- The document discusses Mercury's rotation period and orbital period, and how they relate to other astronomical cycles and periods in the solar system, including Pluto's orbital period. Complex equations are presented relating distances, velocities, and time periods of various astronomical bodies. The author's conclusion is that a light beam traveling before planets created the dimensions of matter, space and time, and planets follow these dimensions like a "slave follows its master."

1. 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
Mercury Rotation Period Creation
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 –20th
November 2020
Abstract
- 142984 km = 13.1 km/sec (Jupiter velocity) x10921 seconds, this equation tells
that, Jupiter needs to move during 10921 seconds to pass a distance =142984 km
(= Jupiter diameter). But on the other side, the equation tells, the Earth moon needs
to rotate 13.1 times around its axis to pass by rotation a distance = 142984 km (the
Earth moon circumference 10921 km) (The moon needs 27.3 days to rotate around
its axis and by that the 13.1 rotations needs 13.1 x 27.3 = 358 days = 1 year)
- 51118 km = 6.8 km/sec (Uranus velocity)x 7511 seconds, this equation tells that,
Uranus needs to move during 7511 seconds to pass a distance =51118 km
(=Uranus diameter). But on the other side, the equation tells, Pluto needs to rotate
6.8 times around its axis to pass by rotation a distance = 51118 km (Pluto
circumference 7511 km) (Pluto 153.3 hours to rotate around its axis and by that the
6.8 rotations needs 6.8 x 153.3 = 1042.5 hours)
- 51118 km = 4.7 km/sec (Pluto velocity) x 10921 seconds, this equation tells that,
Pluto needs to move during 10921 seconds to pass a distance =51118 km
(=Uranus diameter). But on the other side, the equation tells, the Earth moon needs
to rotate 4.7 times around its axis to pass by rotation a distance = 51118 km (The
moon needs 27.3 days to rotate around its axis and by that the 4.7 rotations needs
4.7 x 27.3 = 128.3 days).
- 142984 km = 4.7 km/sec (Pluto velocity) x 30589 seconds, this equation tells
that, Pluto needs to move during 30589 seconds to pass a distance =142984 km
(=Jupiter diameter). But on the other side 30589 days = Uranus orbital period.
Equation No. (1)
- The moon (rotation) to move (51118 km) needs 128.3 days BUT Pluto (rotation)
to move (51118 km) needs 1042.5 hours (=43.32 days) i.e. (128.3/43.3) = 3
Equation No. (2)
- The moon (rotation) to move (142984 km) needs 1year days BUT Pluto (rotation)
to move (142984 km) needs 30589 seconds (suppose days) (30589 /346.6) =88

2. 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
2
- The 30589 seconds are used as 30589 solar days in Equation no. (2), let's left the
reason behind this using till the end of the discussion.
- What have we here?
- Mercury day period (175.9 days) = 3 x 58.66 days (Mercury rotation period),
where no. (3) is equation No. 1 result
- Mercury orbital period= 88 days (Equation no. 2 result)
Notice No. (1)
- 346.6 days (the nodal year) x 88 days (Mercury orbital period) = 30501 days and
- 30589 days (Uranus orbital period) = 30501 days + 88 days (accurately)
- This is astronomical cycles behaviors which proves that, Uranus orbital period
depends on (the nodal year and Mercury orbital period)
- How to prove that, this is astronomical behavior? Because we have seen it before
- 6939.75 days (Metonic Cycle period) =6585.3 days (Saros Cycle) +354.36 days
(Lunar synodic year), that means, the greater cycle is consisted by addition one
cycle to another smaller cycle – So Metonic is produced by addition of saros plus
one synodic year – here Also, Uranus orbital period is produced by addition the
value 30501 days (=346.6 x 88) to the value 88 days which is Mercury cycle
- Also we have seen that the cycle (2737 years) depends on smaller cycle (2736
years) where this last one (2736 years) depends on the moon metonic cycle (where
2736 = 144 x 19) and by that the cycle 2737 is produced by addition only 1 year.
- The Analysis Results
- We have 88 days (Mercury orbital period) and we have the rate (3) which is the
rate between Mercury day period and its rotation period – but we don't have any of
these 2 periods! How to create them
Equation No. (3)
- 90560 days (Pluto orbital period) = 2 x 778.6 mkm x 58.66 days
- 58.66 days = Mercury rotation period
- 778.6 mkm = (Jupiter Orbital Distance) (2 because the equation use the orbital
diameter which is 1557 mkm = 2 x 778.6 mkm).
- (Question No.1) Why Mercury rotation period is created based on Pluto orbital
period? Because Pluto orbital period is the solar system original period! How?
Equation No. (4)
- 90560 days (Pluto orbital period) = 1461 days x 2π3

3. 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
3
- We know (1461 days = 365+365+365+366 days), but this period should be
considered as the basic period in the solar system because this value is produced
directly by a light motion as we may remember
o 86400 seconds x 0.3 mkm /sec (light known velocity) = 25920 mkm
o 17.75mkm (planets motions distances daily total) x 1461 days = 25920 mkm
So
o Light motion distance in 1 solar day = planets motions distances total during
1461 days – which makes this cycle (1461 days) as first cycle produced in
comparison with light motion – that makes it an original cycle period -
because of that the solar system depends on it to define Mercury rotation
period and based on the 2 equations (no. 1 and 2) mercury orbital period and
day period are produced.
- (Question No.2) Why does the equation (no.3) depends on Jupiter orbital diameter
1557.2 mkm to define Mercury cycles periods?
- Jupiter orbital circumference = Jupiter circumference x the moon circumference,
the previous different motions are done to pass distance (= Jupiter diameter) or a
distance (= Uranus diameter) –that shows – the previous motions have relationship
with the used terms in the equation.
- I want to say that, there's some geometrical machine here, this machine uses the
previous planets motions to define these data, so Jupiter orbital distance is defined
to be in consistency with Jupiter diameter which is defined to be in consistency
with the previous planets motions!!
- Who created this consistency – it's not the gravity force- it's a geometrical ability
which can create the matter, space, time and motions dimensions in such great
harmony- but why this harmony is necessary? Why if the moon motion to pass a
distance = Jupiter diameter was defined by another period of time, what would
happen in this case?
- My conclusion is the same one which is
A light beam traveled before the planet, and this light beam created the matter, space
and time dimensions to be in consistency then the planet is the slave follow its master
and perform its instructions.

4. 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
4
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)
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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
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