The Moon Uses Pythagoras Triangle Technique In Its Orbital Motion (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 Uses Pythagoras Triangle Technique In Its Orbital Motion (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 – 15th
December 2020
Abstract
Paper hypothesis:
- The moon uses Pythagoras triangle as one of the moon orbital motion techniques
The hypothesis Explanation:
- The moon uses Pythagoras triangle to do the following job. While the moon moves
a daily displacement =88000 km, The moon real displacement through its orbit can
be less than (88000 km) by using Pythagoras triangle technique.
- The moon needs to decrease its daily displacement (88000 km) through its orbit to
enable the moon to revolve around Earth in more near orbits to Earth
- That means, if the moon displacement (88000 km) can't be decreased and be its
daily orbital displacement through its, the moon would revolve around Earth
through only its orbit apogee (r=0.406 mkm) and can't revolve through any more
near orbits.
- The moon using of Pythagoras triangle in its orbital motion, creates a great
Pythagoras triangle controls the moon orbital motion and its orbit geometrical
structure.
Paper objective:
- The paper proves this fact and discusses the geometrical basics based on which the
moon orbital triangle is created.

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Contents
Subject Page No.
1- The Moon Orbital Triangle Geometrical Basics 3
2- The Moon Orbital Motion Equation 11
3- Jupiter and Uranus effects on the moon orbit 23
4- Appendix No. 1 26
4- References 28

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1- The Moon Orbital Triangle Geometrical Basics
1-1- Introduction
- The moon orbital motion 4 points show the Pythagoras rule using. These 4 points
are:
- Perigee radius (r=0.363 mkm), the most near point the moon can reach to Earth
- Apogee radius (r=0.406 mkm), the most far point the moon can reach from Earth.
- T.S. Eclipse (r=0.373 mkm), the moon creates the total solar eclipse at this radius
or shorter
- Orbital distance (r=0.384 mkm), and this is the moon data registered point as its
orbital distance.
I-Data
- (363000 mkm)2
+ (86000 km)2
= (373000 km)2
- (373000 mkm)2
+ (86000 km)2
= (384000 km)2
(error 0.5%)
- (384000 mkm)2
+ (86000 km)2
= (393500 km)2
- (393500 mkm)2
+ (86000 km)2
= (403000 km)2
(Apogee radius =406000 km creates as error with 403000, equal = 0.8%)
II-Discussion
- The previous data was the first reason to suppose, the moon uses Pythagoras rule
in its orbital motion. the data is questionable because the dimension (86000 km) is
not found in the moon orbital motion data, we know the distance between perigee
(r=0.363 mkm) and apogee (r= 0.406 mkm) = 43000 km but this dimension 86000
km isn't found in the moon orbital motion data?!
- The next question is why the moon uses Pythagoras triangle? The paper hypothesis
tells, the moon does to decrease its daily displacement (88000km) through its orbit,
this answer needs more explanation…
- In point no. (1-2) we will discuss the moon displacement decreasing technique
- And In point no. (1-3) we will discuss why the dimension 86000 km is used to
define the moon orbital motion basic 4 points.

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1-2- The Moon Displacement Decreasing Technique
- 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 orbital horizontal
level, by that the real displacement will be (L= 88000 km cos (θ)) through the
moon orbit, and by that, spite the moon moves 88000 km and the orbital horizontal
distance 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 So the real displacement (L) = 77237 km, So,
if the moon real displacement daily be equal this value during 29.53 days the moon
will move a distance = 2.28 million km and this will be the moon orbital
circumference, BUT for what radius? 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 BUT for what radius? 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving round 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) needs to pay some 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 that means the moon will 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 will be near or at to apogee orbit
o That means, the angle (θ) does change the real displacement (L) and also
change the distance between the moon to perigee or to apogee, shortly, the
angle defines the moon position (as a ship) between 2 river banks….
- In fact the angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation… but here we need to see
the direct effect of the angle (θ) on the moon motion because it's used in the moon
orbital triangle geometrical basics
- Now let's analyze the angle (θ) in some deep
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with the previous dimensions, what's the angle (θ)
value? (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
- i.e., The angle (1.1 deg.) controls the moon motion from perigee to apogee. More
about the angle (θ) we have to discuss after the moon orbital geometrical basics

