The Moon Orbital Motion More Analysis

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
The Moon Orbital Motion More Analysis
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 – 4th
January 2021
Abstract
Paper hypothesis:
The Moon Metonic Cycle (6939.75 solar days) is created because of Uranus
Motion effect on The Earth Moon Motion.
The hypothesis Proves
- (1st
Proof) Uranus Orbital Circumference = 19 Earth Orbital Circumference,
that means, Earth revolves around the sun 19 times (19 sidereal years = 6939.75
solar days) while Uranus revolves around the sun 1 time only, if Uranus has effect
on the Earth and its moon motion, this effect should shows the value (19 sidereal
years).
- (2nd
Proof) Earth Motion Distance During Its Day Period = The Earth Moon
Motion Distance During Its Day Period = Pluto Motion Distance During Its
Day Period (Error 1%)
- (3rd
Proof) Uranus Moves During (1440 Of Its Days Period) A Distance = The
Earth Moon Total Displacement During Metonic Cycle (6939.75 Solar Days)
- (4th
Proof) The moon orbital triangle angles (33 degrees and 37 degrees) show
that the triangle is created by Uranus effect on the moon orbital motion.

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Contents
Subject Page No.
1- The Moon Orbital Motion Analysis 3
2- The Moon Motion Study Tools (New Discovered) 11
3- Metonic Cycle Creation 25
4- Uranus Motion Analysis 40
5- Appendix No.1 55

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1- The Moon Orbital Motion Analysis
1-1 Preface
1-2 Why Does The Moon Use Pythagoras Triangle In Its Motion?
1-3 How Does The Moon Use Pythagoras Triangle In Its Motion?
1-4 The Moon Orbital Motion
1-1 Preface
- The moon uses Pythagoras triangle as one of the moon orbital motion techniques
- 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.
- The moon motion 4 basic points were the method by which I have discovered that
the moon uses Pythagoras triangle in its orbital motion.. these 4 points are
o Perigee radius (r=0.363 mkm), which is the most near point the moon can
reach to Earth.
o Apogee radius (r=0.406 mkm), which is the most far point the moon can
reach from Earth.
o T. S. Eclipse (r= 0.373 mkm), the moon creates A total solar eclipse when
the moon be at this distance from Earth or Shorter.

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o Orbital distance (r=0.384 mkm), this is the registered distance in the moon
data sheet as the moon orbital distance
- These 4 points are defined based on each other by Pythagoras rule:
o (363000 km)2
+ (86000 km)2
= (373000 km)2
o (373000 km)2
+ (86000 km)2
= (384000 km)2
o (384000 km)2
+ (86000 km)2
= (393000 km)2
o (393000 km)2
+ (86000 km)2
= (406000 km)2
(Error 1%)
- Based on this data, the concept is iscovered that, The Moon Uses Pythagoras
Triangle As One Of The Moon Motion Techniques
- But
- Why does the moon use Pythagoras triangle as one of its motion techniques?
Let's discuss how the moon uses this intelligent technique in following…

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1-2 Why Does The Moon Use Pythagoras Triangle In Its Motion?
- Let's try to summarize the answer in following:
o The moon uses Pythagoras triangle basically to decrease its displacement
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
will 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 Let's explain this intelligent technique with some data and details to show
the useful result of using Pythagoras triangle by the moon motion….

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1-3 How Does The Moon Use Pythagoras Triangle In Its Motion?
- The moon moves daily (88000 km) on the right triangle hypotenuse (AC), but the
moon creates an angle (θ) between its motion direction and its orbit horizontal
level, by that the real displacement through the moon orbit will be (L= 88000 km
cos (θ)), and by that, spite the moon moves 88000 km, but the real orbital
horizontal displacement be less than (88000 km) and this is the objective for which
the moon uses Pythagoras triangle –
As an example,
- If (θ) =28.63 degrees, the real displacement (L) = 77237 km, So, if the moon real
displacement daily be (77237 km), during 29.53 days the moon will pass a
distance = 2.28 million km and this will be the moon orbital circumference, where
2.28 mkm = 2π x (0.363 mkm)
- The Moon Orbital Perigee Radius =0.363 mkm
- That means, the moon by a real displacement =77237 km can move around Earth
through the perigee orbit (radius =0.363 mkm), this is the useful result the moon
performs by using Pythagoras triangle,
- Now let's suppose the moon doesn't use Pythagoras triangle, what would happen?
- The moon daily displacement = 88000 km, during 29.53 days the moon moves a
distance = 2.598 mkm where 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving around Earth through its apogee
orbit (or even far from apogee orbit) because the total distance can't be passed
through any more near orbit around Earth…
- The data shows how Pythagoras triangle is so useful for the moon orbital motion.

