangle ACE
- The triangle ACE is A Vertical Triangle (z-axis)
- The triangle ACE depends on its base (AE) (The Green Line),
- The point (A) is the intersection point between the Ecliptic Line and the triangle
Base (EA), where the ecliptic line is above the base (EA)
- There's an angle =0.443 degrees between the ecliptic line and the base (EA)
- Because of that, the base (EA) reaches to the Earth under the Earth Center with a
distance =3460 km (because of the angle 0.443 deg between the ecliptic (above)
and the triangle base EA (under)). (notice, the moon diameter =3475 km)
- The Line (BC) is perpendicular (90 degrees) on the triangle Base (EA), where the
line BC = 86000 km.
- The point (A) is a point on the triangle base and on the Ecliptic Line with a
distance 86000 km from the point B. by that the triangle ABC is an isosceles
triangle.

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The Triangle ACE Data
- BE = 364481.5 km
- BC =86000 km
- BA =86000 km
- BD = DA= 43000 km
- The hypotenuse CD = 96151 km The hypotenuse CZ = 88000 km
- CE = 374490 km (= Saturn Circumference) (1%)
- CA = 121622 km (= Saturn Diameter) (1%)
- The base (EA) (the Green Line ) = 450481.5 km (= Jupiter Circumference).
The Angles
- The angle is Zero between the points (E, B, D and A)
- CEB = 13.276 degrees
- ECB = 76.724 degrees
- BCA = CAB = 45 degrees
- BCD = 26.56 degrees
- ECA =121.724 degrees
- There's an angle 0.443 degrees between the triangle base (EA) (the green line) and
the ecliptic line (notice the ecliptic is above the triangle base)

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- (2nd
) The Triangle AFE
- The Triangle AFE is a horizontal triangle (x-y Plain)
- The Point T is the Earth Center
- The Point E is found inside Earth
- The Point M refers to the Moon Center
- The Red line TM is the moon orbit plane and it has an angle 5.1 degrees with the
ecliptic line (the moon orbital inclination)
- The line MF is the moon equator line and it has an angle 1.543 degrees with the
ecliptic line, the line MF isn't intersected with the ecliptic line at the point (A), the
MF & ecliptic lines will intersect each other beyond the point (A) with a long
distance. Based on that, The triangle ACE is closed triangle but the triangle AFE is
an opened one at the point (A).
- Based on that, between the moon equator line (the blue line) and the triangle base
(AE) an angle =1.1 degrees.
- The Point (F) is the intersection point between the triangle dimension CE and the
moon equator line MA.
- The Point (K) refers to the intersection point between the moon orbit plane (TM)
and the triangle base (the green line) (EA)
- The blue line in Earth is the Earth equator line declines with 23.4 deg
The Triangle AFE Data
- MT = 363000 km = The Perigee Point (the nearest point the moon reach to Earth).
- MB = 31604 km
- MC= 117603 km
- MK = 327180 km but KT = 35820 km
- MF = 542722 km
- CF = 557518 km
- EF = 183028 km

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The Angles
- MTB = 5.1 degrees (the moon orbital inclination)
- CMT = 91.1 degrees
- CMA = 88.9 degrees
- AFE = 12.176 degrees
- FEB = 166.724 degrees
- MKB = 5.543 degrees
- MAB = 1.1 degrees (between the base EA and the moon equator line MA)
Notice No.1
- The moon (center) is found under the ecliptic line with a distance = 32269 km
- But
- The ecliptic line is above the (ACE) triangle base (EA) (the Green line) at this
point with a distance =665 km because of that the distance from the point (B) (on
the triangle Base EA) and the moon (center) = MB = 31604 km
Notice No.2
- The moon equator line will intersect with the ecliptic line beyond the point (A)
with a distance =1198380 km

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- This figure of 2 circles I have brought from internet to use in the Explanation -
- We have supposed, the inner circle is Perigee orbit and the outer circle is apogee
orbit, And we have calculated the tangent DB = 181843 km
- AB = 363686 km (= Perigee Radius Approximately)
- 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 the moon orbital triangle is a similar to this triangle (ODB)
where their dimensions are rated and their angles are equal both are created as a
specific Pythagoras triangle (1, 2 and 51/2
)
- The first question should be (what's the geometrical necessity for which the
specific Pythagoras triangle (1, 2 and 51/2
) is used for the moon orbital motion?)

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The Triangle BCD
- Please remember, the green line (the triangle base EA) has a n angle 1.1 degrees with
the moon equator line, and an angle 0.443 degrees with the Earth Ecliptic
- The triangle (BCD) should be the basic triangle in the moon orbit, because
o The moon in the point M is in Perigee Point (r=0.363 mkm) which is
parallel to the point (B),
o The point (D) is supposed to be the parallel point to the apogee point
(r=0.406 mkm)
o means, the distance (BD) refers to the distance the moon moves from
perigee to apogee.
o The triangle (BCD) is similar to the triangle (ODB) and both are specific
type of Pythagoras triangle (1,2, 51/2
)
o The angle (BCD) = 26.56 degrees
o The angle (CDB) = 63.44 degrees
o The perimeter of triangle (BCD) = 225151 km
o Sin (5.1) x 225151 km x 2 = 40080 km (Earth Circumference)
o (5.1 degrees = The Moon Orbital Inclination).

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The Triangle BCZ
- The triangle BCZ is a specific triangle in the moon orbit because
o BZ = 18586 km
o The Angle BCZ =12.195 degrees
o The hypotenuseCZ = 88000 km = the moon displacement daily
- The data is interesting because it tells that, there's some relationship between the
moon daily displacement (88000 km) and the angle (BCZ =12.195 degrees)
- The angle 12.195 degrees = 13.177 degrees – 0.9856262 degrees
- Where
o 13.177 degrees = The Moon Motion Degrees Daily
o 0.98562 degrees= Earth Moon Motion Degrees Daily
o Because of that
o 12.195 degrees x 29.53 days (the moon day period ) = 360 degrees
o Can we conclude that, the moon daily displacement is defined relative to
this angle (12.195 degrees)? We should discuss this question later.

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The Point (A)
- 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!
- But 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) where no clear reason we have to explain why this point has such
massive importance?!
- The paper answers this question by the hypothesis: (Another force effects on the
moon orbital motion in addition to Earth gravity force and this point (A) refers to
this 2nd
force) –
- The paper tries to prove its hypotheses.
Notice
- In following we need to answer Why & How the moon uses Pythagoras triangle
in its motion, explaining The Moon Orbital Motion Equation… This discussion
is necessary to be done before our try to deepen the moon orbital triangle data
analysis.

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3- The Moon Orbital Motion Analysis
3-3 The Moon Orbital Motion Analysis

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- Let's summarize this question answer in following:
o The moon uses Pythagoras 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 Pythagoras 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 Pythagoras 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 Pythagoras triangle,
- Now let's suppose the moon doesn't use Pythagoras 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 Pythagoras 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

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3-3 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 Pythagoras 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 Pythagoras 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 Pythagoras 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
Orbits 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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3-4-1 The Equation Concept
3-4-2 The Equation Test and Accuracy
3-4-1 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 before which is (the moon uses Pythagoras triangle in its orbital motion)
- The moon uses Pythagoras 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 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? 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 Pythagoras 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 Pythagoras triangle its angle (θ) = 26.93 degrees, 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 (tomorrow) will be at the point 368722 km
and will have the Pythagoras 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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3-4-2 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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3-4-3 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. 6)

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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 more less
enables 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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4-The Moon Orbit Geometrical Design
4-1 Preface
4-2 The Necessity Of Pythagoras Triangle (1, 2, 51/2
)
4-3 Why The Moon Displacement Daily =88000 km?
4-4 The moon motion angle (12.195 deg) Analysis
4-5 Why The Moon Day Period =29.53 days?
4-6 The Perpendicular Line BC (=86000 km)
4-7 The Point (F) Position Analysis

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4-1 Preface
On What Facts This Study Depend?
On The Logical Geometrical Structure
- Please remember, the green line (the triangle base EA) has an angle 1.1 degrees with the
moon equator line, and an angle 0.443 degrees with the Earth Ecliptic
- Example.
- The moon orbital triangle base (The Green Line) (EA) = 450481.5 km
- In this distance, the point (A) I have concluded and was not found in the moon
motion data sheet, so Can be this point (A) a real point, or it's invented one?
o The distance EA causes the distance BD (43000 km) be = DA (43000 km)
o The distance EA 450481.5 km = Jupiter Circumference
o The distance BA = 86000 km = BC
o The triangle BCD is a Pythagoras specific triangle (1, 2, 51/2
)
o The perimeter of the triangle (ECA) = the distance from the point (A) to the
end on the lunar eclipse umbra length (1.392 mkm).
If I have invented the point (A), how can I created these relationships with it, where I
depend on the moon orbital motion real data? The main power behind this analytical
study is The Logical Geometrical Structure Of The Moon Orbital Motion Data.

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4-2 The Necessity of Pythagoras Triangle (1, 2, 51/2
)
(1st
Point) The Apogee Point Definition (r= 406000 km)
- In this moon orbital triangle I have added the line CS to create a total angle =137
degrees – based on that
- The angle ECS =137 degrees
- The distance BS = 150628 km
- The hypotenuse CS = 173450 km
- The perimeter of the triangle CBS = 173450 +150628 +86000 = 410080 km
- The triangle perimeter =410080 km= the apogee radius (406000 km) (error 1%)
- The perigee point radius (r= 363000 km) is defined by Earth gravity, where Earth
centrifugation prevents the moon to be more near than The Perigee Point.
- The idea is that:
o The moon orbital triangle is created based on a logical geometrical structure,
if it's true, The apogee point (r=0.406 mkm) should be defined based on the
triangle geometrical rules and data.
o Specifically the apogee point (r=0.406 mkm) should be defined based on the
angle 137 degrees (ECS). And this angle should have a massive effect on
the moon orbital motion, let's test that in following….

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(2nd
Point) The Rate 0.08
Why Pythagoras 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 x 0.08
(3)
Sin (10.96 degrees) x 406000 km = 77237 km
(4)
Cos (10.96 degrees) 88000 km = 86400 km
(5)
Sin (10.96 degrees) 450481.5 km = 85650 km
II – Discussion
- Why is the Pythagoras 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) based on which
most of the moon orbital motion data will be created.
- Because of that, the rates (1,2,51/2
) was required to produce this rate (0.08).

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- So, based on the angle (CSB =137 degrees), the moon orbital motion receives 2
basic data which are
o The apogee point radius (r=0.406 mkm) which is defined by the triangle
ECS) Perimeter
o And the rate (0.08) which is defined between the tangent DB (181843 km)
and the perigee orbital circumference (2.28 mkm).
o 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.
- 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….
- This explanation is not so correct because the apogee radius is defined before by
the triangle (ECS) Perimeter and (the rate 0.08) is defined based on it because we
use it in the circles figure.
- I try to show that, we deal here with few players are created depending on each
other , all of them has on 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, and by that, I may prove this is Directed Data.