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1-3- The Moon Orbital Triangle Geometrical Structure
Figure No. (1) (my figure)
Let's Review The Moon Orbital Triangle Data
(1st
Point)
- The figure I brought from internet to use in the Explanation -
- We have supposed that the inner circle is Perigee orbit and
the outer circle is apogee orbit – and we have calculated the
tangent DB = 181843 km
- AB = 363686 km (= perigee radius approximately)
- Perigee radius r =0.363 mkm Apogee radius r =0.406 mkm
- Based on that, the triangle (ODB) is a specific Pythagoras
triangle (1, 2 and 51/2
)
- The triangle (ODB) angles are 26.564 deg. and 63.435 deg.

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(2nd
Point) The Moon Orbital Triangle Data Correction
- EB = Perigee radius = 363000 km
- ED = Apogee radius = 406000 km
- EA= (Jupiter Circumference) =449197 km
- AC = (Saturn diameter) =121620 km (error 1%)
- ES = total solar eclipse radius = 373900 km (error 1%)
(EC = 373000 km = Earth moon distance at T. Solar eclipse, BUT point C is NOT
the moon position in T. solar eclipse, because the distance BC= 86000 km but the
distance between perigee point and total solar eclipse point = 11000 km)
- CX= =87521 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 km CY= 97766 km
- BA = BC = 86000 km
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km
- BY = = 46475 km
- SZ = 7665 km ZF= 2414 km
- DY = 3475 km BX= 16203 km
THE ANGLES
- The angle between the black and red lines (under E) = 1.1 degrees
- (E) = 13.33 degrees (C)= 121.67 degrees (A) = 45 degrees
- (ECB) = 76.67 degrees (BCA) = 45 degrees
- (BCS = 7.23 deg) (BCZ = 12.195 deg) (BCF = 13.72 deg) (BCD = 26.564 deg)
(ACD = 18.435 deg)
- (BSC = 82.7 deg) (BZC = 77.8 deg) (BFC = 76.82 deg) (BDC = 63.434 deg)
- (CSA =97.23 deg) (CZA =102.195 deg) (CFA= 103.7 deg) (CDA = 116.564 deg)
- (CYA = 118.3 deg)
- BCY = 28.39 degrees ECZ= 88.9 degrees
- XCE = 66 degrees
- CZS = 77.8 degrees
- CZF =102.195 degrees
- XCB = 10.67deg
- (Uranus Axial Tilt = 97.8 degrees = FSC 97.2 degrees + 0.6 degrees) (i.e. the
angle under FSC)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

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The Moon Orbital Triangle Details Discussion
- How to draw The Moon Orbital Triangle….?
- The first horizontal black thick line which is under all triangle details and has zero
angle with the horizontal level this black line is the Moon Axial Tilt (6.7 degrees)
- The triangle base (red) thick line declines on the horizontal level (the black line)
with an angle =1.1 degrees
- Point E represents the Earth
- Point B represents Perigee radius (r=0.363 mkm)
- Point D represents Apogee radius (r=0.406 mkm)
- Point A represents a point in space far from Apogee radius with 43000 km at the
same horizontal level, means no angle between these points (E,B,D,A)
- The Ecliptic Line which is seen in the triangle has an angle = 0.5 degrees between
it and the moon orbital triangle base (The red line), why?!
- Because 1.6 degrees is found between the Earth Ecliptic & The Moon Axial Tilt
- The moon orbital motion is ranged between the point (B) (Perigee radius r=0.363
mkm) and the point D (Apogee radius =0.406 mkm).
- We will discuss the triangle details in full analysis one after one – but – at first
- Our basic discussion triangle is the triangle BCD because it contains the moon
orbital motion from perigee (Point B) to apogee (Point D)
- Please Note, the triangle (BCD) is a similar to the general triangle we have
discussed separately in page no.6 (the triangle DOB) where the dimensions are
rated (406000km , 363000 mkm and 181843 km) and (96151 km, 86000km and
43000km), because of that the angles are equal, which makes both triangles are
similar, both are typical to Pythagoras triangle (1,2, (5)1/2
)
- Our consideration now should be directed to the line BC =86000 km, this is the
value which we have found in the moon motion 4 points definition and we have
asked why all points use this dimension (86000 km) which is not found in the
moon orbital motion data, let's consider it in following

