Jupiter Has A Cycle Of 8 Days (II) (Revised)

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
Jupiter Has A Cycle Of 8 Days (II) (Revised)
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 – 26th
December 2020
Abstract
Paper hypothesis:
Planet day period is defined depends on other planets motions
Paper objective:
The paper discuss if this fact supports the hypothesis (Planet Motion Depends On Light Motion)
Paper contents
1- Planet Velocity And Diameter Relationship
2- Planet Day Period Definition
3- The Moon Day Period Discussion

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1- Planet Velocity And Diameter Relationship
1-1 Data 1-2 Discussion
1-1 Data
Table No.1 Planet Velocity And Diameter Relationship
Planet Mercury
4879 km
Venus
12104 km
Earth
12756 km
Mars
6792 km
Moon
10921
Jupiter
142984km
Saturn
120536 km
Uranus
51118 km
Neptune
49528 km
Pluto
7511 km
Mercury 47.4 k/s 231265 573730 604634 321941 517655.4 6777442 5713407 2422993 2347627 356021
Venus 35 km/s 170765 423640 446460 237720 382235 5004440 4218760 1789130 1733480 262885
Earth 29.8 km /s 145394 360700 380129 202402 325446 4260923 3591973 1523316 1475934 223828
Moon 27.78 km/s 135539 336249 354362 188682 303386 3972096 3348490 1420058 1375888 208656
Mars 24.1 km /s 117584 291706 307419.6 163687 263196 3445915 2904918 1231944 1193625 181015
Jupiter 13.1 km /s 63915 158563 167104 88975 143065 1873091 1579022 669646 648817 98394
Saturn 9.7 km /s 47326.3 117409 123733 65883 105934 1386945 1169199 495845 480422 72857
Uranus 6.8 km /s 33177 82307 86741 46186 74263 972292 819645 347603 336791 51118
Neptune 5.4 km /s 26347 65362 68882.4 36677 58974 772114 650895 276037 267451 40560
Pluto 4.7 km /s 22931.3 56889 59953 31922.4 51329 672025 566519 240255 232782 35302
Notice The marked numbers are the used numbers because they refer to distances passed by 2 different planets velocities during
2 defined periods of time (as we will discuss in following)

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More Data (1)
(1)
117584 km = 12104 seconds x 9.7 km/s = 4879 seconds x 24.1 km /s
(2)
432640km = 12104 seconds x 35 km/s = (17.2 h x 3600 s) x 6.8 km /s
(3)
336249 km =12104 seconds x 27.78 km/s = 49528 seconds x 6.8 km /s
(4)
160592 km =12104 seconds x 13.1 km/s = π x 10921 seconds x 4.7 km /s (error 1.2%)
(5)
65362 km =12104 seconds x 5.4 km/s = 6972 seconds x 9.7 km /s
(6)
446460 km =12756 seconds x 35 km/s = π x 10921 seconds x 13.1 km /s
(7)
380129 km = 12756 seconds x 29.8 km/s = 10921 seconds x 35 km /s = (10.7 h x 3600) x 9.7 km/s (error 2%)
(8)
(9)
307420 km = 12756 seconds x 24.1 km/s = 10921 seconds x 27.78 km /s (error 1.3%)
(10)
167104 km = 12756 seconds x 13.1 km/s = 6792 seconds x 24.1 km /s (error 2%)
(11)
60000 km = 12756 seconds x 4.7 km/s
(12)
312941 km = 6792 seconds x 47.4 km/s = 10921 seconds x 29.8 km /s = (16.1 h x 3600s) x 5.4 km/s
(13)
263196 km = 10921 seconds x 24.1 km/s = 7511 seconds x 35 km /s

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(14)
(15)
4260923 km = 142984 seconds x 29.8 km/s = 120536 seconds x 35 km /s
(16)
1873091 km = 142984 seconds x 13.1 km/s = 2 x 943817 seconds
(17)
(18)
(19)
1420058 km = π2
x 142984 seconds = 51118 seconds x 27.78 km /s
(20)
1523316 km = π2
x 49528 seconds = 51118 seconds x 29.8 km /s
(21)
669646 km = 51118 seconds x13.1 km /s = 49528 seconds x 2 x 6.8 km /s
(22)
40080 km = 7511 seconds x 5.4 km /s
(23)
35302 km = 7511 seconds x 4.7 km /s
Notice
4879 km = Mercury Diameter 51118 km = Uranus Diameter Mercury Velocity =47.4 km/s
12104 km = Venus Diameter 49528 km = Neptune Diameter Venus Velocity = 35 km/s Saturn Velocity = 9.7 km/s
12756 km = Earth Diameter 6792 km = Mars Diameter Earth Velocity = 29.8 km/s Uranus Velocity = 6.8 km/s
142984 km = Jupiter Diameter 120536 km = Saturn Diameter the moon Velocity = 27.78 km/s
7511 km = Pluto Circumference Mars Velocity = 24.1 km/s Neptune Velocity = 5.4 km/s
10921 km = the Earth moon Circumference Jupiter Velocity = 13.1 km/s Pluto Velocity = 4.7 km/s