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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…..
- Where
o The moon orbital circumference at apogee radius (r=0.406 mkm) equals
only 2.55 mkm and this distance is short!
o Because
o The moon daily displacement =88000 km and during 29.53 solar days the
total displacements will be = 2.598 mkm …..if this distance be the moon
orbital circumference the radius will be = 0.413 mkm
o Means, the apogee radius will not be 0.406 mkm but 0.413 mkm !
o Which proves the paper claim, that, the moon uses Pythagoras triangle in its
motion,
o But
o Why the moon orbital circumference at apogee is not = 2.598 mkm? Why
we have 2 values contradict each other (which are the moon daily
displacement and the moon orbital circumference at apogee),
- Equation No. (4) tells us this story clearly, where the apogee orbital permits for a
daily displacement =86400 km and NOT 88000 km
- Please note, this is a theoretical analysis and not practical, the moon could use
88000 km as its displacement without using Pythagoras triangle technique in any
days during the month BUT NOT Every Day.
- The question still the same
- Why the apogee orbital circumference doesn't equal 2.598 mkm?
- Let's analyze the moon different motions in following to see this data as clear as
possible

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More Data
(A)
The moon orbital circumference at apogee point = 2.55 mkm (100 %)
The moon moves per solar day a distance = 2.5734 mkm (101%)
The moon total displacements during 29.53 days = 2.598 mkm (102%)
Pluto motion distance during its day (153.3 h) = 2.5938 mkm (102%)
(B)
137 =95.1 x 1.44
More Discussion
Data No. A
- The first and third distances are the moon motion distances, where, it’s the moon
orbital circumference (2.55 mkm) and its total displacements (2.598 mkm)…
- The second distance is the moon motion distance also, because the moon moves
per solar day a distance equal earth motion distance per solar day perfectly
otherwise the moon and Earth will be separated in the motions course.
- We have 3 motions are arranged in (100%, 101%, 102%) all of the are done by the
moon motions – There must be a geometrical mechanism behind this order-
- We deal with some gears, and these gears are required to be rated to each other to
enable to do their jobs –
- i.e.
- The moon orbital circumference at apogee (2.55 mkm) is NOT short distance, it's
created for some geometrical necessity to enable the machine of gears to work
- Nothing more can be told here, this data should be discussed with the answer of
(Why the moon daily displacement =88000 km? (the next point "point 4-3")
- I have brought Pluto motion also here to show that the machine of gears isn't
limited to the moon and Earth motion but other planets are players in this machine.

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Equation No. (B)
137 =95.1 x 1.44
- We still don't know why this angle 137 degrees has so massive effect on the moon
orbital motion…?
- Equation no. (B) may help us let's discuss it
o 95.1 degrees = 90 degrees + 5.1 degrees (the moon orbital inclination)
o 1.44 degrees = the moon orbit regression degrees per month
- So, the angle 137 degrees, is created by the moon orbit motion effect,
- 2 features of the moon orbit motion unifies together to produce this angle (137
degrees) which is the origin of the moon motion distance from perigee to apogee..
which are
o The moon orbital inclination 5.1 degrees
o The moon orbit regression 19 degrees per year
- These 2 features of the moon orbital motion creates together the angle 137 degrees
as their platform to create the moon orbital motion in harmony with these 2
features…
Notice
- 180 degrees -137 degrees = 43 degrees
- If 1 degree =1000 km, so
- The value 43 degrees expresses the distance 43000 km which is the distance
between Perigee and apogee….
- The data tells that, a geometrical mechanism is found behind it creates this data
based on each other geometrically

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(3rd
Point) The Angle 10.96 degrees
By this triangle we follow the moon motion data based on the angle 10.96 degrees.
We start from apogee radius (r=406000 km) on the AC as following
(1)
- AC =406000 km the angle C= 10.96 deg what's BC?
- BC = 398595 km
(2)
- BC = 391324 km
(3)
- BC = 384186 km
(4)
- BC = 377179 km
(5)
- BC = 370300 km
(6)
- BC = 363546 km
The 4 distances (in blue color) are the moon motion basic 4 points.
The moon motion depends on the angle 10.96 degrees.

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Equation No. (5)
Sin (10.96 degrees) 450481.5 km = 85650 km
- Equation no. (5) tries to help the explanation,
o The distance 85650 km is very near to the line BC =86000 km (error 0.4%)
o Also the distance 86000 km = 2 x 43000 km ( Perigee apogee distance)
o But
o The distance 450481.5 km is created based on the point (A) which is created
and not found in the moon data sheet….
o By what geometrical mechanism the angle 10.96 degrees uses the distance
450481.5 km to produce the line BC 86000 km?! it still needs to answer
with more analysis but the point is that, the data (450481.5 km) is a real one
and isn't invented …. Also the line BC (86000 km) is real data
o The angle (10.96 degrees) proves that, because, the moon orbital motion
data depends on this angle and because of the invented data by the moon
orbital triangle also depend on this same angle…

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4-3 Why The Moon Displacement Daily =88000 Km?
Earth motion distance during its day period = the moon displacements total
during its day period = Pluto motion distance during its day period (error 1%)
I - Data
(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
2(4)
Uranus Orbital Circumference = 18048.449 mkm
Where 153.3 hours = Pluto day period 29.53 days = the moon day period
The moon daily displacement =88000 km
II - Data Analysis
(4) – (1) = 189.124 mkm
(4) – (2) = 47.879 mkm
(4) – (3) = 14.171 mkm
(3) – (1) = 174.953 mkm
(3) – (2) = 33.708 mkm
(2) – (1) = 141.245 mkm
- Sin (17.2) x 47.879 mkm = 14.171 mkm
- Tan (10.96) x 174.953 mkm = 33.708 mkm
- Tan (13.3) x 141.245 mkm = 33.708 mkm (error 1%)
- 0.8 x 174.953 mkm = 141.245 mkm (error 1%)
- Sin (4.63) x 174.953 mkm = 14.171 mkm
o 0.8 degrees = Uranus Orbital Inclination. Sin (4.63) =0.08
o 17.2 degrees = Pluto Orbital Inclination
o 13.3 degrees = The angle of (E) in the moon orbital inclination
o 10.96 degrees (Cos 10.96 degrees x 88000 km = 86400 km)

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III –Discussion
- The previous 4 angles are the basics data for their planets, let's try to show that
o 0.8 degrees = Uranus Orbital Inclination
o 122.5 deg (Pluto Axial Tilt) x 0.8 = 97.8 deg (Uranus Axial Tilt)
o Pluto orbital inclination 17.2 degrees = 0.99 x 17.4 deg (The inner planets
orbital inclinations total) … also
o Pluto orbital inclination 17.2 deg x 7.1 = 122.5 deg (Pluto Axial Tilt)
o 13.3 degrees is the angle of point (E) (Earth) in the moon orbital triangle
(Earth Orb. Period 365.25 d = The moon Orb. Period 27.3 d x 13.3)
o The angle 10.96 degrees is the valuable angle we have discussed deeply
where (Cos 10.96 degrees x 88000 km = 86400 km).
o Sin (4.63) = 0.08 This rate effects on the moon orbit geometrical design
There's an interaction occurred here between these 4 planets (Uranus, Pluto, Earth and
its moon), and in this interaction, these 4 basic values are created and based on these 4
values many other data of these planets is created … means, this interaction forms the
geometrical structure of these planets motions …. And if we limited our discussion
for the moon orbit structure, that lead us to conclude that, the moon orbit geometrical
structure is effected by these 4 planets motions interaction as seen in the data.
i.e.
These 4 planets motions interaction effects on the moon orbital motion and causes to
create Metonic Cycle.
That supports the 2nd
hypothesis (Metonic Cycle is a proof of Uranus motion effect on
the moon motion.)
(This discussion is extended with full detail in Metonic Cycle discussion Point No. 6)

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4-4 The Moon Motion Angle (12.195 degrees) Analysis
I-Data
(I)
And
13.177 degrees – 0.98562 degrees = 12.195 degrees
(II)
(10.96 degrees) + 1.25 degrees =12.195 degrees
Where
13.177 degrees = the moon daily motion degrees
0.98562 degrees = Earth daily motion degrees
0.8 degrees = Uranus Orbital Inclination
II- Discussion
- The Apogee Orbit (r=0.406 mkm) permits a displacement =86400 km only based
on the valuable angle (10.96 degrees), as maximum displacement during 29.53
days because (86400 km x 29.53 days = 2.55 mkm = 2π x 0.406 mkm)
- But
- What about the actual displacement 88000 km, which angle expresses it?
- The data shows that, the angle 12.195 degrees can define this displacement (88000
km) relative to the radius (407300 km) which is very near to apogee radius =
(406000 km) (error 0.3%).
- Equation No (II) tells that, Uranus orbital inclination 0.8 degrees is used as (1/0.8),
i.e.
- The angle (10.96 degrees) + (1/0.8 degrees) = 12.195 degrees
- The data shows Uranus effect on the moon orbital motion

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NOTICE
The following explanation shows a new geometrical technique is using in the moon
geometrical structure, it's just example using the angle 12.195 deg in this technique
I-Data
- In the triangle ABC
- AB = 12.195 km
- AC = 2 x 29.53 km
- The Angle A = 78.081
- The Angle C = 11.919 degrees
- But
- Cos (12.195 degrees) x 12.195 degrees = 11.919 degrees
1- How This Triangle Is Created?
- The geometrical structure uses the angle 12.195 degrees as a distance= 12.195 km,
and creates the angle (C) depends on the angle 12.195 degrees as the data shows
- So this triangle is created depending on the angle 12.195 degrees
2- This Triangle Purpose
- The triangle aims to create the hypotenuse AC = 59.06 km = 2 x 29.53 km
3- Why This Triangle Is Created?
- To create the value (29.53 km) depends on the value 12.195 degrees geometrically,
both data is the moon motion data, but the triangle tries to connect both data
geometrically, why? because Nothing is independent (the geometrical concept),
because of that, the new data should be created based on the old data, and by that
there's always one line connecting all data
- This simple example is for this technique explanation.. and the rate (1km=1degree)
is used here only and not a general rate, although the value (2x 29.53) is used more
widely than (29.53) in all data. (For example, Earth during 59 days moves a
distance = its orbital distance "Error 1%" ).