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The Dimension 86000 km
- The moon orbital triangle is a vertical triangle, the line BC is perpendicular on
the base EA (=449197 km)
- By that
- While the moon motion is done from perigee (B) to apogee (D) on (x-y plain) the
line BC is found on (z-axis) perpendicular on the base EA.
- Based on that,
- The line CE =373000 km = The Total Solar Eclipse Radius …… BUT
- The line CE Is NOT the Total Solar Eclipse Radius Because
- The line CE is found vertical level (z=axis) while the moon moves on (x-y plain)
- Shortly
- The moon orbital triangle is a Pythagoras triangle found on the vertical level
(z=axis) and this triangle defines the moon orbital motion points using Pythagoras
rule….
- The dimension 86000 km is found on the vertical level (z-axis)…
- What does that tell us?
- The distance EC =373000 km has an angle =13.33 degrees with the horizontal
base (EA) because the point (C) is on the vertical axis (BC) (z-axis) but when this
angle 13.33 deg be not found, the distance EC =373000 km on the horizontal level
will = the total solar eclipse radius..
Means
- The moon orbital triangle (Pythagoras triangle) defines the moon orbital motion
points vertically but the moon uses the (vertical) definition by its horizontal motion
an by that, the points definition which is done by the vertical triangle is used by the
moon horizontal motion… we should discuss this point more deeply with the
moon orbital motion equation (Point No. 2) of this current paper.

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The Point (A)
- The moon orbital triangle geometrical structure depends on 3 points (E, C and A),
E is Earth (by its gravity the moon revolves around it) and C is a vertical point
found by geometrical necessity because the Pythagoras triangle is vertical on the
base (EA), I want to say, if the Pythagoras triangle technique using by the moon
for its orbital motion necessities this triangle to be vertical on the bases (EA) that
necessitates to create the point (C) on a vertical axis (z-axis) as it's now, means,
this point (C) is found by geometrical necessities. But
- What's the point (A)? how this point can be created and can effect on the moon
orbital 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 Ecliptic Line
- The ecliptic line is seen in the figure creates an angle = 0.5 degrees with the
triangle base (red line), because the moon axial tilt declines on the Earth ecliptic
with (1.6 degrees).
- Please Note, this angle (0.5 degrees) is its right triangle hypotenuse =396800 km,
so its dimension will be =3475 km (the moon diameter), but if its right triangle
hypotenuse =1.392 mkm (lunar umbra length), so its dimension will be =12104 km
(Venus diameter).

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2- The Moon Orbital Motion Equation
2-1- 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 2.58 million km with an angle declines on the horizontal level
0.98562 degrees as typical as 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 because Lorentz Length Contraction effects on the moon motion daily distance
(2.58 mkm) with a rate 1.0725 and causes this distance to be contracted (2.41
mkm)
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1, page
26 of this current Paper)
- 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 2.58 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 n 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

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- Because of that, the moon moves its daily displacement (88000 km) depends on
Earth gravity force, but the different distance (0.17 mkm) to be covered still needs
the moon to move one more displacement (= 88000 km)
- 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 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 (2.58 mkm) which saves the moon and Earth motion 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 the actual motion distance (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.41 mkm) and Earth motion distance (2.58 mkm), so
the moon needs to move this distance perfectly, but if it's used as a full
displacement the moon would be prisoner in the apogee orbit (r=0.406 mkm)
always as we have discussed before, because of that, the moon creates Pythagoras
triangle technique by which the moon moves actually 88000 km 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 and second to
move in near orbits to Earth, that shows the intelligent moon motion techniques…
- But
- How the moon defines its motion boundaries! means, How the perigee and apogee
radiuses are defined? Let's try to answer in the following point…