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1-2 Discussion
- What does this data tell us?
- Let's see how this analysis started
o 142984 km = 13.1 km/s x 10921 seconds, this equation we have seen before…. It tells that, Jupiter (whose velocity
13.1 km/s) moves during 10921 seconds a distance = 142984 km = Jupiter Diameter
o But 10921 km = The Earth Moon Circumference
o That means, Jupiter velocity uses the moon circumference (10921 km) as a period of time and during it (10921 seconds)
Jupiter moves a distance = 142984 km = Jupiter diameter! why?
o Jupiter is found before the Earth moon creation, means, this previous equation has 2 values as ancient values (142984
km and 13.1 km/s) and only the moon circumference 10921 km which is used as a period of time (10921 seconds) is the
new value in this equation because the moon is created after Jupiter Creation
o We have another similar equation … let's see if in following
o 51118 km = 6.8 km/s x 7511 seconds, this equation also we know …. It tells, Uranus (whose velocity 6.8 km/s) moves
during 7511 seconds a distance = 51118 km = Uranus Diameter
o But 7511 km = Pluto Circumference
o It's a typical equation to the previous one… but we still don't understand why the planets behave by such behavior…
o We accept that, the planet (Jupiter or Uranus) uses its diameter and velocity to create a period of time (10921 seconds in
Jupiter equation and 7511 seconds in Uranus equation) and we discovered that (10921 km = the moon circumference
and 7511 km = Pluto circumference), still we don't realize what's going here….?!

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- Let's consider the next suggestion:
o Jupiter diameter 142984 km and velocity 13.1 km/s are known before the moon creation, so Jupiter uses its own data to
create a value (10921 seconds) which can be in harmony in Jupiter motion, and later Jupiter uses this value (of time
10921 seconds) to creates a planet circumference (10921 km) and by that this new planet motion will be in harmony
with Jupiter motion
o So, the production of the value 10921 seconds from Jupiter data (142984 km and 13.1 km/sec) is a similar to a birth
process to produce a new planet be its motion in harmony with the other planets motion, to perform this harmony
Jupiter creates the moon data to be in harmony with Jupiter motion and because Jupiter motion is in harmony with the
other planets motions the new born planet motion will be in harmony with all other planets motion – it's a giving birth-
o By this same description Uranus gave birth to Pluto
o This giving birth guarantees the general harmony of motions
Now …
- Let's suppose that, This suggestion is a correct one, What we will expect based on it? we should expect that, by this same
method all planets are created based on reach other, we have discussed this description before, let's remember it here
o We should refuse the description that, the solar planets are rigid bodies separated from each other revolves around the
sun by the masses gravity effect
o Instead we have to consider The solar planets as points found on the same trajectory of energy, or as puppets connected
together in one thread or as one body and each planet is a member in this same body.

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o By this description, each planet data is created relative and complementary to the other planets data… now if we accept
this conclusion…. The Birth of the Earth moon from Jupiter data will show the geometrical mechanism by which the
planets data is created depending on each other… by this same method Pluto circumference (7511 km) is created by
Uranus motion data (Uranus diameter and velocity)…
o If this conclusion also is acceptable
o So, We should expect that the table no. (1) should show the connections between the planets by which each planet is
created depending on another one (simply the table shows the network based on which the planets data is created)
o So, the equations (from 1 to 23) try to show the table details in separated equations to help our investigation
- That tells us we need to move with each equation individually because it's a point in the network and we need to move
through the whole network to see how it's created and how it effects on the planets motions.
- The explanation is complex in both cases theoretically or practically let's use one example only
o 4.7 km/s (Pluto velocity) x 10921 seconds = 51118 km = Uranus Diameter
o 4.7 km/s (Pluto velocity) x 51118 seconds = 2 x 120536 km = Saturn Diameter
o The previous data shows the accumulation effect, by that, the Earth moon circumference, Uranus diameter and Saturn
diameter all connected together strongly by one motion velocity (Pluto velocity 4.7 km/sec)
o Why this is useful and effective on the planets motions? Let's try to answer in following

