The Moon Orbital Motion Description (2nd 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 (2nd
Revised)
The Author Authorized To Be Used By
Mr. Gerges Francis Tawdrous
A Student–Physics Department- Physics
& Mathematics Faculty –
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Dr. Budochkina, Svetlana Aleksandrovna
Associate Professor (Mathematical Analysis
and Theory of Functions Department)
Peoples' Friendship University of Russia
(RUDN University) – Moscow – Russia
Phone +201022532292
E-Mail: mrwaheid@gmail.com
Curriculum Vitae http://vixra.org/abs/1902.0044
Phone +7 (495) 952-35-83
E-Mail: budochkina-sa@rudn.ru, sbudotchkina@yandex.ru
Website
http://web-local.rudn.ru/web-local/prep/rj/index.php?id=2944&p=19024
The Assumption Of S. Virgin Mary -Written in Cairo –Egypt –29th
January 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
19
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?
35
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
56
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)
61
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
82
8- Appendix No.1 103

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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
- In following we discuss how this triangle is created

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The Moon Orbital Triangle Building
(1st
Point) The Earth Position (Point E)
- The Point (T) refers to The Earth Center
- The Point (M) refers to The Moon Center
- The Points (T, Q and Y) are on the Ecliptic Line
- The Red Line (TM) is the moon orbit plane with an inclination 5.1 degrees on the
ecliptic.
- The Green Line (BE) is the moon triangle base, I choose the distance BE to be =
363000 km and accordingly I have to define the point (E) position.
- The line BC is a perpendicular on the triangle base (BE), its length =86000 km
- The line EC is the triangle dimension to connect the points (E and C).
- The points (Q and P) are the intersection points of CE with the ecliptic and the
moon orbit plane respectively.
- The line TX is a perpendicular from the Earth Center on the base BE
- The point K is the intersection point between the moon orbital triangle base (BE)
(the Green Line) and the moon orbit plane (the Red Line)
- The angle is Zero between the points ( A, B , K , X and E).

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Notice
- The Line MB is Perpendicular on the Moon Orbital Triangle Base (BE)
- And
- MB has an angle =90.443 degrees with the Ecliptic Line……….because
- The moon orbital triangle base is under the ecliptic with an angle =0.443 degrees
- Let's define the Earth Point in following:
o The Triangle TMY
o The distance TM = 363000 km (The perigee radius)
o The distance TY = 361313 km
o The distance MY = 32269 km
o The angle MTY = 5.1 degrees (the moon orbital inclination)
o The Triangle TXK
o The distance KX = 35640 km
o The distance KT =35807 km
o The distance TX = 3460 km
o The angle XKT = 5.543 degrees
o The Triangle BMK
o The distance KM = 327193 km
o The distance KB =325663 km
o The distance MB = 31604 km The distance YB = 665 km
o The angle KMB = 84.9 degrees
o The distance XB = KX + KB = 35640 +325663 = 361303 km …But
o The triangle base (BE) I choose to be =363000 km, based on that the
distance XE = 1697 km

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o The Triangle TXE
o The distance TX = 3460 km
o The distance XE =1697 km
o The distance TE =3854 km
o The angle XET = 26.12 degrees But
o The triangle angle CEB = 13.328 degrees, because of that the triangle
dimension CE connects the point (E) under the hypotenuse ET.
o The Triangle PET
o The distance PE = 6244 km
o The distance PT =2640 km The angle ETP = 148.4 degrees
o The Triangle KPE
o The distance KE =37337 km
o The distance KP = 26554 km
o The distance PE =11152 km
o The angle EPK = 161.12 degrees
Data Analysis
(1)
o The Triangle TXE
o The distance TX = 3460 km The distance XE =1697 km
o The moon diameter =3475 km and the moon radius =1737.5 km, both are
equal the triangle 2 dimensions (error around 2%). That shows geometrical
interaction in this distances definition.
(2)
o The line MY is perpendicular on the moon orbital triangle base (BE) and has
an angle 90.443 degrees with the Ecliptic Line. And
o There's an angle = 0.443 degrees between the ecliptic line and the moon
orbital triangle base (EB).

