Does Planet Day Period Depend On Its Circumference (VI)

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
Does Planet Day Period Depend On Its Circumference? (VI)
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 –16th
November 2020
Abstract
Paper hypothesis
- The Moon Orbital Distance Is Proportional To Its Parent Circumference
- This hypothesis is suggested by the previous observation, which are
o Saturn moves during its day period a distance = its circumference
o Jupiter moves during its day period a distance = its circumference (+4%)
o Neptune moves during its day period a distance = 2 x circumferences
o Uranus moves during its day period a distance = 2.6 x circumferences
- This observations lead us to conclude that, Planet motion may depend on its
circumference or the motion distance be function of the planet circumference..
- Then, the moon orbital radiuses analysis lead us to conclude this hypothesis,
this paper tries to prove it
- By the hypothesis prove, many other old claims can be supported in addition,
let's write them down here:
o The solar system is one geometrical building, each planet is a part of it.
o The angle 63.7 degrees (The Sun Declination) is created by the solar
planets motions effect
o Pluto was the Mercury Moon
o Pluto had migrated and extort Neptune order in the solar system, and
created an interaction between its motion and Neptune motion, by which
Neptune defines its own orbital distance (because of that Titus Bode law
couldn't predict of Neptune orbital distance).

I do this research
2nd
2
1- Introduction
In his introduction I refer to the solar system geometry study basic difficulty…
- Light Motion Effects On A Planet Motion By Undefined Force
- The previous sentence summarizes the situation clearly… The equations aren't
working because this undefined force ….
- For that
- Newton gravitation equation works only with Jupiter and Saturn, But all other
planets orbital distances are defined out of the equation, clearly the greatest
planet mass (Jupiter) is found in order No. 5 while Mercury (0.000173 Jupiter
mass) is the nearest planet to the sun. also Uranus is nearer to the sun than
Neptune (greater mass)…
- To what extent the light effect on planet motion disproves our explanations
and equations?
- For example
- We accept that the Earth moon was created by "Giant-impact hypothesis".
And my theory of "Mars Migration" solves many questions in the Giant-
impact hypothesis and gives a clear vision how the moon was created by
collisions done between Mars and Venus then Mars and Earth – and the moon
is created from these collisions debris by Earth Gravity – The total idea solves
all questions and explains how the moon was created clearly.
- BUT
- The moon diameter (3475 km) is created by an effect of light motion
- If we refuse this conclusion, so how to explain the following relationships:
o 10921 km (the moon circumference) x 27.3 days = 300000 km, this
equation tells, if the moon rotates around its axis once per solar day, it
would pass (during its orbital period) a distance =0.3mkm = light motion
distance per second
o 10921 km (the moon circumference) x 86400 s = 940 mkm, this
equation tells, if Earth revolves around the sun a complete revolution in
one solar day only, the moon circumference will be = a distance of Earth
motion during 1 second
o 10921 km (the moon circumference) x 13.1 = 142984 km, this one
tells, Jupiter needs to move by its real velocity (13.1 km/sec) during a
period= 10921 seconds , to pass a distance = its diameter, the equation
tells also, the moon has to rotate around its axis 13.1 times to pass a
distance = Jupiter diameter …..
o 10921 km (the moon circumference) x 449197 km = 4900 mkm, this
one tells, Jupiter orbital circumference (4900 mkm) , depends on Jupiter
circumference (449197 km) and the Earth moon circumference.
o 10921 km (the moon circumference) /(Uranus Circumference) = (The
Moon Axial Tilt) / Uranus Axial Tilt
- These are few of hundreds which can be created only by light motion effect!