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2-2- Perigee and Apogee Radiuses Definition
- During 29.53 days Earth moves 29.2 deg, because Earth moves daily 0.9856 deg
- During 29.53 days the moon moves 389.2 deg, because the moon moves daily
13.18 deg (389.2 deg =360 deg +29.2 deg)
- The data shows clearly the Earth and moon motions harmony
But
- (389.2 degrees /29.2 degrees) = 13.328
Please remember
- If
- BC =363000 km (Perigee Radius)
- AB = 86000 km so
- AC = 373000 km and
- (θ) = 13.328 degrees
- This is our first triangle in the moon orbital motion because the dimension (BC)
=363000 = Perigee radius, and the moon can't be more near to Earth than perigee
radius…
- Of course the rate (389.2 deg/29.2 deg) = 13.328, is a rate and the angle (θ) is a
value defined in degrees units, but that can't disprove the claim, because the
geometrical interaction in the moon orbital triangle can simply uses the rate 13.328
to produce the value 13.328 degrees
- I want to say that, the moon orbital perigee radius is defined based on the rate
(389.2/29.2 =13.328), because this point (perigee radius =0.363 mkm) is defined
by an interaction of gravities of the sun and Earth effect on the moon motion…this
interaction is seen in this rate between it expresses about the harmony of Earth and
moon motions…

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- Shortly,
- The sun and earth gravities effect on the moon motion prevents the moon to be
more near to Earth than the perigee radius (r=0.363 mkm), this effect which
prevent the moon to be limited to perigee point, this same effect create the deep
harmony between the moon an Earth motion because of that (389.2/29.2 =13.328)
and (365.25 days Earth orbital period /27.32 days the moon orbital period =13.37).
Apogee Radius Definition
- The apogee radius (r=0.406 mkm) is defined by some different method because
o The moon moves daily 88000 km and this distance is fixed and can't be
changed because the moon needs it to cover the different distance with Earth
motion
o During 29.53 days, by this motion distance (88000km) the moon moves a
distance = 2.598 mkm = 2θ x (0.413 mkm)
o This value (413000 km is different from apogee 406000 km with 2%)
o Regardless the error, the point is, the daily distance 88000 km defined this
orbital radius (r=0.406 mkm), because it's the suitable radius to perform the
moon motion daily displacement –
Shortly
- Where perigee radius (r=0.363 mkm) is defined by the sun and Earth gravities
interaction effect on the moon motion, but apogee radius (r=0.406 mkm) is defined
as a result for the moon daily displacement where the radius r=0.406 mkm is
suitable to be passed through to perform the required displacement 88000 km
- The moon motion (neglects) the error (2%) because the moon doesn't use its
displacement 88000 km as a real displacement every solar day, because the moon
depends on Pythagoras triangle technique, by that the moon can move through
orbits more near to earth than the apogee orbit and even can reach to perigee
radius, so to use the displacement 88000 km as a real displacement every day
during 29.53 days is somehow impossible choice, So the error 2% isn't effective.

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2-3 The Moon Orbital Motion Equation
- Based on our previous explanation we may discuss here the moon orbital equation,
The moon orbital motion equation
θ1 = the angle (θ) for today
θ0 = the angle (θ) for yesterday
0.98562 deg = Earth motion per solar day (= the moon motion per solar day
because both motions must be typical).
- We have discussed the great effect of the angle (θ) on the moon motion, for that,
the equation depends on the angle (θ) definition because this angle defines the
moon real displacement (L = 88000 Cos (θ)) and the distance between the
moon and perigee radius (r=0.363 mkm) and by that defines where's the moon is
the distance between perigee and apogee.. by using this technique the moon
position can be defined clearly day by day
- The equation is interesting one but still it has difficulties in the moon position
definition with accurate range, to see the depth behind we need use some data as
examples to show how this equation works and how the moon real motion works
and why the moon position definition still needs more accurate tool
- Let's analyze some data in following…

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The Equation Using Practice
- The angle (θ) can be greater than 28.63 degrees, let's see why
- (L = 88000 Cos (θ) = 77237 km) , this is the real displacement..
- During 29.53 days, the total distance will be 2.2809 mkm = 2π x 363000 km
- That's why The angle (θ) can't be greater than (28.63 degrees) because the moon
can't be more near to Earth than perigee radius (r=0.363 mkm)
- Let's move day after day
o 28.63 degrees – 0.98562 degrees = 27.649 deg, its (L) = 77951 km
o During 29.53 days = 2.30 mkm = 2π x 366357.6 km
o During 29.53 days = 2.322 mkm = 2π x 369606 km
o 22.7209 degrees – 0.98562 degrees = 21.73528 deg, its (L) = 81743.6 km
- The previous explanation uses the moon motion equation with some data to show
how the equation woks…

