The Moon Orbit Design (II) (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 Orbit Design (II) (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 –3rd
February 2021
Abstract
The moon uses Pythagorean triangle as one of its motion techniques
- The moon displacement = 88000 km, and during 29.5 days the total displacements
will =2.598 mkm, which should be = the moon orbital circumference.
- (2.598 mkm = 2π x 413000 km) (the moon apogee radius =0.406 mkm) (1%)
- According to this data, the moon would revolve around Earth through its apogee
orbit only during a month.
- The moon solves this dilemma by creating an angle (θ) between its displacement
motion direction and its orbit horizontal level, by that, the real displacement
through the orbit be (L =88000 km cos (θ)), where (L) will be less than 88000 km
and the moon can revolve around Earth through more orbits than its apogee orbit
- The moon using of Pythagorean triangle technique creates the moon orbit in a
triangle form.
- The using of Pythagorean triangle provides us 2 tools can be used in the moon
orbital motion study and analysis which are (the moon orbital triangle and the
moon orbital motion equation).
- Because the moon apogee radius (r=0.406 mkm) and doesn't =0.413 mkm, that
proves the moon using of Pythagorean triangle as one of its motion technique.
- But
- Why the moon apogee orbital circumference=2.55 mkm and doesn't = 2.598 mkm?

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- The moon uses the difference between these 2 distances (2.598 mkm & 2.55 mkm)
to create its orbital inclination with an angle (5.1 degrees).
- But
- Why the moon motion shows a deep love for Geometry?!
- Because the moon orbital motion is effected by many planets motions effects, and
to create a balance for the moon orbital motion the geometrical rules using is a
necessary option to create balancing points between these planets effects.
Paper hypotheses
(1st
) There's a 2nd
force effects on the moon motion in addition to Earth gravity
(2nd
) Uranus Motion effects on the moon orbital motion and cause Metonic Cycle
Paper Conclusion
- The moon orbit regression (19 degrees per year) is created as a result for the moon
orbital inclination creation.
- Earth Cycle (365+365+365+366 days) is created as a result for the moon obit
regression.

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Contents
Subject Page N
1- Introduction 4
2- The Moon Orbital Triangle Description
2-1 Preface
2-2 The Moon Orbital Triangle Description
2-3 The Moon Orbital Triangle Data Analysis
6
3- The Moon Orbital Motion Analysis
3-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
3-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
3-3 The Moon Orbital Motion Analysis
3-4 The Moon Orbital Motion Equation
27
4- The Moon Orbital Inclination Creation
4-1 Preface
4-2 The Moon orbital inclination creation geometrical process
4-3 Planets motions effect on the moon orbital inclination creation
4-4 The Moon Orbit Regression
4-5 Planets motions cause The Moon Orbit Regression
4-6 The Moon Orbit Regression Effect on The Earth Motion
43
5-The Moon Orbit Geometrical Design
5-1 Preface
5-2 The Necessity Of Pythagorean Triangle (1, 2, 51/2)
5-3 Why the moon orbital circumference at apogee doesn't = 2.598 mkm?
5-4 The moon motion angle (12.195 deg) Analysis
5-5 Why The Moon Displacement Daily =88000 km?
5-6 The angle 71.9 degrees
5-7 Why The Moon Day Period =29.53 days?
5-8 The Perpendicular Line BC (=86000 km)
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6- The Moon Orbital Triangle Benefits
6-1 Preface
6-2 The Moon orbital triangle shows that (2nd
force effect on the moon motion)
6-3 The Moon orbital triangle shows that (There's 2nd
Orbit for the moon motion)
6-4 The Moon orbital triangle shows that Uranus effects on the moon motion
96
7- Metonic Cycle Is A Proof of Uranus Effect On The Moon Motion
7-1 Preface
7-2 Uranus Effect On The Moon Orbital Motion
7-3 Uranus, The Moon And Pluto Motions Interaction
7-4 The Moon Orbital Triangle Angles Discussions
7-5 Moon Day Period Analysis (29.53 Solar Days)
101
8- Uranus Motion Analysis
8-1 Uranus Motion During 1440 Of Its Days Period
8-2 Uranus Motion During 8 Pluto Days period
8-3 Uranus 144 Days Cycle
8-4 The Interaction angle 71.9 degrees (continued)
8-5 The Moon Diameter Creation.
121
8- Appendix No.1 142

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1- Introduction
- The figure is used for the explanation…
- The discussion in the moon orbital triangle depends on the idea that, many planets
motions effect on the moon orbital motion, the question is … How can the far
planets effect on the moon orbital motion? for example the paper hypothesis
tells (Uranus motion effects on the moon orbital motion and causes to create
Metonic Cycle), by what force Uranus can do that? because Newton gravitation
equation prevents us to claim of such effect because of the huge distances between
these planets and the moon.
- How can these far planets effect on the moon orbital motion?
- Imagine we have a triangle, whatsoever this triangle dimensions lengths, but its
angles total is =180 degrees, So if one angle =120 degrees and found at a distance
1 million km that doesn't effect, the rest 2 angles total should be =60 degrees.
- The idea depends on the space nature, it supposes that, the solar planets are found
inside Space created geometrically- as seen in the figure – The space is created
based on geometrical design, So when any planet does any motion the whole space
be effected by it. it's not a force which causes the effect – but the space
geometrical structure – the space is built by geometrical structure, that means, each
motion of any planet should be controlled by geometrical rules.

