The Moon Orbital Triangle (General discussion) (I)

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 Triangle (General discussion) (I)
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 –26th
November 2020
Abstract
Paper hypothesis
- The moon has 2 orbits for its motion
- The second orbit is declined by 0.8 degrees on the 1st
orbit (the moon orbital
triangle)
- The 2 orbits are neighbor and define together the lunar eclipse umbra length (1.392
mkm)
- Venus axial tilt is an effective player in the moon orbital triangle structure
Paper argument
- The paper analyze the moon motion data to conclude this 2nd
moon orbit place and
direction.

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1- The Moon Orbital Triangle Data
Let's review The Earth Moon Orbital Triangle because we use it
Figure No. (1) (my figure)
Please Note
(1) SZ = 7665 km ZF = 2414 km
- CZS = 77.8 degrees CZF =102.195 degrees
(2) DY = 3475 km BCY = 28.39 degrees
(3) XB = 16203 km XCB = 10.67
- XCE = 66 degrees CX = 87513 km

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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.
(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 = 373000 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)
- BS= (the moon Circumference) =10921 km
- BZ = 18586 km BF =21000 km
- BD = DA = 43000 km (BY =46475 km)
- BA = BC = 86000 km
- CS = = 86690 km
- CZ= (the moon daily displacement) =88000 km
- CF= 88526.8 km CD =96150.9 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 = 118.92 deg) (
- (Uranus axial tilt = 97.8 degrees = FSC 0.6 degrees)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

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Figure No. 2 (The Moon Diameter Division)
- The moon diameter is divided into small diameters as following:
- The Blue Circle its diameter R1= 1390 km and r1 =695 km
- The Red Circle its diameter R2= 2085 km and r2 =1042.5 km
- The Black Circle its diameter R3= 2185 km and r3 =1092.5 km
- The Brown Circle its diameter R4= 3208 km and r4 =1604 km
- The Orange Circle is the moon itself its diameter R5= 3475 km
- The red horizontal line is the moon orbital triangle base (AE = 449197 km)
- The blue vertical line is Uranus axial tilt (97.8 degrees), and in the center of the
moon diameter is found the moon axial tilt (6.7 degrees), i.e. the angle in the
center is 91.1 degrees and the value 1.1 degrees is consumed in the moon diameter
(0.5 degrees) and in the green box (0.6 degrees) (the green box shows a
cooperation between Jupiter and Saturn to produce the moon orbit)
- The angle at (S) = 90 degrees (Perpendicularity).
- The angle above the moon diameter = 90.6 degrees
- The moon consumes 0.5 degrees for its diameter (3475 km) for that, I have cut its
diameter into smaller diameters to decrease the consumed value (0.5 degrees), for
that, at R1 = 1390 km, the consumed angle will be 0.2 degrees i.e. the angle above
The Blue Circle = 90.9 degrees.

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- R2= 2085 km (60% of the moon diameter) and because of that, the consumed angle
will be only 0.3 degrees, means, the angle above The Red Circle =90.8 degrees
i.e.
- 90.8 degrees = 90 degrees + 0.8 degrees (Uranus orbital inclination), by that the
rest degree of Uranus axial tilt = its orbital inclination vertically.
Notice No. 1
- 149.6 mkm (Earth Orbital Distance) = 3475 km (the moon diameter) x 43000 km
(the distance from perigee to apogee)
- 149.6 mkm (Earth Orbital Distance) = 2085 km (R2) x 71492 km (Jupiter Radius)
Notice No. 2
- We will need these 2 figure in our following investigation because of that I insert
them here to be our discussion reference – let's start our discussion.

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2-Venus Effect On The Moon Orbital Triangle
2-1 The Triangle BCF Analysis
- BC =86000 km
- BF = 21000 km
- CF =88527 km
- BCF = 13.7 degrees
- BFC = 76.3 degrees
- The line CB is perpendicular on FB (90 degrees) but
- The line CB has an angle 91.1 degrees on the black line
base (under the blue circle "the moon")
Equation No. (1)
1259.3 degrees = 91.9 degrees x 13.7 degrees
- The angle 13.7 degrees (=BCF)
- We have no 91.9 degrees! We have only 91.1 degrees and that means there's
another angle is added here (0.8 degrees) what's this angle? This is our
(1st
Question)
- We have 2 points here to investigate… which are
o (1st
point) To analyze the triangle BCF because of the angle 13.7 degrees
And
o (2nd
point) To analyze the angle 1259.3 degrees other using…
(1st
point)
The triangle BCF
- Its perimeter = 195527 km but (2x 195527 = 391054 km) …..And
- BF = 21000 km and EB = perigee radius = 363000 km means EF = 384000 km
- (384000 km)2
+ (86000 km)2
= 393512 km
- The difference between the 2 values = 0.6% or around 2500 km while the
distance from perigee to apogee =43000 km, means the rate around 5.7%.