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- Let's use the first example in more details
o During 29.53 days = 2.30 mkm = 2π x 366357.6 km
- The moon (before motion) was found on point (363000 km = Perigee radius) and
the moon will move during 1 solar day and that decrease the angle (θ) with the
value 0.9856262 degree…
- So, in this figure, the moon before motion is in point B (perigee point=0.363 mkm)
- The value 0.9856262 degree change the angle (θ) and caused a new (real
displacement (L = 77951), so the moon will move (77951 km) and reach to the
point (A) (= 366357.6 km), here we have a clear direction of motion
- The distance (AB =77951 km)
- The angle BCA (θ) (in the figure) = 12.195 degrees why
- Because the moon move per solar day (13.18 degrees) and we have consumed
0.98562 degrees to define the displacement (77951 km) and the rest is 12.195 deg
- The explanation shows that, how the equation is a useful tool to define motion
daily and its position between perigee and apogee…
The Equation Disadvantages
- The equation is inaccurate in the moon position definition…
- By using real data the equation accuracy in the moon position definition can't be
more than 65%
- The basic difficulty is that
o The equation supposes the steady motion and creates a different in motion
per solar day around 3500 km on the distance from perigee to apogee but
o The moon real motion makes jumps and the moon may jump 6000 km in
one solar day while the equation defines only (3500 km) and that creates the
great error in it.

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2-4 2nd
Force Effects On The Moon Orbital Motion
- As we have discussed in the Point (A) (page no. 10),
- This point (A) is one pillar in the moon orbital triangle but there's no explanation
how this point can be created… it's found basically for geometrical necessities
- I suggest that, there's another force effects on the moon orbital motion and this
point (A) is a proof for this second force existence
- There's one more reason to suppose this (2nd
) force …. Because, the moon motion
needs 2 displacements (2 x 88000 km) to cover the distance (0.17 mkm), but the
moon moves one displacement seen by us through its orbit, where's the other
displacement be passed?
- That creates 2 reasons support the same conclusion that, there's one more force
effect on the moon orbital motion…
- This 2nd
force I claim, is found by interaction between the sun and Jupiter masses
gravities effect on the moon motion, because the sun gravity on the moon is
greater than Earth gravity effect on the moon, and Jupiter effect here creates an
interaction with the sun gravity to create specific effect on the moon motion.. this
interactive effect is effected also by Earth gravity… shortly.. that creates 2 points
of gravities total, Earth and its moon (these are the 2 points) and these 2 points are
effected by (the sun & Jupiter) gravities interactive effect, by that, another force is
created and be effective on the point (A) found on 43000 km from apogee radius
(0.406 mkm).
Please remember
- 149.6 mkm (Earth orbital distance) = 1047 x 142984 km (Jupiter diameter),
this equation shows that (the sun /Jupiter) masses rate effects on the Earth orbital
distance definition and that tells Jupiter effects on Earth orbital distance, and by
this same effect Jupiter effect on the moon orbital motion and that causes the
distance (EA) to be = 449197 km = Jupiter Circumference

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2-5 2nd
Orbit Is Found For The Moon Motion
- The moon needs to move one more displacement =88000 km, to cover the
distance 0.17 mkm.
- The moon moves its first displacement (88000 km) through its orbit around Earth,
but where the moon moves the second displacement?
- There must be 2nd
orbit for the moon motion, but where this orbit?
- We know that, the lunar eclipse umbra length = 1.392 mkm = the sun diameter
- And
- The distance EA = 449197 km = Jupiter diameter
-
- The distance from the point (A) to the end of the lunar umbra length = 942803 km
- But
- The Triangle EAC Perimeter = 942803 km
- That tells us, the point (A) separates between 2 equal values
- The first value is the triangle EAC perimeter and the second value is the distance
from the point (A) to the end of lunar umbra length…
- That tells us, the moon 2nd
orbit is a neighbor one to the first orbit, simply the point
(A) separates between the moon 2 orbits…
- But how to understand that?
- Can the moon be out of apogee radius (0.406 mkm)?! Because Earth gravity
prevents the moon to move out apogee radius… but
- The 2nd
orbit position is defined clearly by the previous data analysis, and we have
to accept that this definition is a correct one and there's one more orbit found
beyond the point (A), even if the moon can't move through this orbit
- So we have some dilemma her because the data tells a second orbit must be found
for the moon motion but there's no way to use this second orbit by the moon
because the moon is a prisoner behind the apogee radius (r=0.406 mkm)