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- To make this picture more clear, we may imagine that, the planets orbital
inclinations total around the Earth Ecliptic can't be greater than 360 degrees. This
limitation isn't created by our geometrical calculations but created by the space
itself and by that no planet can create its inclination violating the rule but planets
orbital inclinations total will not be more than 360 degrees.
- It's A Feature Of This Space
- Any planet motion will be expressed in angle and this angle will effect on the other
planets angles and the motions general harmony will be created when all motions
will be in harmony – that forces planets data also to be in harmony with each other
Let's explain this paper contents in following
- In Point no. 2 the paper introduces The Moon Orbital Triangle
- In Point no. 3 the paper discusses why & how the moon uses Pythagorean rule
- In Point no. 4 the paper discusses how the moon orbital inclination is created
- In Point no. 5 the paper discusses the moon orbital triangle geometrical design.
- In Point no. 6 the paper discusses the moon orbital triangle benefits
- In Point no. 7 the paper discusses Uranus effect on the moon (Metonic Cycle)
- In Point no. 8 the paper analyzes Uranus Motion

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2- The Moon Orbital Triangle Description
2-1 Preface
2-3 The Moon Orbital Triangle Data Analysis

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2-1 Preface
- The moon orbital triangle is created by its base (EA) creation between the Earth
Ecliptic & the moon equator lines, where there's an angle 1.543 degrees is found
between these 2 lines, the triangle base (EA) is created between these 2 lines,
above it the ecliptic with and angle 0.443deg and under it the equator with 1.1 deg.
- Then the perpendicular line BC is created perpendicular on the base (EA)
- The perpendicular line (BC) is created above the moon position, and because the
moon moves from perigee to apogee, this line BC should be used 2 times one on
the Perigee Point and one on the apogee point.
- By that we have 2 forms of the moon orbital triangle, we should discuss them as 2
cases for the same triangle, these 2 cases will be discussed individually.
- The 2 cases are inserted here for reference.
1st
Case
2nd
Case

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2-2-1 The 1st
Case (The Perigee Point).
- This is the suggested moon orbital triangle for the 1st
case
- 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 (M1) refers to The Moon Center (The moon in Perigee Point).
- 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
Earth ecliptic line.
- The Green Line (BE) is the moon triangle base, the distance BE = 363000 km, I
choose it and accordingly I have to define the point (E) position.
- The line BC is a perpendicular on the triangle base (BE), its length =86000 km
- The line BC is perpendicular on the triangle base (BE) on the moon perigee point.
(The 1st
Case)
- The angle CBE =90 degrees but the angle CYT = 89.557 degrees.
- The points (Q and P) are the intersection points of CE with the ecliptic and the
moon orbit plane respectively.
- The line TX is a perpendicular from the Earth Center on the base BE
- K is the intersection point between the triangle base (BE) & the moon orbit plane.
- The angle is Zero between the points ( A, B , K , X and E).
- The line EC connects between the points C & E where BC =86000 km and BE =
363000 km (As The Triangle Creation Requirements).

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

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

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

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

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

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

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

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2-2-2 The 2nd
Case (The Apogee Point).
- The Change Is In The Following:
- The line BC is perpendicular on the triangle Base (AE) on the point (D) which is
parallel to the point (Apogee) of the moon motion. (apogee r =406000 km).
- The 2nd
Case causes no more changes in the moon orbital triangle. The only
change is that, the perpendicular line position is changed from perigee point (in the
1st
Case) to the apogee point (in this 2nd
Case).
- In following we should discuss the changed data in the triangle as a result to
change the line BC position.

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(Point No. I) The Triangle ABC
Under the Ecliptic
- The angle CDA = the angle EDB =90 degrees
- The angle DCA = the angle DCB =26.46 degrees
- The angle ACB = the angle DCB =52.92 degrees
- The angle CAD = the angle CBD =63.54 degrees
- The line CD = 86000 km
- AD = DB = 43000 km (equal to M1R =42800 km error 0.4%).
- AC =AB = 96151 km
- The distance EA =449197 km
The Ecliptic
- The angle ydC = 89.557 degrees
- The angle DAd =0.443 degrees
- The distance Dd =333 km
- The distance BY =665 km
- The distance By =740 km
- The distance Cy =95411 km
- The distance dy =42664 km
- The distance Ay = 85350 km
- The perimeter of the triangle ACB = 278302 km

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(Point No. II) The Triangle CBX
Under the Ecliptic
- The distance DX = BX 361300 km + BD 43000 km =404300 km
- The distance AX = DX 404300 km + AD 43000 km =447300 km
- The line BC = 86000 km
- The hypotenuse CX = 413345 km
- The angle CXB = 12 degrees
- The angle BCX = 51.44 degrees
- The angle CBE =116.46 degrees
The Ecliptic
- The distance AT = 447313 km
- The distance yT = 361963 km
- The distance Tq = 12491 km
- The distance yq = 349472 km
- The distance Td = 404630 km

