Kepler Was Right (Planet Density Is A Function Of Its Distance)

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
Kepler Was Right (Planet Density Is A Function Of Its Distance)
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
Paper hypothesis
Planet Density Is A Function Of Its Orbital Distance
As a result
Planet Velocity Is A Function Of Its Diameter
The hypothesis Explanation
- The solar planets data shows that, planet velocity is defined as a function of its
diameter,
- This interesting observation we have caught before in Saturn motion, Because
Saturn moves during its day period a distance equal its circumference,
- Based on this observation, the conclusion was, (Planet day period is defines based
on some function between the planet velocity and its diameter)
- But – another observation has been caught – which is
- The outer planets days periods are proportional with their masses, and because of
that Uranus day period is longer than Neptune's because its mass is less –
- Newton tells us that, planet velocity depends on its masses gravity
- This data total leads to conclude the paper conclusion..
Paper conclusion
- Planet velocity is created as a function of its diameter
- The paper tries to prove this fact.

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1- Planet Velocity Is A Function Of Its Diameter (Proves Discussion)
Pluto Data Analysis
I-Data
Group No. 1
I. 51118 km Uranus diameter =
(a) = 4.7 km/sec (Pluto velocity) x 10921 seconds
(b) = 6.8 km/sec (Uranus velocity) x 7511 seconds
(c) = 2.58 mkm / (7.1)2
II. (0.406 mkm per day /51118) = (346.6/43.3) =8
III. (Uranus diameter / the moon diameter) / (the moon diameter / Pluto diameter) =C
But (Mercury velocity / Pluto velocity) = C
IV. (71492/4879) =(51118/3475) = (757350/51118) = (366500/49528) = (4.87/0.33) =
(86.8/5.97) = (97.8/6.7) (26.7/1.8) = (19/1.3) . (will be explained in the discussion)
V. 1206 mkm (Mars Saturn distance) = Uranus circumference x 7511 km
Group No. 2
- 2.58 million km = Pluto motion distance during its day period (153.3 hours)
- 2.58 million km = Earth motion distance during its day period (24 hours)
- 2.58 million km = the Earth moon motion distance during its day (708.7 hours)
II-Discussion
- The data is so puzzled let's start step by step
o (1st
Step), the Data in Group No. 2 tells us, we have 3 planets moves equal
distances in their days periods –which are Earth, Its Moon And Pluto
o These 3 planets move equally (2.58 mkm) for each during its day period
o Why? (1st
question)
o Still the situation needs explanation – because
o Uranus diameter is the secret word here – the data shows that clearly
o The distance 2.58 million km = 7.1 x 0.363 mkm, and this distance 0.363
mkm is the distance between Earth and its moon at perigee point – from that
we have learnt the rate 7.1 is used geometrically here
o 0.363 mkm = 7.1 x 51118 km (Uranus Diameter), so the secret word is
seen here – Why the distance (2.58 mkm) divided by (7.1)2
produces Uranus
diameter (equation no. c) ?
o Or to whom this distance (2.58 mkm) is belonged?! (2nd
question)

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o (2nd
Step), Pluto moves during 10921 seconds a distance = 51118 km =
Uranus diameter (equation no. a) and we know that 10921 km = the moon
circumference – but
o Uranus moves during 7511 seconds a distance=51118 km= Uranus diameter
(equation no. b) and we know that 7511 km = Pluto circumference..
o All equations have other sides – for example
o Pluto has to rotate around its axis (6.8 times) to move (by rotation) a
distance = 51118 km = Uranus diameter, but the rotation for (6.8 times)
around Pluto axis needs a period = (6.8 x 153.3 h = 1042.5 hours)
o Also the Earth moon needs to rotate around its axis 4.7 times to move by
rotation a distance = 51118 km = Uranus Diameter, but the moon rotation
for 4.7 times needs a period (4.7 x 27.3 days = 128.3 solar days)
o To solve this dilemma we need to see Equation no. (III) where
o Equation (III) (Uranus diameter / the moon diameter) / (the moon diameter
/ Pluto diameter) = C - But (Mercury velocity / Pluto velocity) = C
o Pluto and the moon diameters are created as function in Uranus diameter by
one rate (C), by this same rate Mercury velocity is rated to Pluto velocity!
Why? (3rd
question)
o The rate (C=10.08), this is so important rate because it's a basic number in
different data rates – and frequently we have seen the rate (10) is used in
different values (for example 1.3 deg (Jupiter orbital inclination) x 10 = 13
where Jupiter velocity =13.1 km/sec) –
o Let's suggest the main idea, (The Planets Diameters Are Created Based
On Their Velocities To Perform Some Geometrical Interaction By
Their Diameters Through Their Motions) – the idea is proved in the data
o Please note that, Mercury is the greatest velocity and Pluto is the smallest
one – by that – The Rate Controls All Other Velocities Between – and
that controls all planets velocities.
o That may explain old data, Let's remember it (Mercury velocity daily x
Ceres velocity daily = Venus velocity daily x Mars velocity daily = Earth
velocity daily x its Moon velocity daily) (daily = per solar day).
o But we have 2 rates, the first one is 14.7 (Uranus diameter / moon diameter)
and the rate (1.47 "error 1%") – the rate between them is 10 (or 10.08)
o The rate 14.7 controls the equation no. (IV), many different values are rated
based on this rate… the rate 14.7 in not a common in the planet data but it's
a cornerstone in the solar system geometrical structure – let's explain this
equation terms in following…

