Does Planet Day Period Depend On Its Circumference (II)

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
Does Planet Day Period Depend On Its Circumference ? (II)
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 – 14th
November 2020
Abstract
Paper hypothesis
The Planet Day Period Definition
The Period Through Which The Planet Moves A Distance = Its Circumference.
The hypothesis Explanation
- Saturn and Jupiter velocities support the paper hypothesis
- But
- The other planets don’t follow this same rule ….Why?
- Because
- The motion is transported through the (outer planets) as motion of gears, i.e. Each planet defines the next planet motion.
- Then
- The outer planets motions total is balanced with the inner planets motions to create a general harmony for the solar system..
the motion transportation provides a strong proof that, the solar system in one geometrical building and no planet can be
considered as a separated point from the whole trajectory of energy which created the solar system.

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1st
Table (The Planet Circumference And Its Day Period) (An Old Table)
Planet and its day
period
Its Circumference Its Motion distance
during its day period
The rate
Mercury 175.94 days 15334 km 720.7 million km 47000
Venus 116.75 days 38041 km 352.4 million km 9280
Earth moon 29.53 days 10921 km 71.16 million km 6510
The Earth 24 hours 40080 km 2.58 million km 64.3
Mars 24.7 hours 21346.6 km 2.1467 million km 100.5
Jupiter 9.9 hours 449197.5 km 466884 km 4%
Saturn 10.7 hours 378675 km 373644 km 1.3%
Uranus 17.2 160656.5 km 421056 km 2.62
Neptune 16.1 155660 km 312984 km 2
Pluto 153.3 7511.4 km 2.59 million km 345.3
Total 1277423 km 1153.15 million km 902.7
- Please Note
- The Paper Discussion Will Be Started After The Tables Of Data (2 Tables Only)

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Table (Planet motion degrees relative to different periods of time)
- Planet motion distance degrees per a solar day period (P. M. S) Planet days periods during its orbital period (P.D.O)
- Planet rotation periods during its orbital period (P.R.O) Planet motion distance per its day period (P. M. Dist)
- Planet motion distance degrees per its day period (P. M. D) Planet motion distance per its rotation (P. M. R. Dist)
- Planet motion distance degrees per its rotation period (P. M. R)
Planet P. M.S. P.D.O P.R.O (P. M. Dist) P.M.D. P. M. R. Dist P.M.R.
Mercury 4.095 mkm = 4.09 deg 0.5 days 1.5 rotations 720.7 m km 720.7 deg 240 mkm 240 deg
Venus 3.02 mkm = 1.6 deg 1.929 days 0.924 rotations 352.4 mkm 186.56 deg 733.8 mk 388.5 deg
Moon 88000 km = 13.18 deg 13.2 days 0.924 rotations 71.16 mkm - 65.8 mkm -
Earth 2.58 mkm = 0.985deg 365.25 days 366.7 rotations 2.58 mkm 0.98562 d 2.56 mk 0.9817 deg
Mars 2.082 mkm = 0.524 deg 667.5 days 670.2 rotations 2.146 m km 0.539 deg 2.138 m 0.5371 deg
Jupiter 1.1318 mkm = 0.083 deg 10495 days 10495 rotations 466884 km 0.0343 d 466884 km 0.0343 d
Saturn 0.838 mkm = 0.03349 24113.8 days 24113.8 rotation 373644 km 0.014929d 373644 km 0.014929d
Uranus 0.587 mkm = 0.01176 42880.3 days 42880.3 rotation 421056 km 0.00839 421056 km 0.00839
Neptune 0.466 mkm = 0.00602 90276.2 days 90276.2 rotation 312984 km 0.0039877 312984 km 0.0039877
Pluto 0.406 mkm = 0.003972 14324.3 days 14324.3 rotation 2.59 m km 0.025132 2.59 mkm 0.025132
Total (7.20 +0.1382 =7.342 deg) 183138 days 183129.84 1153.15 m 907.2+1.6067 1048.5 628.5+1.605
(940 mkm /71.16 mkm)= 13.2 =27.3 degrees
The moon moves per solar day 3 distances (1st
) 2.58 mkm (2nd
) 2.41 mkm (3rd
) 88000 km
- During 27.3 days , the moon moves (1st
) 71 mkm (2nd
) 65.8 mkm (3rd
) 2.41
- During 29.53 days , the moon moves (1st
) 76.2 mkm (2nd
) 65.8 mkm (3rd
) 2.58 mkm
- The moon orbital circumferences are (940 mkm, 2.58 mkm, 2.41 mkm and 2.28 mkm).
- The degrees (940/2.58= 0.985 deg) (940/2.41= 0.9365 deg) (940/0.00088 =0.0337), (2.58/0.88 =12.286 deg) (2.28/0.88=13.9)

