Planet velocity is defined as a function of its diameter

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
Planet Velocity Is Defined As A Function Of Its Diameter
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
November 2020
Abstract
Paper hypotheses
(1)
Planet velocity is defined as a function in its diameter (or circumference)
(2)
Planets diameters and circumferences are used as periods of time for their
motions.
(3)
Light Motion Is A Necessary Contributor In Planet Motion.
Paper discussion
- Uranus effect on Pluto and the Earth moon creation and motion proves the paper
hypotheses.

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1- Uranus effect on Pluto and Earth Moon Creation and Motion.
I- Data
(Group No. 1)
Equation No. 1
51118 km Uranus Diameter = 4.7 km/sec (Pluto velocity) x 10921 seconds
10921 km x π = 4.7 km/sec (Pluto velocity) x 7511 seconds
Equation No. 2
2x 378674 km = 4.7 km/sec (Pluto velocity) x 160592 seconds
Equation No. 3
2π x 51118 km Uranus Diameter= 2π x 4.7 km/s (29.8 km/s Earth velocity) x 10921s
(29.8 times x 29.53 days =880 days= 10 x 88 days)
Equation No. I
4.7 x π = 14.7 (where 14.7 = Uranus diameter / the moon diameter)
(Group No. 2)
During 5040 seconds
- Mercury moves a distance = 238896 km = 2 Saturn diameters (1%)
- Venus moves a distance = 176400 km = 2 x 88200 km
- Earth moves a distance = 150192 km = 43.2 x the moon diameter 3475 km
- The moon moves a distance = 140112 km = 10921 km x12.83
- Mars moves a distance = 121464 km = Saturn diameters
- Jupiter moves a distance = 66024 km
- Saturn moves a distance = 49528 km = Neptune diameter (1.3%)
- Uranus moves a distance = 34309km=π x 10921 km (Earth Circumference)
- Neptune moves a distance = 27216 km = 2.5 x 10921 km
- Pluto moves a distance = 23688 km = π x 7511 km (Pluto circumference)
Notice No. (1)
5040 seconds are required by Mercury day period to be =176 solar days
Notice No. (2)
The 3 equations are part of equations group which is inserted after the discussion to
be a reference for it – and I didn't insert it here to avoid the confusion
Let's start our discussion

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II- Discussion
Equation No. 1
10921 km x π = 4.7 km/sec (Pluto velocity) x 7511 seconds
- Equation No. (I) shows that, Pluto used its circumference (7511 km) as a
period of time (7511 seconds) to move a distance =34309 km= 10921 km x π
(where 10921 km = the moon circumference)
- Then Pluto used the moon circumference (10921 km) as a period of time
(10921 second) to move a distance = 51118 km = Uranus Diameter
Please Note
- Uranus moves during 5040 seconds a distance =34309 km= 10921 km x π
(please review Data group No. 2)
- That shows some rate between Uranus motion velocity and the moon diameter,
How such rate can be created?
Equation No. I
4.7 x π = 14.7 (where 14.7 = Uranus diameter / the moon diameter)
How can 2 diameters rate effect on a planet velocity? (Pluto velocity = 4.7 km /s)
- Let's try to solve this question in following…
o Planet diameter and circumference is used as period of time so Earth
circumference =40080 km but can be used as 40080 seconds (according
to the paper hypotheses)
o i.e. Planet Diameter = A Period Of Time
o What does mean a rate of 2 diameters accordingly?
o A rate of 2 diameters means (a period of time / a period of time)! And
when we find that? it's the acceleration because the velocity is a distance
per time, and the acceleration is a velocity per time – so – 2 diameters
rate is equivalent to acceleration.
o 14.7 (= acceleration) = π x 4.7 km/sec (velocity) how to define π in this
case?! π is the circle circumference (i.e. π is a distance)
o But we have a velocity 4.7 km/sec produced from acceleration (14.7), so
what's π in this case? It's a period of time (specifically π =1/t) because
without π we have acceleration but with π we have a velocity
o How π can be = 1/t?
o Let's remember - Light motion is accompanying with planet motion, light
moves along the circle circumference but its follower planet moves along
the circle radius.
o So, π is a distance for light motion
o We see it as a time period (because x = ct and when c=1 so x=t)
o i.e. π =1/t, because there's a light motion with plant motion

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to see the previous equation significance we need to see the influence range of this
rate 14.7 in following…
Equation No. II
14.7 =
= Uranus Diameter 51118 km / The Earth Moon Diameter 3475 km
= 2 Saturn circumferences / Uranus Diameter 51118 km
= Jupiter Radius / Mercury diameter 4879 km
=Uranus mass /Earth Mass
= Venus Mass / Mercury Mass
= Uranus Axial Tilt 97.8 deg / the Earth Moon Axial Tilt 6.7 deg
= Saturn Axial Tilt 26.7 deg / Neptune orbital inclination 1.8 deg
= 19 degrees /Jupiter orbital inclination 1.3 deg
= 366500 km /Neptune diameter
(366500 km =the outer planets diameters total)
- The rate 14.7 is so effective on the planets different data.
let's ask
- What we try to do?
- Pluto velocity (4.7 km/sec) is created as a function in this rate 14.7 between
Uranus diameter and the moon diameter- I try to prove this fact by analyze their
data as possible – we need to see the geometrical mechanism by which Pluto
velocity depended on this rate to be defined as 4.7 km/sec –
- 1st
Question why this rate is so effective?
- 2nd
Question why many data uses 2 values of Saturn diameters (or
circumferences) (please review equation no.2, and the distance passed by
Mercury during 5040 seconds and also Equation no. II)
1st
Question why this rate is so effective?
Equation No. III
97.8 seconds x 0.3mkm /sec (light known velocity) =29.4 mkm = 2 x 14.7
- Equation no. (III) tells that, light moves during 97.8 seconds a distance = 29.4
mkm … but
- Uranus orbital distance (2872.5 mkm) depends on this rate because 2872.5
mkm = 97.8 x 29.4 that means Uranus orbital distance is crated by light motion
during a defined period. (97.8 seconds)2
x 0.3 mkm/s (light known velocity).
Means light depends on squared value of (97.8 seconds) to pass Uranus orbital
distance – and by this light motion the rate 29.4 is produced and the periods of
time (97.8 seconds) is seen by us as distance (97.8 mkm) which is equivalent to
97.8 degrees = Uranus axial tilt) -
2nd
Question why many data uses 2 values of Saturn diameters
- The original rate =29.4 = 2 x 14.7
- That explains the ability to produce the double value