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o (71492 Jupiter radius /4879 Mercury diameter) =(51118 Uranus
diameter /3475 moon diameter) = (757350 (two Saturn circumferences
/51118 moon diameter) = (366500 outer planets diameters total /49528
Neptune diameter) = (4.87 Venus mass /0.33 Mercury mass) = (86.8
Uranus mass /5.97 Earth Mass ) = (97.8 Uranus axial tilt /6.7 moon axial
tilt) (26.7 Saturn axial tilt /1.8 Neptune orbital inclination ) = (19 degrees the
moon regression yearly/1.3 deg Jupiter orbital inclination).
o The previous explanation is a part only because Jupiter data does the same
play! So
o 142984 km = 13.1 km /sec x 10921 seconds (Equation no. A), this
equation tells Jupiter needs to move during 10921 seconds to pass a distance
= 142984 km = Jupiter diameter and ON THE OTHER SIDE
o The moon has to rotate 13.1 times around its axis to move a distance =
142984 km= Jupiter diameter (13.1 x27.32 days =357.9 days around 1 year)
o But
o 142984 km = 4.7 km /sec x 30589 seconds (Equation no. B), this equation
tells by Pluto velocity (4.7 km/sec), Pluto needs to move during 30589
seconds to pass a distance = 142984 km = Jupiter diameter (error 0.5%)
o But 30589 Solar Days = Uranus Orbital Period
o Uranus Orbital Period = 7.1 x Jupiter Orbital Period (error 0.5%)
o Equation no (IV)-
o 1206 mkm (Mars Saturn distance) = Uranus circumference x 7511 km
o Equation (IV) tells that Uranus and Pluto circumference define Mars Saturn
distance, this distance is so specific because Mars orbital circumference =
Saturn orbital distance
o Please Note, the rate 14.7 is confirmed to be so specific because, 14.7 x 71
= 1047 and this rate = (The Sun Mass / Jupiter Mass) , where 71 is found
as a contraction rate.
For more clear vision – I add to this discussion –the moon diameter analysis
discussion to see that the data which is used as distance is used again as periods of
time
THE CONCLUSION
Planet velocity is created a function of its diameter
Because
Planet density is a function of its orbital distance.

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Please Note
It's an old discussion inserted completely without change
It's published under title (Uranus Is Perpendicular On Earth Moon Orbit (VI))
1- The Moon 2 Orbits Proves Discussion
1-1 The Moon Diameter Analysis
Figure No. (1)
- This figure I have created as a part of the moon orbital triangle
- The figure provides a clear vision for what's happening inside the moon diameter
to create Uranus orbital inclination
- In this figure we use the moon orbital triangle data, where I have inserted this
triangle in the paper end to be our discussion reference –
Let's start our discussion
- We woks basically with the line (BTM L), it's Uranus axial tilt line (97.8 degrees),
- The line KL is the Earth Moon Axial Tilt (6.7 degrees)
- The angle KLM = 91.1 degrees which supposed to be equal the angle in (the moon
center IMT)
- The angle 1.1 degrees is consumed in the moon diameter (3475 km = 0.5 degrees)
and the green rectangle =0.6 degrees (created by Saturn and Jupiter interaction)
- The angle UTB = 90 degrees +0.6 degrees = 90.6 degrees
- We search for an angle =90.8 degrees (Uranus orbital inclination 0.8+90)
- The moon diameter consumes 0.5 degrees, so we will cut the moon great diameter
to a smaller one which consume only 0.2 degree, this small moon r = 695 km..
- So the small moon (its diameter = 695 km x 2= 1390 km) consumes 0.2 deg
- The (Uranus axial tilt line) angle on this small moon surface = 90.9 degrees
- This 90.9 degrees = 90.8 degrees + 0.1 degree
- Now let's try to see how this analysis will wok in following ….