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4-5 Why the moon day period =29.53 solar days?
I-Data
Equation No. (A)
Tan (12.195 deg) x 708.7 hours (the moon day period) = 153.3 h (Pluto day period)
Tan (13.177 deg) x 655.7 h. (the moon rotation period) = 153.3 h (Pluto day period)
- The angle 12.195 deg. is the moon angle (12.195 deg. = 13.177 deg. - 0.9856 deg),
Based on this angle the moon & Pluto days periods are defined relative to each
other… Why?
- The angle 13.177 degrees is the moon motion daily angle (360 =13.17 deg x 27.3)
and based on this angle the moon & Pluto rotations periods are defined relative to
each other… Why?
- Why the moon day period =29.53 solar days? Because the moon day period is
created in proportionality with Pluto day period and both are created relative to
each other…..But the better question is ….
Why Earth day period =24 hours?
Equation No. (B)
Tan (8.9 deg) x 153.3h (Pluto day period) = 24 hours
- The angle 8.9 degrees =98.9 degrees – 90 degrees (discussed in page no. 35)
- By this angle Earth and Pluto days periods are created relative to each other!
- Pluto, Earth and the moon motions are interacted because of their motion distances
relative to Uranus orbital circumference (discussed in page no. 37) means this data
is a point of a sea of data which we have to discuss in Metonic Cycle discussion
Shortly
- The moon day period (= 29.53 solar days) because it's created by 2 motions effect
on the moon orbital motion (Earth & Uranus motions) through the 4 planets
motions interaction. (Metonic Cycle is discussed in Point No. 6)
- (In that discussion we should discuss, Why "Earth velocity/ Pluto velocity" = Pluto
day period / Earth day period?).

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4-6 The Perpendicular Line BC (=86000 km)
- Let's summarize how this triangle idea is created in following:
o Uranus Axial Tilt =97.8 deg and the Earth Moon Axial Tilt =6.7 deg. So
between them (97.8 – 6.7 = 91.1 degrees)
o The number 91.1 degrees gives a reference for some perpendicularity
between the moon axial tilt and Uranus axial tilt, but there's 1.1 deg!
o So, the solution was to decline the triangle base (EA) with 1.1 degrees on
the horizontal level and by the Uranus axial tilt will be perpendicular on the
triangle base if this triangle based depends on the moon axial tilt…
o This is the original idea of this triangle
o For that reason the line BC is perpendicular on the moon orbital triangle
- I want to say that….
- The line BC expresses Uranus effect on the moon orbital motion and we see
clearly how this dimension is so effective on the moon orbital triangle.

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4-7 The Point (F) Position Analysis
- Let's summarize the idea in following…
- The angle MTB =5.1 degrees (the moon orbital inclination), and based on the
distance (363000 km = Perigee radius), the distance MB = 31604 km ….. BUT
- The other triangle MBA (which is opened one), its angle is 1.543 degrees (or even
1.6 degrees), and when we try to know where the ecliptic line will intersect with
the moon equator line, we will find that at distance around 1.19 million km from
the point (A)
- I tried to find better calculations for this distance, so I suggest the following:
o Let's suppose the angle (BTM) be used as 5.6 degrees (in this case the moon
orbital inclination will be measured above the moon diameter which
consumes 0.5 degrees)
o The distance MB will be = 34757 km
o The angle (1.6 deg) also will use an additional (0.5 deg) = 2.1 deg.
o In this case the distance to the intersection point will be = 966676 km
beyond the point (A)

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o The Base EA =450481.5 km, means the total distance from Earth to the
intersection point between the Earth ecliptic line and the moon equator line
will be = 1417157.5 km
o Where
o The lunar eclipse umbra length =1392000 km
o The error is around (1.8%)
That creates a geometrical structure defines the lunar umbra length.
Notice
- The previous suggestion doesn't change the triangle, it's a suggestion for the lunar
umbra length definition geometrical rules… that means, this triangle is correct as
it, if this calculation is correct that means more details should be added to this
same triangle to create this result.

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5- The Moon Orbital Triangle Geometrical Benefits
5-1 Preface
5-2 The Moon orbital triangle shows that (2nd
5-3 The Moon orbital triangle shows that (There's 2nd
5-1 Preface
- The moon orbital triangle geometrical analysis provides a new and effective idea
let's try to summarize it in following
o The moon orbital triangle shows that many forces effect on the moon orbital
motion because of that many geometrical rules are used in this motion to
define each force balancing points
o I refer to Earth gravity force effect on the moon motion as 1st
force
o I refer to all other planets effects on the moon motion as 2nd
force
o The sun gravity force is considered to be including into both forces
- The triangle shows that, many forces (or motions) interaction effects on the moon
motion and by that the moon orbit geometrical design became a specific one,
showing these forces effects.
- The triangle analysis depends on the Logical Geometrical Analysis, for that, the
absent data can be concluded and (more important) the forces created this data can
be discovered
- Based on that, Jupiter and Uranus (in addition to other planets) have effects on the
moon orbital motion. this conclusion can be formed by the moon orbital triangle
data analysis.
- This analysis can support the paper hypotheses which are: (1st
) (There's 2nd
force
effects on The Earth Moon Orbital Motion (2nd
) (Uranus Motion effects on the
Earth moon orbital motion and creates Metonic Cycle)

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5-2 The Moon orbital triangle shows (2nd
(The Triangle Data Analysis In Discussed In Point No. 1 Of This Current Paper)
- What Proves Can Be Provided For The 2nd
Force Hypothesis?
o (1st
Proof) The Point (A) In The Moon Orbital Triangle
o (2nd
Proof) The 2nd
Displacement 88000 km
o (3rd
Proof) Metonic Cycle Creation…. Let's discuss them in following:
(1st
Proof)
- The moon orbital triangle causes to raise the question, because the Point (A) is one
of its 3 basic points and no force we know can create this Point (A) which is found
far from apogee radius (r=0.406 mkm) with a distance =43000 km, because of that
the distance EA =449197 km
- So how this point is found and effect on the moon orbital triangle? We have no
answer except that 2nd
force is found effects on the moon orbital motion, this 2nd
force effects on the Point (A). So Earth gravity force effects on the moon motion
on one side and this 2nd
force effects on the moon motion on other side to create
general balancing of the moon motion.
- Although no clear definition for the force creates the point (A), this force is still
fact because of the geometrical massive significance of the point (A).
- means, the point (A) should be considered as a proof for this force existence

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(2nd
Proof) The 2nd
Displacement 88000 km
- The moon orbital motion story tells us, the moon contracted distance (2.399 mkm)
needs (0.17 mkm) to be = Earth motion distance (2.573 mkm) per solar day, and
the moon has to move this additional distance (0.17 mkm) on the same solar day,
But the moon daily displacement =88000 km, means, the moon has to move one
more displacement (88000 km) which we don't see…
- If this story is real, and the distance 0.17 mkm should be passed, and if 1 force
only effects on the moon, so this force should cause the moon to move 0.17 mkm
completely…but the moon displacement is only (a half) of the required distance…
that tells us there are 2 forces causes 2 equal displacements (regardless our
observation for them).
- The argument here depends on the moon basic motion (2.573 mkm) which creates
the moon daily displacement (88000 km), if the connection between these 2
distances is a real one, so the 2nd
displacement must be a fact and that necessitates
to find 2nd
force effects on the moon orbital motion.
(3rd
Proof) Metonic Cycle Creation.
- Uranus Orbital Circumference =19 Earth Orbital Circumference …… means
- While Uranus revolves around the sun one revolution, Earth (and its moon)
revolve around the sun 19 revolutions (19 years =6939.75 solar days)
- If Uranus motion effect on the Earth moon motion, the period 19 years should be
seen in this effect data because it’s the basic rate between the 2 orbits
- The moon Metonic Cycle (6939.75 solar days=19 years) tells that, there's a
possibility of Uranus motion effect on the moon motion..
- The point is, if Uranus really effects on the moon orbital motion to create Metonic
Cycle, so this will be a solution for the question (What's this 2nd
force effects on
the moon orbital motion), or at least will give us a light to see other players effect
on the moon orbital motion in place of the one planet gravity effect vision.

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5-3 The Moon orbital triangle shows (There's 2nd
I- Data
(1)
The moon orbital triangle (ECA) Perimeter = 943817 km
The Lunar Eclipse Umbra Length = 1.392 mkm
The distance (EA) = 449197 km + (The Perimeter) 943817 km = 1.392 mkm
II- Discussion
- The Point (A) divides (the lunar eclipse umbra length) into 2 equal parts, after the
Point (A) this part is seen in the triangle perimeter (ECA) and
- Before the Point (A) this part is seen in the distance from the Point (A) to the end
of The Lunar Eclipse Umbra Length
- Can This Be A Proof?
- The geometrical division is a proof, because the moon orbit data is created based
on geometrical interactions for that reason the moon orbital triangle shows these
geometrical interactions and rules, and these geometrical rules tell, many players
are interacted here –for that reason, the triangle (ECA) perimeter has a relationship
with The Lunar Eclipse Umbra Length (Where the geometrical necessity of this
relationship still need to be caught, but the mere existence of this relationship is a
proof for different player effect on the moon orbit geometrical creation).
- I want to say, the moon orbit is NOT a trajectory of a rigid body revolves around
Earth, instead, it's a network of forces lines and the moon moves through this
networks taking into consideration these forces lines effects AND shows that in its
motion data.

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- Let's review the triangle concept in following:
o The moon orbital triangle is a vertical triangle effects on the moon orbit,
where the line (BC) is perpendicular on the moon orbital triangle base (EA)
and because of that the point (C) is found on (z-axis), where the moon
orbital motion is done on (x-y plain)
o How That Can Be Possible?
o I supposed Uranus Axial Tilt (97.8 degrees) is the line (BC), the moon axial
tilt =6.7 degrees and the difference =91.1 degrees, for that reason the moon
orbital triangle declines on the moon equator line with 1.1 degrees and the
line (BC) is perpendicular (90 deg) on the moon orbital triangle base (EA).
o I have designed this triangle basically based on this data and the triangle is
used sufficiently for the moon real motion and data.
o Uranus indeed effects on the moon orbital motion in different features, not
only in Metonic Cycle, but also by Uranus axial tilt effect on the moon axial
tilt, not that only…
o Earth moves during its day period a distance = The moon displacements
total during its day period = Pluto motion during its day period, (error 1%),
This feature also is found by Uranus effect on the moon orbital motion
o The moon day period (29.53 solar days) is a piece of gold because this
period of time shows that it's created by 2 motions effect on the moon
orbital motion – shortly – Earth and Uranus motions effect on the moon
orbital motion, forcing the moon day period to be 29.53 solar days.
o This discussion should be completed with Metonic Cycle Discussion (Point
No. 6 of this paper).

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6- Metonic Cycle Is A Proof of Uranus Effect On The Moon Motion
6-1 Preface
6-2 Uranus Effect On The Moon Orbital Motion
6-3 The 4 Planets Motions Interaction

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6-1 Preface
Paper 2nd
hypothesis
- The Earth Moon Metonic Cycle (6939.75 Solar Days) is created by Uranus
Motion Effect On The Moon Orbital Motion.
The Hypothesis Proves
o (1st
Proof) Uranus Orbital Circumference =19 Earth Orbital Circumference,
So while Uranus revolves around the sun 1 complete revolution the Earth
(with its moon) revolve around the sun 19 revolutions..
o If Uranus Motion effects on the moon orbital motion, the number 19 should
be seen in this effect data because it’s the rate between both orbits.
(Sub-Point 6-2)
o (2nd
Proof) Earth Motion Distance During Its Day Period = The Moon Total
Displacements During 29.53 solar days (The Moon Day Period) = Pluto
Motion Distance During 153.3 hours (Pluto Day Period) – this feature of
motion is created by Uranus motion effect on the 3 planets.
(Sub-Point 6-3)
o (3rd
Proof) Uranus Moves During (1440 Of Its Days Period) A Distance =
The Earth Moon Total Displacements During Metonic Cycle (6939.75 Solar
Days)
(Point No.7 "Uranus Motion Analysis")
o (4th
Proof) The Moon Orbital Triangle Data Shows Uranus Effect On The
Moon Motion.