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1- The Motion Transportation
1-1 The Outer Planets Motions Transportation 1-2 The Outer Planets Motions Total is balanced with Mercury Motion
1-3 Venus & The Moon Motions Interaction 1-4 Jupiter Is A Solar Planets Source Of Energy
1-1 The Outer Planets Motions Transportation
I- Data analysis
In The Table Number 2, we see
o Jupiter Motion Degrees During One Solar Day = 0.083 deg = Uranus Motion Degrees During Its Day (0.00839)
o Saturn Motion Degrees During One Solar Day = 0.03349 deg = Jupiter Motion Degrees During Its Day (0.0343)
(error 2.3%)
o Pluto Motion Degrees During 1 Solar Day =0.003972 deg = Neptune Motion Degrees During Its Day (0.0039877)
- Neptune moves during its day period a distance = 2 Neptune Circumferences
- Where 2 Neptune Circumferences = (Jupiter diameter + Saturn diameter+ Neptune diameter)
- Note Please
- 2 Neptune Circumferences = light known velocity (0.3 km/sec) motion during 1 second and 0.04 second
II- Discussion
- The table tells us simply there's some transportation of motion is done between Jupiter, Saturn and Uranus, this
conclusion can be reach clearly based on the data no (1 and 2)….. Also there's another transportation of motion is done
between Neptune and Pluto as shown in Data no. (3)
- Now we need only one transportation between the 2 groups to create a general transportation of motion though the
outer planets – and this one transportation we know it perfectly… let's remember it
- Jupiter energy is sent to Pluto and then reflected from Pluto to Neptune where Neptune used 14% of Jupiter Total
Energy to create its orbital circumference, that may tells Neptune motion depends on Jupiter motion….
- There's one more proof here, because Neptune motion distance during its day period (=312984 km) = (light 0.3 mkm/sec
motion during 1 second of time with error (4%), but we know that, Jupiter moves during its day period a distance = its
circumference with error (4%), and Saturn moves during its day period a distance = its circumference (1.3%), means

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Jupiter causes the motion transpiration to Neptune and even the error (4%) which was found in Jupiter motion is
transported also to create the error (4%) between Neptune motion during its day period and light motion for 1 second
- The previous analysis tells that, the 4 inner planets create some transportation of motion between each other….
- Note Please
o Light motion for 1 second which is compared with Neptune motion during its day period is a basic motion in the
solar system, because Jupiter energy was sent in light beam form from Jupiter to Pluto, that supports our old
hypotheses which are (1st
) (Planet motion depends on light motion) and 2nd
(light motion for 1 second cause
planet motion for 1 day)
o A basic point here is that, because Neptune orbital circumference is created by Jupiter energy, and Jupiter motion
during its day causes the transportation motion to Neptune motion per its day, we need to find some connection
between Neptune orbital circumference and Neptune circumference to support this claim – and we have 2 proves
o (1st
) Neptune orbital circumference x π = C2
for 1 second (C2
=90000 million km) (error 1.3%)
And 2 x Neptune circumferences = light motion for 1 second (error 4%)
o (2nd
) Neptune orbital circumference = Neptune circumferences x 181843 km (Where the value 181843 km is
the basic value in the moon orbital triangle data (Please review it in point no. 2 of the current paper).
i.e.
- The previous analysis tries to prove 2 claims
- (1st
) the planet moves during its day period a distance = its circumference (the ideal rule)
- (2nd
) not all planets follow this ideal rule because the motion is transported from planet to another to perform the
same result of this ideal rule in the planet motion actual situation – based on that there's a clear motion is
transported from planet to another through the outer planets
So, let's ask, what the task which should be done by this transportation of motion??
Let's answer this question in following

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1-2 The Outer Planets Motions Total is balanced with Mercury Motion
I-Data
(1)
The outer planets motions distances during their days periods = 4.16 mkm
(Mercury Motion Distance During 1 Solar Day = 4.095 mkm (error 1.6%)
(2)
175.94 days (Mercury Day Period) x π = 552.7 days BUT
511.1 degrees = 1.0725 x 552.7 degrees (511.1 degrees = The Planets Axial Tilts Total)
(3)
The 4 outer planets motions distances total during their days periods =1.57 mkm (without Pluto)
Their days total = 53.9 hours Pluto day period 153.3
(4)
(153.3/53.9) x 1.57 mkm = 1.0725 x 4.16 mkm
II- Discussion
- The previous data tells the following:
1. Pluto motion during its day is defined according to the 4 other outer planets motions distances total as rated to their days
periods one another –i.e.– the total value 4.16 mkm is defined based on the (Total Day Period) of the 5 outer planets.
2. The 5 outer planets motions distances during their days periods = Mercury motion distance during a solar day (1.6%)
3. but
4. Mercury during its day period (4222.6 hours) moves a distance =720.7 mkm, so these 5 outer planets motions distances
can't be balanced with Mercury Motion because Mercury day =175.94 solar days
5. Equation No. (2) tells that the planets axial tilts total (511.1 degrees) creates this required balance for both motions
6. That means, the planets axial tits total =511.1 degrees to create this balance…
7. The previous explanation shows that the motion transportation is used usefully to perform the geometrical necessary
tasks of the solar system motion which prove that, this motion transportation is a real one.