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- The calculation is clear and puzzled, because the double value of the triangle
perimeter = The Triangle Base + Perigee Radius
- What geometrical reason behind this calculation, because it tells us that, the
triangle perimeter (knows!!) the perigee radius r= 363000 km, which is not part of
this triangle at all-
- Any way the dimension (86000 km) is part of the triangle but how this dimension
tells to the triangle perimeter about the perigee radius?!
- What's common between the triangle EBC (which has the perigee radius as one
dimension) and the triangle BFC (whose its double perimeter is produced by
Pythagoras rule? (Why does the moon use Pythagoras rule, Do we remember?)
what's the common value between these 2 triangles ( EBC and BFC)??
o 13.328 degrees the angle (E) in the triangle EBC
o 13.7 degrees the angle (BCF) in the triangle BCF
Equation No. (2)
I - (365.25 days /354.39 days) = 13.736 /13.328
II- (2.6 degrees /2.5 degrees) = 13.7 /13.18
- Equation (2) (I), shows that, the required rate is used between the lunar synodic
year and the sidereal year
- Equation (2) (II), shows that, 2.5 degrees = Saturn Orbital Inclination) deals with
(2.6 degrees = 180 deg – 177.4 deg (Venus Axial Tilt). We know that, the moon
daily motion =13.18 degrees.
- What's the useful result for this equation no. (2)? Why we searched for it? the
triangle (BCF) tells us that, its perimeter (or circumference) knows perigee radius
dimension which is not included in this triangle, we don't know how this triangle
catch this information – so we have searched for something common between the
2 triangles, the most near value is the angles (13.328 and 13.7), and we analyze
them here to know how these 2 angles connects the triangle BCF with the great
triangle EBC?

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- Equation no (20 (I) tells that the value (13.7 deg) depends on the value (13.3)! if
365.25 days depends on 354.39 days or vice versa?
- We accept that the moon motion depends on earth motion and that may cause the
rate (365.25/354.39) is created by some dependency…
- That means, there's some hidden machine connects the angle 13.3 and 13.7 degrees
and that causes perigee radius to be included in the triangle BCF perimeter, may
later we can discover this geometrical machine.
(2nd
Point)
Let's analyze this value angle (1259.3 degrees)
- 177.4 deg (Venus axial tilt) +2.5 deg (Saturn orbital inclination) = 179.9 degrees
- 179.9 degrees x 7 deg (Mercury orbital inclination) =1259.3 degrees2
- We know this value, let's remember it here
o Because of the planets (Mars And Pluto) migration, the solar system basic
geometrical structure has changed, and because of that, the value 180
degrees is destroyed in the solar system
o Venus and Saturn together unified their powers to create another (180 deg)
but
o The available value was only 179.9 degrees, so the solar system has to use
this value 179.9 degrees in the geometrical process in place of 180 degrees
o As a proof for that
o 180 degrees x 0.8 =225 but 179.9 degrees x 0.8 = 224.7 degrees (224.7 days
= Venus Orbital Period)
i.e.
o The solar geometrical structure depends on 179.9 degrees in place of
180 degrees.