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- How to solve this dilemma?
- The power which is provided by the moon orbital triangle is the geometrical rules,
the triangle shows that the moon orbital motion is done based on geometrical rules
perfectly an by that, the geometrical structure shows many data about this motion
- So, what advantage can be provided by the moon orbital geometrical structure
- Let's imagine a simple description in following
o Imagine a car connected by a chain of steel with another car both are the
same type and manufacturer and production date, simply 2 typical cars
o One moves by its motor and the other doesn't
o Both cars are on 2 different traces as 2 prisoner cars, no one can be out of its
track
o The only available motion is to forward… now the working car moves by its
motor and can pass the track but the other can't move …. But because they 2
cars are connected with chain of steel the working car pulls the other and
both move equal distances through the 2 tracks… that perform 2 distances
by one car motion
o What we need to perform this experiment?
o We need a suitable geometrical structure only…
o The moon doesn't move beyond apogee radius (r=0.406 mkm) but the other
displacement (88000 km) is passed through the other orbit (behind the point
A) how? Because the moon orbital geometrical structure provides this
chance for the moon orbital motion and this is the basic positive result of the
complex geometrical structure of the moon orbital triangle
o Shortly
o The second force which effects on the moon orbital motion, effects to create
a geometrical structure interactive with the moon orbital motion and create a
parallel displacement in the (2nd
orbit) as a result for the moon displacement
in its orbit around Earth and by that the moon moves both displacements and
creates the total distance (0.17 mkm) to cover the difference.

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2-6 More Basic Comments concerning the moon orbital triangle
Please review this triangle details in pages (6-7) of this current paper..
- The ecliptic line creates an angle (0.5 degrees) with the triangle base (EA) but the
triangle base is declined on the horizontal with (1.1 degrees), the total =1.6 deg
because the moon axial tilt declines on the Earth ecliptic with 1.6 degrees, but
there's 0.5 degrees is consumed in the moon diameter as we will discuss in point
(no.3) that means, the angle between the moon and the triangle base (EA) =0.6
degrees and that means the angle between the moon and the ecliptic =1.1 degrees
- We remember that, by Pythagoras triangle, the angle 1.1 degrees defines the moon
orbital distance from perigee to apogee, and that means the angle 1.1 degrees
allows the moon to move through this distance from perigee to apogee but because
we know that perigee radius (r=0.363 mkm) is defined by the gravity interaction
effect and the apogee radius (r=0.406 mkm) is defined to be suitable for the moon
daily displacement, that means, the angle 1.1 degrees is related to the moon daily
displacement definition.

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- In angle (BCD =26.6 degrees) defines the distance from perigee to apogee, as seen
in the triangle..
- That means, a deep dependency is found between these 2 angles (1.1 and 26.6 deg)
- Notice (1)
- In this triangle
- AC =406000 km (apogee radius) and AB =86000 km ….So
- BC = 396800 km
- But
- Tan (0.5 deg) x 396800 km = 3475 (The Moon Diameter)
- Also
- Tan (0.5 deg) x 1.392 mkm (the lunar umbra length) = 12104 (Venus Diameter)
Notice (2)
- 181843 km sin (1.1 degrees) = 3475 (The Moon Diameter)
o The value (181843 km) we have discussed before where
o (406000 km)2
= (363000 km)2
+ (181843 km)2
o Also
o The triangle BCS perimeter =181843 km (error 1%)
Notice (3)
- In the triangle BCZ, the hypotenuse CZ = 88000 km while the angle BCZ =12.195
degrees! Where we know that (13.18 deg – 0.98562 deg = 12.195 deg), so the
angle 12.195 degrees defines the moon daily motion after remove (0.98562 deg)
but the distance 88000 km is the moon daily displacement… the data tells that they
are defined based on geometrical rules and that means, the orbital triangle data
takes into consideration the moon daily displacement in the triangle data
definition.

