The Moon Orbital Motion Rules

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
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 Rules
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 – 22nd
December 2020
Abstract
Paper hypothesis:
- The Earth moon is found inside an interaction occurred between Mercury and
Jupiter Motions, because of that, the moon orbital motion can't be explained
independent from Mercury and Jupiter Motions.
The hypothesis Explanation
- The hypothesis tells that, Mercury and Jupiter motions effect on the moon motion
and creates features can't be explained independent of Mercury and Jupiter
Motions effect.
- The claim refers to that, the moon orbital motion is done based on geometrical
rules that means, the moon moves by gravity force by not in a direct effect but
through geometrical rules, to explain that, let's imagine a car moves on a track and
this track has walls on both sides, so the car has one option only to move forward,
so this motion of car is done by its motor but if the walls aren't found the car could
move in different directions but now the car has only one option – similar to that-
the moon moves by gravity forces but in a geometrical designed track and because
of that the moon has only limited options to do its motion- by that – the motion
doesn't depend on gravity forces only but also on the track geometrical design.
Paper objective
- The paper analyzes the moon orbital motion track geometrical design and rules.

I do this research
2nd
2
1- Introduction
- There are 3 new tools can be used for the moon orbital motion analysis and to
define the moon position in its orbit, let's remember these 3 tools in following:
1. The Concept (the moon uses Pythagoras triangle in its orbital motion)
2. The Equation (θ1 = θ0 + 1.7 degrees)
3. The Moon Orbital Triangle
- We have discussed these 3 tools in the previous paper, and we have answered why
the moon needs to use Pythagoras triangle in its orbital motion, where the moon
daily displacement (88000 km) is so long distance and during 29.53 days the total
distance will be 2.59 mkm which means, if the moon moves by this displacement
as a real displacement through its orbit the moon would revolve around Earth
through only its apogee orbit (r=0.406 mkm).
- For that reason, the moon creates an angle (θ) between its motion direction and its
orbital horizontal level and by this angle the real displacement through the moon
orbit will be (L = 88000 cos (θ)), makes (L) less than (88000 km) and enables the
moon to revolve round Earth through more near orbits than apogee orbit.
- Because the angle (θ) defines the moon orbital motion (all) features, the equation
(the moon orbital motion equation) uses this angle (θ1 = θ0 + 1.7 degrees) where
this the 2nd
tool to define the moon position in its orbit – an the 3rd
tool which is the
moon orbital triangle is inserted in this paper, because we need it in our discussion
- This paper discusses one more feature of the moon orbital motion, which is the
geometrical rules by which the moon move.. The paper claims, in addition to the
previous tools, there are rules control the moon orbital motion, and these rules are
defined by effect of Jupiter and Mercury because (as the paper claims) the moon is
found inside an interaction occurred between Jupiter and Mercury Motions

I do this research
2nd
3
2 – Jupiter and Mercury Effects On The Moon Orbital Motion
2-1 The Moon Orbital Triangle Perimeter Analysis
2-2 Jupiter Motion Effect On The Moon Orbital Motion
2-1 The Moon Orbital Triangle Perimeter Analysis
I-Data
(1)
136.3 km /sec x 6939.75 seconds = 943817 km
(2)
0.3 mkm /sec x 6939.75 seconds = 2082 mkm
(3)
11.77 mkm/day x 6939.75 days = 2082 mkm x (2π)2
(4)
11.77 mkm/day x 175.94 days = 2082 mkm
(5)
3.02 mkm /day x 687 days =2082 mkm
(but)
3.02 mkm /day x 365.25days =1103 mkm = 2π x175.94 mkm
II-Discussion
Equation No. (1)
136.3 km /sec x 6939.75 seconds = 943817 km
- 136.3 km /sec = the total velocities of (Mercury + Venus + Earth + Mars)
- 943817 km = The Perimeter Of The Moon Orbital Triangle (ACE)
- Equation no. (1) tells that, the inner planets motions distances total during 6939.75
seconds = the perimeter of moon orbital triangle (ACE) (this is the greatest triangle
of the moon orbit, review (The Moon Orbital Triangle Geometrical Structure)
- What's the treasure equation no. (1) gives us?! It's the period of time (6939.75 s)
- The equation tells that, the moon orbital triangle perimeter creates a period of time
for the 4 inner planets motions… the positive result is the period of time. Why this
is a useful result? Try to know in following discussion….