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Equation No. (3)
2094 degrees x 0.6 degree = 1259.3 degrees2
- 0.6 degrees is the angle created by the green box in figures (no. 1 and 2), it's found
above the moon and under the moon orbital triangle base (AE) = 449197 km -
- 2094 mkm = Jupiter Uranus Distance
- We know that 1 mkm = 1 degree, where Mercury orbital circumference =360 mkm
= 360 degrees, and the solar system is one machine, so any planet use this rate.
- i.e. 2094 mkm = 2094 degrees
- But
- The distance 2094 mkm (Jupiter Uranus Distance) is so important in our
discussion about the moon orbital triangle, because the data tells that this distance
effects on the triangle data- but how – because it's just a distance?! Let's review its
analysis as possible to see what effect this distance (2094 mkm) can practice on
the moon orbital motion…
o Light known velocity (0.3 mkm/sec) travels this distance in 6939.75 seconds
o The value 6939.75 seconds = 71 x 97.8 seconds
o We know that 71 is Lorentz Length Contraction Effect Rate
o 97.8 degrees = Uranus Axial Tilt, the blue line in figure no. 2, and we
know that, light motion can use the time periods as distances (because x =ct
and if c=1 so x=t) (where 1 mkm=1 degree, so 97.8 s= 97.8 mkm =97.8 deg)
where (97.8 s x 0.3 mkm/s =29.33 mkm and 29.33 mkm x 97.8 = 2872.5
mkm = Uranus Orbital Distance – that means Uranus orbital distance also is
created base on (97.8s) and not only Uranus axial tilt.
Also
o 2094 seconds x 0.3 mkm /sec (light known velocity) =627 mkm = Earth
Jupiter distance – that means, by using this distance (2094 mkm) as a period
of time (2094 seconds) the light travels from Jupiter to Earth

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Notice No. 1
o Based on that, the light almost travels from Earth to Jupiter and then to
Uranus, it's one light beam starts from Earth to Jupiter and then to Uranus,
how we know this is the direction and not the reversed one? because 1.2
mkm = 40080 seconds x 29.8 km/sec (Earth velocity), where Earth saves a
value 40080 and rest (1.16 mkm/sec) Earth sent to Jupiter, this is Jupiter
light supposed velocity (1.16 mkm/sec), that means Earth motion is Jupiter
source of energy…
Notice No. 2
o Earth orbital distance (149.6 mkm) = Jupiter diameter (142984 km) x 1047
(where 1047 = the sun mass / Jupiter mass), that shows Earth orbital
distance is defined by Jupiter mass effect on the sun gravity
Also
o 14298 km (Jupiter diameter) = 13.1 x 10921 seconds, the Equation tells that
Jupiter velocity 13.1 km/sec needs to move during 10921 seconds to pass a
distance = 142984 km, (where the Earth moon circumference =10921 km).
Notice No. 3
o Please remember in the moon diameter division we have found 2 diameters
(R2) =2085 km and we know that the moon orbital diameter =1 mkm
because of that this value 2085 km can be 2085 million km and be the
distance which we deal with here… Please remember, above this diameter
(2085 km) the blue line has an angle =90.8 degrees accurately which =
Uranus orbital inclination vertically (0.8 deg + 90 deg) – can be there a
geometrical connection between these 2 values 2085 mkm and 2085 km?
Notice No. 4
o The distance (2094 mkm) is very puzzled one, because it has different
values as (2094 mkm –2088 mkm- 2082 mkm – 2107 mkm), the differences

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are so small (less than 1%) but these are different values because they are
used by these values in very accurate equations….
o For example
o Light known velocity (0.3 mkm/sec) during 6939.75 s moves 2082 mkm
But
o Light supposed velocity (1.16 mkm/sec) during 3600 s moves 2088 mkm
o Because of light motion, I prefer to use the value 2082 or 2088 mkm.
But
- 2073 degrees = 2073 mkm = 0.99 x 2094 mkm
o 2073 mkm = 1153 mkm (the planets motion distances total during their
days periods) x 1.8 deg (Neptune orbital inclination)
o 2073 mkm = 175.94 mkm x 11.8 deg (11.8 deg=6.7 deg the moon axial tilt
+ 5.1 deg = the moon orbital inclination) but (175.94 days = Mercury day
period, where Venus moves during 365.25 days a distance=2π x 175.94mkm
o 2073 degrees = 1440 x 1.44 deg (the moon orbit regression monthly)
o 2073 degrees=12.195 degrees (the moon daily motion deg13.18– 0.985 deg)
x 170 where (170 deg = 6.7 deg x 25.2 deg Mars Axial Tilt = 11.8 deg (the
moon axial tilt + orbital inclination) x 14.4 degrees (in the moon orbital
triangle this angle = E angle (13.3 deg +1.1 deg) the angle under E
- Please remember, Uranus axial tilt (97.8 degrees) has an angle (91.1 degrees)
with the moon axial tilt (6.7 deg).. but Uranus is perpendicular on the moon orbital
triangle based (EA =449197 km), because Uranus creates an angle 1.1 degrees
between the triangle base and the horizontal level (where this horizontal level is
the moon axial tilt 6.7). and this angle (1.1 degrees) is found under (E) as shown in
the moon orbital triangle. So (E angle =13.3 degrees +1.1 = 14.4 degrees)