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6-2 Uranus Effect On The Moon Orbital Motion (1st
Proof)
In this figure
- The Red Ball Shows Earth
- The Yellow Ball shows The Earth Moon
- The Blue Ball shows Uranus
- (S) is the Sun
- The figure suggests that, a triangle contains these 3 planets together in their
revolutions around the sun
- Let's suppose the three planets, Earth, its moon and Uranus move in parallel to
each other in their revolutions around the sun, and to guarantee this parallelism
between them the figure provides a triangle contains these 3 planets -
- Uranus orbital circumference = Earth orbital circumference x 19
In accurate calculations
- Uranus (18048 mkm) = Earth (940 mkm) x 19 (error 1%)
- This data means, while Earth revolves around the sun 19 times, Uranus revolves
around the sun 1 time only

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- If the 3 planets move in parallel to each other, that means, Uranus will divide its
revolution trajectory around the sun into 19 parts, and each part will be a qualified
for one Earth orbital circumference (difference 1%)
- Uranus motion trajectory effect is observed on the Earth moon motion trajectory,
let's show how that happens:
- The moon moves through its orbital circumference revolving around the Earth
(while the masses gravity forces imprison the moon inside the range from perigee
(0.363 mkm) to apogee (0.406 mkm) and prevents the moon to move out of this
motion range).
- But
- Uranus motion effects on the Earth moon motion (inside its prison) and forces the
moon to change its motion trajectory through 19 years. Because of that the moon
doesn't move through the same point 2 times during 19 years (6939.75 solar days),
that creates Metonic Cycle, that happens because the moon motion reflects Uranus
Motion Effect revolving around the sun, where Uranus moves on a trajectory
doesn't pass through the same point 2 times during (19 years) (according to the
moon time) similar to that the moon moves through its orbital circumference
doesn't pass through the same point 2 times during 19 sidereal years.
- Shortly
- Metonic Cycle Is Created By Uranus Motion Effect On The Moon Orbital Motion.

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6-3 The 4 Planets Motions Interaction (2nd
Proof)
(We us a discussion of the moon orbit design – Point no. 4-3)
Earth motion distance during its day period = the moon displacements total
during its day period = Pluto motion distance during its day period (Error 1%)
6-3-1 The 4 Planets Motions Interaction Analysis
I - Data
(1)
(2)
(3)
(4)
II - Data Analysis
(4) – (1) = 189.124 mkm
(4) – (2) = 47.879 mkm
(4) – (3) = 14.171 mkm
(3) – (1) = 174.953 mkm
(3) – (2) = 33.708 mkm
(2) – (1) = 141.245 mkm
- Sin (17.2) x 47.879 mkm = 14.171 mkm
- Tan (10.96) x 174.953 mkm = 33.708 mkm
- Tan (13.3) x 141.245 mkm = 33.708 mkm (error 1%)
- 0.8 x 174.953 mkm = 141.245 mkm (error 1%)
o 0.8 degrees = Uranus Orbital Inclination.
o 13.3 degrees = The angle of (E) in the moon orbital inclination
o 10.96 degrees (Cos 10.96 degrees x 88000 km = 86400 km)

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III –Discussion
- The previous 4 angles are the basics data for their planets, let's try to show that
o 122.5 deg (Pluto Axial Tilt) x 0.8 = 97.8 deg (Uranus Axial Tilt)
o Pluto orbital inclination 17.2 degrees = 0.99 x 17.4 deg (The inner planets
orbital inclinations total) … also
o Pluto orbital inclination 17.2 deg x 7.1 = 122.5 deg (Pluto Axial Tilt)
o 13.3 degrees is the angle of point (E) (Earth) in the moon orbital triangle
(Earth Orb. Period 365.25 d = The moon Orb. Period 27.3 d x 13.3)
o The angle 10.96 degrees is used to define the moon orbital apogee radius
(r= 0.406 mkm) because (86400 km x 29.53 days = 2π x0.406 mkm). The
apogee orbit doesn't permits for a daily displacement greater than 86400 km,
where (Cos 10.96 degrees x 88000 km = 86400 km).
There's an interaction occurred here between these 4 planets (Uranus, Pluto, Earth and
its moon), and in this interaction, these 4 basic values are created and based on these 4
values many other data of these planets is created … means, this interaction forms the
geometrical structure of these planets motions …. And if we limited our discussion
for the moon orbit structure, that lead us to conclude that, the moon orbit geometrical
structure is effected by these 4 planets motions interaction as seen in the data.
i.e.
These 4 planets motions interaction effects on the moon orbital motion and causes to
create Metonic Cycle.
That supports the hypothesis (Metonic Cycle is a proof of Uranus motion effect on the
moon motion.)

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6-3-2 The Interaction Angle 71.9 Degrees
I- Data
1- The Moon Orbital Circumference at apogee radius = 2550973 km (100%)
2- Earth Daily Motion Distance = 2573483 km (101%)
3- Pluto moves during 153.3 hours =2593836 (102%)
4- The displacements 88000 km total during (29.53 days) = 2598693 km (102%)
5- Uranus motion distance (during 378675 seconds) = 2574990 km (101%)
5-1 = +24017 km
5-2 = +1507 km
5-3 = - 18846 km
5-4 = - 23703 km
4-1 = 42863 km
4-2 = 25210 km
4-3 = 5867 km
3-2 = 20353 km
3-1 = 41853 km
2-1 = 22510 km
II- Data Analysis
(I)
Cos (71.9) x 18846 km = 5867 km
Sin (71.9) x 23703 km = 22510 km And (Cos (71.9) = tan (17.25))
- The angle (71.9 degrees) I call (The Interaction Angle)
- This angle connects 5 basic values which are:
o 17.2 deg (Pluto orbital inclination)
o - 18846 km = the difference (Uranus motion & Pluto motion)
o – 23703 km = the difference (the moon displacements & Uranus motion)
o 22510 km = the difference (the moon orbit & Earth motion)
o 5867 km = the difference (the moon displacements & Pluto motion).

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- It's a clear interaction between the 4 planets motions, because it's directed data….
This data is not random but directed, because of that the same angle (71.9 degrees)
is used frequently Because It's Found In The Interaction Point.
(II)
2 x 71.9 degrees = 12.195 degrees x 11.8 degrees
Where
12.195 degrees = The moon motion angle (13.177 deg – 0.98562 deg)
11.8 degrees =6.7 degrees (moon axial tilt) + 5.1 deg (moon orbital inclination)
Why does the data use double values (2 x 71.9 deg)??
(III)
122.5 = 71.9 degrees x 1.7
Where
122.5 degrees = Pluto Axial Tilt
1.7 degrees = The moon motion equation constant ((θ1= θ0 + 1.7 degrees)
- Why does the equation use 1.7 degrees for moon motion daily? (this question is
asked in the moon motion equation discussion), the data tells that the angle 71.9
degrees (the interaction angle) has an effect to do that
- So, the constant (1.7 deg) depends on the interaction angle (71.9 deg) and Pluto
Axial Tilt (122.5 deg)…
BUT
- (122.5 deg -71.9 deg) x 2 = 101.2 degrees
- In the distances data Earth motion distance daily (2573483 km) is considered as
(101%), If there's a relationship between this 101% and the value 101 deg, we may
conclude, this value also refers to the using of (2 x 71.9 degrees)! Why?
ALSO
- 71.9 degrees / 101.2 = 0.712 we remember θ1= θ0 + 1.7 degrees where 1.7 deg =
0.98562 deg +0.712 deg, it's another proof that, the constant (1.7 deg) is produced
by the planets interaction (specifically between Pluto and the moon motion).

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(IV)
- 14 degrees x 5.1 degrees (the moon orbital inclination) =71.4 degrees
- 71.9 degrees = 71.4 degrees + 0.5 degree (the moon angular diameter)
- And
- 14 degrees = (5.1 degrees (the moon orbital inclination) + 8.9 degrees)
Let's remember 8.9 degrees
o 95.6 deg + 1.1 deg = 96.7 deg
o 96.7 deg + 1.1 deg = 97.8 deg
o 97.8 deg + 1.1 deg = 98.9 deg
o (Please review page no. 36)
Where
o 95.6 deg = 90 deg + 0.5 degrees + 5.1 deg (The Moon orbital inclination)
o 96.7 deg = 90 deg + 6.7 deg (The Moon Axial Tilt)
o 97.8 deg = Uranus Axial Tilt
o 96.7 deg = 90 degrees + 8.9 degrees
o 1.1 deg = the angle of the moon triangle base (EA) & moon equator line.
(V)
- 63.7 degrees = (71.9 deg – 8.9 deg) + (71.9 deg – 8 x 8.9 deg)
- Where
- 63.7 deg = The Sun Declination
- Equation no. (V) tells a very important information, which are:
o (1) The interaction angle (71.9 deg) is used in double Value (2 x 71.9),
because of a geometrical necessity.
o (2) The (8 days) Cycle, we have discovered in Jupiter & Uranus motions, is
used here to define the interaction angle based on which the most of the
moon data is created – i.e. the cycle (8 days) effects on the moon motion
- The cycle (8 days) is discussed with many details in Point No. (7) (Uranus Motion
Analysis).

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II- Data Analysis (Continued)
(VI)
Tan (29.53) x 18846 km = 23703 km
o - 18846 km = the difference (Uranus motion & Pluto motion)
o – 23703 km = the difference (the moon displacements & Uranus motion)
- We have found the moon day period (29.53 days), it's created as an angle (29.53
degrees) in this same interaction ….
- The moon orbit regresses 19 degrees per year and causes to change the eclipse
calendar by 19 days by this regression , showing that, 1 degree = 1 day
- By this data we can explain some other important data, let's remember them
Old Data
Tan (12.195 deg) x 708.7 hours (the moon day period) = 153.3 h (Pluto day period)
Tan (13.177 deg) x 655.7 h. (the moon rotation period) = 153.3 h (Pluto day period)
- We have discussed this data with our discussion for the question (Why the moon
day period = 29.53 solar days?) (Point 4-5) (Page no. 42).
- This old data told us, the moon day =29.53 solar days because the moon an Pluto
days are created relative to each other (depends on the angle 12.195), and here we
catch the interaction point on which these 2 days periods are created relative to
each other…
- Where
- 29.53 degrees x 12.195 = 360 degrees.