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(2nd
Point) The Moon Motion (From Perigee To Apogee)
- The moon moves on its orbit planet (MT) with an inclination 5.1 degrees on the
ecliptic Line.
- The moon moves from perigee point (r=363000 km) to the apogee point
(r=406000 km)
- The Point M1 refers to the perigee Point, and M2 refers to the apogee point.
- The distance M1 M2 = 43000 km (=The Perigee Apogee Distance)
- The distance between perigee and apogee =43000 km.
- M1B and M2D are perpendicular on the moon orbital triangle base (EA) (the
Green Line) …… Based on that
- The distance BD is parallel to M1R, where this distance (M1R) is the adjacent in
the triangle M1M2N in which the distance M1M2 expresses the perigee apogee
distance, by that, the distance BD expresses the moon motion from perigee to
apogee and vice versa.
- The basic reference in this triangle is that, the 2 distances M1B and M2D are
perpendicular on the moon orbital triangle base (AB) and Not on the Ecliptic Line,
where the moon orbital triangle base (AB) has an angle 0.443 degrees with
ecliptic, i.e., the triangle base (AB) passes under the ecliptic with 0.443 degrees.

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- The blue line is the moon equator line, where an angle =1.543 degrees is found
between this line and the ecliptic line.
- Let's consider this triangle data in following:
- I- Data
o The angle M1 T Y = 5.1 degrees (the moon orbital inclination)
o The angle M2 M1 R = 5.543 degrees
o The angle M1 M2 R = 84.457 degrees
o The angle B M1 R = 90 degrees
o The angle T M1 B = 84.457 degrees
o The angle T Y B = 90.443 degrees
o The angle B N M2 = 88.9 degrees
o The angle M1 N V = 91.1 degrees
o The distance M1 Y = 32269 km But
o The distance B Y = 665 km
o The distance M1 B = 31604 km
o The distance M2 D = 36092 km But
o The distance D d = 333 km
o The distance M2 D = 35760 km
o In the Triangle RM1M2
o The distance RM2 = 35760-31604 = 4156 km
o The distance M1M2 = 43000 km
o The distance M1 R = 42800 km = The distance BD

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(3rd
Point) The Point (A)
- The blue line is the moon equator line and it has an angle =1.543 degrees with the
Ecliptic Line
- The Point (A) is a point on the Ecliptic Line I have choose and caused an angle
=0.443 degrees under the ecliptic line and created the Base AB =86000 km.
- That means, the triangle base (AB) depends on the ecliptic line only
- But
- Because of the angle 1.543 degrees between the ecliptic & the moon equator lines,
the triangle base (AB) gets an angle =1.1 degrees with the moon equator line (blue
line)
- i.e. above the triangle base (AB) there's an angle with the Ecliptic = 0.443 deg and
under the triangle base (AB) there's an angle with the moon equator =1.1 deg.
- The triangle ABC is a closed triangle where the point (A) is the intersection point
between the ecliptic line, the triangle base AB and the triangle dimension AC

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- But
- The moon equator line (the blue line) doesn't interest neither with the ecliptic nor
the moon orbital triangle AB on the point (A),
- The moon equator line (the blue line) will interest the ecliptic line beyond the
point (A) with a long distance
- Let's define this intersection point position in following:
o The moon orbit plane declines on the Ecliptic line with 5.1 degrees, means,
far distance be found between the Earth and moon will cause longer
perpendicular distance between the moon center and the ecliptic line
o For that, we use the moon distance on a apogee because it's the most far
point the moon can reach from Earth
o ON APOGEE …
o Earth moon distance on apogee point = 406000 km
o The perpendicular distance from the moon center to the ecliptic line = 36091
km, because of the moon orbital inclination (5.1 degrees)
o But
o The angle between the ecliptic line and the moon equator line =1.543 deg
o So these 2 lines will be intersected each other at a distance =1340318 km
o i.e.
o The ecliptic line will intersect with the moon equator line after the apogee
point with a distance =1340318 km
o but the distance from perigee to apogee =43000 km
o i.e. The ecliptic line will intersect with the moon equator line after the
perigee point with a distance =1383318 km
o Notice, the lunar eclipse umbra length =1392000 km (error 0.6%)
The Useful Result :
The triangle base (AE) has an angle = 1.1 degrees with the moon equator line.