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Also
Equation No. (4)
2073 degrees/1259.3 degrees = 327.6 /540
- 327.6 days = the lunar sidereal year
- 540 seconds x 1.16 mkm/sec (light supposed velocity) = 627 mkm = Earth Jupiter
distance
- Equation no. (4) makes the days are equivalent to seconds, it's an application my
research 2nd
hypothesis which is (light motion for 1 second cause planet motion for
1 solar day) –
- I try to show that, it's a complex machine by which the planets move, i.e. planet
motion can't be considered as result of rigid body mechanics, because the light
motion is part can't be departed from planet motion and planet motion depends on
light motion…
Please Note.
- The Equation (2073 degrees = 2073 mkm = 0.99 x 2094 mkm) gives us again the
rate 0.99 which we have found frequently in different data and I supposed this rate
is found by light motion effect- now this puzzle is solve because
o Uranus Axial Tilt =97.8 degrees
o Sin (97.8) = 0.99
o That means, all values used the rate 0.99 are effected by Uranus Axial Tilt,
and we know that Uranus axial tilt is so strong value and it's produced
directly by light motion as we remember (6939.75 =71 x 97.8)

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2-2- The Angle 1259.3 Deg
We still discuss equation no. 1 but before let's take look on one more equation
Equation No. (5)
- 95.6 degrees =90+5.6 degrees and 5.6 degrees = 0.5 degree +5.1 degrees (the
moon orbital inclination), and that means, the value 5.6 degrees is measured above
the moon diameter (the moon diameter consumes 0.5 degree).
- 13.18 degrees = The Moon Motion Daily
- Equation no. (5) tells that, the value 1259.3 degrees is defined based on the moon
motion data!
- But
- We know that 1259.3 deg =7 deg x 179.9 deg and 179.9 deg =177.4 deg+2.5 deg,
so it's some strange to define this value by the moon motion data, because the
value (1259.3 deg) is consisted of many players and how can be defined only by
the moon motion data?!
Let's now return to Equation no. (1) it may help our investigation greatly
Equation No. (1) (continued)
- What's the key of this equation? It's the value 91.9 degrees
- We have no this value in the moon orbital triangle, Uranus Axial Tilt (97.8 deg)
creates and angle =91.1 degrees with the Moon Axial Tilt (6.7 deg) and this angle
91.1 deg is decreased by the moon diameter with 0.5 degree and by the green box
(in the figure) by 0.6 degrees and then the angle between the triangle base (EA)
and Uranus axial tilt will be 90 degrees
- So how to find this angle 91.9 degrees we need 0.8 deg +91.1 deg to produce this
angle 91.9 degrees? So from where we can find this angle (0.8 degree )
- From the 2nd
orbit of the moon motion
Let's summarize this idea in following:

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The 2nd
Orbit For The Moon Motion:
- The moon does 2 displacement each equal (88000 km) so the total displacements =
2 x 88000 km (why?)
- One displacement we see because the moon moves it through its orbit very near to
Earth, but the other displacement the moon moves on earth orbital circumference
revolving around the sun, so we don't see it
- So another displacement is passed by the moon in addition to the one which we see
(88000 km).
- This second displacement is moved in 2nd
orbit for the moon motion –
- So where is the orbit – spite we don't see the displacement motion we can conclude
the 2nd
orbit position based on the moon orbital triangle
- So where is this 2nd
orbit …?
- The 2nd
moon orbit direction is defined by Equation no. (1)
Equation No. (1) (continued 2)
- The angle 91.9 degrees =91.1 degrees +0.8 degrees
- We have the angle 91.1 degrees and then will add also 0.8 degree and this will
be the moon orbit direction …
- The 2nd
orbit should be a neighbor for the 1st
one – simply the 2 moon motion
orbits together define the lunar eclipse umbra length (1.392 mkm)
- In the moon orbital triangle, after the point(A) in the opposite direction to of Earth,
the distance from this point (A) to the end of the lunar eclipse umbra length
defined the second orbit…
- Shortly
- The distance from the point (A) to the End of the lunar eclipse umbra = 1.392
mkm – 449197 km = 942803 km
- And this value 942803 km = the moon orbital triangle (ABC) Perimeter,
- That means the point (A) separates between 2 orbits of the moon motion,