I do this research
2nd
3
2- The Moon Orbital Distance Is Proportional To Its Parent Circumference
2-1 The Earth Moon Orbit
I-Data
- 0.406 mkm (apogee radius) x 2π = 40080 km (Earth Circumference) x 63.7
- The moon orbital circumference at a apogee (r=0.406 mkm) which is the most
far point the moon can reach far from Earth = 2.55 million
- This value 2.55 million =Earth Circumference x 63.7
- Shortly
- The rate (63.7) controls the relationship between the moon orbital distance and
its parent circumference – this same result we can find in many other planets…
2-2 Saturn Moon Orbit (Titan)
I-Data
- 1.22 mkm (Titan Semi major) x 2π = 7.66 mkm
- 7.66 mkm = 120536 km (Saturn diameter) x 63.7
2-3 Mercury Moon Orbit (Ancient Pluto)
I-Data
- 15534 km (Mercury Circumference) x 63.7
- = 2π x 312984 km (Neptune motion distance during its day period)
II-Discussion
- Neptune moves during its day period a distance = 2 Neptune circumferences,
(312984 km)
- Pluto was the Mercury moon and had migrated, so the distance between
Mercury and its ancient moon (Pluto) is defined in Pluto data which is seen in
Neptune motion during its day! Why?!
- Because
- Neptune orbital period 58900 solar days = 90560 Neptune days, but Pluto
orbital period = 90560 solar days – it's clear interaction of 2 motions
Before to complete our discussion, we need to remember the special case contains
these 3 planets orbital distances and diameters where
o Mercury Orbital Distance = Mercury Diameter x 1092
o Earth Orbital Distance = Earth Diameter x 1092
o Saturn Orbital Distance = Saturn Diameter x 1092
i.e.
- These 3 planets orbital distances are defined based on the rule (d= R x 1092
)
and that creates their moons orbital distances dependency on the value 63.7.
And
- Because many other planets don't follow this same rule (d= R x 1092
), so their
moons don't follow the new rule and also their days periods don't follow the
rule (planet moves during its day period a distance = its circumference), there's
on force behind connects these 3 rules together.

I do this research
2nd
4
2-4 Pluto Moon Orbit (Charon)
I-Data
- 7511.4 km (Pluto Circumference) x 63.7
- = π x 155597 km (Neptune Circumference) = 488822 km
II-Discussion
- This distance 488822 km is not Chiron orbital circumference but it=123097 km
mans this value 488822 km = 3.97 x 123097 km
- This is done (again) because of the interaction between Pluto and Neptune
motions – let's provide some data to prove that in following
o Saturn diameter (120536 km) x Neptune diameter (49528 km) = 5906
mkm (Pluto Orbital Distance)
o Saturn diameter (120536 km) x 0.5 Neptune diameter (24764 km) = 2985
km (Saturn Neptune Distance)
- The interaction in motions between Pluto and Neptune is so deep in different
data- and because of that, Pluto different distances are defined effected by
Neptune motion interaction.
Please Note
- Venus Circumference (38041 km) x 63.7 = 2.42 mkm which is the Earth moon
orbital circumference at average radius (r= 0.384 mkm) – means – Venus deals
with the Earth moon as its own moon and used its data for own motion effect.
More Discussion
- Why many moon distances are relate to the value 63.7?
- In the solar system there only one angle =63.7 degrees, which is the sun
declination…. The data provides a claim that (the sun declination is crated as a
result of planets motions total effect)
- Such claim is hard to be acceptable, because we consider the solar planets as
separated masses points move by the gravity force.
- The point is,
- Light motion on which planet motion depends, frequently violates our physics
laws and prove that, our description for Planet motion as only A Rigid Body
Motion must be mistaken

I do this research
2nd
5
3- Jupiter and Uranus identical behaviors (Data For Explanation)
1-1 For Jupiter
- Let's summarize the data in following:
o Jupiter Velocity =13.1 km/sec, i.e. Jupiter has to move during a period
10921 seconds to pass a distance =142984 km = Jupiter diameter
o Because the Earth moon circumference = 10921 km
o The moon has to rotate around its axis 13.1 times to pass a distance =
142984 km = Jupiter diameter! (13.1 rotation x 27.3 days =A Year)
o 1.16 mkm = 13.18 x 88000 km (the moon displacement per solar day)
o 13.18 degrees = the moon motion degrees per a solar day (=360/27.3)
o Jupiter moves per a solar day a distance = 1.1318 mkm (difference only
2.5% with the value 1.16 mkm)
o 1.16 mkm x1 mkm = 449197 days x 2.58 mkm (Earth velocity daily)
where 449197 km = Jupiter Circumference!
o Also
o 13.18 - 13.1 = 0.08 but 0.08 x 2.28 mkm =181843 km
o Where 2.28 mkm =the moon orbital circumference at perigee radius and
181843 km is the main distance defined in the moon orbital triangle
(Please review in the end of the paper)
NOW
o The moon defines the period of time (A Year=365.25 solar days) as a
result of the moon and Jupiter motions interaction
o Mars moves during A Year=365.25 solar days a distance =760 mkm
(Result No. 1)
1-2 For Uranus
- Uranus Velocity =6.8 km/sec, i.e., Uranus has to move during a period 7511
seconds to pass a distance =51118 km = Uranus diameter
- Because Pluto circumference = 7511 km
- Pluto has to rotate around its axis 6.8 times to pass a distance = 51118 km =
Uranus diameter! (6.8 rotation x 153.3 hours =1042.5 hours)
- 0.406 million km (Pluto velocity per solar day) = 6.8 km/sec x 59800 seconds
(but 59800 days = Neptune orbital period) – (Please note that- Neptune orbital
period 59800 solar days = 90560 Neptune days but 90560 solar days = Pluto
orbital period)
- Pluto moves during the period 1042.5 hours a distance = 17.6 million km
(Result No. 2)