I do this research
2nd
4
Equation No. (2)
0.3 mkm /sec x 6939.75 seconds = 2082 mkm
- During this same period (6939.75 seconds) the light known velocity (0.3 mkm/se)
travels a distance =2082 mkm = Jupiter Uranus Distance
- So, the moon orbital triangle perimeter defines the period of time depends on the 4
inner planets velocities total, and this same period of time (6939.75 sec) is used by
light known velocity to pass the distance from Jupiter to Uranus
- So, by this same period of time (6939.75 seconds) we have 2 motions (the first by
4 planets velocities) and (the second by light motion)
- We accept that, if a light motion is found, that can create relativistic effects, and
that causes to create new rates of time and by that we can suppose that 1 second of
light motion is equivalent to 1 solar day of planet motion and based on this
hypothesis we will use the period 6939.75 days (Metonic Cycle period) as a period
of these same 4 inner planets motions to see what useful result we may receive
Equation No. (3)
11.77 mkm/day x 6939.75 days = 2082 mkm x (2π)2
- 11.77 mkm/day = the daily velocities total of (Mercury + Venus + Earth + Mars)
- Equation No. (3) tells that, the 4 inner planets motions distances total during
(6939.75 days) = light motion distance during 6939.75 seconds x (2π)2
- To make this equation more easy, so 6939.75 days = (2π)2
x 175.94 days (Mercury
Day Period) that means ….
- The 4 inner planets motions distances total (during Mercury day period175.94
days) = light motion distance during (6939.75 seconds)
- The previous explanation tells that, there's a dependency or interaction between
Metonic Cycle Period (6939.75 days) and Mercury Day Period 175.94 Days.
- Please remember, the moon orbital triangle is used to define a period of time
(6939.75 seconds), means, the moon data here is used to create a period of time

I do this research
2nd
5
Equation No. (5)
3.02 mkm /day x 687 days =2082 mkm
- Equation No. (5) tells that, the 4 inner planets motions distances total during
175.94 days (Mercury day period) = the distance Venus moves during Mars orbital
period (687 days), that creates an interaction between Mercury day period and
Mars orbital period and based on that with Metonic Cycle….let's see the following
o (Mercury velocity / Pluto velocity) = (6939.75 days / 687 days)
o This equation is so important because the rate of velocity between Mercury
and Pluto controls all planets velocities because Mercury is the greatest and
Pluto is the smallest velocities, by that, the rate between them control all
planets velocities and motions and based on that the rate between Metonic
Cycle period (6939.75 days) and Mars orbital period (687 days) should be
considered a general rate in the solar system motion.
Equation No. (5) (Part 2)
3.02 mkm /day x 365.25days =1103 mkm = 2π x175.94 mkm
- The Equation tells Venus moves during (365.25 day) a distance = 2π x175.94 mkm
but Mercury day period =175.94 solar days…That means
- The distance 2082 mkm = 1103 mkm x 1.9 (Mars orbital inclination) and because
of that (687 days /365.25 days) =1.9
o We have here 2 motions done by Venus
o (1st
) Venus moves during (365.25 days) a distance = 2π x175.94 mkm
o (2nd
) Venus moves during (687 days) a distance = 2082 mkm
2082 mkm = 2π x 175.94 mkm x 1.9
But 175.94 solar days = Mercury Day Period
the period of time (175.94 days) is used as a distance (175.94 mkm)
and this is a usual relationship between Mercury and Venus because
(Mercury moves during its rotation period 58.66 days a distance =243
mkm But Venus rotation period =243 solar days)

I do this research
2nd
6
- The useful result of the previous analysis is that
o The moon orbital triangle perimeter is used to define a period of time
(6939.75 seconds)
o The planets motions interaction is seen by Venus Motion
o By Venus Motion during (365.25 days) and (687 days), Mars orbital
inclination 1.9 degrees is created.
- We leave the data at this point and will start another point of data analysis…. The
general discussion will use all of them

I do this research
2nd
7
2-2 Jupiter Motion Effect On The Moon Orbital Motion
I-Data
(6) (a)
- 943817 km = 27.78 km/sec x 33968 seconds
(b)
- 449197 km = 13.1 km /sec x 33968 seconds
(7)
- 4900 mkm = 10921 km x 449197.5 km
(8)
- 142984 = 13.1 km /sec x 10921 seconds
(9)
- 203.7 km /sec x 6939.75 seconds = 1392000 km (1.6%)
II-Discussion
Equation No. (6) (a)
943817 km = 27.78 km/sec x 33968 seconds
- 27.78 km/sec = The Moon Velocity, which = (Earth velocity /1.0725)
- 943817 km = the perimeter of the moon orbital triangle (ACE)
- Equation no. (6) (a) tells that, the moon moves during 33968 seconds a distance =
943817 km = its orbital triangle perimeter…. Why?!
- As in the previous discussion, the moon data is used to define a period of time
(33968 seconds).
- Please observe that clearly, the moon data is used to define periods of time for
other planets motions
Equation No. (6) (b)
- 449197 km = 13.1 km /sec x 33968 seconds (error 1%)
- Jupiter moves during (33968 seconds) a distance = Jupiter circumference
- So the connection between Jupiter motion and the moon motion is the period of
time (33968 seconds)