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(VII)
(42863/25210) =1.7 (The constant =1.7 deg)
(42863/5867) =7.25 (an angle in the moon triangle its dimension =10921 km)
(42863/22510) = 1.9 (Mars Orbital Inclination = 1.9 deg)
- The distance 42863 km should get our attention because many of the moon data
depends on it
o 42863 km= (the moon displacements total) – (the moon orbit circumference)
o We should consider this relationship as deep as possible because it may
answer the question, Why the moon apogee orbit circumference=2.55 mkm?
This is the question we need to answer… because the moon apogee orbit
circumference is smaller than the moon displacements total during 29.53
days which equal (2598693 km) where the apogee orbit (2550973 km) and
the difference between them is 42863 km BUT the distance from perigee to
apogee =43000 km…… So, let's ask, Is the difference between the moon
displacements total and its apogee orbital circumference is found to create
the distance 43000 km (perigee apogee distance)?
o Shortly
o Is this difference found for the geometrical necessity required to create the
distance between perigee and apogee?
- The previous interaction effects clearly on the interacting planets data, let's see the
proportionality in Earth and Pluto Data to prove this claim.
Earth and Pluto Data
1- (Pluto day period / Earth day period) = (Earth velocity / Pluto velocity)
2- Pluto orbital distance 5906 mkm= Earth orbital Circumference 940 mkm x 2π
3- Pluto orbital period 90560 solar days= 1461 solar days x 2π3

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4- Pluto moves during 365.25 solar days a distance = 149.6 mkm = Earth orbital
distance.
5- Pluto orbital inclination 17.2 deg = 99% the inner planets orbital inclinations
total (17.4 deg)
6- Earth Axial Tilt 23.4 deg= 99% the outer planets orbital inclinations total (23.6
deg)
Notice
- The Interaction Angle Data is so rich data but we can't extend our discussion here,
because we need before to analyze Uranus motion, for that reason, we have to
complete this discussion with the Point no. (7) (Uranus Motion Analysis) under the
same title ("The Interaction Angle 71.9 degrees" Continued)

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The Moon Orbital Triangle Data Shows Uranus Effect On The Moon Motion.
(4th
Proof)
We analyze here 2 angles (32.967 degrees & 36.912 degrees)
Point No. (A) (The angle 32.967 degrees)
- This is the moon orbital triangle, I have added the triangle CUB
- BCU = 32.967 degrees
- CU = 102500 km
- BU = 55756 km
Let's analyze this data in following
- BU = 55756 km = 43000 km +12756 km (Earth Diameter)
- CU= 102500 km = 2 x 51118 km (Uranus Diameter)
- The angle BCU = 32.967 degrees where 32.967 deg x 0.8 = 26.36 degrees
o 26.6 degrees = the angle controls the moon motion from perigee to
apogee as we have seen in the moon orbital triangle original form (BCD),
but the angle (BCD) = 26.56 degrees (Error 1%)

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- The Idea Summary:
o Uranus motion effects on the moon motion revolving around Earth, this is
one of the paper hypotheses – Earth force imprison the moon inside the
range (Perigee and Apogee distance =43000 km) – so – any effect of Uranus
motion on the moon motion will be acceptable if it be in the range (Perigee
And Apogee Distance).
o Uranus Motion creates The Red Line BC in The Moon Orbital Triangle
o Let's suppose, the line BC receives Uranus Motion effect and provides it to
the moon motion.
Let's imagine how that's doing
o The Line BC moves in angle (32.967 degrees), that means, the line BC
changes the angle (BCU) from Zero degree to (32.967 degrees) and then
return to Zero Again
o It's a cycle, but the line BC moves in its opening for the angle from Zero to
(32.967 degrees) in some way and NOT return through this same way when
the opened angle (32.967 degrees) be closing to be Zero
o That creates a motion of cycle of this line CB (column).
o This motion is A Waving Motion (going and return back but NOT through
the same way).
o By this motion the line BC effects on the moon motion revolving around
the Earth… now the line BC should be considered as a column built on the
moon body or is connected by it – and that means- if this line BC moves (by
angle opening or closing) the moon will move with it or effected by it.
o The angle is (BCU =32.967 degrees), but the moon doesn't reach to this
angle range for 2 reasons (1st
) Because the moon can't move beyond apogee
radius (0.406 mkm) (2nd
) Because of Uranus Orbital Inclination effect.
o (32.967 degrees) x 0.8 = 26.36 degrees
o That means, this angle (32.967 deg) is seen in the moon orbital motion as
(26.36 degrees) because of Uranus orbital inclination effect on this angle.

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o The angle = (BCD) =26.6 degrees
o The angle 26.36 degrees controls the moon motion during a distance 42500
km (i.e. From perigee to Before apogee point with 500 km) (error 1%).
o Error 1% is found frequently in Uranus effect on Earth & moon motions.
Notice
- Uranus effect is seen strongly in the data for example
o CU = 102500 km = 2 Uranus Diameters
o BU = 43000 km (perigee apogee distance) + 12756 km (Earth Diameter)
o AU = 30589 km (error 0.4%) (where 30589 days = Uranus orbital period)
More Data
- (BCU) = (32.967 degrees) x 3 = 98.9 degrees
- Where
o 98.9 degrees = 97.8 degrees (Uranus Axial Tilt) + 1.1 degrees
o 97.8 degrees (Uranus Axial Tilt) = 96.7 degrees + 1.1 degrees
o (96.7 degrees =90 degrees +6.7 deg The Moon Axial Tilt).

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Point No. (B) (The angle 36.912 degrees)
- In This moon orbital triangle I have removed all details and added the line CX
- The angle BCX = 36.912 degrees
- BX = 64600 km
- CX = 107560 km
- The angle ECX = 113.58 degrees
- Note Please
o Cos (36.912 degrees) = 0.8
o Tan (36.912 degrees) = 0.7511
I-Data Analysis
- (97.8 degrees /122.5 degrees) = Cos (36.912 degrees)
o 97.8 deg = Uranus Axial Tilt
o 122.5 deg= Pluto Axial Tilt
o Uranus Orbital Inclination = 0.8 degrees
o Also Cos (36.912 degrees) = 0.8
The data tells that, the angle (36.912 degrees) is used in Uranus & Pluto motions
interaction data

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- (13.177 degrees /17.4 degrees) = Tan (36.912 degrees)
o 13.177 degrees = The Moon Motion Per Solar Day
o 17.4 degrees = The Inner Planets Orbital Inclinations Total
The data tells that, the angle (36.912 degrees) is used for The Moon Daily Motion
- (17.4 degrees/ 23.45 degrees) = Tan (36.6 degrees) (error 1%)
o 23.45 degrees = Earth Axial Tilt
o (36.6 degrees) is different with (36.912 degrees) with 1%
The data tells that, the angle (36.912 degrees) is used for Earth Axial Tilt (Please
remember Earth data has always an error =1% concerning Uranus effect).
- (26.3 degrees/ 32.96 degrees) = Cos (36.912 degrees)
o 23.45 degrees = Earth Axial Tilt
o (36.6 degrees) is different with (36.912 degrees) with 1%
The data tells that, the angle (36.912 degrees) is used for The moon motion angle
(26.6 deg) From Perigee To Apogee.
o The angle = BCX
o 29.53 days = the moon day period ……………. Also
o 29.2/29.53 = 0.99
The data tells, the angle (36.912 degrees) is used for The Moon Day Period (29.53
solar days).

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o The angle (A) =45 degrees
o The triangle base (EA) is declined with 1.1 degrees on the horizontal level,
so the total angle will be 45 deg +1.1 deg = 46.1 degrees ….. So
o The angle (BCX) (36.912 deg.) / the total (46.1 deg.) = Cos (36.912 deg.)
The data tells, the angle (36.912 degrees) is used for The angle (A) in the Moon
Orbital Triangle
Notice (1)
o 29.53 days = the moon day period
o 29.2/29.53 = 0.99
o Earth moves during 29.53 solar days (29.2 degrees) but the moon moves
during the same period (389.2 degrees = 360 deg +29.2 degrees)…
Notice (2)
- 37 x π2
=365.25
- This data shows the massive importance of the angle (36.912 degrees) (BCX).
- Please Note
o Uranus, Pluto and the moon data is controlled by this angle (36.912 deg)
A Conclusion
o It's the same angle (36.912 deg) is used for Uranus, Pluto, the moon and
Earth motions data showing that this angle (36.912 deg) is created inside the
interaction of these planets motions

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The Triangle (ECX) Analysis
- The angle ECX = 113.6 degrees = 90 deg +23.6 degrees
Where
- 23.6 deg (The Outer Planets Orbital Inclinations Total) x 0.99 =23.45 deg (Earth
Axial Tilt)
- 17.4 deg (The Inner Planets Orbital Inclinations Total) x 0.99 =17.2 deg (Pluto
Orbital Inclination).
- The Right Triangle Hypotenuse (CX) = 107560 km = 51118 km +56382 km
- 51118 km = Uranus Diameter
- 56382 km = the distance BU (55756 km) (error 1%)
Note Please
- The Point X divides the distance BA into BT = 3 and TA =1
- means, BX = 43000 km +21500 km
- and , XA =21500 km
- (21500 km = Mars Circumference)
A Comment
- The angle and triangle analysis shows that, Uranus data is used strongly in the
moon orbital triangle, in addition to many other planets, as the distance 449197 km
= Jupiter circumference, or the distance 21500 km = Mars circumference, BUT
- Uranus data is used dominantly along the moon orbital triangle data specially
through the angle (36.912 deg) which should be origin point from which different
data is created and Uranus axial tilt based on which the moon orbital triangle is
created
- The angle ECX =113.6 degrees tells us that, Earth axial tilt (23.45 degrees) is
created based on the moon orbital triangle geometrical structure.