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2-3 The Moon Orbital Triangle Data Analysis (1st
Case)
- 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
)
- In the triangle data analysis we should answer the question (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
1. The distance BD refers to the moon motion distance from perigee to apogee.
That tells this triangle expresses the basic part of the moon orbital motion.
2. The triangle (BCD) is similar to the triangle (ODB) and both are specific
type of Pythagorean triangle (1,2, 51/2
)
Data
- The angle (BCD) = 26.46 degrees, the angle (CDB)= 63.54 degrees
- The hypotenuse CD = 96151 km the distance AC = 121622 km
- The distance BD = 85600km (86000 km) where AD=BD =42800 km
Data Analysis
o The perimeter of triangle (BCD) = 225000 km
o Sin (5.1) x 225000 km x 2 = 40000 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 claims that (Another force effects on the moon orbital motion in
addition to Earth gravity force and this point (A) refers to this 2nd
force) –

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The Moon Orbital Triangle Data Analysis (2nd
Case)
Data Analysis
- The Triangle ACB
- The perimeter of the triangle ACB = 278302 km = π x 88600 km (error 0.6%)
Accurately
o 88000 x π =86000 km +2 x 95230 km
o The hypotenuse AC = AB =96151 km
o 1% accurately the difference between 95230 km and 96151 km
o The data tells that, the moon daily displacement (88000 km) must be defined
depending on this triangle ABC (Data).
- The Triangle TdM2
- The distance T M2 = 406000 km (the apogee radius)
- The distance M2 d = 36092 km
- The distance T d = 404630 km
- The angle dTM2 = 5.1 degrees (the moon orbital inclination)
- The angle TM2d = 84.457 degrees
- The angle M2dT = 90.443 degrees
- (Please remember 943819 km = the perimeter of the triangle ACE in the 1st
case)
- The perimeter of the triangle TM2d = 846722 km

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Notice (1)
o 943819 km cos (26.22 degrees) = 846722 km
o By the angle 26.22 degrees (in the triangle ABD), the hypotenuse will be
=95900 km where 95900 km x cos (23.4 deg) = 88000 km.
Notice (2)
o (88000 x 29.53) + 61400 km = 846722 km x π
o 60800 km =1/2 the distance AC (121622 km) in the 1st
case triangle.
o This data tells that, there's 2 values of this triangle perimeter 846492 km…
Why? because the data uses a half of distance AC 121622 km?!
o 846492 km x 2 = 1.69 mkm ……………. If 1 degree = 1 mkm
o So this value 1.69 mkm is very near the value 1.7 degrees which is used in
the moon orbital equation…. That supports the claim (there are 2 values of
this triangle perimeter 846492 km).
o Note, 846722 km = π2
x 85790 km (equal 86000 km error 0.2%).
- The previous analysis is a simple analysis for the triangle data….
- But
- In the triangle design analysis we have to consider 2 basic questions, one question
for each triangle case …..
- For the 1st
Case the question is:
- Why the Pythagorean Triangle (1,2 and 51/2
) is a necessary tool for the moon
orbital motion?
- For the 2nd
Case the question is:
- Why the moon orbital circumference at apogee radius doesn't Equal 2.598 mkm
but is 2.55 mkm which is shorter than the moon displacements total during 29.53
days?

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The Moon Orbital Triangle Design Analysis
- The previous description of the moon orbital triangle in its 2 cases tried to
summarize the basic data found in this triangle
- This triangle should be the basic tool used by the moon in its orbital motion, for
that reason, we need to analyze this data as deep as possible
- Because of that, the paper dedicates the Point No. 4 to analyze the moon orbital
triangle data and to see its effect on the moon orbital motion
- But before to analyze this triangle data…
- We have to discuss how the moon uses Pythagorean triangle rule in its orbital
motion and how this using can be useful to produce the moon orbital motion
equation and then we have to test this equation accuracy with the moon motion
real data, this process we have to do in the point No. (3) of this paper (the next
Point) and then we'll discuss the moon orbital inclination geometrical mechanism
in point No. (4) and then we should return to the moon orbital triangle design
analysis in the point No. (5)

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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 Pythagorean triangle basically to decrease its displacement
daily through its orbit
o The moon daily displacement = 88000 km and the moon has to move this
distance every day without any decreasing (later we will know why!)
o But
o If the moon moves by this displacement as its orbital displacement the moon
would revolve around Earth through its apogee orbit only (r=0.406 mkm)
o For that reason
o The moon creates an angle between its motion direction and its orbit
horizontal level to create a displacement through its orbit less than (88000
km)
o As a result of this technique, the moon can revolve around Earth through
more near orbits than apogee orbit (r=0.406 mkm)
o Simply, because the moon uses this technique the moon can revolve around
Earth through perigee orbit (r=0.363 mkm)
o Let's explain this intelligent technique with some details to show the useful
result of using Pythagorean triangle by the moon orbital motion….

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

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The Angle θ
- The angle (θ) should get our attention for its specific effect…let's summarize the
idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and by that the moon can be near or at
Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon can be near or at apogee orbit
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees),
but before we need to analyze the moon orbital motion

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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 Pythagorean triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagorean triangle
in its motion, because the moon can't decrease its daily displacement (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.399 mkm) and Earth motion distance (2.573 mkm),
So the moon needs to move this displacement perfectly, but if it's used as a
displacement through the moon orbit, the moon would be always a prisoner in the
apogee orbit (r=0.406 mkm) as we have discussed before, because of that, the
moon creates Pythagorean triangle technique by which the moon moves actually
88000 km daily but the real displacement through the moon orbit became less (L =
88000 Cos θ) and by that the moon can achieve 2 objectives, First to pass the
required distance (88000 km) and Second to move in near orbits to Earth, that
shows the intelligent moon motion technique…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1)