I do this research
2nd
8
Equation No. (7)
- 4900 mkm (Jupiter orbital circumference) = 10921 km x 449197.5 km
o 10921 km = The Moon Circumference
o 449197.5 km = Jupiter Circumference
- So, why Jupiter orbital circumference is defined based on the moon circumference
and Jupiter circumference? Because
o Jupiter moves a distance = its circumference in a period (33968 seconds)
and
o The moon moves during this same period (33968 seconds) a distance = the
perimeter of the moon orbital triangle (ACE)
o (Please note EA = Jupiter Circumference) …But …
Equation No. (8)
- 142984 = 13.1 km /sec x 10921 seconds
- Jupiter moves during 10921 seconds a distance = Jupiter diameter but the moon
circumference = (10921 km)
- That means
- The Period (33968 seconds = π x 10921 seconds) (error 1%)
i.e.
- The moon uses period = (10921 seconds x π) for its motion to pass a distance = the
perimeter of the moon orbital triangle (ACE)
- Again
- The moon data is used to define periods of time
Equation No. (9)
- 203.7 km /sec x 6939.75 seconds = 1392000 km (1.6%)
- 203.7 km/sec = The Solar Planets Velocities Total
- Equation no. (9) tells that, during the same period (6939.75 seconds) all planets
motions distances total = the sun diameter (error 1.6%) but we know that
- The sun diameter = Jupiter diameter x π2
(error 1.4%)

I do this research
2nd
9
The Moon Orbital Triangle Perimeter Analysis (More Analysis)
Equation No. (10)
943817 km x 2 = 86000 km x 21.8
- 943817 km = The perimeter of the moon orbital triangle (ACE)
- 86000 km = The Vertical Line BC
- 21.8 = (Jupiter Mass / Uranus Mass)
- Equation no. (10) tells, there are 2 values of (943817 km), and based on these 2
values the vertical line BC (86000 km) is created as a function of the masses rate
between Jupiter and Uranus
- Equation no. (10) tells that there are 2 orbits of the moon motion (as we have
concluded before) and these 2 orbits cooperate together to create the vertical line
BC (86000 km) based on the masses rate…. So the equation tells that these 2
planets (Jupiter and Uranus) effect on the moon orbital geometrical structure by
their masses rate.
- Also it tells that, the gravity forces effect inside a geometrical structure and not by
a direct effect and because of that the rate of masses is seen behind other data

I do this research
2nd
10
4- The Moon Orbital Triangle Discussion
4-1 The Moon orbital triangle Data (Revision)
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.

I do this research
2nd
11
(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)

I do this research
2nd
12
4-2 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 in the small figure separately (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

I do this research
2nd
13
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
and by that, the points definition which is done by the vertical triangle is used by
the moon horizontal motion

I do this research
2nd
14
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 moon orbital triangle (Pythagoras
triangle) is a vertical triangle on the base (EA), because of that the point (C) is
found on (Z-axis) by geometrical necessities and form one of the triangle basic
points. BUT
- What's the point (A)? how this point can be created and can effect on the moon
orbital triangle and motion?! Where this point is far from apogee radius with
43000 km and the moon can't move beyond the apogee radius, and can never reach
to this point (A). So how this point is found and why it has an effect on the moon
orbital triangle and motion?! So this point (A) raises an essential question in the
moon orbital triangle geometrical structure analysis.
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) and we have to find the reason by which this point (A) is created here
and causes such massive effect on the moon orbital triangle and motion.
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), If 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).

I do this research
2nd
15
2nd
Force Effects On The Moon Orbital Motion
- We have discussed the Point (A) before
- 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 effect 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..
Shortly.. 2 points of gravities total are created, 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 effects on the moon orbital motion and that causes the
distance (EA) to be = 449197 km = Jupiter Circumference
1047 = (The Sun Mass /Jupiter Mass)

I do this research
2nd
16
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 (2 equal distances)
- 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 of 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 here 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)

I do this research
2nd
17
- 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 Orbit 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 tracks as 2 prisoners 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 the 2
cars are connected with chain of steel the working car pulls the other one
and both move equal distances through the 2 tracks… that perform 2
distances by one car motion.
o What do 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
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.

I do this research
2nd
18
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

The Moon Orbital Motion Rules

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to The Moon Orbital Motion Rules

Similar to The Moon Orbital Motion Rules (20)

More from Gerges francis

More from Gerges francis (20)

Recently uploaded

Recently uploaded (20)

The Moon Orbital Motion Rules