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- That also explains why this triangle (ABC) perimeter x 2 =1.88 mkm which is
equivalent to 1.9 degrees (Mars orbital inclination).
- But why the value 1259.3 degrees is defined by the moon motion?
Equation No. (6)
- 180.8 degrees – 91.9 degrees = 88.9 degrees
- 180.8 degrees = 177.4 deg (Venus axial tilt) + 3.4 deg (Venus orbital inclination)
- 91.9 degrees our angle (Equation n. 1)
- 88.9 degrees is the angle ECZ
- Why this angle 88.9 deg is distinguished? Because the hypotenuse CZ =88000 km
the moon daily displacement –
- That tells, Venus data (177.4+3.4) is created in this interaction under the moon
orbital triangle – basically Venus diameter =1.774 x Mars diameter and Venus
axial tilt be 1.774 x 100 = 177.4 deg because of the interaction occurred under the
moon orbital triangle and by that Venus axial tilt became 1.774 and because of the
Green box in the figure (shows Jupiter and Saturn Interaction), because of that, a
deep interaction is happened between Saturn and Venus based on it the value 179.9
deg is created (177.4 deg +2.5 deg) – and the moon orbital inclination be created in
consistency with this data 180-177.4 =2.6 x 2=5.2 (where 5.1 the moon orbital
inclination + 0.1 degree is consumed for mars orbital inclination which was 1.8
degrees and became 1.9 degrees)
Conclusion
- The moon has 2 orbits for its motion
- The second orbit is declined by 0.8 degrees on the 1st
orbit (the moon orbital
triangle)
- The 2 orbits are neighbor and define together the lunar eclipse umbra length (1.392 mkm)
- Venus axial tilt is an effective player in the moon orbital triangle structure

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The Moon Orbital Motion (Revision)
- Please remember why we need the moon orbital triangle….
o The moon daily displacement =88000 km but the moon doesn’t use it as its
real displacement but instead the moon uses Pythagoras triangle to define its
real displacement
o Based on that
o The moon uses the right triangle dimension (L= 88000 km Cos θ) where
this (L) is the moon real displacement through its orbit daily
o The angle (θ) is the smallest angle in the right triangle, and it effects on the
moon real displacement and its height in motion above perigee radius!
o Why?
o Because the displacement 88000 km during 29.53 days is a great distance
can be provided only by the apogee orbit whose radius (r=0.406 mkm) so if
the moon uses only 88000 km as a real displacement daily, the moon would
move only through apogee radius
o But
o Because the moon uses real displacement technique (L= 88000 km Cos θ)
so the moon has the ability to move through lower orbits with the Earth, and
based on that, when the angle θ be smaller the real displacement be greater
and needs more wide orbit to be performed which force the moon to move
in high orbits above perigee radius (r=0.363 mkm).
o Then based on that I have suggested the moon motion equation which is
Gerges Equation For The Moon Orbital Motion
θ Per Solar Day = θ Of The Previous Day + 0.985 degrees
o Then by more analysis, we have discovered that, a 2nd
force effects on the
moon orbital motion and this force effects on the point (A) in the moon
orbital triangle – where this point is an essential part of the triangle while it's
far from apogee radius with 43000 km
o The 2nd
force is a result of interaction gravity forces between the sun, Earth
and Jupiter on 2 points (Earth and its moon), and because of this interaction
Jupiter causes some gravity force (10% of Earth gravity force) to be effected
on the point (A) and causes the moon motion to apogee radius.
o Then Uranus axial tilt perpendicularity effects analysis gives us the
suggestion that another orbit must be found for the moon motion and this
orbit is found under the first one as described in the following figure.

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References
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
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
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 Triangle (General discussion) (I)

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