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1-3 Venus & The Moon Motions Interaction
I-Data
(1) (The Moon Rotation Period 27.3 Solar Days / The Moon Day Period 29.53 Solar Days) = 0.924
(2) (Venus Orbital Period 224.7 Solar Days / Venus Rotation Period 243 Solar Days) = 0.924
(3) 25.2 days = 27.3 days x 0.924
II- Discussion
- Equations No. (1 and 2) shows that, The Earth Moon and Venus Cycles Periods are rated to each other by the same rate,
but this same rate (0.924) is so specific one – because it's resulted after the contraction –
- The contraction rate (1.0725) causes the value 1 to be 0.924
That means,
- The moon rotation period is equal the moon day period =29.53 solar day but we see the first one under the contraction
rate (1.0725) and so we see it as (27.3), that's perfectly done for Venus, it's orbital period is a resulted under the
contraction for the rotation period 243 days = 1.0725 x 224.7 days
- That explains more old data, because, Mercury moves during its rotation period (58.66 solar days) a distance = 243mkm
and also Mars moves during Venus day period (116.7 solar days) a distance = 243mkm but this distance is seen wirth
Venus as 243 days (Why?) because it's a light motion (x =ct when c =1 so x =t) then the distance is used as a period of
time because it's a light motion (or high velocity motion) and another proof for this high velocity motion is the
contraction with rate (1.0725) which is done for Venus and the moon, means
- Venus and the earth moon motions are under the contraction
- Venus and the moon motions are interacted
- Equation no. (3) tells that, a third value (25.2) is produced under this same contraction from the moon rotation period (or
orbital period), the contraction created the value 25.2 (where 25.2 degrees = Mars axial tilt), that explains the interaction
of motion between Venus, Mars and the Earth moon (For example 687 days "Mars Orbital Period" =25.2 x 27.3), (also
13.18 deg the moon motion daily = 25.2 deg /1.9 deg "1.9 deg = Mars orbital inclination").

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1-4 Jupiter Is A Solar Planets Source Of Energy
I-Data
Equation No. (I)
1153.15 million km = 4900 days x 0.466884 mkm (error 1%) (4900 Jupiter days where each day =9.9 hours)
II-Discussion
Equation No. (I)
1153.15 million km = 4900 days x 0.466884 mkm (Jupiter motion during its day)
1153.15 mkm = The Planets Motions Distances During Their Days Periods
4900 mkm = Jupiter Orbital Circumference
- Equation no. (I) tells that, The Planets Motions Distances During Their Days Periods = The Distance passed by Jupiter
during 4900 days of Jupiter days
- What's interesting data here?
- The period 4900 (Jupiter) days is used as 4900 mkm = Jupiter orbital circumference …! How to understand that?
- We have only Jupiter data but used in different units why?? because it's light motion, we remember that…
- A light with supposed velocity (1.16 mkm/sec) travels during 4222.6 seconds and pass a distance = 4900 mkm, this
distance = Jupiter orbital circumference, but the period 4222.6 seconds was reached to Mercury and used it as 4222.6
hours (Mercury day period)
- The light motion connects between Jupiter and Mercury and because of that, Mercury moves during its day period a
distance = 720.7 mkm= Mercury Jupiter distance = 1.0725 x 778.6 mkm (Jupiter orbital distance), that means, Jupiter
orbital distance = Mercury Jupiter distance but the last distance is seen under contraction, so it's seen shorter
- Shortly
- I try to show that, this is a light motion and this motion uses the distance value as a time period, and
- All values are Jupiter data because Jupiter is the solar system energy source – because of that- its motion through light
beam equal all the planets motions distance- we may remember also that
- 4900 mkm (Jupiter orbital circumference) = the inner planet orbital circumferences total + Mars orbital circumference

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2- More Analysis In The Moon Orbital Triangle
Basic Data Correction And Revision

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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)
- 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.

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(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

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References
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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