The Moon Orbital Triangle Analysis (V)

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 Analysis (V)
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 – 9th
November 2020
Abstract
Paper Claim
- The moon orbital geometrical structure uses the distance between perigee and
apogee (43000 km) as a building unit
- The moon orbit uses Pythagoras rule with this distance 43000 km to create all
other distances found in the moon orbit
- The moon uses Pythagoras rule in its motion to be enable to change its daily
displacement
- Because the moon daily displacement 88000 km isn't used as a real displacement
totally, Where the moon uses Pythagoras triangle to create another real
displacement which can be shorter than 88000 km, this technique enables the
moon to move through orbits more near to the Earth than apogee orbit (r=0.406
mkm)
- The moon orbital geometrical structure depends on the distance 43000 km to
concentrate the energy into the moon orbit.

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1- The Moon Orbital Triangle Analysis
1-1 The Moon Orbital Geometrical Structure
Figure No. (1)
(my figure)
Figure no. (2) For Geometrical Explanation
What do we try to do here??
- We try to find the small circle tangent (AB) in the moon orbital figure –
o We deal with the moon orbit as a circle and not as an ellipse (for simplicity)
o OB = apogee radius (r=0.406 mkm) and OD = perigee radius (0.363mkm)
o (DB)2
= (OB)2
- (OD)2
= (0.406 mkm)2
– (0.363 mkm)2
= (0.18184 mkm)2
o DB = 181840 km ….. AB = 363000 km = Perigee Radius
o Means, The Tangent AB = Perigee Radius = 0.363 Mkm! Why?
o How This Orbit Is Created?
o The triangle ADO is a specific triangle of Pythagoras rule its dimensions are
(1, 2 and (5)1/2
) and its angles are (90, 63.4 and 26.6 degrees)
o i.e., The value (181840 km) is the moon orbital geometrical structure unit

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Now let's analysis this value 181840 km
o (43000 km)2
+(86000)2
= (96000 mkm) but
o 181840 km = 86000 km + 96000 km How To Explain This Result?
o the distance 181840 km is build as a function in the distance 43000 km
o The distance from perigee to apogee = 43000 km
i.e.
o The moon orbit geometrical structure is built based on 43000 km! Why?
Conclusions:
- The moon orbital geometrical structure is built based on 2 factors
o (1st
Factor) Pythagoras Rule
o (2nd
Factor) The distance 43000 km
- Based on Pythagoras Rule the distance (43000 km) produced all other distances in
the moon orbit – means – this distance is the build unit.
- The rule is Pythagoras rule simply (43000)2
x2=(61000)2
and (61000)2
x 2= 86000,
it's the method clearly, each distance is produced based on another one…
- (1st
) Why the moon uses Pythagoras rule to define its orbital motion points?
Because by this rule the moon could decrease its daily displacement – let's
remember that-
o The moon daily displacement =88000 km
o The moon doesn't use this displacement totally as a real displacement
through its orbit but uses a part of it only as a real displacement –
o Why do that? because during 29.53 days the moon moves (88000km) a total
distance =2.58 mkm and this distance = the apogee circumference (r=0.406
mkm) , means if the moon uses only the displacement 88000 km it will have
to move only through apogee radius
o But the moon uses Pythagoras rule to decrease the displacement, where the
displacement 88000 km is used as the Pythagoras right triangle hypotenuse
and the real displacement will be define as L (where L =88000 cos θ)
o The angle θ is the smallest angle in the triangle
o The angle θ changes the real displacement (L) and by that forces the moon
to move into a higher orbit with the real displacement increasing.
- Note Please
- DB = 181840 km = The Triangle Perimeter (BSC =86000 km +86693 km
+10921km = 183614 km (error 1%) (10921km =the moon circumference)
Now let's ask
(2nd
) Why the moon orbit depends on the distance 43000 km?