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3- Jupiter and Uranus effects on the moon orbit
3-1 The Moon Diameter Division
Figure No. 2 (my figure) (The Moon Diameter Division)
- Uranus Axial Tilt (97.8 degrees) has an angle (91.1 degrees) with The Moon Axial
Tilt (6.7 degrees), so The blue vertical line is Uranus axial tilt and the angle in the
circles center = 91.1 degrees
- This value 91.1 degrees is decreased by 0.5 degrees which is consumed by the
moon diameter (the circles express the moon diameter with some division), also
0.6 degrees is consumed by the green box (I suppose Saturn & Jupiter motions
interaction effect is done here to create this 0.6 degrees)
- So the angle between the blue vertical line and the horizontal red line 90 degrees
- The red horizontal line is the moon orbital triangle base (AE = 449197 km)
- The blue vertical line angle above the moon directly =90.6 degrees
- But
- In this figure I cut layers from the moon diameter to create smaller moon diameters
to consume smaller angles than (0.5 degrees), and based on that we measure the
angle of The blue vertical line, let's explain this moon diameter division:
- The Blue Circle its diameter R1= 1390 km and r1 =695 km
- The Red Circle its diameter R2= 2085 km and r2 =1042.5 km
- The Black Circle its diameter R3= 2185 km =π x 695 km and r3 =1092.5 km
- The Brown Circle its diameter R4= 3208 km and r4 =1604 km =5040/π

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- The Orange Circle is the moon itself its diameter R5= 3475 km = 5 x 695 km
- The moon consumes 0.5 degrees for its diameter (3475 km) …So
- R1 = 1390 km, the consumed angle will be 0.2 degrees i.e. the angle above The
Blue Circle = 90.9 degrees.
- R2= 2085 km (60% of the moon diameter) and because of that, the consumed angle
will be only 0.3 degrees, means, the angle above The Red Circle =90.8 degrees
- Note Please, 90.8 degrees = 90 degrees + 0.8 degrees (Uranus orbital inclination),
by that the rest degree of Uranus axial tilt = its orbital inclination vertically.
Notice No. 1
- Pluto Axial Tilt 122.5 degrees = 7.1 x 17.2 degrees (Pluto orbital inclination),
- 7.1 a rate created by Lorentz Length Contraction Phenomenon…
- That tells us, the values (122.5 deg and 17.2 deg) are equivalent values, how? It's a
contraction, the energy of (122.5 deg) in contracted by 7.1 and the rest 17.2 deg is
created from this same energy (122.5 deg). they are 2 equivalent values but one of
them passed through a different frame and faced a contraction phenomenon and
because of that its value (122.5 deg) became (17.2 deg)
- This notice is provided here to show that in the solar system there's some way to
create A Planet Axial Tilt = Its Orbital Inclination…. What's the important
result for this equality? We have seen this equality in Uranus & Pluto data, where
we know that (122.5 deg x 0.8 deg = 97.8 deg)– there's some geometrical machine
found behind this equality…
Notice No. 2
- Are these divided diameters real diameters or invented numbers? Let's test them :
(1)
- 149.6 mkm (Earth Orbital Distance) = 3475 km (the moon diameter) x 43000 km
(the distance from perigee to apogee)
- 149.6 mkm (Earth Orbital Distance) = 2085 km (R2) x 71492 km (Jupiter Radius)

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- 149.6 mkm (Earth Orbital Distance) = 1047 x 142984 km (Jupiter diameter) (1047
= The Sun Mass / Jupiter Mass)
(2)
- R4= 3208 km x 466884 km = 149.6 mkm x π2
- 466884 km = Jupiter motion distance during its day period.
(3)
- R3= 2185 km x π x 21346.6 km (Mars Circumference) = 149.6 mkm (2%)
(4)
- R2= 2085 km x 71492 km (Jupiter Radius) =149.6 mkm
(5)
- 4900 mkm = 4222.6 x 1.16 mkm = 10921 km x 449197 km
- 302 mkm = 4222.6 x 71492 km (Jupiter Radius)
- 305 mkm = 88000 km x 3475 km = 25.2 x (3475 km)2
- 177.4 degrees -90 degrees = 87.4 degrees (+0.6 degree = 88 degrees)
- 3475 s x 0.3 mkm/sec =1042.5 mkm but
- 2085 mkm = 3475s x 0.3 mkm/s x2
Discussion
- The points (1-4) show, Earth orbital distance 149.6 mkm, is defined as a function
in the moon diameter, This same feature is used for the smaller diameters, which
proves these diameters are real values
- Point (No.5) shows light motions behind Planet Data – Let's remember 1st
hypothesis (Planet Motion depends On Light Motion), the show some of these
light motions which caused to create the provided data.