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II-Discussion
- I wish to use this triangle data to explain my idea about how planets data is created
….. let's try to do that in following:
o The angle (36.912 degrees) has 2 distinguish values, its (cos) = 0.8 and its
(tan) = 0.7511, where
o 7511 km = Pluto Circumference
o I consider the planets data is created based on each other – let's deepen this
meaning as possible in following
o Uranus (6.8 km/sec) moves during a period (7511 seconds) a distance =
51118 km (Uranus diameter), the value 7511 km = Pluto Circumference, the
data uses this value as a period of time – and because I try to explain how
the planets data is created– I have the charge to explain how the 7511 km
(Matter Dimension) can be used as 7511 seconds (A period of time) – and
based on the known physics theories, I suppose Pluto rotates around its axis
and by this rotation Pluto moves a distance =7511 km – this distance = Pluto
circumference but it's different from Pluto circumference because Pluto
circumference (7511 km) is a matter dimension but the motion distance
7511 km is a space (a distance)-
o And - We know that, light motion can cause time and distance values to be
equivalent because (x=ct) and when c=1 that will cause t=d
o I explained the data supposing that, A light beam moves in accompanying
with Pluto motion and uses the distance Pluto moves during its rotation as a
period of time and produce the equation (6.8 x 7511 s= 51118 km)
o This explanation still needs to be developed
o The angle (36.912 degrees) tells, Pluto (7511) and Uranus (0.8) data is
created based on the same source, and because of that, the interaction of
Uranus and Pluto motions is created by their origin and not as an event

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occurred later, they are created in interaction – because both data has the
same source which is (36.912 degrees)
o The explanation development should tell us how the matter is created…
- I need to explain this meaning as clear as possible – because it's a cornerstone in
the solar system suggested description – so let's summarize the idea in following:
o The solar system is a theater of puppets, all planets are connected with each
other by the same thread, and each planet data is created in harmony of the
general motion of this thread and that forces this data to be complementary
to each other –
o The double production experiment is a good example to explain this idea,
from Gamma ray, electron and positron are created complementary to each
other and so they are equal in mass and opposite in charges
o This is the meaning of (complementary to each other), without observation I
expect that, Gamma rays will produce positron in addition to the electron –
even if I can't catch this positron by observation, simply because of the
charge law conservation –
o It's the concept of the matter creation – the complementary couple – for that
reason – Pluto circumference =7511 km because Uranus velocity =6.8 km/s,
It's a geometrical mechanism connects Pluto with Uranus regardless our
observation –this connection is created based on geometrical rules which
control planet motion and creation data.
o The basic conclusion is that (The solar system is a network of motions)
We release ourselves from rigid bodies motions – we should see the motions as an
independent objective – the motion is not a feature of rigid body in space – the motion
is done in space regardless any rigid body as the sea waves push the ships but found
without the ships – we see the rigid body moving but without this rigid body the
motion is still found – the motion be potential.

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7- Uranus Motion Analysis
7-2 Uranus Motion During 8 Pluto Days period
7-3 Uranus 144 days Cycle
7-4 The Interaction angle 71.9 degrees (continued)
7-5 The Moon Diameter Creation.
(3rd
Proof)
Uranus Moves During (1440 Of Its Days Period) A Distance = The Earth Moon
Total Displacements During Metonic Cycle (6939.75 Solar Days)
I-Data
- Uranus has a cycle with (144 of its days), Where
- Uranus 144 days = 2476.8 hours
- Pluto 16 days = 2452.8 hours
- The difference = 1 Solar Day
- This cycle we should discuss later in details …. Now we try to know if this cycle
effect on the moon motion….
- Uranus moves during 1440 of its days (1440 x 17.2 h = 24768 hours), during this
period Uranus moves a distance = 606.3 mkm
- The Earth moon moves per a solar day a displacement =88000 km,
- During 6939.75 days (Metonic Cycle), the moon moves a distance = 610.7 mkm
And
- Uranus diameter 51118 km x (1092
) = 607.3 mkm
II- Discussion
- The values (606.3 mkm and 610.7 mkm) are different with around (1%)
- The data tells that, the distance Uranus moves during its cycle (1440 Uranus days)
= the moon displacements total during Metonic Cycle, which shows that both
values are related to each other.

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- For more confirmation, Uranus gives us the value (607.3 mkm) as a result of the
equation (51118 km x 1092
), where we remember this equation because
o Mercury orbital distance (57.9 mkm) = Mercury diameter x 1092
o Earth orbital distance (149.6 mkm) = Earth diameter x 1092
o Saturn orbital distance (1433.5 mkm) = Saturn diameter x 1092
- By this same equation Uranus produces the result 607 mkm = the moon total
displacement during Metonic Cycle = Uranus motion distance during 1440 its days
- The data shows, the moon motion is effected by Uranus Motion, Supporting the
hypothesis, (Metonic Cycle is created by Uranus Effect on the moon motion)
Notice
- During (1440 days of Uranus days period) Uranus moves a distance = 606.3 mkm
- 1440 days x 17.2 hours = 24768 hours = 1032 solar days
- The Moon total displacement during (6939.75 solar days) = 610.7 mkm
- i.e.
- Equal distances (error 1%) are passed in 2 different periods of time
o (6939.75 solar days / 1032 solar days) =6.724
o But
o 6.7 degrees = The Moon Axial Tilt (error 0.3%)
o We may remember that, a deep relationship is found between Uranus axial
tilt and the moon axial tilt (97.8 degrees – 6.7 degree = 91.1 degree).
Shortly
Uranus motion effects on the Earth moon motion and forces the moon to move
Metonic Cycle during 19 years

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7-2 Uranus Motion During 8 Pluto Days Period
- Why do we need to remember Uranus 144 days Cycle? We have 2 reasons
o (1st
) Because Uranus moves during 1440 Uranus days (17.2 h) a distance =
The Moon Total Displacements During Metonic Cycle
o (2nd
) Because Uranus & Pluto Motions interaction can be seen clearly in
studying Uranus 144 days Cycle….
- How Uranus Motion Cycle (During 8 Pluto Days) Is Discovered?
o Jupiter (13.1 km/s) moves during its day period (9.9 h) a distance = Jupiter
circumference (449197 km) + 17695 km
o During 8 of Jupiter days periods (9.9 x 8 =79.2 h), Jupiter moves a distance
= 8 Jupiter circumferences + Jupiter diameter (error 1%)
o Because of Jupiter diameter I concluded that, Jupiter has a cycle in 8 days
Then
o Jupiter motion distance during 8 of its days (79.2 h) which = 3735072 km,
this same distance = Saturn motion distance during 10 of Saturn days period
o Means, Saturn moves during 10 of its days a distance = 3735072 km,
o I have concluded that, Jupiter motion energy is transported to Saturn motion
energy by the rate 80% Because of that I expected that, Saturn should
transport its motion energy to Uranus by the same 80% (but it's incorrect!)
o Saturn transported the motion energy with this rate 80% to Neptune and that
means the distance Saturn passes during 8 of Saturn days period, Neptune
passes during 10 days of Neptune days period (error 5%)
The question is why Saturn doesn't transport the motion energy to Uranus?!
- A surprise was in our waiting…
- Uranus (6.8 km/sec) moves during Pluto day period (153.3 h) a distance = Jupiter
motion distance during 8 of its days period (79.2 h) + 17695 km
- That means, during 8 Pluto days period (153.3 h x 8) Uranus moves a distance =
Jupiter motion distance during 64 Jupiter days (64x 9.9 h) +Jupiter diameter (1%)

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- It's very similar to Jupiter 8 days Cycle ….
- Easily we have concluded that, Uranus motion creates a cycle of Pluto 8 days
where Uranus moves during this period a distance = Jupiter motion distance during
64 Jupiter days
- Because Jupiter motion energy is transported to Saturn and from Saturn to Neptune
the distance should be equal for all these planets – by that – the chance if found to
compare between these motions distances - let's write them in following
I- Data
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
5. Jupiter Circumference (449197 km) x 64 = 28.748639 million km
6. Saturn Circumference (378675 km) x 80 = 30.294001 million km
7. Neptune Circumference x 2 (311193.6 km) x 100 = 31.119360 million km
II-Data Analysis
- Let's remember these cycles in following:
- Jupiter moves during its day period (9.9 h) a distance = 466884 km but Jupiter
circumference = 449197 km, the difference = 17687 km
- During 8 of Jupiter days period, Jupiter moves distance = 3735072 km = 8 Jupiter
circumferences + 1 Jupiter diameter (142984 km =8x 17687 km) (error 1%)
- Based on that I have concluded, Jupiter has a cycle in 8 of its days (79.2 hours)
- By this analysis Jupiter 8 days Cycle is discovered
Then

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- Uranus (6.8 km/s) moves during (1 Pluto day period =153.3 hours) a distance =
3752784 km, this value is very near to 3735072 km (Jupiter motion during 8 days)
- But 3752784 km - 3735072 km = 17687 km
- That means, during 8 days of Pluto days period (153.3 h x 8 = 1226.4 hours), this
difference will be = 142984 km (Jupiter diameter) (17687 km x 8)
- By this analysis also, Pluto 8 days cycle is discovered…. We should note that,
Jupiter uses its day period but Uranus uses Pluto day period, so Pluto 8 days cycle
is found by Uranus motion…
- This analysis tells that, 8 days of Pluto days period is a cycle contains 8 cycles of
Jupiter cycle and each cycle is consisted of 8 Jupiter days period, for that reason I
use Jupiter motion distance during 64 Jupiter days as seen in data No (2)
- There's one more noticeable observation, that, Jupiter motion distance during 8 of
Jupiter days period = Saturn motion distance during 10 of Saturn days period, by
this notice I have concluded that, Jupiter motion energy is transported from Jupiter
to Saturn based on a rate 80%
- So, I have supposed that, this is the way by which the motion is transported from a
planet to another, so the motion energy of Jupiter is transported to Saturn with this
rate 80%, based on this supposition, I have supposed that, Saturn motion energy
will be transported to Neptune with this same rate 80%
- For that reason, the distance during 64 of Jupiter days period should be equal the
distance Saturn passed during 80 of its days period (I provided this in data No. 3)
- And based on that, Saturn motion distance during 80 of its days should be equal
Neptune motion distance during 100 of its days period (provided in Data No. 4)
- I have supposed that, a planet circumference is a player effect on this planet
motion produced cycles, I have provided 64 of Jupiter circumferences in Data No.
5 and also provided 80 Saturn Circumferences in data No.6 and then 200 of
Neptune Circumferences in Data No.7 (Neptune moves during its day period a
distance = 2 Neptune Circumference)

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III-Discussion
(1st
Point)
- Uranus motion distance during 8 of Pluto days period (30.022272 mkm) – Jupiter
motion distance during 64 of Jupiter days period (29.880576 mkm) = 142984 km
(= Jupiter diameter) (error 1%)
o Because of Jupiter diameter value, I have concluded that, Uranus & Jupiter
Creates a cycle by their motions interaction.. Where Uranus uses 8 of Pluto
days periods and Jupiter uses 64 of its days period.
(2nd
Point)
- Jupiter motion distance during 64 of its days period (29.880576 mkm) – the total
of 64 Jupiter Circumferences (28.748639 mkm) = 1.1318 m km
o Jupiter moves per a solar day a distance = 1.1318 m km
o That means, these 2 values express 2 motions interacted together to produce
this value (1.1318 m km) which is considered as a value defined based on
cycle period (a solar day) (this value will be analyzed more deep later)
(3rd
Point)
- Saturn motion distance during 80 of its days (=29.891520 mkm) – Jupiter motion
durance during 64 of its days period (= 29.880576 mkm) = 10921 km
o 10921 km = The Earth Moon Circumference
o The 2 values are considered to be equal, and its almost correct because the
difference between both = 0.036%
o But this very small error = the moon circumference
o Please Note 29891520 km = 10921 x 2737
o But 29880576 km = 10921 x 2736
o The difference between both values shows that, these motions are part of
great cycles and interactions –support the claim (new cycles are discovered)

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(4th
Point)
- Neptune motion distance during 100 of Neptune days periods (=31298400 km) –
Saturn motion distance during 80 of Saturn days periods (=29891520 km) =
1406880 km ………. But
o 1406880 km = The Sun Diameter 139200 km (Error 1%)
o The Sun Diameter 139200 km = 49528 km Neptune diameter x 28.1
o Neptune Axial Tilt =28.3 degrees
(5th
Point)
- Neptune motion distance during 100 of Neptune days periods (=31.298400 mkm)
– Uranus motion distance during 8 of Pluto days periods (=30.022272 mkm) =
1.276128 mkm = π x 406000 km
o Earth Moon Distance at apogee radius = 406000 km, where this is the most
far point the moon can reach from Earth.