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- Please Note
- The using of (2 Saturn diameters or circumferences) is found because of a
geometrical necessity where Uranus orbital distance =2 Saturn orbital distance,
that shows the using of 2 values depends on a geometrical mechanism.
Pluto And The Earth Moon Diameters Creation
Equation No. IV
The moon diameter 3475 km = Pluto diameter 2390 x 1.454
- Now we have 2 rates (14.7 and 1.454)
14.7 / 1.454 = 10.1 (Mercury velocity / Pluto velocity)
- The previous data creates one thread in the solar system motion, means, there
are many data use this rate (10.1 or 10) in different using – and because of that
the solar system has one geometrical design extending through the system-
- Let's explain my idea in 2 points
o (1st
Point) The solar system data is created based on a systematic design,
means not any planet data is created by a random process – it's a
systematic design to produce the data depending on other data – so the
rates 10, 14.7 and 1.454 are used frequently in the planets data because
this data is created based on these rates originally
o (2nd
Point) Planets motions, accordingly, are created to perform some
geometrical tasks defined for their diameters – based on that – the
equations aren't strange for us – when we find that, Pluto motion during
(7511 seconds) produce a distance = π x 10921 km and based on this
10921 seconds Pluto (itself) passes another distance = 51118 km =
Uranus diameter – no surprise here – if we find that Uranus moves
during (5040 seconds) a distance = π x 10921 km, why no surprise here?
because the period 5040 s is defined for Mercury day which has a deep
relationship with Pluto – can that be sure??
Equation No. V
4222.6 hours (Mercury day period) = 24.7 hours x 10 x 17.2
Where
24.7 hours = Mars day period and 17.2 degrees = Pluto orbital inclination
The rate 10 is used here also
- The relationship between Mercury day and Mars day depends on Pluto orbital
inclination (17.2 deg)… why this relationship (if found) is significant?
- Because … Mercury moves during its rotation period (58.66 days) a distance =
243 mkm = Mars motion distance during Venus day period (116.7 days)

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- i.e.
- There's some interaction of motions between Mars and Mercury which depends
on Pluto orbital inclination…. Let's see one more equation …
Equation No. VI
86400 million km = 17.2 million km x 5040
- We know that 86400 mkm is sent by reflected by Neptune toward the inner
planets
- 17.2 mkm is equivalent to 17.2 degrees
- 5040 seconds are required for Mercury day to be 176 solar days
- i.e.
- The value 17.2 degrees is found I interaction with the basic energy to produce
the value 5040 seconds – that explains more clear the geometrical mechanism
by which the solar system data is built.
A Conclusion
- Planet Velocity Is Defined As A Function In Its Diameter.
o Why?!
o If the planet velocity depends on the mass gravity force, that means, the
planet density is a function of the planet orbital distance – this idea is
told by kepler before
Otherwise
o light motion causes planet motion and effects on it. In this case the light
creates planet matter dimensions to be in proportionality with its motion
data and velocity.

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The Reference Equations
Equation No. 1
Equation No. 2
Equation No. 3
2π x 51118 km Uranus Diameter = 8 x 40080 km (Earth Circumference)
Equation No. 4
346.6 days (Nodal Year) = 8 x 43.3 days
43.3 days = 1042.5 hours = (153.3 h x 6.8) Because 51118 km = 6.8 km/s x 7511
Equation No. 5
2.58 mkm = 7511 x 346.6 (Error 1%) (346.6/153.3)= 2.26
Equation No. 6
2.58 mkm per solar day x 10 = 2390 km (Pluto diameter) x 10921 (1%)
(2.58 mkm = Earth motion distance per solar day = Pluto motion distance during its
day period = the earth moon motion distance during its day period)
10921 km = 1.0725 x 10182
Equation No. 7
300000 km /s (light velocity) = 29.8 km/s (Earth velocity) x 10067 (10182) (1%)
Equation No. 8
300000 km /sec (light velocity) = 29.8 km/sec (Earth velocity) x 5040 x 2
Equation No. 9
2.58 mkm = 2π x 51118 km Uranus Diameter x 8
2.58 mkm = 40080 km (Earth Circumference) x 82
= 64
Please remember
Equation No. 10
10921 km = 4.7 km/sec (Pluto velocity) x7511 seconds (2.8%)
Equation No. 11
2x 378674 km = 4.7 km/sec (Pluto velocity) x 160592 seconds

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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
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 =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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The Moon Orbital Motion
- 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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A Model For The Moon Motion 2 Orbits

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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

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