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

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I-Data (B)
Why These Distances Are NOT Equal?
1. 0725.1
mkm2.41nceCircumfereOrbitalMoon
mkm2.58MotionDailyEarth
=
2. 1.0725
km)(378500radiusEclipseSolarTotal
km)(406000radiusorbitalApogee
=
3. 0725.1
distanceMercuryJupitermkm720.3
DistanceOrbitalJuppitermkm6.778
= (Error 0.7%)
4. 1.0725
DistanceVenusJupitermkm670
distanceMercuryJupitermkm720.3
=
5. 1.0725
DistanceEarthJupitermkm629
DistanceVenusJupitermkm670
= (0.6%)
6. 1.0725
mkm)(1325.3DistanceVenusSarurn
mkm)(1433.5DistanceOrbitalSaturn
= (0.8%)
7. 1.0725
mkm)(1205.6DistanceMarsSarurn
mkm)(1284DistanceEarthSaturn
= (0.7%)
8. 1.0725
mkm)(2644DistanceMarsUranus
mkm)(2872.5DistanceOrbitalUranus
= (0.7%)
9. 1.0725
mkm)(4495.1DistanceOrbitalNeptune
mkm)(4894nceCircumfereOrbitalJupiter
= (1.5 %)
(10)
I-Discussion (B)
The same rate (1.0725) is used for all equations (around 18 distances = 40% of all
solar system distances) – why?
Suppose the equal distances are produced by light reflection and that cause these
distances to be equal – as I have supposed in the previous point (A).
Now suppose– part of these equal distances – is passed through another frame relative
to us – so this part of distances will suffer from Lorentz Length Contraction Effect
which is seen in the rate 1.0725
(Another frame can be found in the solar system because we deal with light motion) –
This explanation can answer why some distances are equal and others are rated with
the same rate (1.0725) – it's simply a feature of light motion.
0725.1
T.AxailEarth23.4
T.AxailMars25.2
T.AxailMars25.2
T.AxailSatrun26.7
TiltAxailSatrun26.7
TiltAxailNeptune28.3
===

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References
Light Motion Features Are Discovered in Planet Motion
https://www.slideshare.net/Gergesfrancis/light-motion-features-are-discovered-in-planet-motion
or
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-motion-trajectory-analysis-ii
Can Different Rates Of Time Be Found In The Solar System Motion?(II)
https://www.academia.edu/44334645/Can_Different_Rates_Of_Time_Be_Found_In_The_Solar_System_Motion_II_
Does Particle Data Depend on Its Motion? (Lorentz Transformations Analysis)
https://vixra.org/abs/1912.0134
Dr. Budochkina, Svetlana Aleksandrovna
Associate professor - Candidate of physico-mathematical sciences (2005)
http://www.mathnet.ru/eng/person22119
List of publications on Google Scholar
List of publications on ZentralBlatt
https://mathscinet.ams.org/mathscinet/MRAuthorID/757317
http://elibrary.ru/author_items.asp?spin=6087-3245
http://orcid.org/0000-0003-3447-0425
http://www.researcherid.com/rid/G-7453-2014
http://www.scopus.com/authid/detail.url?authorId=6507007003
https://www.researchgate.net/profile/Svetlana_Budochkina
Full list of
publications:
http://web-local.rudn.ru/web-
local/prep/rj/index.php?id=2944&p=15209
Mr.Gerges Francis Tawdrous +201022532292
Physics Department- Physics & Mathematics Faculty
Gerges Francis Tawdrous +201022532292
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
Facebook https://www.facebook.com
Researcherid https://publons.com/researcher/3510834/gerges-tawadrous/
ORCID https://orcid.org/0000-0002-1041-7147
Quora https://www.quora.com/profile/Gerges-F-Tawdrous
Google https://scholar.google.com/citations?user=2Y4ZdTUAAAAJ&hl=en
Academia https://rudn.academia.edu/GergesTawadrous
List of publications http://vixra.org/author/gerges_francis_tawdrous

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