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The Angle θ
- The angle (θ) should get our attention for its specific effect…let's summarize the
idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and by that the moon can be near or at
Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon can be near or at apogee orbit
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees)
- Any way here we need to refer to one important notice in following
Notice
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with dimensions (363000 km, 373000km, 86000 km),
what's the angle (θ)? The angle (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
i.e.,
The angle (1.1 deg.) controls the moon motion from perigee to apogee, we will need
this notice later in our discussion

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1-4 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 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)
- 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 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, 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

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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 motions accompanying.
- Now we have 2 basic information about the moon orbital motion
o (1st
information) the moon uses Pythagoras triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagoras triangle in
its motion, because the moon can't decrease 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 displacement
through the moon orbit, the moon would be a prisoner in the apogee orbit (r=0.406
mkm) as we have discussed before, because of that, the moon creates Pythagoras
triangle technique by which the moon moves actually 88000 km daily but the real
displacement through the moon orbit became less (L = 88000 Cos θ) and by that
the moon can achieve 2 objectives, first to pass the required distance (88000 km)
and second to move in near orbits to Earth, that shows the intelligent moon motion
techniques…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1 of
This Current Paper)

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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 (2.58 mkm) which is contracted to be
(2.41 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 also 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
regardless our observation ability
That 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?
- By the moon orbital geometrical structure. We can discover this second orbit by
the moon orbit geometrical data analysis. This geometrical structure analysis can
lead us to discover the 2nd
orbit position. For that we have to discuss the moon
motion 2nd
orbit with The Moon Orbital Triangle Geometrical Structure Discussion

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2- The Moon Motion Study Tools (New Discovered)
2-1 Preface
2-2 The Moon Orbital Triangle Data
2-3 The Moon Orbital Motion Equation
2-1 Preface
- There are 3 new tools are discovered which can be used to study and analyze the
moon orbital motion, these 3 tools are
o The Concept (The Moon Uses Pythagoras Triangle Technique As One Of
The Moon Orbital Motion Techniques), this concept we have discussed in
point no. (1) of this paper
o The Moon Orbital Triangle
o The Moon Orbital Motion Equation
o In this point (No. 2) we will discuss the rest 2 tools which are (the moon
orbital triangle) and (the moon orbital motion equation)
o Let's start in following

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2-2 The Moon Orbital Triangle Data
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
- Ecliptic Line creates 0.5 degrees with the moon orbital triangle base (EA)

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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… more clear explanation will be provided with the
discussion of (Metonic Cycle Creation Point No. 3)

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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 is used by the moon
for its orbital motion that necessities this triangle to be vertical on the bases (EA)
and 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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2-3 The Moon Orbital Motion Equation
2-3-1 The Equation Concept
2-3-2 The Equation Test and Accuracy
2-3-1 The Equation Concept
The moon orbital motion equation
(θ1= θ0 + 1.7 degrees)
- The moon orbital motion equation is created depending on the concept which we
have discussed in point No. (1)
- The moon uses Pythagoras triangle in its orbital motion, and by this intelligent
technique the moon be under control of the angle (θ) change
- As we have discussed, the angle (θ) defines almost all the moon motion features,
this massive effect of the angle (θ) is found as a result of the moon using for
Pythagoras triangle concept…
- The moon uses Pythagoras triangle because the moon daily displacement (88000
km) is so long and if the moon moves it as its orbit real displacement the total
distance during 29.53 days will cause the moon to revolve around Earth only
through its apogee orbit (radius =0.406 mkm)
- For that reason the moon uses Pythagoras triangle, and by this intelligent technique
the moon moves 88000 km but the moon real displacement (L) through its orbit
will be less than it because (L =88000 Cos θ), by that the moon moves 88000 km
and can revolve around Earth in more near orbits than apogee orbit (radius =0.406
mkm).

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- This technique caused the moon motion be under control of the angle (θ), because,
when the angle (θ) be decreased the real displacement (L) will be longer and that
necessitates the moon to move through orbits far from perigee orbit (r=0.363
mkm) to find more wide orbits to move through (i.e. Orbits more far from Earth)
- When the angle (θ) be greater the real displacement (L =88000 km Cos (θ)) will be
shorter and the moon can be more near to perigee orbit (more near Earth)
- By that, the angle (θ) almost controls the moon motion (all) features
- For that reason the moon orbital motion equation uses the angle (θ) with a constant
- Let's see The Moon Orbital Motion Equation
- θ1= The Pythagoras triangle angle for today
- θ0= The Pythagoras triangle angle for yesterday
- 1.7 degrees = Constant (the angle 1.7 degrees expresses the moon Daily Motion)
- How does this equation work?