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7-3 Uranus 144 days Cycle
- 144 days of Uranus days periods = 144 x 17.2 hours = 2476.8 hours
- 16 days of Pluto days period = 16 x 153.3 hours = 2452.8 hours
- The Difference = 24 hours = 1 Solar Day
- The data shows these are 3 values (144 Uranus days, 16 Pluto days and the solar
day) they are 3 cycles
- We have seen that before
o 6939.75 solar days = Metonic Cycle
o 6585.36 solar days = Saros Cycle
o 354.39 solar days = The lunar year
And
- The data shows that, there's an interaction of Uranus, Pluto and Earth motions
- This data we have discussed before and reach to the following conclusions
o Uranus motion effect on Pluto motion causes Pluto day period to be =153.3
hours where it's so long day period in comparison with the other outer
planets
o This effect is found because Pluto moves during (6939.75 x 153.3 hours) a
distance = Uranus orbital circumference
o This data also we have discussed before

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Uranus & Pluto Motions Interaction
I- Data
(a)
4.7 km / sec x 10921 seconds = 51118 km (Uranus diameter) (error 0.4%)
(b)
4.7 km / sec x 51118 seconds = 2 x 120536 (Saturn Diameter) (error 0.4%)
(c)
4.7 km / sec x 2 x 120536 (Saturn Diameter) =1.1318 mkm (Jupiter daily velocity)
II- Discussion
- The data shows that, the results are created as different values in the cycles of 8
Pluto days cycle and the other planets equal distances
- That tells, these cycles are created by an effect of Pluto motion during different
periods of time
- That shows an effect of Pluto motion more extending than our expectation for its
effect on the solar system motion…
More Data shows Uranus and Pluto Motions Interactions
Equation No. (d)
90560 days = 13.177 x 0.99 x 6939.75 days
- 90560 solar days = Pluto Orbital Period
- Equation (d) shows, Pluto orbital period depends on Metonic Cycle (6939.75 days)
- and on the value 13.177, where the moon moves per solar day 13.177 degrees
Equation No. (e)
Pluto during 6939.75 days moves a distance = 2815 mkm
- The data tells that, Pluto uses also Metonic Cycle (6939.75 solar days) and moves
during this period a distance = Mercury Uranus Distance
Equation No. (f)
21.8 x 0.8 degrees (Uranus orbital inclination) = 17.4 degrees
21.8 = Jupiter Mass / Uranus Mass

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7-4 The Interaction Angle 71.9 Degrees
We complete the discussion in (6-3-2) (Metonic Cycle Discussion).
I- Data
1- The Moon Orbital Circumference at apogee radius = 2550973 km (100%)
2- Earth Daily Motion Distance = 2573483 km (101%)
3- Pluto moves during 153.3 hours =2593836 (102%)
4- The displacements 88000 km total during (29.53 days) = 2598693 km (102%)
5- Uranus motion distance (during 378675 seconds) = 2574990 km (101%)
5-1 = +24017 km
5-2 = +1507 km
5-3 = - 18846 km
5-4 = - 23703 km
4-1 = 42863 km
4-2 = 25210 km
4-3 = 5867 km
3-2 = 20353 km
3-1 = 41853 km
2-1 = 22510 km
II- Data Analysis
- This data is analyzed in point (6-3-2), we use this data here again for a new point.
- We will add Uranus motion distance differences together as following:
- The total will be = (5-1) + (5-2) +(5-3) +(5-4) = -17025 km
- What does this value (-17025 km) means?!
- It's Uranus effect on the 3 Planets (Earth, its moon and Pluto), Why?
- Because these 3 planets move during their cycles periods a distances = Uranus
orbital circumference as we have discussed in that point (6-3-2), because their
(Metonic) Cycles depends on their motions distances which equal Uranus Orbital

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Circumference, that shows Uranus effect on these 3 planets, where their motions
are interacted because of that.
- For that reason, the distance (-17025 km) should show Uranus effect on these 3
planets!!
- What's This Distance (-17025 km)?!
- 17025 km x 2 = 34309 = π x 10921 km (the moon Circumference). (Error 0.7)
More Data
(a)
5040 sec. x 6.8 km /s (Uranus velocity) = 34309 km (= 10921 km x π)
(b)
34309 sec x 13.1 km/s (Jupiter velocity) = 449197 km = Jupiter Circumference
(c)
34309 sec x 4.7 km/s (Pluto velocity) = 160592 km = Uranus Circumference
(d)
34309 sec x 27.78 km/s (the moon velocity) = 943817 km
(943817 km = the perimeter of the moon orbital triangle ACE) (error 1%)
(e)
10921 sec x 35 km/s (Venus velocity) = 120536 km (Saturn diameter) (error 1%)
(f)
10921 sec x 29.8 km/s (Earth velocity)=321183 km (2 Uranus Circumferences) (1.3%)
(g)
1153 sec x 29.8 km/s (Earth velocity)= 34309 km = 5040 sec. x 6.8 km /s
III- Discussion
- The data shows clearly that the value (34309) is used widely in the solar planets
motions data, showing that there's a deep effect practiced from this interaction on
almost all solar planet… but why the used value is a double value?
- We have found that before, the angle 71.9 degrees is used by its double value!!

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- Let's remember that in following….
(a)
2 x 71.9 degrees = 12.195 degrees x 11.8 degrees
Where
12.195 degrees = The moon motion angle (13.177 deg – 0.98562 deg)
11.8 degrees =6.7 degrees (moon axial tilt) + 5.1 deg (moon orbital inclination)
Why does the data use double values (2 x 71.9 deg)??
(b)
(122.5 deg -71.9 deg) x 2 = 101.2 degrees
Where
122.5 degrees = Pluto Axial Tilt
(c)
63.7 degrees = (71.9 deg – 8.9 deg) + (71.9 deg – 8 x 8.9 deg)
It's usual using of the double value, Why?! because
(Old Data)
(1)
(2)
(3)
(4)
Data Analysis
The period (6939.75) x (29.53 solar days) of the moon = 204931 days = 561.07 years
But
30589 days (Uranus Orbital period) = 27.3 days (the moon orbital days) x 1120
561.07 years x 2 = 1123

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- 2 moon Cycles are required to produce the rate (1123) which is very near to the
rate between both cycles (1120)… these (1123) and (1120) aren't 2 random
number found by chance, these are 2 geometrical values are produced by one
machine depends on the planets motions as motions of gears, for that reason,
double Cycle is required, and the moon data will show its cycle (561 years) but
many other planets motions will deal with the double cycle (1123 years), this also
we have seen in the angle 12.195 degrees analysis where the triangle hypotenuse
was = 2 x 29.53 km, and the value 29.53 solar days is the moon day period, but the
value 59 days is used more widely, for that, the Earth moves during 59 days a
distance = its orbital distance.
(The angle 12.195 analysis is in the moon orbit geometrical design)

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7-5 The Moon Diameter Creation.
I-Data
- In this figure I cut layers from the moon diameter to create
smaller moon diameters
- he Blue Circle its diameter R1= 695 km and r1 =347.5 km
- The Red Circle its diameter R2= 1390 km and r2 =695 km
- The Black Circle its diameter R3= 2085 km and r3 =1042.5 km
- The Brown Circle its diameter R4= 2780 km and r4 =1390 km
- The Orange Circle its diameter R5= 3208 km r5 = 1604 km = (5040/π)
- The moon diameter R6= 3475 km r6 = 1737.5 km
II- Data Analysis
- Let's suppose that, (the moon diameter is created as a difference between 88000
km and 86400 km = 1600 km)
- Why this hypothesis is logical? Because the moon orbit apogee circumference
(2.55 mkm) is sufficient for a daily displacement =86400 km, as we discussed
before, 86400 km x 29.53 days = 2.55 mkm
- But the actual moon displacement =88000 km
- The difference =1600 km
- Where 5040 km = π x 1604 km
- Mercury day period needs 5040 seconds to be =176 solar days,
- To use 1 km equivalent to 1 second as a using found frequently…
- (for example, The moon day period =29.53 solar days =2.55 million seconds and
the moon orbital apogee circumference =2.55 mkm "Zero error")
- The data provides an interaction between (distance & time period) values showing
that, their interaction (may) cause the matter creation and define its dimensions,

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based on that I suppose the moon diameter may be created based on this difference
between the 2 values (1600 = 88000 km – 86400 km)
- Where the value 86400 km is (in fact) a solar day (86400 seconds) which is used
here as a distance for a geometrical necessity, because the moon apogee orbital
Circumference = 86400 km x 29.53 days, i.e. the value 86400 is created depends
on the moon day period and the moon apogee orbital circumference, because the
facts tell that, the moon can't move beyond apogee radius (r=406000 km) so we
have to suppose that the value 86400 is a distance 86400 km.
- Now we have 2 points to analysis in this investigation…(1st
point) to analyze how
the period of time can be used as a distance (2nd
) to define the reason for which the
moon diameter is cut in these 5 parts (5 smaller diameters), let's do that in
following…

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(1st
Point) Time and Distance Values Equivalence.
I-Data
(1)
116.75 solar days (Venus day period) x π = 366.7 Solar Days
(Earth orbital period =366.7 Earth rotation)
(2)
1461 solar days x 0.08 = 116.75 solar days (1461 days =365+365+365+366)
(3)
29.2 deg x 4 = 116.75 degrees
(4)
2.55 mkm = 116.75 x 2 x 10921 km
And
2.55 mkm = 116750 km x 21.866
(5)
Mercury moves during its rotation period (58.66 days) a distance = 243 mkm
II-Discussion
- The data is a clear, Equations no. (1 and 2) tries to create a proportionality
between Earth orbital period and Venus Day period, this proportionality depends
(almost) on the rate 0.08 which is an effective rate and discussed deeply with the
moon orbital triangle design (Point .4) and based on this rate the moon angle
(10.96 deg) is created
- That tells, there's a relationship between Earth orbital period and Venus Day
Period and this relationship depends on the moon orbital motion between the
orbital period 365.25 days is a cycle for Earth and for the moon also
- Equation no. (3), we know that, Earth moves during (29.53 days) a value (29.2
degrees) and the moon during the same period move (360 deg + 29.2 degrees)
- Equation no. (3) tells that, Venus day period (116.75 deg) depends on this
(common) value (29.2 deg) by the rate (4), where Uranus diameter =4 Earth
diameter.. Equation no. (3) refers to Uranus effect to produce this result, the
equation no. (4) tells how that's happening, let's seen it here

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Equation no. (4)
2.55 mkm = 116750 km x 21.866 where, 21.86 = Jupiter Mass / Uranus Mass,
- Equation no. (4) tells that, Jupiter Uranus masses rate effects on the moon apogee
orbital circumference (2.55 mkm) to produce the value 116750 km
- This is the secret value behind Venus day period (116.75 days)
- Because
- Based on this value (IF) 1000 km = 1 day
o 116750 km be = 116.75 days (Venus day Period) ….. also
o 88000 km be = 88 days (Mercury orbital Period)
o 176000 km be = 176 days (Mercury day Period 175.94 days)
- The distance value is used as a period of time simply by the same rate…
Equation no. (5)
Mercury moves during its rotation period (58.66 days) a distance = 243 mkm
- Venus rotation period =243 days
- Again the distance value is used as a period of time by a different rate ..
- No invented idea we provide here, it's the data which suggest this same idea,
almost there's a mechanism to use the periods of time as a distance values and vice
versa.
Even Venus day period is created based on the Venus diameter, the data shows that
- 115 x (180/177.4) = 116.75
o 115 = (The Sun Diameter / Venus Diameter)
o 177.4 degrees = Venus Axial Tilt
o 116.75 solar days = Venus Day Period.