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

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- This figure of 2 circles I have brought from internet to use in the Explanation -
- We have supposed, the inner circle is the Perigee orbit and the outer circle is the
apogee orbit, And we have calculated the tangent DB = 181843 km
- 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 Pythagorean triangle (1, 2 and 51/2
)
- The first question should be (what's the geometrical necessity for which the
specific Pythagorean 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 distance BD refers to the moon motion distance from perigee to apogee
o The triangle (BCD) is similar to the triangle (ODB) and both are specific
type of Pythagorean 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 hypotenuse CZ = 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 Pythagorean
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 Pythagorean Triangle (1, 2, 51/2
)
4-3 The moon motion angle (12.195 deg) Analysis
4-4 Why The Moon Displacement Daily =88000 km?
4-5 The angle 71.9 degrees
4-6 Why The Moon Day Period =29.53 days?
4-7 The Perpendicular Line BC (=86000 km)
4-8 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) = 449197 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 449197 km = Jupiter Circumference
o The distance BA = 86000 km = BC
o The triangle BCD is a Pythagorean 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 Pythagorean Triangle (1, 2, 51/2
)
(1st
Point) The Moon Motion Limits Definition
- In this moon orbital triangle I have added the line CS to create a total angle =137
degrees – based on that
(A)
- The angle ECS =137 degrees
- The distance BS = 150628 km
- The distance SA = 64628 km
- The hypotenuse CS = 173450 km
- The perimeter of the triangle BCS = 173450 +150628 +86000 = 410080 km
- The triangle perimeter (BCS) =410080 km= the apogee radius (406000 km)
(error 1%)
(B)
- The perimeter of the triangle (ACS) = 121622 + 173450 +64628 = 359700 km
- Perigee radius = 363000 km (error 1%)
A Conclusion
- The triangle BCS defines the moon motion limits from perigee to apogee by a
geometrical mechanism depends on The angle 137 degrees……. Why & How?

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(2nd
Point) The Rate 0.08
Why Pythagorean Triangle (1,2, 51/2
) Is Required?
This figure is discussed before.
- The inner circle refers to the perigee orbit
- The outer circle refers to the apogee orbit
- OB = 406000 km = Apogee Radius
- OR = 363000 km = Perigee Radius
- DB = 181843 km
- Perigee Orbital Circumference = 2.28 mkm
- Apogee Orbital Circumference = 2.55 mkm
I - Data
(1)
(DB / Perigee Orbital Circumference) = (181843 km/2.28 mkm) = 0.08
(2)
10.96 = 137 (The basic Angle) x 0.08
(3)
Sin (10.96 degrees) x 406000 km = 77237 km
(4)
Cos (10.96 degrees) 88000 km = 86400 km
(5)
Sin (10.96 degrees) 449197 km = 85403 km
II – Discussion
- Why is the Pythagorean triangle (1,2,51/2
) required for the moon orbital motion?
- Because, the rate (0.08) is required to create interaction with the angle (137 deg),
and based on this interaction, the valuable angle (10.96 degrees) will be created,
and based on this angle (10.96 degrees) most of the moon orbital motion data will
be created.

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- That answers the question why the rates (1,2,51/2
) were required necessary for the
moon orbital motion? because based on these rates the rate (0.08) will be produced
which will be used to produce the angle (10.96 degrees)…… So
- Based on the angle (CSB =137 degrees), the moon orbital motion receives 3 basic
data which are
o The apogee point radius (r=0.406 mkm) which is defined by the triangle
BCS) Perimeter
o The Perigee point radius (r=0.363 mkm) which is defined by the triangle
ACS) Perimeter
o And the rate (0.08) which is defined between the tangent DB (181843 km)
and the perigee orbital circumference (2.28 mkm)…….. then
o 10.96 = 137 x 0.08
o The valuable angle (10.96 degrees) is created.
Equation No. (3)
- This equation tells the story in more clear way….
- The value 77237 km is very important…. If the moon moves daily a displacement
= 77237 km, during 29.53 days, the total distance will be = 2.28 mkm = the moon
orbital circumference at perigee orbit (r= 363000 km)
- Means,
- The perigee orbital circumference = 29.53 displacements each =77237 km, that
tells the value (77237 km) is defined by perigee radius (r=0.363 mkm) and the
moon day period (29.53 solar days), whatsoever the moon apogee radius be ….
Now the angle (10.96 deg) is defined before (10.96 = 137 x 0.08), and by that the
apogee radius is defined….
- This explanation is not so correct because the apogee radius is defined before by
the triangle (BCS) Perimeter and (the rate 0.08) is defined based on it because we
use it in the circles figure.