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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 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 through its orbit. For day motion the
equation uses 1.7 degrees, that means, the moon on perigee uses
Pythagoras triangle with angle (28.63 degrees) and during one solar day the
moon uses - 1.7 degrees and by that the angle will be (26.93 degrees)…...
The angle 1.7 degrees expresses The Moon Daily Motion
o By using Pythagoras triangle its angle (θ) = 26.93 degrees, the displacement
(88000 km) will create a real displacement through the moon orbit = 78454
km and the moon will finish its motion today at a distance 368722 km

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means the moon is far from perigee radius with (368722 km-363000 km
=5722 km )
o So, the moon tomorrow will be at the point 368722 km and will have the
Pythagoras triangle its angle 26.93 degrees.
The Descending Motion
o When the moon moves from apogee (0.406 mkm) to perigee (0.363 mkm),
so the angle (1.7 degrees) will be positive (+1.7 degrees) because the angle
(θ) is increased and the real displacement (L = 88000 Cos (θ)) be shorter.
So
o If the moon in apogee radius (r=0.406 mkm), what's the angle (θ)?
o The apogee orbital circumference = 0.406 mkm x2π =2.55 mkm = 29.53
days x 86400 km, the angle (θ) = 11 degrees
o The moon moves from apogee to perigee (descending motion)
o (θ1= θ0 + 1.7 degrees) means (θ1= 11 degrees + 1.7 degrees) = 12.7 deg.
o L = 88000 Cos (12.7 degrees) = 85847 km
o During 29.53 days so (85847 km x 29.53 days = 2.535 mkm)
o 2.535 mkm = 2π x 403467 km
So
o After one day the moon will be on 403467 km far from apogee (406000 km)
with 2540 km
Now let's see this equation test and efficiency in following

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2-3-2 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 sufficiency
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- Metonic Cycle Creation
3-1 Preface
3-2 Uranus Effect On The Moon Orbital Motion (1st
Proof)
3-3 Uranus, The Moon And Pluto Motions Interaction (2nd
Proof)
3-4 The Moon Orbital Triangle Angles Discussions (4th
Proof)
3-1 Preface
- The paper hypothesis tells (The Moon Metonic Cycle 6939.75 solar days) is
created because of Uranus Motion effect on the Earth Moon Motion.)
- What proves can be provided for this hypothesis?
- (1st
Proof) Uranus Orbital Circumference = 19 Earth Orbital Circumference,
that means, Earth revolves around the sun 19 times (19 sidereal years = 6939.75
solar days) while Uranus revolves around the sun 1 time only, if Uranus has effect
on the Earth and its moon motion, this effect should shows the value (19 sidereal
years)
- (2nd
Motion Distance During Its Day Period = Pluto Motion Distance During Its
Day Period (Error 1%) (This data proves the paper hypothesis)
- (3rd
Earth Moon Total Displacement During Mitonic Cycle (6939.75 Solar Days)
(3rd
proof is discussed in "Uranus Motion Analysis" (Point No. 4)
- (4th
Proof) the moon orbital triangle angles (33 degrees and 37 degrees) show
that the triangle is created by Uranus effect on the moon orbital motion.
- The hypothesis refers to that, Planet mass gravity effects more extending than the
range defined by the gravitation equation based on (1/r2
), showing that the law
may need to be revised

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3-2 Uranus Effect On The Moon Orbital Motion (1st
Proof)
In this figure
- The Red Ball Shows Earth
- The Yellow Ball shows The Earth Moon
- The Blue Ball shows Uranus
- (S) is the Sun
- The figure suggests that, a triangle contains these 3 planets together in their
revolutions around the sun – that means
Paper hypothesis
- (The Moon Metonic Cycle 6939.75 solar days) is created because of Uranus
Motion effect on the Earth Moon Motion.)
The hypothesis Explanation
- Let's suppose the three planets, Earth, its moon and Uranus move in parallel to
each other in their revolutions around the sun, and to guarantee this parallelism
between them the figure provides a triangle contains these 3 planets -
- Uranus orbital circumference = Earth orbital circumference x 19
In accurate calculations
- Uranus (18048 mkm) = Earth (940 mkm) x 19 (error 1%)