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Equation no. (12)
312941 km = 6792 seconds x 47.4 km/s = 10921 seconds x 29.8 km /s = (16.1 h x 3600s) x 5.4 km/s
- Neptune moves during its day period (16.1 hours) a distance = 312941 km ……But
o 312941 km = 6792 seconds x 47.4 km/s = 10921 seconds x 29.8 km /s
o Mercury (whose velocity 47.4 km/s) moves during 6792 seconds (but 6792 km =Mars diameter) a distance=312941 km
and Earth (whose velocity 29.8 km/s) moves during 10921 s (10921 km = the moon circumference) a distance = 312941
km – So –
o This same distance 312941 km is passed by 3 planets (Neptune, Earth and Mercury) during 3 different periods, 2 of
them are (mars diameter and the moon circumference), that makes, Neptune day period a value rated to these planets
diameters values! Why?
o We may remember that, Pluto (whose velocity 4.7 km/s) moves during 7511 seconds a distance = 35302 km, where
7511 km = Pluto circumference, now, If we use the produced distance 35302 km as a period of time 35302 seconds so
this value = Jupiter day period (9.9 hours) with error (1%)
o Shortly, the planets days periods are created to be in consistency with the planets diameters or circumferences if the day
period is used as a distance value (1 second= 1 km) or if the planet diameter or circumference is used as a period of time
(1 km = 1 second).
- Let's try to find a general conclusion for this discussion… The data tries to create a rate between the planets diameters and
their days periods…. This process is used for the outer planets only Why?

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- Because
- The outer planets move during their days periods distance rated to their circumferences
o Saturn moves during its day period a distance = its circumference (error 1.2%)
o Jupiter moves during its day period a distance = its circumference (error 4 %)
o Neptune moves during its day period a distance = 2 its circumferences
o Uranus moves during its day period a distance = 2.6 it circumference
But
- The inner planets move during their days periods distances rated to their orbital circumferences… Why? Because the inner
planets define their orbital circumferences to be equal their distances to Jupiter, so the inner planets orbital circumferences are
created by a direct effect of Jupiter because of that their days periods are so long and doesn't related to their circumferences
but their orbital circumferences….. so
- That divides the solar planets into 2 groups, the inner planets whose days periods motions distances are rated to their orbital
circumferences and the outer planets whose days periods motions distances are rated to their circumferences…
- Still the inner planets velocities are rated to other planets diameters and circumferences
- It's a great and deep network, which is seen clearly in the table no. (1)

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2- Planet Day Period Definition
2-1 Jupiter Day Period 9.9 hours
(a)
4.7 km/s x 7511 seconds = 29.8 km /s x 1195 seconds = 3600 x 9.9 (error 1%)
- We have discussed Jupiter day period before, let's remember this discussion
o Pluto (whose velocity 4.7 km/s) moves during 7511 seconds a distance = 35302 km, if we use this distance as a period
of time (33502 seconds), it = 9.8 hours but Jupiter day = 9.9 hours, So the error is 1%
But
o Earth (whose velocity 29.8 km/s) moves during 1195 seconds a distance = 35640 km, if we use this distance as a period
of time (35640 km seconds), it = 9.9 hours = Jupiter day = 9.9 hours accurately without error
i.e.
o Earth motion during 1195 seconds moves the accurate distance which can be used as Jupiter day period and the error
1% is created between Pluto motion during (7511 seconds) and Earth motion during (1195 second), means, Jupiter day
is defined by Earth motion but the difference 1% between Earth and Pluto motions is found because of geometrical
necessity…
Notice
o 7511 km = Pluto Circumference
o 1195 km = Pluto Radius, the confusion in equations is found because (Earth velocity / Pluto velocity) = (Earth day/
Pluto day).