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- 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 A Directed Data.
Equation No. (4)
Cos (10.96 degrees) 88000 km = 86400 km
- The analysis is still complex and we need to consider it deeply in following…..
- 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 Pythagorean triangle in
its motion,
o But
o Why the moon orbital circumference at apogee is not = 2.598 mkm? Why
the moon orbital circumference at apogee =2.55 mkm and les with (1%)
than the total displacements during 29.53 days?
- Equation No. (4) tells us this story clearly, where the apogee orbit permits for a
moon daily displacement =86400 km and NOT 88000 km
Notice
- This is a theoretical analysis and not a practical one, the moon could use 88000 km
as its displacement without using Pythagorean triangle technique for any days
during the month BUT with a condition that, the total distance isn't grater than
2.55 million km (= the moon apogee orbital circumference).

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- 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
More Data
(A)
The moon orbital circumference at apogee point = 2.55 mkm (100 %)
The Earth 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
- This discussion should be completed with answering of the question (Why the
moon daily displacement =88000 km? (its discussion is in "point No. 4-4")

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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….
- Also, the triangle (ACS) Perimeter =359700 km = 360000 km
- If 1 degree =1000 km, so this value 360000 km will be equivalent to 360 degrees.
- 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) 449197 km = 85403 km
- Equation no. (5) tries to help the explanation,
o The distance 85403 km is very near to the line BC =86000 km (error 0.7%)
o Also the distance 86000 km = 2 x 43000 km ( Perigee apogee distance)
o But
o The distance 449197 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
449197 km to produce the line BC 86000 km?! The data tells that the
distance (449197 km) is a real one and isn't invented …. Also the line BC
(86000 km) is real data.
o That means, the moon orbital triangle is discovered and not invented.
o And the data which is concluded by its as real as the moon registered data
by observation.

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4-3 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
NOTICE (1)
Uranus effect on the moon orbital motion will be discussed in the next point (no. 4-4,
why the moon daily displacement =88000 km?)

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NOTICE (2)
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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4-4 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 (Old Data)
The moon orbital circumference at apogee point = 2.55 mkm (100 %)
The Earth 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%)
II - Data (New 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
(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
III - 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

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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)
VI –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. (this discussion should be completed with Metonic Cycle
Discussion Point No. 6)

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4-5 The angle 71.9 degrees
- 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
The angle 71.9 degrees is an angle created by the interaction between the 4 planets
Motions (Earth, its moon, Pluto and Uranus).
(Why we need to discuss this angle 71.9 degrees?)
Because this angle can answer why the moon orbital motion equation uses the
constant 1.7 degrees for the moon daily motion (θ1= θ0 +1.7 degrees).
The Figure Description
- In this moon triangle, I added CM, where the angle ECM= 49.77 degrees
- And the angle MCA = 71.9 degrees
- The angle M1 N M2 =88.9 degrees
- AM = 129630 km
- CM = 96434 km
- EM = 319370 km
- The Perimeter of the triangle (MCA) = 347684 km

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I- Data
(1)
The angle M1 N M2 =88.9 degrees
88.9 degrees – 71.9 degrees = 17 degrees
(2)
(17 degrees /0.8) = 21.25 degrees
(3)
21.25 degrees x 0.08 = 1.7 degrees (the moon motion equation constant)
(4)
17 degrees x 1.7 degrees = 29 degrees
(5)
23.4 deg = 1.8 deg x 13.177 deg x 0.98562 deg
II- Discussion
- Equations (from 1 to 3) give us a simple geometrical method to change the value
17 degrees into 1.7 degrees, but why this method is useful?
- Because the value 21.25 degrees is one of the moon motion angles which is
- 21.25 degrees = 11.8 degrees x 1.8 degrees
- Where
- 11.8 degrees = 5.1 deg (the moon orbital inclination) +6.7 deg (the moon axial tilt)
- But what's 1.8 degrees?! Let's discover in following…
o The moon moves from perigee to apogee and return back during its orbital
period.
o The distance from perigee to apogee on the moon orbital triangle (BD)
controlled by the angle (BCD =26.56 degrees)
o The moon go and return during the cycle (26.56 degrees x 2 = 53.12 deg)
o (53.12 degrees /29.2 solar days) =1.8 degrees
o Why I divide this angle 53.12 degrees on 29.2 days?
o Because