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

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3-3 Uranus, The Moon And Pluto Motions Interaction(2nd
Proof)
(2nd
Motion Distance During Its Day Period = Pluto Motion Distance During Its Day
Period (Error 1%)
Let's analysis this data as deep as possible in following:
Earth Moves Per A Solar Day A Distance = 2.5734 Mkm
But
The Earth moon daily displacement =88000 km and during 29.53 solar days the total
distance = 2.598 mkm (+1%)
And
Pluto moves during its day period (153.3 hours ) and distance = 2.598 mkm
(+1%)
Means
- During 6939.75 solar days (day =24 h), Earth moves a distance = Uranus Orbital
Circumference, but with less (1%) of Uranus Orbital Circumference
- During 6939.75 Pluto days (day =153.3 h), Pluto moves a distance = Uranus
Orbital Circumference, (Zero error)
- During 6939.75 The moon day period (day =29.53 solar days), the moon moves a
distance = Uranus Orbital Circumference, (Zero error)
- Why?? Because…
- Earth And Its Moon Motions Are Effected By Uranus Motion and because Urnus
orbital circumference = 19 Earth orbital circumference, because of that, The moon
Metonic Cycle be (19 Sidereal Years = 6939.75 solar days) …… But
- Pluto motion distance = Uranus Orbital Circumference because of Uranus and
Pluto motions interaction, by that, the 3 planets (Earth, its moon and Pluto) be
effected by the same one source (Uranus motion) and because of that they move
during their days periods equal distances – let's see the data with some details in
following…

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I-Data
(i)
4.7 km/s x 153.3 h x 3600 x 6939.75 = 18050 mkm (Zero error)
(ii)
29.8 km/s x 24 h x 3600 x 6939.75 = 18050 mkm (-1%)
(iii)
88000 km x 365.25 days x 1123 = 18050 mkm x 2 (Zero Error)
(iv)
90560 days = 13.177 x 0.99 x 6939.75 days
(v)
Pluto during 6939.75 days moves a distance = 2815 mkm
II-Discussion
Equation No. (i)
4.7 km/s x 153.3 h x 3600 x 6939.75 = 18050 mkm
Equation No. (ii)
29.8 km/s x 24 h x 3600 x 6939.75 = 18050 mkm
- The Distances Are Equal = Uranus Orbital Circumference 18055 mkm
- The days number (6939.75 days) are equal
- Means, the distances are equal and the days numbers are equal!
- What's the different??
- The velocities are different….. so the days periods are created based on the rate of
the velocities (That tell Why "Earth Velocity / Pluto Velocity" = "Pluto Day
Period / Earth Day Period")
- The data tells how that happens
- Earth and its moon motions are effected by Uranus, and because Uranus orbital
circumference = 19 Earthg orbtial circumference, Metonic Cycle became 19 years
- But Pluto is connected by its motion interaction with Uranus Motion…

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- That creates a connection between 4 Planets
- The simple proof is that
- Earth motion distance during 19 sidereal years = Uranus orbital circumference but
with less 1%
- This 1% is not found neither in Earth moon motion distance nor in Pluto motion
distance.
- So this error (1%) can be used as a clear proof for the paper hypothesis..
Notice
- Because Pluto in a partner in Metonic Cycle by its motion interaction with Uranus
motion, many of Earth and Pluto data be very similar as seen in data no. (iv) & (v),
where many other data shows that between the Earth and Pluto (for example Pluto
orbital distance 5906 mkm = Earth orbital circumference 940 mkm x 2π)
Equation No. (iii)
88000 km x 365.25 days x 1123 = 18050 mkm x 2
- The moon moves a displacement = 88000 km per solar day, and during 365.25
solar days the total displacements = 32.14 mkm
- To pass Uranus orbital circumference (18050 mkm), the moon needs 561.7
sidereal year (561.7 x 365.25 days) …..But
- To pass double Uranus orbital circumference (18050 mkm x 2), the moon needs
1123 sidereal year (1123 x 365.25 days)
- I have used 2 values of Uranus orbital circumference because of the number 1123
- Where …. (30589 solar days /27.32 solar days) =1120
- 30589 days = Uranus Orbital Period and 27.32 days = The Moon Orbital Period
- It's almost the same rate ….. 1123 years = 19 years x 59 Where ( The moon day
period =(59/2) =29.53 days) and (59 days x 2.573 mkm = 151.8 mkm (Earth
orbital distance) (error 1%).
- The data tells, There's greater cycle of 38 years (basically between Uranus and the
moon motions) in addition Metonic Cycle (19 years =6939.75 solar days)

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3-4 The Moon Orbital Triangle Angles Discussions (4th
Proof)
Point No. (A) (The angle 32.967 degrees)
- This is the moon orbital triangle, but I have removed all data and added only the
new line (CW) as seen in the figure
- The angle BCW = 32.967 degrees
- CW = 102500 km
- BW = 55756 km
Let's analyze this data in following
- BW = 55756 km = 43000 km +12756 km (Earth Diameter)
- CW= 102500 km = 2 x 51118 km (Uranus Diameter)
- The angle BCW = 32.967 degrees where 32.967 deg x 0.8 = 26.36 degrees
o 0.8 degrees = Uranus Orbital Inclination
o 26.6 degrees = the angle controls the moon motion from perigee to
apogee as we have seen in the moon orbital triangle original form (BCD), in
our investigation this angle be = 26.6 degrees (Error 1%)
- Before to move into the details, let's explain the general idea in following:
o Uranus motion causes the moon motion revolving around the Earth, this is
the paper hypothesis– Earth force imprison the moon inside the range
(Perigee & Apogee) – so – any effect of Uranus motion on the moon motion
will be acceptable if it be in the range (Perigee Apogee).