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Equation No. 1
1.8 deg (Neptune orbital inclination) x 0.8 deg (Uranus orbital inclination) =1.44 deg2
- 1.44 degrees = the moon orbit regresses degrees monthly
- We have 0.8 degrees in the previous figure, but where's 1.8 degrees? How to find
Neptune orbital inclination in the previous figure,
- Because
- The red line (SB) declines on the horizontal level (KL = the moon axial tilt) with
only 1.1 degrees – so how to create 1.8 degrees in this figure
(Question No. 1)
Equation No. 2
The greater circle r = 1042.5 km (3475 km -1390 km = 2 x 1042.5 km)
Where
51118 km = 6.8 km/sec x 7511 seconds
- Let's remember this beautiful equation …
- Equation no. (2) tells that,
o Uranus needs to move 7511 seconds to pass a distance = its diameter
o Where 7511 km = Pluto circumference….
o On the other side
o Pluto has to rotate around its axis 6.8 times to move a distance = 51118 km=
Uranus diameter,
o but
o Pluto rotates around its axis each 153.3 hours, so 6.8 x 153.3 = 1042.5 h
- This interesting data will be more attractive when we remember Jupiter case
o 142984 km (Jupiter diameter) = 13.1 km/sec x 10921 km
o Jupiter needs to move 10921 seconds to pass a distance = its diameter
o Where 10921 km = The Earth moon circumference….
o On the other side
o The moon has to rotate around its axis 13.1 times to move a distance
=142984 km= Jupiter diameter,
o but
o the moon rotates around its axis each 27.3 days, so 13.1 x 27.3 = a year
- It's hard to consider both cases as pure coincidences….
- Any way we have found the number 1042.5 inside the moon diameter, and we as
we have seen in 2 examples, to use the distance value as a time period is usual
using in the plants data…!
- How to explain this data
(Question No. 2)

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Equation No. 3
91.7 degrees = 63.7 degrees x 1.44 degrees
- 63.7 degrees is the sun declination
- 1.44 degrees = the moon orbit regression monthly
- But
- What's this value 91.7 degrees??
-
- In the previous figure, the total decline angle between the moon triangle base (the
red line BS) on the horizontal level (the moon axial tilt) only 1.1 degrees and the
total angle between Uranus axial tilt and the moon axial tilt =97.8- 6.7 =91.1 deg.!
- So
- How to create this number 91.7 degrees? We need 91.1 deg +0.6 deg
- How to create this strange value (97.1 =91.1 deg +0.6 deg)??
(Question No. 3)
Equation No. 4
1.9 degrees = 1.1 degrees + 0.8 degree
- 1.9 degrees = Mars orbital incaution
- 0.8 degree = Uranus orbital incaution
- 1.1 degrees = (91.1 degrees -90 degrees) (i.e. the total decline angle)
i.e.
- If we consume 0.8 degrees from the total decline angle (1.1 degrees) how to add
the same angle (1.1) again to this total?
Also
- In Neptune orbital inclination 1.8 degrees
- We need
- 0.1 degrees works negatively against (1.9 degrees) (on the opposite direction!) so
(1.9 degrees – 0.1 degree = 1.8 degrees)
- That tells w have 2 value of the same total angle (1.1 degrees), and they may be in
opposite to each other!
Equation No. 5
1.44 degrees = 1.9 degrees x 0.378 degree x 2
- 1.44 degrees = the moon orbit regression per month
- 1.9 degrees = Mars orbital inclination
- 0.378 mkm = The Earth Moon Distance at total solar eclipse radius
- Equation No. (5) tells us, there's some reason to create an interaction between the
moon orbit regression and Mars motion! can really such interaction be created?
(Question No. 4)