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

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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 Pythagorean triangle in its orbital motion)
- The moon uses Pythagorean triangle and by this intelligent technique the moon be
under control of the angle (θ) change
- The angle (θ) defines almost all the moon motion features.…
- The moon uses this technique, aiming to create a real displacement shorter than its
actual displacement (88000 km) based on the equation (L =88000 cos (θ)) and by
that while the moon moves a displacement =88000 km but the real displacement
(L) through its orbit be shorter than 88000 km and by that the moon can revolve
around Earth through more near orbits than its apogee orbit (r=0.406 mkm).
- The moon orbital motion equation depends on this concept and, the equation
uses (the constant) 1.7 degrees as the moon daily motion degrees, and the equation
uses the previous day angle (θ0) to produce the today angle (θ1)
- We have 3 questions in this equation 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 Pythagorean triangle with
angle (28.63 degrees) and during one solar day the moon uses - 1.7 degrees
and by that the angle will be (26.93 degrees)…... The angle 1.7 degrees
expresses The Moon Daily Motion
o By using Pythagorean triangle its angle (θ) = 26.93 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

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

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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)

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

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4- The Moon Orbital Inclination Creation
4-1 Preface
4-2 The Moon orbital inclination creation geometrical process

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4-1 Preface
- The moon daily displacement =88000 km, during the moon day period which is
29.53 solar days, the total displacements be 2598640 km.
- The moon orbital circumference at apogee = 2550973 km
- The apogee orbit is the most far point from the Earth which the moon can reach
- The moon orbit circumference is shorter than the total displacements during 29.53
days with a distance = 47667 km
- That proves (The moon uses Pythagorean triangle rule as one of the moon
orbital motion techniques)
- Then the question be
- Why the moon orbital circumference doesn't = 2598640 km
- The answer is …
- Because the moon uses this difference (47667 km) to create its orbital inclination
angle (5.1 degrees)…
- That means, if the moon orbital circumference be =2598640 km, in this case the
moon orbital inclination will be = Zero
- More conclusions can be raised based on this description, the most important one
is that….
- The moon orbit regression is caused by the same geometrical mechanism by
which the moon orbital inclination is created
- That tells the moon orbital inclination creation process causes also the moon orbit
to be regressed (1.44 degrees) per month.
- These conclusions tell, the process by which the moon orbital inclination is created
is a very effective and significant process found in the moon orbital motion
This point tries to analyze this process details to see how the moon orbital inclination
is created and why does the moon orbit regress?, in addition to we should check the
claim (Earth Cycle 1461 days is created as a result for the moon orbit regression 1.44
degrees per month).

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4-2 The Moon Orbital Inclination Creation Geometrical Process
In this triangle
- ab = 0.232 mkm
- ac = 0.2608975 mkm
- bc = 2.598 mkm (the moon displacements total)
- The angle c = 5.1 degrees (The Moon Orbital Inclination)
- This figure tells us that
o To create the angle 5.1 degrees we need 2 distances (1st
distance) the moon
displacements total during 29.53 days (2.598 mkm) and this distance is
found and defined by the moon daily displacement.
o Also we need the distance ab =232000 km
o This is the factor based on which the moon orbital inclination will be created
o So, we need to produce this distance (232000 km)… so let's try to do
Equation No. (1)
Cos (10.96 degrees) x 2598640 km = 2550973 km
- The angle 10.96 degrees is the most valuable angle in the moon orbital triangle we
should discuss its origin in the moon triangle geometrical design (Point No. 5)
- The distance 2598640 km =the moon displacements total during 29.53 days
- The distance 2550973 km = the moon apogee orbital circumference
- Equation No. (1) tells, the moon apogee orbital circumference 2.55 mkm is created
depending on the moon displacements total by the angle 10.96 degrees
- The difference = 2598640 km – 2550973 km = 47667 km
- Then
- From this difference 47667 km we need to create the distance 232000 km
- How??

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- The moon displacements total during 29.53 days = 2598640 km
- And
- The Earth moves during 29.53 days a value 29.2 degrees
- The moon moves during 29.53 days a value (360 + 29.2 degrees)
- Let's create another triangle, its base = 47667 km and its angle 29
- In this triangle
- The BC = 47667 km
- The angle A = 29 degrees based on that
- The hypotenuse AC = 98321 km
- The distance AB = 86000 km
- The triangle perimeter = 231982 km
- The input data is 47667 km and the angle 29.2 deg (is used as 29 deg)
- The output is the perimeter of triangle (ABC) =231982 km
- Tan (5.1) x 2598640 km (the displacements total) =232000 km
- The previous explanation shows the geometrical mechanism by which the moon
orbital inclination (5.1 degrees) is created depends on the difference between the 2
distances (2598640 km and 2550973 km).

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The Moon Orbital Inclination (5.1 degrees)
- The moon orbital inclination (5.1 degrees) is a geometrical value
- For example
- Sin (5.1) x (180/π) =5.1
o The moon orbital inclination (5.1 degrees) is created based on geometrical
calculations and because of that this value has its geometrical power (as all
the moon other data)
o The planet is a geometrical structure as one building, based on this idea, the
planet data is created based on each other geometrically…. That means, no
data is found without geometrical reason otherwise the building will be
useless – imagine one building is built and has no a door or stair how to use
it- the building is built based on a geometrical concept and similar to that the
plant data is created based on a geometrical concept.
o The moon orbital inclination (5.1 degrees) is created with some geometrical
interaction to cause the moon orbit regression 1.44 degrees per month
o The moon orbit regression is created by the geometrical mechanism based
on which the moon orbital inclination is created, that means the moon
orbital inclination creation process contains both features the inclination
degrees 5.1 degrees and the regression effect 1.44 degrees per month….
o The angle 137 degrees which we will discuss in the moon orbital
geometrical design shows this fact (137 =95.1 degrees x 1.44 degrees),
telling that, from one process the 2 features are created.