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1-2 Why does the moon orbit depend on the distance 43000 km?
I-Data
1- 43000 km = 1.0725 x 40080 km
2- 43000 km x 27.3 x 0.99 = 1.16 mkm
• 10921 km x 27.3 = 0.3 mkm
3- Earth moves a distance 43000 km during 24 minutes
• 2.58 mkm = 43000 km x 60
• 2.41 mkm = 40080 km x 60
4- Pluto moves a distance 43000 km during 9124 seconds
• Light known velocity (0.3 mkm/sec) moves during 9124 seconds moves
a distance =2737 million km
• Pluto during 17.75 hours moves 300000 km
• 4222.6 hours =17.75 hours x 237
5- 181840 km = 43000 km x 4.2226
6- 43000 km = 4437 seconds x 9.7
7- 43000 km = 243 km x 176.95
8- 2737 = 243mkm x2 x 5.6
9- 43000 km = 29.34 x 1461
10- 43000 mkm = 37000 sec x 1.16 mkm /sec
II-Discussion
Equation No. 1
43000 km = 1.0725 x 40080 km (Earth Circumference)
- Earth rotates around its axis once per solar day, and in its rotation moves a distance
= 40080 per solar day, so the previous equation shows 2 distances, the distance
between perigee and apogee (43000 km) and Earth rotation distance (40080 km)
- 1.0725 We have discussed this rate frequently before, this rate is created by
Lorentz Length contraction effect for the distance 43000 km (as done for around
40% of all solar system distances – as we discussed)
- What news we have from this equation?
- Equation No. (1) tells that, Earth Circumference is a result and not a reason, i.e. it's
produced from some other value (43000 km)
- i.e. the distance 43000 km was found before the Earth Circumference Creation! If
this distance (43000 km) is the moon free distance and the moon is the follower to
Earth –How Earth Circumference can be a result from this distance 43000km?

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Equation No. 2
(a)- 43000 km x 27.3 x 0.99 = 1.16 mkm
And
(b)- 10921 km x 27.3 = 0.3 mkm
- Equation No. (2) tells that the matter is created based on its motion – or as Lorentz
had taught us, Particle Properties Are Created As A Function Of Its Motion,
Because of that, Particle length and mass are changed with this particle high
velocity motion.
- Equation No. (a) shows the motion from which the moon is created.. it's a light
motion done by a light supposed velocity (1.16 mkm/sec), this light beam travels
for 1 second to create the cycle 27.3 days through the distance 43000 km
- The moon inherited the motion and its cycle (27.3 days)
- Equation No (b) we know perfectly, it tells, if the moon rotates around its axis one
time per a solar day, so the moon during its orbital period (27.3 days) will move a
distance = 0.3mkm = light motion for 1 second (light known velocity 0.3 mkm/s)
- Equation No. (2) tells that, light motion created the moon based on its motion, But
- How the light beam with supposed velocity (1.16 mkm/sec) became light known
velocity (0.3mkm/sec)? how to explain that?!
- The moon creation depends on light supposed velocity (1.16 mkm/sec) but the
moon motion depends on light known velocity (0.3 mkm)
- The moon creation tells some miracle, light supposed velocity 1.16 mkm/sec,
became 0.3 mkm/sec – It's One Direct Result Found By The Moon Creation-
- By the light supposed velocity 1.16 mkm/sec the universe is created before the
moon creation, after the moon creation one more light is born (0.3 mkm/sec) –
- Equation No. (2) tells basically that, the matter is created because of the motion
which is found before any matter creation- it was the light motion based on which
the universe of matter is created – the light provides the motion… only one feature
of life "The Motion" and by this motion is matter is born. For the moon the
inherited motion is found in its orbital period (27.3 days), it's the life feature the
moon inherited from the light and lives by it till now