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o Uranus Motion causes the line BC in the moon orbital triangle
o Let's suppose the line BC expresses Uranus Axial Tilt, We need here is to
imagine that, This line (BC) receives the motion from Uranus and provides
it to the moon – let's try to see how that happens….
o The Line BC moves in angle (32.967 degrees), that means, the line BC
changes the angle (BCW) from Zero degree to (32.967 degrees) and then
return to Zero Again
o It's a cycle, but the line BC moves in its opening for the angle from Zero to
(32.967 degrees) in some way and NOT return through this same way when
the opened angle (32.967 degrees) be closing to be Zero
o That creates a motion of cycle of this line CB (column).
o This motion is A Waving Motion (going and return but not through the
same way).
o By this motion the line BC causes the moon to revolve around the Earth…
now the line BC should be considered as a column built on the moon body
or is connected strongly by it – and that means- if this line BC moves (by
angle opening or closing) the moon will move with it
o The angle is (BCW =32.967 degrees), but the moon doesn't reach to this
angle range for 2 reasons (1st
) Because the moon can't move beyond apogee
radius (0.406 mkm) (2nd
) Because of Uranus Orbital Inclination on this
angle, and that means, Uranus motion defines the moon range to be (32.967
degrees) but in this definition Uranus motion took into consideration Uranus
orbital inclination effect and based on that the rest angle is 26.36 degrees
o In fact, the angle = (BCD) =26.6 degrees
o The angle 26.36 degrees controls the moon motion during a distance 42500
km (i.e. From perigee to Before apogee point with 500 km) (error 1%).
o Error 1% is found frequently in Uranus effect on Earth & moon motions.

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Notice
- Uranus effect is seen strongly in the data for example
o CW = 102500 km = 2 Uranus Diameters
o BW = 43000 km (perigee apogee distance) + 12756 km (Earth Diameter)
o AW = 30589 km (error 0.4%) (where 30589 days = Uranus orbital period)
More Data
- (BCW) = (32.967 degrees) x 3 = 98.9 degrees
- Where
o 98.9 degrees = 97.8 degrees (Uranus Axial Tilt) + 1.1 degrees
o And
o 97.8 degrees (Uranus Axial Tilt) = 96.7 degrees + 1.1 degrees
o Where (96.7 degrees =90 degrees +6.7 deg The Moon Axial Tilt).

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Point No. (B) (The angle 36.912 degrees)
- In this figure I use again the moon orbital triangle after I have removed all details
and added the black line CT
- The angle BCT = 36.912 degrees
- BT = 64600 km
- CT = 107560 km
- The angle ECT = 113.58 degrees
- Note Please
o Cos (36.912 degrees) = 0.8
o Tan (36.912 degrees) = 0.7511
I-Data Analysis
- (97.8 degrees /122.5 degrees) = Cos (36.912 degrees)
o 97.8 deg = Uranus Axial Tilt
o 122.5 deg= Pluto Axial Tilt
o Uranus Orbital Inclination = 0.8 degrees
o Also Cos (36.912 degrees) = 0.8
The data tells that, the angle (36.912 degrees) is used in Uranus & Pluto motions
interaction data

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- (13.177 degrees /17.4 degrees) = Tan (36.912 degrees)
o 13.177 degrees = The Moon Motion Per Solar Day
o 17.4 degrees = The Inner Planets Orbital Inclinations Total
o 17.2 degrees = Pluto Orbital Inclination
The data tells that, the angle (36.912 degrees) is used for The Moon daily motion
- (17.4 degrees/ 23.45 degrees) = Tan (36.6 degrees) (error 1%)
o 23.45 degrees = Earth Axial Tilt
o (36.6 degrees) is different with (36.912 degrees) with 1%
The data tells that, the angle (36.912 degrees) is used for Earth Axial Tilt (Please
remember Earth data has always an error =1% concerning Uranus effect).
- (26.3 degrees/ 32.96 degrees) = Cos (36.912 degrees)
o 23.45 degrees = Earth Axial Tilt
o (36.6 degrees) is different with (36.912 degrees) with 1%
The data tells that, the angle (36.912 degrees) is used for The moon motion angle
(26.6 deg) From Perigee To Apogee.
o The angle = BCT
o 29.53 days = the moon day period ……………. Also
o 29.2/29.53 = 0.99
The data tells, the angle (36.912 degrees) is used for The Moon Day Period (29.53
solar days).