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The using of the moon orbital triangle
- let's use the moon orbital triangle…
- The triangle CDb is our triangle, because its perimeter = 232000 km
- The distance CD =86000 km
- The distance Db = 47667 km (please remember DB =42800 km)
- The distance DX = 2598640 km (the moon displacements total)
- The distance bX = 2550973 km (the moon orbital circumference at apogee)
- The angle CXD =1.89 degrees
- The angle DCb =29 degrees
So
- The Perimeter of triangle CDb =232000 km
- The Distance DX = 2598640 km (the moon displacements total)
- Tan (5.1) x 2598640 km = 232000 km
- By that the moon orbital inclination is created by the proportionality between the
perimeter of triangle CDb and the distance DX.

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I-Data
Mars Velocity (24.1 km/sec) = Pluto Velocity (4.7 km/sec) x 5.1
II-Discussion
- The data tells that, Mars and Pluto motions effect on the moon orbital motion and
causes to create the moon orbital inclination =5.1 degrees
- That means, the geometrical process which we have studied in the previous point
was the geometrical mechanism by which the moon perform the effect of these 2
planets on the moon orbital motion.
- That explains many data has no explanation before but now we may explain it, for
example
o Pluto moves during a solar day a distance =406000 km = The Earth moon
distance at apogee radius. That tells us Pluto effect on the previous process
is found in the moon orbital circumference creation (at apogee orbit whose
radius =0.406 mkm and circumference =2.55 mkm).
o Mar moves during a solar day a distance =2.082 mkm = 0.8 x 2.609 mkm
(This distance is the length of the hypotenuse CX in the moon orbital
triangle (in the previous page) and = the hypotenuse ac in the discussion
triangle for point (2-1). The data tells that, Mars motion depends on the
moon displacements total –
o These are 2 forces, Pluto works for the moon orbital circumference (2.55
mkm) and Mars works for the moon displacements total and the balance
between these 2 forces create the moon orbital inclination 5.1 degrees.
So
o The moon orbital inclination is created by Pluto and Mars Motions effect
on the moon orbital motion
o If the moon orbit regression is done by the same process by which the moon
orbital inclination is created, one of these 2 planets must be a player.

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(1st
Point)
- Let's return to this triangle again
- The hypotenuse ac =2608975 km
- The moon apogee orbital circumference = 2550973 km
- The difference =58000
- Sin (1.3) x 2550973 km =58000 km
- The data leads us to the angle 1.3 degrees! Why? because
- 8 deg = 1.3 deg (Jupiter orbital inclination) + 6.7 deg (the moon axial tilt)
- The data tells that,
- The moon axial tilt is created with Jupiter orbital inclination in the same process
based on the valuable value (8 degrees)
- 8 degrees expresses Uranus orbital inclination 0.8 degrees, because Uranus uses
this value in different forms as 0.08 or 8
(2nd
Point)
- The previous discussion still has benefits for our analysis….
- The value 47667 km tells us the following
o Tan (1.44 degrees) x 47667 = 1195 km (Pluto Radius)
o Tan (1.44 degrees) x 69118 = 1737.5 km (The Earth Moon Radius)
o But what's the value 69118 km?

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Equation No. (2)
Cos (13.244 degrees) x 2598640 km = 2529522 km
- The difference 2598640 km – 2529522 = 69118 km
- The moon daily motion =13.177 degrees (error 0.5%)
The data tells that,
- Pluto radius is created by the distance 47667 km based on the angle 1.44 deg.
And
- The moon radius is created by the distance 69118 km based on the angle 1.44 deg.
- The moon and Pluto motions interaction should be discussed in Uranus motion
effect on the moon orbital motion – what we need here from this data is the angle
1.44 degrees –
- The moon orbital circumference 2.55 mkm is created based on the moon total
displacements during a month (29.53 solar days) and because of that the angle 1.44
degrees is used because the moon orbit regresses 1.44 degrees Per Month
- The total = 47667 km + 69118 km = 116785 km = 2 x 58000 km (error 0.6%)
where the value 58000 is used with Jupiter orbital inclination creation.
Equation No. (3)
(π)1/2
x 1.44 degrees x 2 = 5.1 degrees
- The moon orbital inclination is a geometrical value, as we have discussed before,
and the value 1.44 degrees is created based on a geometrical interaction as the
equation shows.
- The data tells that, the moon orbit regression (1.44 degrees) is created as a feature
with the moon orbital inclination (5.1 degrees), both are created in the same
process.