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o The Earth moves during 29.53 solar days 29.2 degrees
o But
o The moon moves during 29.53 solar days (360 degrees + 29.2 degrees)
- The previous explanation shows that, the angle 21.25 degrees is used in the moon
orbital motion because it depends on 2 angles (11.8 deg) and (1.8 deg) are used in
the moon day motion. based on that, the interaction between the angle 17 degrees
and 21.25 degrees can be created because both angles are used in the same motion
- Then
- The last step is to change the angle 21.25 degrees into 1.7 degrees as following
- 21.25 degrees x 0.08 = 1.7 degrees
- We remember this rate (0.08) based on which the valuable angle (10.96 deg) is
created.
Notice
- The most 3 basic values in the moon motion are (137 deg, 10.96 deg and 0.08)
- As the valuable angle (10.96 deg) is created based on this rate (0.08), the moon
orbital motion equation angle (1.7 deg) is created based on it….BUT
- Why the data shows that, Uranus orbital inclination (0.8 degrees) is used in this
process? The data uses (17 degrees /0.8 degrees) = 21.25 degrees, showing clearly
the using of Uranus orbital inclination (0.8 degrees) Why? because the data tries to
show Uranus effect on the moon orbital motion…. the next points supports it.
Equation No. (4)
17 degrees x 1.7 degrees = 29 degrees
- We know both angles 17 and 1.7 degrees but what's this 29 degrees?!
- The major lunar standstill can be +28.5 = (23.4 deg + 5.1 deg)
- The moon angular diameter = 0.5 degrees, that means, when the moon orbital
inclination is measured above the moon diameter it will be =5.6 degrees
- So the angle 28.55 degrees +0.5 degrees = 29.05 degrees

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- That shows Uranus effect on the moon motion during Metonic Cycle, which effect
on the moon daily orbital motion and effect on the moon motion equation by the
constant (1.7 degrees)
Equation No. (5)
23.4 deg = 1.8 deg x 13.177 deg x 0.98562 deg
Where
23.45 deg = Earth Axial Tilt 0.98562 deg = Earth motion daily degrees
13.177 deg = the moon daily motion degrees
1.8 degrees = is the angle we have discussed in the previous equation (no.3), the
angle of the moon motion from perigee to apogee during its day period (52.6/29.2)
Equation no. (6) shows that Earth axial tilt is created depending on the moon motion.
Planets Effect on Data
(1)
- Uranus (6.8 km/sec) moves during 51118 seconds a distance = 347603 km
- This distance = the triangle (MCA) perimeter accurately, showing Uranus effect
on the moon orbital geometrical structure by using the angle 71.9 degrees.
(2)
- Tan (71.9 deg) x 43000 km = 129630 km (error 1.5%)
- Where 43000 km = Perigee Apogee Distance and AM =129630 km
(3)
- 17 degrees = 0.99 x 17.2 degrees (Pluto orbital inclination)
- But , 17.2 degrees = 0.99 x 17.4 degrees (the inner planets orbital inclinations total)
- Also, 23.4 degrees = 0.99 x 23.6 degrees (the outer planets orbital inclinations total)
Notice
The angle 71.9 degrees is a very rich angle and the previous discussion is a small part
of it, for that, this discussion should be completed with Meronic Cycle Discussion
Point .6 and Uranus Motion analysis Point no. 7.
under the title ("The Interaction Angle 71.9 degrees" Continued)

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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 that Uranus axial tilt will be perpendicular on the
triangle base (AE) 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
- Based on this description
- The line BC shows Uranus motion effect on the moon orbital motion.
- In Metonic Cycle Discussion we should discuss more effects done by this line BC
on the moon orbital motion trying to prove that Uranus Motion effect on the orbtial
motion is a real effect.

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4-7 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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4-8 The Point (F) Position Analysis
- In the triangle PM2F
o The angle PM2F = 6.643 degrees
o The angle M2PF = 161.129 degrees
o The angle F = 12.228 degrees
o The distance PM2 = 403360 km
o The distance M2F = 615895 km
o The distance PF = 220305 km
o The distance EF = 214061 km

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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
6-4 The Moon Orbital Triangle Angles Discussions

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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 use here a discussion from the Point no. 4-5 page 49 of this paper)
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)
Earth moves during (6939.75 solar days) a distance = 17859.325 mkm
(2)
Pluto moves during (6939.75) x (153.3 hours) a distance = 18000.57 mkm
(3)
The moon displacements total during (6939.75) x (29.53 days) = 18034.278 mkm
(4)
Uranus Orbital Circumference = 18048.449 mkm
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%)
o 0.8 degrees = Uranus Orbital Inclination.
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 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).

I do this research
2nd
70
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.

The Moon Orbital Motion Description (2nd Revised)

The Moon Orbital Motion Description (2nd Revised)

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