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

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

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

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

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

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- For more confirmation, Uranus gives us the value (607.3 mkm) as a result of the
equation (51118 km x 1092
), where we remember this equation because
o Mercury orbital distance (57.9 mkm) = Mercury diameter x 1092
o Earth orbital distance (149.6 mkm) = Earth diameter x 1092
o Saturn orbital distance (1433.5 mkm) = Saturn diameter x 1092
- By this same equation Uranus produces the result 607 mkm = the moon total
displacement during Metonic Cycle = Uranus motion distance during 1440 its days
- The data shows, the moon motion is effected by Uranus Motion, Supporting the
hypothesis, (Metonic Cycle is created by Uranus effects on the moon motion)
Notice
- During (1440 days of Uranus days period) Uranus moves a distance = 606.3 mkm
- 1440 days x 17.2 hours = 24768 hours = 1032 solar days
- The Moon total displacement during (6939.75 solar days) = 610.7 mkm
- i.e.
- Equal distances (error 1%) are passed in 2 different periods of time
o (6939.75 solar days / 1032 solar days) =6.724
o But
o 6.7 degrees = The Moon Axial Tilt (error 0.3%)
o We may remember that, a deep relationship is found between Uranus axial
tilt and the moon axial tilt (97.8 degrees – 6.7 degree = 91.1 degree) and
based on this relationship the moon orbital triangle is designed because the
moon orbital triangle base (EA) declines on the horizontal level with (1.1
deg) which enable the line BC to be perpendicular on the triangle base EA
(= 449197 km).

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- let's summarize the whole idea in following to see the left questions:
o Uranus motion effects on the Earth moon motion and forces the moon to
move Metonic Cycle during 19 years
Shortly
o Metonic Cycle is created because Uranus orbital distance = 19 Earth orbital
distance and because Uranus effects on the moon motion, Uranus revolution
around the sun effects on the moon motion trajectory creating Metonic
Cycle
o This data is so puzzled because Uranus is so far from Earth and the distance
prevents Uranus gravity force to effect such clear effect
o The data arrangement provides a real puzzle needs to be solved
o The next question ..,
o How does Uranus motion effect on the moon motion?
o The data tells that, Uranus axial tilt has a strong effect on the moon axial tilt,
that means, Uranus motion effects on Uranus axial tilt and Uranus axial tilt
effects on the moon axial tilt which effects on its orbital motion …
o So
o We have 4 pieces of motion, from motion to axial tilt to axial tilt to motion,
how that is happened? We still need to deepen our analysis as possible to
see how this is done….

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II-More Data Analysis
(1)
97.8 degrees = 19 x 5.1 degrees (the moon orbital inclination) (error 1%)
Where, (97.8 degrees =Uranus axial tilt)
- Error 1% is used usually in Uranus motion effect on Earth and its moon motion
- 19 = (Uranus orbital distance/ Earth orbital distance)
- The previous equation tells that, Uranus axial tilt effects by (97.8 degrees) during
19 years (with error 1%), If so, as a result Uranus axial tilt will effect with a value
5.1 degrees per a year
- If so, the conclusion will be, The Moon Moves During 1 Year Based On Its
Orbital Inclination 5.1 Degrees…
How?!
(2)
5.1 million km x 2π = 32.05 million km
And
88000 km x 365.25 solar days = 32.05 million km
- The data tells that,
- Because of the moon orbital inclination (5.1 degree) the moon daily displacement
=88000 km
- We know, 1 degree =1 mkm because Mercury orbital circumference =360 degrees
= 360 mkm.
- The data is clear and understandable but no geometrical mechanism is seen in it,
because how the displacement 88000 km depends on the moon orbital inclination
5.1 degrees?!
- Let's try to analyze that as deep as possible in following