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A Discussion
- There are many simple and direct questions are needed to be solved – for example
- Why does the moon orbital motion depend on geometrical rules?
- Because many planets effect on the moon orbital motion and the geometrical rules
create balancing points between these planets motions effects, as the paper
supposes that, 2nd
force effect on the moon orbital motion and Uranus motion
effect on the moon orbital motion and causes to Create Metonic Cycle… the idea
is that …. If many planets gravities effect on the moon orbital motion the
geometrical rules will be a necessary tool to create a balancing for the moon orbital
motion.
- The next question is
- How can many planets motions effect on the moon orbital motion?
- We see that in the planet data, for example
- (Jupiter Mass / Earth Mass) x 142984 km = 149.6 mkm (Earth orbital distance)
(where 142984 km = Jupiter diameter)
- The data tells that, Earth orbital distance definition is effected by Jupiter mass rate
to Earth mass, that shows an effect of Jupiter on Earth motion
Also
- The moon apogee orbital circumference 2550973 km = 21.86 x 116785 km
- Where 21.86 = (Jupiter mass / Uranus mass)
- And 116785 km = 47667 km + 69118 km
- We have used this distance 116785 km in the previous page explanation to show
how the moon motion data depends on the angle (1.44 degree = the moon orbit
regression per month)
- This data shows the force behind causes the interaction from which the data is
created, this force is the masses rate between Jupiter and Uranus which are the
most 2 planets effective on the moon orbital motion data.

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I-Data
(No. 1)
6.8 km /sec (Uranus Velocity) = 4.7 km /sec (Pluto Velocity) x 1.44
II-Discussion
- The moon orbit regresses 1.44 degrees per month
- The data tells that
- Mars and Pluto motions interaction cause the moon orbital inclination to be 5.1
degrees
- And
- Uranus & Pluto motions Interaction cause the moon orbit to regress 1.44 degrees
per month.
- That makes Pluto motion as a central point of both planets (Mars and Uranus)
motions effect, and Pluto transports this effect to the moon orbital motion
- Let's discuss more data in following

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II-More Data
Uranus Data
(a)
97 degrees = 19 x 5.1 degrees (the moon orbital inclination)
(97 degrees = 97.8 deg Uranus axial tilt -0.8 deg Uranus orbital inclination)
(b)
Uranus orbital distance =19 x Earth orbital distance
(c)
95.1 degrees +1.1 = 96.7 degrees
96.7 degrees +1.1 = 97.8 degrees (Uranus Axial Tilt)
97.8 degrees +1.1 =98.9 degrees
(Please note, 97.8 - 6.7 =91.1 degrees =90 +1.1 degrees)
(97.8 deg = Uranus Axial Tilt 6.7 deg = The Moon Axial Tilt)
(d)
122.5 deg ( Pluto Axial Tilt) x 0.8 deg (Uranus orbital inclination) = 97.8 deg
(e)
122.5 degrees = 95.1 degrees +27.4 degrees
(95.1 deg = 90 deg +5.1 deg (the moon orbital inclination))
(f)
176.4 deg = 122.5 deg ( Pluto Axial Tilt) x 1.44
( 177.4 = Venus Axial Tilt)

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The Moon Orbit regression 19 degrees per year
- The idea is a simple one
- Uranus Orbital Distance =19 Earth Orbital Distance
- So, if Uranus motion effects on the Earth moon orbital motion, the number 19
years should be seen in this effect, that lead us to conclude, (Uranus motion effect
on the moon orbital motion causes to create the moon Metonic Cycle)
- The next question should be,
- Is this 19 created by Uranus /Earth orbits rate or it's an independent number?
- Because
- 97 degrees = 5.1 degrees x 19
- 97 deg (= 97.8 deg Uranus Axial Tilt – 0.8 deg Uranus orbital inclination)
- 5.1 deg (= the moon orbital inclination)
- if Metonic Cycle is created by Uranus effect so the number 19 is the orbits rate but
why we see it between the axial tilt and orbital inclination?!
- The data tells that,
- Uranus orbital distance must be created depending on its axial tilt, and Earth
orbital distance is created depending on its axial tilt and Earth axial tilt is rated to
the moon orbital inclination!
- The data my show that in following
More Data
(j)
(149.6/23.45) = (29.8/4.7) = (153.3/24) = (5906/928)
149.6 mkm = Earth Orbital Distance 23.4 deg = Earth Axial Tilt
(29.8 /4.7) = Earth Velocity / Pluto Velocity
(153.3/24) = Pluto day period / Earth day period
5906 mkm = Pluto orbital distance
928 mkm = Earth Jupiter distance when they be on 2 different sides from the sun

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(k)
119.2 deg = 5.1 deg x 23.4 deg
- Earth moves during 29.53 solar day (29.2 degrees) (29.53 x 0.9862 deg)
- The moon moves during 29.53 solar day (360 deg + 29.2 deg) (29.53 x13.17 deg)
- 29.2 + 90 = 119.2 degrees
- So
- The angle which leads both motions is created depending on Earth axial tilt and the
moon orbital inclination.

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- Let's remember an old question,
- If the Earth is a fixed point in Space and the moon orbit regresses 19 degrees per
year, can the distances between the moon and Earth be without change although
the moon orbit regression?
- The answer is Not
- If the Earth is a fixed point in space the distances between the Earth and the moon
must be changed by the moon orbit regression
- That means,
- Earth does some motion as a result for the moon orbit regression to save the
distances between the Earth and moon without changes
- The next question is,
- Does Earth do any motion as a result of the moon orbit regression?
- The answer is yes
- It's the Earth Cycle (365 +365 + 365 +366 days) which is 1461 days
- This cycle is created as a result of the moon orbit regression 19 degrees per year
- Let's try to prove this claim by data analysis in following…
Equation No. (4)
1.461 x 0.98562 degrees = 1.44 degrees
- 0.98562 degrees = The Earth motion degrees daily
- 1.44 degrees = The moon orbit regression per month
- 1.461 = 1461 days /1000
Also
- 95.1 degrees = 23.41 deg x 4 +1.44 degrees
o 23.45 degrees = Earth Axial Tilt
o If Earth motion for 1 year depends on its axial tilt 23.45 degrees, so this
equation shows 4 years Cycle. (similar to that, the moon displacements per
year depends on its orbital inclination 5.1 deg)