I do this research
2nd
6
Jupiter orbital distance 778.6 million km
778.6 mkm = 760 mkm (Mars Motion) +17.6 mkm (Pluto Motion)
- Let's remember our hypothesis
o (1st
) Planet motion depends on light motion
o (2nd
) light motion for 1 second causes planet motion for 1 day
o The Motion Rule ( Light moves by a circle circumference but its
follower planet moves along its radius)
i.e.
- Planets motions are found based on light motions, so I inserted in this paper in
No. 2 a part of light motion network found behind the planets motions
- Jupiter orbital distance is created as 2 parts (778.6 mkm = 760 mkm +17.8
mkm) why?
- Because Mars and Pluto were migrated from their original orbital distances and
Jupiter created its orbital distance to establish connection points with the 2
migrants planets to support the solar system geometry
- How to define these 2 parts?
o Part No. 1 (760 mkm), the Earth moon defines the period (1 year = solar
365.25 days) and Mars does the motion (2.082 mkm per solar day)
during this period, and by Mars Motion the distance 760 mkm is created
o Part No. 2 (17.6 mkm), Uranus defines the period (1042 hours = 43.14
solar days) and Pluto did the motion (0.406 mkm per solar day)
- The previous explanation shows a deep dependency between the planets
motions which can be understood through the light motions interactions which
are found behind the planets motions…
Note Please
- 17.2 degrees = Pluto orbital inclination = 0.99 x 17.4 degrees (the inner planets
orbital inclinations total)
- 0.18 = 2.58 mkm Earth motion velocity per solar day / 0.466884 mkm Jupiter
motion during its day
- 160591.9 km (Uranus circumference) = 13.18 x 12104 km (Venus diameter)
- 142984 km (Jupiter diameter) = 27.78 km/sec (the moon velocity) x 5127 s
- (5127 mkm = Jupiter Pluto Distance)

I do this research
2nd
7
4- Light Motions Behind Planets Motions
I- Data
- Mercury is created by a period of time =50 seconds
- Light beam with velocity =1.16 mkm/sec travels for 50 seconds and perform a
distance =58 mkm = Mercury Orbital Distance
- Mercury orbital circumference (58 x 2π =360 mkm)
- 360 mkm because of high velocity motion – is used as 360 seconds
- Light beam with velocity (0.3 mkm/sec) during 360 seconds travels a distance =
=108 mkm = Venus Orbital Distance
- 216.4 mkm (Venus orbital diameter) is a distance passed by light beam its velocity
0.3mkm/sec during a period =720 seconds
- 720 seconds (because of high velocity motion) is used as a distance =720 mkm =
Mercury Jupiter distance –
- Mercury Jupiter diameter (720 mkm x 2) needs a period = 720 seconds x 2 which
needs a distance = 216.4 x 2 mkm
- (the distance 216.4 mkm x 2) will be used as a time period in the following
equation 216.4x 2 x 1.16 = 500 mkm
- 500 mkm is used as a time period = 500 seconds
- Light beam ( 0.3 mkm/sec) needs 500 seconds to pass Earth orbital distance (149.6
mkm)
- Still the distance 720 mkm is produced by a light beam (1.16 mkm/sec) during a
period = 627 seconds
- Light its velocity (0.3mkm/sec) needs 2090 seconds to pass 627 mkm where 2090
seconds is used as distance 2090 mkm = Jupiter Uranus Distance
- But 6939.75 seconds x 0.3mkm/sec = 2090 mkm – means light with c velocity
travels during (6939.75 seconds) a distance = Jupiter Uranus Distance (6939.75
days =Metonic Cycle).
- 108 mkm = Venus Orbital Distance – So Venus Orbital Circumference =680
mkm but Venus Jupiter distance = 670.4 mkm (Venus & Jupiter positions are
defined before by their distances to Mercury and Earth – that means – the distance
670.4 mkm we didn't bring it from the planets data sheet but we define it relative
to Earth & Mercury positions to Jupiter - i.e. 670.4 mkm is not a new data but a
concluded data)
- light beam with velocity 1.16 mkm passes during 670.4 seconds a distance =778.6
mkm (Jupiter orbital distance)
- Also 670.4 mkm x2π = 4222.6 mkm (high velocity motion uses this value as time)