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Equation No. 3
Earth moves a distance 43000 km during 24 minutes
(c ) 2.58 mkm = 43000 km x 60 (2.58 mkm =Earth velocity per solar day)
(d) 2.41 mkm = 40080 km x 60 (2.41 mkm = moon velocity per solar day)
- Equation No. (3) tells that, in 24 minutes Earth moves 43000 km!!
- Earth motion distance per solar day (2.58 mkm) = 60 x 43000 km But
- The moon motion per solar day (2.41 mkm) = 60 x Earth Circumference!!
Also
- The moon orbital circumference at (r= 0.384 mkm) = (2.41 mkm)
- Simply That Because
- 2.58 mkm =1.0725 x 2.41 mkm and 43000 km = 1.0725 x 40080 km
- How to understand that?
- 60 x 2.58 mkm (Earth motion distance per solar day) = 154.4 mkm (Note Please,
Earth ancient orbital distance was 153 mkm, the equation has error 1%)
- 60 x 2.41 mkm (The moon motion distance per solar day) = 144 mkm (Mars
Displacement from 84 mkm to 227.9 mkm)
- Why Is There A Rate 60 Between Different Distances?
- Because the distances are used s periods of time, and because of that, the distances
distribution is created based on the rate 60 because it's the rate between time units
(60 seconds =1 minute, 60 minutes = 1 hour).
- It's one more feature of light motion discovered in the planets motions, because the
distance using as period of time is a light motion feature where (x =ct, when c=1
so x=t)… this feature of light motion is used in planets motions and because of that
the distances are distributed based on this rate to save its ability to be used as
periods of time…
A Conclusion
- The solar system distances are distributed to be used as periods of time, as we have
seen the distance (25920 mkm) which is used as (25920 years = the precession
Cycle)… because this feature is a basic geometrical feature of the solar system
distances, these distances are divided by the rate 60 to be qualified to be used as
periods of time
- i.e. the solar system distances distribution shows one more feature of light motion
discovered in planet motion.

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Equation No. 4
Pluto moves a distance 43000 km during 9124 seconds
(e) Light known velocity (0.3 mkm/sec) moves during 9124 seconds moves a distance
= 2737 million km
(f) Pluto during 17.75 hours moves 300000 km
(g) 4222.6 hours =17.75 hours x 237
- Equation No. (4) tells that, Pluto moves 43000 km during 9124 sec!!
- Equation (e) tells that, during 9124 sec Pluto moves 43000 km but the light known
velocity (0.3 mkm/sec) moves 2737 million km – (2 motions are done during the
same period of time)!
- The distance 2737 mkm starts from the point (135.5 mkm) from the sun to Uranus
point (Uranus orbital distance =2872.5 mkm)!
- What a big deal if we have 2 motions are done in the same period of time? Why
this is specific geometrical feature? 9124 sec = 2.534 hours …. If 1 hour = 1 mkm!
So this 2.534 mkm = the moon orbital circumference at apogee radius (error 0.6%)
o What is the useful result here?
o 2 motions are done during the same period of time
o If this period of time is used as distance value, it will be equal the moon
orbital circumference at apogee radius …. Why is this useful??
Pluto and light motions are done to create the moon orbital circumference
o Because of that
o Light motion effect is discovered in the moon orbital motion (Point 1-3)
o Pluto motion effect is discovered in the moon orbital motion
How??
o By what mechanism can Pluto effect on the moon orbital motion?!
Let's see the next equation
(f) Pluto during 17.75 hours moves 300000 km
(g) 4222.6 hours =17.75 hours x 237
- Equation (f) tells that, Pluto moves during 17.75 hours a distance = light motion
per a second 0.3 mkm (light known velocity 0.3 mkm/sec)…!
- We need Equation (f) because 2 motions were done in the previous equation by
light (0.3mkm/sec) & Pluto during the same period of time- so we need to know
how this process is done, so Pluto needs 17.75 hours to cover 1 second of light
motion distance…
- 17.75 this value we know as a distance value (17.75 mkm), let's remember that,
25920 mkm = 17.75 mkm x 1461 days where 17.75 mkm = the planets motions
distances total per a solar day – the distance 25920 mkm is passed by light known
velocity (0.3mkm/sec) motion during 1 solar day (86400 km), this green equation
tells that, the distance passed by light in 1 solar day, will be passed by all planets