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5-The Moon Orbit Geometrical Design
5-1 Preface
5-2 The Necessity Of Pythagorean Triangle (1, 2, 51/2
)
5-4 The moon motion angle (12.195 deg) Analysis
5-5 Why The Moon Displacement Daily =88000 km?
5-6 The angle 71.9 degrees
5-7 Why The Moon Day Period =29.53 days?
5-8 The Perpendicular Line BC (=86000 km)

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5-1 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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5-2 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)
Sin (10.96 degrees) x 406000 km = 77237 km
- 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 less 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 greater 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 them are done by
the moon– 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 the next point (4-3) because more data
analysis may help us greatly. We should do after the data discussion completion.

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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 are unified 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 1.44 degrees per Month.
- 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 (Part I)
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 invented
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 it as real as the moon registered data by
observation.

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(4th
Point) The Angle 10.96 degrees (Part II)
- In This Triangle
- ab = 88000 km
- bc = 449197 km
- ac= 457735.6 km
- The angle acb = 11.084 degrees.
- Tan (11.084 degrees) x 449197 km = 88000 km
- Why have I created this triangle?
- I suppose that, the moon daily displacement 88000 km may be created depending
on the moon triangle base (AE =449197 km) which gives some release from the
full dependency on the angle 10.96 degrees… I try to know if the moon
displacement 88000 is created by any other factor than the angle 10.96 degrees.
- The question starts with the equation no. 5 (Sin (10.96 degrees) x 449197 km =
85403 km – this equation causes disappointment for the investigation because
neither the value 88000 km nor 86000 km is created based on the triangle base
(EA=449197 km) based on our valuable angle (10.96 deg), so, that tells something
must be un-understandable!
Shortly
How that is happened? As following:
o 137 degrees x 0.08 = 10.96 degrees (our angle)
o (137 degrees +1.543 degrees) x 0.08 =11.084 degrees
o (137 degrees -1.543 degrees) x 0.08 =10.836 degrees
Based on that
o Tan (11.084 degrees) x 449197 km = 88000 km
o Tan (10.836 degrees) x 449197 km = 86000 km

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- Both values (88000 km and 86000 km) are defined based on the triangle base
(EA=449197 km) based on both angles (11.084 and 10.836 degrees) where these 2
angles are created by the original angle 137 degrees (as our angle 10.96 deg).
- But
- The angle 1.543 degrees (found between the ecliptic line an the moon equator line)
effects on our angle (10.96 degrees) to produce these 2 new angles (11.084 and
10.836 degrees) where these 2 angles should be considered as similar forms for our
angle (10.96 degrees).
- The data proves the existence of the hypotenuse ac= 457735.6 km
- Where the moon triangle base (EA =449197 km) is used as a adjacent in all
equations..
- Please note this data importance because the base EA =449197 km = Jupiter
Circumference, because of that, this data may refer to Jupiter effect on the moon
orbital motion.
Notice
- Tan (10.836) x 29.2 = 5.6
- Where
- Earth moves during 29.53 solar days a value 29.2 degrees but the moon moves
during this same period (360 deg + 29.2 deg)
- 5.6 degrees = 0.5 deg +5.1 deg = that means, when the moon orbital inclination be
measure above the moon diameter the value will be 5.6 degrees
- That tells us, the moon orbital inclination is rated to the Earth and moon motions
during 29.53 days by this angle (10.836). That means these 3 values are created
rated to each other.

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(Point No. I)
- In the triangle DCX the hypotenuse CX = 413345 km
- Let's remember
o The moon displacement for a solar day = 88000 km
o During 29.53 solar days the total will be = 2.59864 mkm
o 2.59864 mkm = 2π x 413560 km
o means, the moon orbital triangle data considers the distance 2.598 mkm and
uses it in its geometrical structure but for some geometrical necessity the
moon orbital circumference at apogee doesn't=2.598 mkm BUT= 2.55 mkm.
BUT
- What's this geometrical necessity for which the moon orbital circumference at
apogee radius be 2.55 mkm in place of 2.598 mkm? Let's try to answer in
following….
Notice
o Because XE =1700 km the hypotenuse CE will be =415000 km.

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(Point No. II) The Moon total displacements 2.598 mkm
- In this moon orbital triangle I have used the distance 2.598
- Where
- The hypotenuse CE2 = 2598640 km = the moon total displacements during 29.53
solar days
- The point E2 is defined by this hypotenuse on the triangle base (AE)
- No more changes are done …. Let's consider what's happening as a result
o The angle E2 = 1.896 degrees
o The angle E2 CB = 88.1 degrees
o The hypotenuse CE2 = 2598640 km
o The distance DE2 = 2597217 km
o But
o BD = 42800 km
o BE2 = 2554417 km = 2π x 406550 km
o The moon apogee radius = 406000 km
- The data tells that
o (1st
) The moon displacements total (2.598 mkm) is considered as the basic
value in the moon orbital triangle because it's used as the hypotenuse
o (2nd
) the moon orbital apogee radius (406000 km) is produced based on the
moon orbital triangle geometrical interaction.
o But
o For what geometrical interaction the apogee circumference be 2.55 mkm?!