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(3)
The Moon Orbital Motion Equation
θ1 = θ0 + 1.7 degrees
- Why does the equation use 1.7 degrees as the moon daily motion value?
- 5.1 degrees (the moon orbital inclination) =3 x 1.7 degrees
- But
- 98.9 degrees = 32.967 x 3 (where 98.9 deg = 1.1 +97.8 deg Uranus Axial Tilt)
- There's some geometrical mechanism for this data explanation
(4)
449197 km = 88000 km x 5.1
- This is the equation which defines the moon daily displacement….
- The Moon Orbital Triangle Base (EA =449197 km) is the main line in the moon
orbital triangle and it contains the distance from Earth to the moon in perigee and
in apogee and to the Point (A) (where EA = 449197 km)
- The equation tells that, by some geometrical rule the moon orbital inclination
causes the moon daily displacement to be = 88000 km
(5)
5.1 mkm = 2 x 2.55 mkm
- The moon orbital circumference = 2.55 mkm, but
o Earth moves per solar day a distance = 2.573 mkm (+1%)
o The moon moves during its day period a distance = 2.598 mkm (+2%)
- The data tells, the moon orbital circumference is created depending on the moon
orbital inclination (accurately) ….. but because the moon day period (29.53 solar
days) = 2.55 million seconds, that tells us, the moon orbital circumference and its
day period are built depends on the moon orbital inclination …..But
- Why 5.1 mkm =2.55 mkm x 2 ??
- Why the moon orbital circumference doesn't = 5.1 mkm?!

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- Because
o 29.53 solar days x 2 = 59 days
o 59 days x 2.573 mkm (Earth motion daily) = 151.8 mkm (Earth orbital
distance) (1%)
o Also
o (59)2
=3475 km = The Moon Diameter
(6)
449197 km = 1.0725 x 421056 km
- 449197 km = The Distance (EA) = Jupiter Circumference
- 421056 km = Uranus Motion distance during its day period (17.2 mkm)
- 1.0725 = The Contraction Rate
- We remember that,
- The moon moves per a solar day a motion typical to Earth Motion, means the
moon moves during a solar day a distance =2.573 mkm with an angle 0.98562
degrees on the horizontal level revolving around the sun on the Earth orbital
circumference
- If there's no contraction effect on the moon motion, the moon motion trajectory
should be shown as a parallel line to Earth motion trajectory
- But
- Because of the contraction effect with rate (1.0725) on the moon daily motion
(2.573 mkm) and contracted its to be (2.41 mkm), creating a different distance =
0.17 mkm where this different distance would cause to separate the moon and
Earth during their motions
- Because of this risk, the moon has to move its daily displacement (88000 km)
depends on Earth gravity forces and it's not enough to recover the different
distance where the other inner planets with Jupiter help the moon to create an
equal distance in parallel by its orbital motion to make total displacement (88000
x 2 = 0.17 mkm) and to recover the different distance

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- Why does the moon motion distance suffer from contraction phenomenon?
- It's a result of Jupiter and Uranus motions interaction effect on the moon orbital
motion.
- Jupiter different distances are suffered from the contraction phenomenon, let's
remember that inflowing:
o 778.6 mkm (Jupiter Orbital Distance) = 1.0725 x 720.7 mkm
o 720.7 mkm (Mercury Jupiter Distance) = 1.0725 x 671 mkm
o 671 mkm (Venus Jupiter Distance) = 1.0725 x 629 mkm
o Where 629 mkm (Jupiter Earth distance)
- The data shows, the contraction phenomenon is used for Jupiter different distances
and because Jupiter & Uranus motions interaction effect on the moon orbital
motion, the moon motion distance is suffered also from the contraction
phenomenon.
(7)
(327.6 days/88 days) = (19 degrees/5.1 degrees)
- Mercury orbital period (88 solar days) is rated to the moon sidereal year (327.6
solar days) with the rate between 19 and 5.1 degrees why?
- 19 degrees the moon regresses per a year
- 5.1 degrees = the moon orbital inclination
- We remember that, the moon daily displacement (88000 km) depends on the moon
orbital inclination (5.1 degrees),
- That means, the rate (19/5.1) expresses the moon motion per year
- So the year is found in the equation (327.6 solar days)
- But Why 88 solar days?
- What's the relationship between Mercury orbital period (88 solar days) and the
moon sidereal year (327.6 solar days)??