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motions distances total during 1461 days, in this equation we knew that the
distance period 25920 mkm can be used as time period (25920 years = the
precession Cycle) because light motion features are found in the planet motion.
So
- 17.75 mkm = the planets motions distances total per solar day
- Pluto uses this value as a period of time! (17.75 hours!) What does that mean??
- It's a collective motion… let’s imagine that 2 cars one moves by 200 km /hour and
the other moves by 300 km/ hour and both are connected together to move with
each other always, what will happen? Both will move by 250 km/hour, why? it's a
collective motion…. That's what happens Pluto uses all planets motions and add
to its velocity to produce a total motion distance = light motion – but in this case
Pluto doesn't use the planets motions distances as distances to be added to each
other but as period of time to define its velocity rate to the light velocity – that
produces a collective motion…
- i.e.
- The Light Beam Motion Is Produced By All Planets Motions Total
- That happened before in the green equation (17.75 mkm x 1461 days =25920
mkm) the distance 25920 mkm is light motion and can be produced by all planets
motions total – and that what Pluto did in equation no. (f) Pluto during 17.75
hours moves 300000 km – Pluto used the planets motions distances total as a
period of time–It's a collective motion produces light motion distance for 1 second.
- Why Pluto did that?
(g) 4222.6 hours =17.75 hours x 237
- This is an interesting equation, Mercury day period 4222.6 hours is rated to the
period 17.75 hours (Pluto Period) - with the rate 237! What's this?
- Venus Rotation Period / Mars Rotation Period = 237
- Let's remember (light motion for 1 second causes planet motion for 1 solar day)
this hypothesis created the 1st
rate of time (1 second is equivalent to 1 solar day)
this rate which is used frequently through the solar system motion …. But
- In the moon orbit this rate was changed because Venus and Mars rotation periods
created the rate 237 based on Venus and Mars motions interaction, so the 1st
rate
which is (1s is equivalent to 86400 s) changed to be (1s is equivalent to 365.25 s)
this is the 2nd
rate which is used in the moon orbit and based on this 2nd
rate the sun
rays is created by the planets motions energies total
- Can we conclude equation no. (g) meaning now? Not yet … first let's remember
- 4222.6 million km = 2π x670 mkm (Venus Jupiter Distance)
i.e.
o If Pluto uses the distance 17.75 mkm as time period =17.75 hours where
1mkm = 1 hour …..

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o A similar process is done by Mercury, where Mercury used 1 hour for 1
mkm perfectly as Pluto does
o We may remember that, a light beam with supposed velocity travels during
4222.6 seconds a distance = 4900 mkm (= Jupiter orbital circumference),
and this light beam is the mercury , Venus, earth and Mars source of energy,
as we have discussed before – so the light energy during 4222.6 seconds is
transported to the inner planets in different forms for Mercury the form is
4222.6 hours but for Venus was 4222.6 mkm- to explain why we need to
discover the light motion effect details on the planets motions …
o I want to say that
o Pluto motion shows a very similar motion to the inner planets motions and
so to have an interaction between the 2 periods of time (4222.6 hours and
17.75 hours) it can be possible – let's now try to discover the equation
meaning
- Equation no (g) 4222.6 hours =17.75 hours x 237, the equation tells that, Venus
and Mars motions interaction effect which create the rate 237 by which the 1st
rate
of time is changed to the 2nd
rate of time (1sec is equivalent to 365.25 sec), this
interaction between Mars & Venus, was supported by Mercury and Pluto motions
where Pluto motion was supported by all planets motions
- i.e. this process was on collective motion, although we see only Venus and Mars
motions interaction effect on the moon orbit – but this effect was supported by all
planets motions total through the direct effect found by Pluto…
- That explains Pluto great effect on the Earth moon orbital motion which we have
seen frequently – for example – Pluto moves during a solar day a distance =
apogee radius =0.406 million km
Equation No. 5
181840 km = 43000 km x 4.2226
- Let's remember the moon orbital triangle where The Tangent AB = 363000 km
and the distance (DB) = 181840 km, and we have been astonished because the
triangle is one specific of the Pythagoras triangles, where ( (1) 181840 km and (2)
363000 km (Perigee) and (5)1/2
= 406000km (apogee)) and then we have
discovered that the distance 181840 mkm is created as a function of the distance
43000 km (perigee apogee distance) – based on that we have concluded that (the
moon orbital geometrical structure is build depending on the distance 43000 km)
- Equation no. (5) confirms this same meaning, the value 181840 km is created as
function of 43000 km based on Mercury day period ((4222.6 hours) /1000).. it's
the meaning which I can provide now, but the deep meaning still needs more
analysis …. Because Mercury orbital period =88 days and the moon daily
displacement = 88000 km, if this is done for one time we my consider as pure
coincidence.. but Mercury moves during its rotation period (58.66 days) a distance
= 243 mkm where Venus rotation period =243 days… (Also Mars moves during