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Equation No. 6
5.6 degrees = 1.1 degrees x 5.1 degree
- The value 1.1 degrees is the decline angle total
- 5.1 degrees = the moon orbital inclination
- 5.6 degrees = the moon orbital inclination + the moon diameter angle (5.1+0.5)
i.e. the moon orbital inclination is measure above the moon diameter
- Equation No. (6) tells that, the moon orbital inclination (5.1 deg) is produced as a
result of Uranus perpendicularly on the moon orbit and specifically as a result of
the decline total angle (1.1 degrees) in the moon orbital triangle
How to solve the
previous
questions?
A CONCLUSION
THERE'S ONE MORE ORBIT IS FOUND UNDER THE HORIZONTAL
LEVEL (AS SHOWS IN THE FIGURE)
This conclusion can solve the questions
- The value 1.9 degrees can be produced So 1.1 degrees (1st
orbit) +0.8 deg ( 2nd
orbit)
- Also (1.9 deg – 0.1 deg) = 1.8 deg where the value (0.1 deg) effects in opposite
direction to the first one (1.9 deg)
- Also
- The value 91.7 degrees = 63.7 x 1.44 degrees can be easily produced, because the
value 91.1 degrees (1st
orbit) +0.6 degrees (2nd
orbit)

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1-2 Pluto Effect On The Moon Orbital Triangle
We discuss here the interesting equation No.2
Equation No. 2 (Revision)
The greater circle r = 1042.5 km (3475 km -1390 km = 2 x 1042.5 km)
Let's add more data to extend our discussion
Equation No. 7
(a) 1042.5 km x 2 = 6.01 x 346.6
(b) 8 degrees = 1.9 degrees + 6.1 degrees
(c) 346.6 days (the nodal year) = 8 x 43.3 days but (1042.5 hours = 43.3 days)
How to understand this equation, let's move step by step:
- 1042.5 km = the greater circle radius (R = 1042.5 km) in our figure No. 1
- So the diameter = 1042.5 km x 2 =2085 km
- We remember the story concerning the equation no. (2), that Pluto needs to move
during 1042.5 hours to rotate around its axis 6.8 times…. But (1042.5 h =43.3 day)
- The hours (1042.5) causes to create the days (346.6 days) based on the rate 8
- 8 = 1.9 deg (Mars orbital inclination) + 6.1 degrees
- But we have only 6.01 and how to create 6.1? (6.01 /0.1) =6.1 degrees
- Equation no. (b) tells that, the value (0.1 degree) is owned by Mars and it uses
freely as here, proving that (1.9 deg= 0.1 deg +1.8 deg Neptune orbital inclination)
- What conclusions here?
- Mars and Pluto are 2 players in the nodal year creation (346.6 days)
- Because of that
o (687 days mars orbital inclination + 2 x 3.1) = 2 x 346.6 days
(3.1 degrees = Jupiter Axial Tilt which is used here in days unit)
o 90560 days (Pluto Orbital Period) = 346.6 days x 261.3
o 261.3 x 0.1 = 26.13 = (5.1)2
(5.1 degrees = The moon orbital inclination)

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Equation No. 8
180.8 degrees = 179.9 degrees +0.9 degrees
- 180.8 degrees = 177.4 deg (Venus axial tilt) +3.4 deg (Venus orbital inclination)
- But
- 179.9 degrees = 177.4 deg (Venus axial tilt) +2.5 deg (Saturn orbital inclination)
- 0.9 degrees is the total decline angle above the moon small diameter (r=695 km)
And
- We know that the green triangle in the figure (mo.1) is created to consume 0.6 by
interaction between Jupiter and Saturn (that proved by the equations 2.5 deg = 0.6
deg +1.9 deg and 1.9 deg = 0.6 deg +1.3 deg)
- So the interaction is done inside the moon body –
- Simply above a radius =695 km, the interaction is done between Venus axial tilt
(177.4 deg) and Saturn orbital incaution (2.5 degrees) =179.9 deg, then add to that
the value 0.9 degrees which is the total decline angle above the small moon
diameter (695 km) – to create the total 180.8 degrees based on which Venus orbital
inclination is created.
- To prove this story… is there any relationship between 695 km and Venus?
Equation No. 9
12104 km (Venus diameter) = 695 km x 17.4
- 17.4 degrees = the inner planets orbital inclination total
(But 17.2 deg Pluto orbital inclination = 17.4 x 0.99)
- 23.6 degrees = the outer planets orbital inclination total
(But 23.4 deg Earth axial tilt = 23.6 x 0.99)
- Equation no. ( 9) tells that there's a strong relationship between Venus diameter
and the small moon radius (695 km)
- But we know that
o Venus circumference x π = Saturn diameter
o (Venus circumference)2
= Saturn orbital distance (1%)
i.e.
- Based on the radius 695 km both planets are interacted and created their data to be
in consistency with this dame distance

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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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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

Kepler Was Right (Planet Density Is A Function Of Its Distance)

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