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4-2 Uranus Motion During 8 Pluto Days period
- Why do we need to remember Uranus 144 days Cycle? We have 2 reasons
o (1st
) Because Uranus moves during 1440 Uranus days (17.2 h) a distance =
The Moon Total Displacements During Metonic Cycle
o (2nd
) Because Uranus & Pluto Motions interaction can be seen clearly in
studying Uranus 144 days Cycle….
- How Uranus Motion Cycle (During 8 Pluto Days) Is Discovered?
o Jupiter (13.1 km/s) moves during its day period (9.9 h) a distance = Jupiter
circumference (449197 km) + 17695 km
o During 8 of Jupiter days periods (9.9 x 8 =79.2 h), Jupiter moves a distance
= 8 Jupiter circumferences + Jupiter diameter (error 1%)
o Because of Jupiter diameter I concluded that, Jupiter has a cycle in 8 days
Then
o Jupiter motion distance during 8 of its days (79.2 h) which = 3735072 km,
this same distance = Saturn motion distance during 10 of Saturn days period
o Means, Saturn moves during 10 of its days a distance = 3735072 km,
o I have concluded that, Jupiter motion energy is transported to Saturn motion
energy by the rate 80% Because of that I expected that, Saturn should
transport its motion energy to Uranus by the same 80% (but it's incorrect!)
o Saturn transported the motion energy with this rate 80% to Neptune and that
means the distance Saturn passes during 8 of Saturn days period, Neptune
passes during 10 days of Neptune days period (error 5%)
The question is why Saturn doesn't transport the motion energy to Uranus?!
- A surprise was in our waiting…
- Uranus (6.8 km/sec) moves during Pluto day period (153.3 h) a distance = Jupiter
motion distance during 8 of its days period (79.2 h) + 17695 km
- That means, during 8 Pluto days period (153.3 h x 8) Uranus moves a distance =
Jupiter motion distance during 64 Jupiter days (64x 9.9 h) +Jupiter diameter (1%)

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- It's very similar to Jupiter 8 days Cycle ….
- Easily we have concluded that, Uranus motion creates a cycle of Pluto 8 days
where Uranus moves during this period a distance = Jupiter motion distance during
64 Jupiter days
- Because Jupiter motion energy is transported to Saturn and from Saturn to Neptune
the distance should be equal for all these planets – by that – the chance if found to
compare between these motions distances - let's write them in following
I- Data
1. Uranus (6.8 km/s) moves during 8 days of Pluto days period (1226.4 hours) a
distance = (4415040 seconds x 6.8 km/s) = 30.022272 million km
2. Jupiter (13.1 km/s) moves during 64 days of Jupiter days period (64 x9.9 h =
633.6 hours) a distance = (2280960 seconds x 13.1 km/s) = 29.880576 million km
3. Saturn (9.7 km/s) moves during 80 days of Saturn days period (80 x10.7 h = 856
hours) a distance = (3081600 seconds x 9.7 km/s) = 29.891520 million km
4. Neptune (5.4 km/s) moves during 100 days of Neptune days period (100 x16.1 h =
1610 hours) a distance = (5796000 seconds x 5.4 km/s) = 31.298400 million km
5. Jupiter Circumference (449197 km) x 64 = 28.748639 million km
6. Saturn Circumference (378675 km) x 80 = 30.294001 million km
7. Neptune Circumference x 2 (311193.6 km) x 100 = 31.119360 million km
II-Data Analysis
- Let's remember these cycles in following:
- Jupiter moves during its day period (9.9 h) a distance = 466884 km but Jupiter
circumference = 449197 km, the difference = 17687 km
- During 8 of Jupiter days period, Jupiter moves distance = 3735072 km = 8 Jupiter
circumferences + 1 Jupiter diameter (142984 km =8x 17687 km) (error 1%)
- Based on that I have concluded, Jupiter has a cycle in 8 of its days (79.2 hours)
- By this analysis Jupiter 8 days Cycle is discovered
Then

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

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

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

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4-3 Uranus 144 days Cycle
- 144 days of Uranus days periods = 144 x 17.2 hours = 2476.8 hours
- 16 days of Pluto days period = 16 x 153.3 hours = 2452.8 hours
- The Difference = 24 hours = 1 Solar Day
- The data shows these are 3 cycles (144 Uranus days, 16 Pluto days and the solar
day) they are 3 cycles
- We have seen that before
o 6939.75 solar days = Metonic Cycle
o 6585.36 solar days = Saros Cycle
o 354.39 solar days = The lunar year
And
- The data shows that, there's an interaction of Uranus, Pluto and Earth motions
- This data we have discussed before and reach to the following conclusions
o Uranus motion effect on Pluto motion causes Pluto day period to be =153.3
hours where it's so long day period in comparison with the other outer
planets
o This effect is found because Pluto moves during (6939.75 x 153.3 hours) a
distance = Uranus orbital circumference
o This data also we have discussed before

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

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Equation No. (f)
21.8 x 0.8 degrees (Uranus orbital inclination) = 17.4 degrees
21.8 = Jupiter Mass / Uranus Mass
Equation No. (g)
17.2 degrees (Pluto Orbital Inclination) x 1.1 deg = (18.92 degrees)
Equation No. (h)
27.32 = 122.5 degrees (Pluto Axial Tilt) – 95.18 deg (= 90 + 5.1 the moon orbital
inclination)
Equation No. (i)
90560 days (Pluto orbital period) = 29.53 days (the moon day period) x 3061
(3061 mkm = Saturn Neptune Distance)

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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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