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Venus day period 116.7 solar days an equal distance =243 mkm). The using of
time period as distance an vice versa is usual using by Mercury and Venus (and
mars sometimes) –So if Venus rotation period depends on Mercury motion- that
means – the moon daily displacement (88000km) must be created depending on
Mercury motion.
Equation No. 6
43000 km = 4437 seconds x 9.7 km/sec (Saturn velocity)
- 4437 mkm = Mercury Neptune Distance, the distance is used as a time period
- Equation No. (6) tells, during 4437 seconds, Saturn moves a distance = 43000 km
- What's interesting in this equation? Saturn uses the distance between Mercury and
Neptune as a period of time! Why?
- (1) Neptune orbital distance = Saturn orbital distance x π
- (2) Mercury day period needs 5040 seconds to be 176 solar days… but during
5040 seconds Mercury moves a distance = 2 Saturn diameters (error 1%)
- There's some deep relationship between Mercury, Saturn and Neptune… that's
seen in their data… shortly… there's a geometrical machine under these 3 planets
which creates the distance 43000 km…. means the distance 43000 km is a result of
cooperation point – but the mechanism still needs deep analysis…
Equation No. 7
43000 km = 243 km x 176.95
- We know …
- 243 mkm = Mercury moves during its rotation period (58.66 days) but Venus
rotation period = 243 days
- 175.94 mkm x 2π = Venus moves during (365.25 days) but Mercury day period =
175.94 days)
Equation No. 8
2737 mkm = 243mkm x2 x 5.6
- We know that…
- 243 mkm = Mercury moves during its rotation period (58.66 days) but Venus
rotation period = 243 days
- 5.6 degrees = 5.1 degrees (the moon orbital inclination) +0.5 degrees (the moon
angular diameter)
- 2737 mkm (a distance from Uranus to the point 135.5 far from the sun), the light
known velocity (0.3mkm/sec) travels this distance in 9123 seconds
i.e.
- 2737 mkm = Light Motion Energy
- 243 mkm = Mercury or Mars motion energy
- The interaction between 2 motions energies produced the moon orbital inclination
which is measured above the moon diameter
- But why use 2 values (2 x 5.6 degrees), this manner is repeated frequently…

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Equation No. 9
43000 km = 29.34 x 1461
- 1461 million km = 1.1318 mkm /day x 1290 days , Jupiter moves during 1290
days a distance =1461 mkm which we see as 1461 days !
- 1290 days, we know this period perfectly, let's remember it,
o The moon daily displacement = 88000 km
o Suppose the moon moves only vertically from perigee to apogee (43000km)
and return back, consuming 86000 km (so the rest is 2000km)
o Suppose the moon moves this 2000 km on the horizontal level through its
orbit
o The moon orbital circumference at apogee (r=0.406 mkm) = 2.58 mkm and
if the moon daily distance =2000 km so the moon needs 1290 days to
revolve its orbit
- 97.8 seconds x 0.3 mkm (light known velocity) = 29.34 million km
- 43000 mkm= 29.34 mkm x 1461 mkm , how this distance became 43000 km only?
Because
- The moon orbital diameter =1 mkm … so
- 43000 km x 1 mkm = 29.34 mkm x 1461 mkm
Equation No. 10
43000 mkm = 37000 sec x 1.16 mkm /sec
- 37000 mkm = Pluto orbital circumference
- Light supposed velocity 1.16 mkm/sec travels during 37000 seconds a distance =
43000 mkm
Shortly
- The distance 43000 mkm express the solar system distances, because Pluto is the
greatest orbital circumference in the solar system, this greatest distance should be
considered the most great cooperation can be done by the planets motions (matter
motions)… the light uses this greatest distance as time period to create even
greater distance which is 43000 mkm
- 43000 mkm = 1 million km (the moon orbital diameter) x 43000 km (the distance
between perigee and apogee) that tells the moon orbit contains the most greatest
energy found in the solar system