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(2nd
Point) The Benefits Of The Moon Diameter Division
- Let's review the triangle concept in following:
- Uranus Axial Tilt =97.8 degrees
- The moon axial tilt 6.7 degrees +90 = 96.7 degrees
- The difference =1.1 degrees
- The moon angular diameter = 0.5 degrees that means, the angle 1.1 degrees is
divided into 2 parts (0.5 degrees inside the moon diameter) + 0.6 degrees
- Let's study these 2 angles individually
The Angle 0.6 Degrees
- I claim The angle 0.6 degrees is used to create Mars orbital inclination (1.9 deg) by
interaction done by Jupiter and Saturn… the following data shows that clearly
Jupiter And Saturn Interaction
- 1.3 deg (Jupiter orbital inclination) +0.6 deg = 1.9 deg (Mars orbital inclination)
- 1.9 deg (Mars orbital inclination) + 0.6 deg = 2.5 deg (Saturn orbital inclination)
- The data shows that, Mars orbital inclination is created based on this 0.6 degrees
which is found by an interaction between Jupiter and Saturn
- Where Mars orbital inclination is created based the interaction between Uranus &
the moon axial tilts that explain the similarly of Mars and the moon motions data,
as following…
o 1.9 deg (Mars orbital inclination) x 13.177 deg = 25.2 deg (Mars Axial Tilt)
(13.177 degrees = The moon daily motion degrees)
o (13.177 deg / 0.524 deg) =25.2 deg (Mars Axial Tilt)
(0.524 degrees = Mars Motion Daily Degrees)
o 687 days (Mars orbital period) =27.3 days (the moon orbital period) x 25.2
o The moon day period (708.7 h) = Mars day period (24.7 h) x 2π
- The Moon And Mars Motions Data almost rated to each other completely.

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The Angle 0.5 Degrees
- This angle is consumed inside the moon diameter (91.1 deg -0.5 deg =90.8 deg)
- Uranus orbital inclination =90.8 degrees, so the difference =0.2 degrees
- I have cut the moon diameter into smaller diameters to find this angle 90.8 degrees
- Let's do in following…
- The Blue Circle its diameter R1= 695 km and r1 =347.5 km
- The Red Circle its diameter R2= 1390 km and r2 =695 km
- The Black Circle its diameter R3= 2085 km and r3 =1042.5 km
- The Brown Circle its diameter R4= 2780 km and r4 =1390 km
- The Orange Circle its diameter R5= 3208 km r5 = 1604 km = (5040/π)
- The moon diameter R6= 3475 km r6 = 1737.5 km
- Based on this description, the angle above the black circle inside the moon
diameter (R3 = 2085 km) above this circle the angle = 90.8 degrees
- So to find this angle we have to cut from the moon diameter 2 layers (695 x 2)
means we move a distance 1390 km from the moon surface into its center and in
this point the angle will be =90.8 degrees.
- That shows the reason of using the rate (0.8) frequently in the moon motion data,
specially the rate (0.08) which is the origin based on which the valuable angle
(10.96 degrees) is created.
- Also that shows Uranus orbital inclination effect on the inner planets (specially on
Venus and Mercury) as proved by different data for example:
o 180.8 deg= 177.4 deg (Venus axial tilt) + 3.4 deg (Venus orbital inclination)

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o 180.8 deg = 180 deg + 0.8 (Uranus orbital inclination)
Also
o 97.8 (Uranus Axial Tilt) = 90 deg + 7.8 deg
o 7.8 deg = 7 deg ( Mercury orbital inclination) + 0.8 (Uranus orbital inclination)
In following we have to discuss the basic geometrical feature of Uranus angle 90.8
degrees on the moon orbital triangle,

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Uranus Angle 90.8 Degrees Geometrical Effect
Data
- 29 deg + 90 deg = 28.2 deg + 90.8 deg
Where
- 90.8 degrees is our investigation angle (Uranus Angle)
- 28.3 degrees = Neptune Axial Tilt
- 29 degrees, we know this angle…
o 28.5 deg +0.5 deg (the moon angular diameter)
o 28.5 The moon declination during the major stand still (+28.5 and -28.5)
- The data shows that, Neptune Axial Tilt, is created by an effect on Uranus angle
(90.8 degrees) (this angle 90.8 is called Uranus angle because = 90 +0.8 deg)
- The data tells that, some interaction contains Uranus and Neptune together is
found in the moon orbit ….
- This data is supported by another one which is
- 1.44 deg = 0.8 deg x 1.8 deg
o 1.44 deg = the moon orbit regression angle per month
o 0.8 deg = Uranus Orbital Inclination
o 1.8 deg =Neptune Orbital Inclination
Notice
- I provide different data to support the claim, because, I try to show that, the data is
created by a geometrical reason, and even if we can't catch all geometrical
mechanism details but the data shows this mechanism from different sides, so I try
to disprove the claim of pure coincidence of numbers, instead, I try to prove that
there's a geometrical mechanism behind, as we have seen in Pythagoras triangle
discussion, the data (3730002
= 860002
+3630002
) could be considered as
coincidence of numbers, but the deep analysis shows that it's created by using of
Pythagoras rule in the moon orbital motion…similar to that, Uranus and Neptune
interaction effect greatly on the moon orbital motion.

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Appendix No.1
Is There Lorentz Length Contraction Effect In The Solar System?
i.e.
(Are There Relativistic Effects In The Solar System?)
Lorentz Length Contraction Effect is the near possible answer to explain the planets
data, in following I provide one example of such planets data to prove that, this
conclusion is the most near one to explain it.
I- Data (A)
Why These Distances Are Equal?
(1)
Saturn Orbital Distance = Saturn Uranus Distance
= Mars Orbital Circumference
= Pluto Neptune Distance
= Pluto eccentricity Distance
= Neptune Orbital Distance/π
= Uranus Orbital Distance /2
= Mercury Jupiter Distance x 2
(2)
Mercury Neptune Distance = Saturn Pluto Distance
Jupiter Pluto Distance = Uranus Neptune Circumference
Earth Neptune Distance = Mercury Saturn Circumference (0.5%)
(3)
Jupiter Mercury Distance = 2 Mercury Orbital Circumference
Jupiter Venus Distance = Venus Orbital Circumference (1.5%)
Jupiter Earth Distance = Earth Orbital Circumference (1.2%)
(Earth and Jupiter at 2 different sides from the sun)
(4)
Jupiter Mercury Distance = Mars Orbital Distance x π (0.6%)
Jupiter Uranus Distance = Venus Jupiter Circumference (0.8%)
Pluto Orbital Distance = Earth Orbital Circumference x 2π
II- Discussion (A)
The previous distances form around 50% of all distances found in the solar system
(All orbital and internal distances)… Why These Distances Are Equal One Other?
We may notice that – the distances equality can be produced more easily by light
motion than the rigid body motion - for example – when we push a ball toward a wall
the ball after collision with the wall will return a distance (NOT) equal the original
one - because the collision causes to decrease the ball motion momentum – but the
light can be reflected at equal distances easily – means – equal distances can be
produced by light motion more easy than the Rigid Body Motion.

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I-Data (B)
Why These Distances Are NOT Equal?
1. 0725
.
1
mkm
2.41
nce
Circumfere
Orbital
Moon
mkm
2.58
Motion
Daily
Earth
=
2. 1.0725
km)
(378500
radius
Eclipse
Solar
Total
km)
(406000
radius
orbital
Apogee
=
3. 0725
.
1
distance
Mercury
Jupiter
mkm
720.3
Distance
Orbital
Juppiter
mkm
6
.
778
= (Error 0.7%)
4. 1.0725
Distance
Venus
Jupiter
mkm
670
distance
Mercury
Jupiter
mkm
720.3
=
5. 1.0725
Distance
Earth
Jupiter
mkm
629
Distance
Venus
Jupiter
mkm
670
= (0.6%)
6. 1.0725
mkm)
(1325.3
Distance
Venus
Sarurn
mkm)
(1433.5
Distance
Orbital
Saturn
= (0.8%)
7. 1.0725
mkm)
(1205.6
Distance
Mars
Sarurn
mkm)
(1284
Distance
Earth
Saturn
= (0.7%)
8. 1.0725
mkm)
(2644
Distance
Mars
Uranus
mkm)
(2872.5
Distance
Orbital
Uranus
= (0.7%)
9. 1.0725
mkm)
(4495.1
Distance
Orbital
Neptune
mkm)
(4894
nce
Circumfere
Orbital
Jupiter
= (1.5 %)
(10)
I-Discussion (B)
The same rate (1.0725) is used for all equations (around 18 distances = 40% of all
solar system distances) – why?
Suppose the equal distances are produced by light reflection and that cause these
distances to be equal – as I have supposed in the previous point (A).
Now suppose– part of these equal distances – is passed through another frame relative
to us – so this part of distances will suffer from Lorentz Length Contraction Effect
which is seen in the rate 1.0725
(Another frame can be found in the solar system because we deal with light motion) –
This explanation can answer why some distances are equal and others are rated with
the same rate (1.0725) – it's simply a feature of light motion.
0725
.
1
T.
Axail
Earth
23.4
T.
Axail
Mars
25.2
T.
Axail
Mars
25.2
T.
Axail
Satrun
26.7
Tilt
Axail
Satrun
26.7
Tilt
Axail
Neptune
28.3
=
=
=

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References
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-trajectory-analysis-ii
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944p=15209
Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics Mathematics Faculty
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Facebook https://www.facebook.com
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJhl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

The Moon Orbital Motion Description (Revised)

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