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(Point No. III) The Moon & Mars Motions Interaction
- The basic result of this triangle is the angle 1.89 degrees based on which the
hypotenuse CE2 =2.598 mkm is created.
- 1.9 degrees = Mars Orbital Inclination
- The moon works as a gear to connect Earth Motion with Mars Motion, in more
clear words, Venus & Earth Motions interaction effect on Mars Motion and this
effect is done by the moon motion effect on Mars Motion! this idea we have to
analyze as deep as possible, let's see the data in following….
Notice
- The perimeter of the triangle CDE2 = 5281856.6 km
- The perimeter of the triangle CBE2 = 5249007.6m
- The difference = 32849 km
- The distance from the moon center and the ecliptic line will be =32849 km when
the moon be far from Earth with a distance = 369530 km (this value less 1% of the
distance CE =373000 km).

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The Moon & Mars Motions Interaction Analysis
I-Data
(a)
1.883 deg = 1.44 deg + 0.443 deg
25.3 = 1.9 +23.4
(b)
137 deg -25 deg = 113.44 deg -1.44 deg
(c)
80 x 1.44 deg = 115.2 deg =17.4 +97.8 81 x 1.44 deg =116.6 deg
(d)
23.36 = 29.2 x 0.8
2.082 = 2.598 x 0.8
2.41 =3.02 x 0.8
1.44 = 1.8 x 0.8
(e)
1.44 x 17.4 = 25.06 1.44 x 12 = 17.34
II-Discussion
- The interaction between Mars and the Earth Moon Motions is known and can be
proved clearly from their data …. For example
o Mars orbital period 687 days = the moon orbital period 27.3 days x 25.2
o The moon daily motion (13.177 deg) / Mars daily motion (0.524 deg) =25.2
o Mars orbital period 687 days = 2 x 343.5 days (the nodal year =346.6 days)
o The moon day period (708.7 h) = Mars day period (24.7 h) x 2π
o Mars orbital period 687 days = Earth orbital period 365.25 days x 1.9
o (1.9 deg = Mars orbital inclination) (25.2 deg = Mars Axial Tilt)
o The data shows the interaction between Mars on one side and the Earth with
its moon on the other side

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But, 2 questions are raised accordingly …
o (1st
question) what's the origin point of this interaction?
o (2nd
question) Why this interaction is seen more clear than Earth and its
moon interaction with Venus where Venus is more near than Mars?!
Let's consider the first question now, and the second we should interest for later
(1st
question) what's the origin point of this interaction?
Equation No. (a)
1.883 deg = 1.44 deg + 0.443 deg
1.9 degrees = Mars orbital inclination,
- This value is created by a direct effect of the moon orbit regression… where the
moon orbit regresses 1.44 degrees per a month and this value is added to the
difference 0.443 degrees (which is found between the moon orbital triangle and the
ecliptic line) to produce the value 1.883 degrees which can be considered as the
moon orbital inclination – this is the angle we have found in the triangle
(CE2D)…
- This angle is created based on the moon displacements total during a month…
the word month should be kept with us because it tells some very important
geometrical reference….
- If so, does the value 1.9 degrees (mars orbital inclination) is created based on a
month period? But Mars daily motion =0.524 degrees where
o (1/0.524 degree) =1.9 degrees
o For Mars the value 1.9 deg is used for the daily motion! it's also a period of
time, but why? what's for Month in the moon motion can be for a day in
Mars motion? how to understand that?!
o The value 1.9 degrees still needs more analysis.. Let's do it in following..

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More Data (old)
1.3 degrees (Jupiter orbital inclination) = 1.9 degrees - 0.6 degrees
2.5 degrees (Saturn orbital inclination) = 1.9 degrees + 0.6 degrees
- Let's remember how the moon orbital triangle originally is created…
- Uranus axial tilt =97.8 degrees and the Earth moon axial tilt =6.7 degrees, the
difference between both =91.1 degrees
- The perpendicularity between Uranus and the moon axial tilt faces the problem of
the angle 1.1 degrees, how to produce this perpendicularity between them?
- The solution was to raise the triangle base with and angle 1.1 degrees and by that
Uranus axial tilt will be perpendicular on the triangle base if this base depends on
the moon axial tilt and has an angle 1.1 degrees with it
- By this description the moon orbital triangle is created and developed
- That means, under the triangle bases there's 1.1 degrees and the moon diameter
consumes 0.5 degrees (the moon angular diameter) by that, the rest angle should
be =0.6 degrees
- Which we see in the data…
- By this data I suppose that Jupiter and Saturn do some interaction for this value 0.6
degrees and this interaction depends on the angle 1.9 degrees which will be used as
Mars orbital inclination
So
- The angle 1.89 deg in the triangle CDE2 is found as one form of this same
interaction done by Jupiter and Saturn based on the angle 0.6 degrees where the
moon uses the other part 0.5 degrees for the total 1.1 degrees.
- That tells
- This interaction of Jupiter and Saturn effect also on the moon motion which is
absolute true, because under the triangle base we have a network of motions
interactions effect.

The Moon Orbit Design (II) (Revised)

The Moon Orbit Design (II) (Revised)

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The Moon Orbit Design (II) (Revised)