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WHY THE MOON ORBITAL GEOMETRICAL STRUCTURE IS
BUILT BASED ON THIS DISTANCE (43000KM)??
Because
- Based The Planets Motions Interaction
- The distance 43000 km contains the greatest energy found in the solar system –
and this energy is stored in 2 distances which are 43000 km (perigee apogee
distance) and 1 mkm (the moon orbital diameter)
- And
- The Greatest Distance Is 43000 km x 1 mkm = 43000 mkm
- The distance 43000 mkm = 21.5 % of all Jupiter energy and greater than
Neptune used energy (14%) of Jupiter total energy…

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1-3 Light travels a distance 2737 mkm during 9123 seconds
In Equation No. 4
Equation No. 4
Pluto moves a distance 43000 km during 9124 seconds
We have found that,
(e) Light known velocity (0.3 mkm/sec) moves during 9124 seconds moves a
distance = 2737 million km
- The distance 2737 mkm starts from the point (135.5 mkm) from the sun to Uranus
point (Uranus orbital distance =2872.5 mkm)!
i.e.
- Light motion passed 2737 mkm from Uranus to the point 135.5 mkm far from the
sun, and this point is so specific in space for 2 reasons
The First Reason
1- Uranus effects on the inner planets and as we have discussed before, Uranus
axial tilt (97.8 degrees) is perpendicular on the moon orbit and creates an
angle (91.1 degrees) with the moon axial tilt
2- Also Uranus axial tilt (97.8 degrees) creates an angle 7.8 degrees (+90 deg)
with Mercury where 7 degrees (= Mercury orbital inclination) + 0.8 degrees
(= Uranus orbital inclination)
- I want to say that, this light motion through 2737 mkm from Uranus to the point
135.5 mkm from the sun, is a real motion and because of that we see Uranus data
effect on the inner planets data through perpendicularity
- For example
- 180.8 degrees = 177.4 deg (Venus axial tilt) +3.4 deg (Venus orbital inclination)…
o But what's this value 180.8 degrees??
o 0.8 degrees (Uranus orbital inclination) + 90 degrees +90 degrees
The Second Reason
- As we know, Jupiter energy is the solar system source of energy.. Jupiter sent its
energy to Pluto, where Pluto reflected it totally to Neptune, then Neptune seized
(consumed) 14% of this energy to built its orbital circumference and reflected the
rest of energy in 2 trajectory of energy, each trajectory contains 86400 mkm
(Space = Energy)
o The 1st
Trajectory of energy is sent to Mercury
o The 2nd
Trajectory of energy is sent to Venus and Earth
- These 2 trajectories of energy is sent to the inner plants almost in the point (135.5
mkm) far from the sun) as a middle point for all inner planets.
- Mars lost its source of energy because of its migration and needs specific support
from Saturn & Jupiter.

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Please remember this triangle data
Let's start our analysis
- EB = Perigee radius = 363000 km
- ES = total solar eclipse radius = EC = 373000 km
- ED = Apogee radius = 406000 km (405500 km)
- EA= (Jupiter Circumference) =449197 km
- AC = (Saturn diameter) =121620 km (error 1%)
- EC = (Saturn circumference) = 373000 km
- SB= (the moon Circumference) =10921 km
- CS = = 86693 km
- CF= (the moon daily displacement) =88000 km
- CD = = 95930 km
- CB= AB = 86000 km
THE ANGLES
- The angle between the black and red lines (under E) = 1.1 degrees
- BCS = 7.25 degrees Angle E = 13.33 degrees
- BCF = 12.195 degrees ECB =76.7 degrees
- BCD= 26.3 degrees
- BCA = 45 degrees
- CSB = 82.75 degrees CSF = 97.25 degrees
- CFB = 77.8 degrees CFD = 102.195 degrees
- CDF= 63.7 degrees CDA= 116.3 degrees
- ECD= 103 degrees

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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
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Full list of
publications:
http://web-local.rudn.ru/web-
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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
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