I do this research
2nd
8
- Light beam with velocity 1.16 mkm travels during 4222.6 seconds a distance
=4900 mkm = Jupiter orbital distance
- Note Please (Mercury Day =4222.6 hours approximately- means the light motion
period is transferred to Mercury motion but the rate of time is changed from 1
second to 1 hour- that's similar to Metonic Cycle 6939.75 days which we have
seen before where the 6939.75 seconds in transferred into the moon motion in
form 6939.75 days)
- Light beam with velocity 1.16 mkm travels during 4900 seconds a distance
=5678.1 mkm = Mars Pluto Distance (6585.39 mkm= 1.16x 5678.1mkm) (where
6585.39 days = Saros Cycle)
- 670.4 mkm (Venus Jupiter Distance) = 1.0725 x 627 mkm (Earth Jupiter distance)
- Earth orbital circumference =940 mkm – which is used as 940 seconds
- A Light beam (0.3mkm/sec) during 940 seconds passes a distance 282 mkm
- A light beam (1.16 mkm/sec) during 282 seconds passes 327.6 mkm (which we
see as lunar sidereal year 327.6 days)
- Light beam (0.3 mkm) during 327.6 seconds pass distance =98.7 mkm (Uranus
axial tilt =97.8 degrees)
- Light beam (1.16 mkm) during 97.8 seconds pass distance =113.45 mkm (where
1mkm=1deg means 113.45 mkm = 113.45 deg = 90 +23.45 deg. Earth axial tilt)
- 149.6 mkm x 2 (Earth orbital diameter) is used as time value – so light with
velocity (1.16 mkm/sec) during this period 149.6 sec x 2 a distance =346.6mkm
where 346.6 days – the nodal year.
- 3717 mkm (Jupiter Neptune Distance) us used as time so –light with velocity
0.3mkm/sec travels during 3717 seconds a distance = Jupiter Mars distance (1.2%)
- 5127 mkm (Pluto Jupiter distance) is used as 5127 seconds where a light with
velocity 1.16 mkm/sec travels during 5127 s a distance = Pluto orbital distance.
But why Mars is exceptional always?!
- 655 mkm (Jupiter Saturn distance) (655 mkm will be used as 655 sec) x 1.16
mkm/sec = 760 mkm (will be used as 760seconds)
- Light beam (0.3mkm/sec) travels during 760 seconds a distance = 227.9 mkm =
Mars orbital distance
Please review
A light beam with velocity (1.16 mkm per sec), creates the solar system
https://www.academia.edu/42069495/A_light_beam_with_velocity_1.16_mkm_per_sec_creates_the_solar_system
or
https://www.slideshare.net/Gergesfrancis/a-light-beam-with-velocity-116-mkm-per-sec-creates-the-solar-system

I do this research
2nd
9
More Analysis In The Moon Orbital Triangle
Basic Data Correction And Revision
Figure No. (1) (my figure)
Please Note
(1) The blue dotted arrow creates a point (Z) between F & S
SZ = 7665 km ZF = 2414 km
CZS = 77.8 degrees CZF =102.195 degrees
(2) The Green arrow creates a point (Y) after the point D
DY = 2513.7 km DYA= 118.92 degrees
Let's review and correct the moon orbital triangle data
(1st
Point)

I do this research
2nd
10
- 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 AB = 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
)
- i.e. the triangle (ODB) angles are 26.564 degrees and 63.435
degrees
NOTE
- for these 2 angles (26.564 deg and 63.435 deg) we have searched, because these 2
angles will correct many data in the orbital triangle.
(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 =47513.7 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) (BYC = 61.08 deg)
- (Uranus axial tilt = 97.8 degrees = FSC 0.6 degrees) and
- (CYA +0.6 deg = 119.5 deg where 119.5 x 0.99 = 118.3 deg (Neptune A. T)
- Angle under (E) = 13.33 degrees 1.1 degrees = 14.43 degrees

I do this research
2nd
11
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

Does Planet Day Period Depend On Its Circumference (VI)

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Does Planet Day Period Depend On Its Circumference (VI)

Similar to Does Planet Day Period Depend On Its Circumference (VI) (20)

More from Gerges francis

More from Gerges francis (20)

Recently uploaded

Recently uploaded (20)

Does Planet Day Period Depend On Its Circumference (VI)