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2-2 Saturn Day Period 10.7 hours
(b)
27.78 km/s x 1390 seconds = 38520 km = 3600 x 10.7 (No Error)
- 27.78 km /s = the moon velocity
- 1390 km = the moon small diameter which we have created by the moon diameter analysis.
- How the moon velocity =27.78 km/sec?
- We know the moon moves a motion typical to Earth motion but a contraction effects on it by a rate 1.0725, so earth velocity
=29.8 km/s and based on that the moon velocity = 29.8 / 1.0725 = 27.78 km/s
- Equation (b) tells that, based on the moon diameter Saturn day period is defined
- We faced a similar data before, where 10747 days (Saturn Orbital Period) = 365.25 days x 29.53 days
- i.e. Saturn Orbital Period is created depending on the moon day period, not only but also Saturn day is created depending on
the smaller moon diameter.

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2-3 Neptune Day Period 16.1 hours
(c)
47.4 km/s x 4879 seconds = 4 x 57960 km = 4 x 3600 x 16.1 (No Error)
- 47.4 km /s Mercury Velocity
- 4879 km = Mercury Diameter
- The distance passed by Mercury during 4879 seconds = 4 folds the value 57960 km, this value if it's used as period of time
(57960 seconds) = 3600 s x 16.1 hours.
2-4 Uranus Day Period 17.2 hours
(d)
9.7 km/s x 6378 seconds = 61920 km = 3600 x 17.2 (No Error)
- 9.7 km /s Saturn Velocity
- 6378 km = Earth Radius
- The distance passed by Saturn during 6378 seconds =61920 km, if this value is used as period of time (61920 seconds) =
3600 s x 17.2 hours.

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2-5 Pluto Day Period 153.3 hours
(e)
5.4 km/s x 51118 seconds x 2 = 551880 km = 3600 x 153.3 (No Error)
- 5.4 km /s Neptune Velocity
- 51118 km = Uranus Diameter
- The distance passed by Neptune during 2 x 51118 seconds =551880 km, if this value is used as period of time (551880
seconds) = 3600 s x 153.3 hours.
2-6 Earth Day Period 24 hours
(f)
6.8 km/s x 12756 seconds = 86400 km = 3600 x 24 (No Error)
- 6.8 km /s Uranus Velocity
- 12756 km = Earth Diameter
- The distance passed by Uranus during 12756 seconds =86400 km, if this value is used as period of time (86400 seconds) =
3600 s x 24 hours.

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2-7 Mars Day Period 24.7 hours
(f)
27.78 km/s x 12756 seconds = 355680 km = 4 x 3600 x 24.7 (No Error)
- 27.78 km /s The moon Velocity
- 12756 km = Earth Diameter
The distance passed by the moon during 12756 seconds =4 x 88920 km, if this value is used as period of time (88920 seconds) =
3600 s x 24.7 hours.
2-8 Venus Day Period 116.75 solar days
(g)
27.78 km/s x 363000 seconds = 10.0872 mkm = 86400 x 116.75 (No Error)
- 27.78 km /s The moon Velocity
- 363000 km = Earth Moon Distance At Perigee Radius
The distance passed by the moon during 363000 seconds = 10.0872 mkm, if this value is used as period of time (10.0872 million
seconds) = 86400 s x 116.75 hours.

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2-9 Mercury Day Period 4222.6 hours
(h)
29.8 km/s x 51118 seconds x 10= 15.201360 mkm = 3600 x 4222.6 (No Error)
- 29.8 km /s Earth Velocity
- 51118 km = Uranus Diameter
The distance passed by Earth during 10 x 51118 seconds = 15.20136 mkm, if this value is used as period of time (15.20136
million seconds) = 3600 s x 4222.6 hours.
2-10 The Earth Moon Day (29.53 solar days)
(h)
6.8 km/s x 378675 seconds = 2.55 mkm = 86400 x 29.53 (No Error)
- 6.8 km /s Uranus Velocity
- 378675 km = Saturn Circumference
- Please consider this data as deeply as possible, the equation is similar to the previous and Uranus motion distance during
(378675 seconds) if be used as a period of time (1km =1second) it will be = 29.53 solar days..
- But
- 2.55 million km = The Moon Orbital Circumference and
- 378675 km = Saturn Circumference = The Earth Moon Distance At The Total Solar Eclipse

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- I kept the moon discussion at the end, because the previous equations are understood simply but the confidence is not so high
because we don't know how the distance can be used by planet low velocity as a period of time
- But the moon data gives us some light to see much better
- The moon day period 29.53 solar days = 2.55 million seconds but the moon orbital circumference =2.55 million km, now we
have 2 real values of planets data show this same idea – the tells us the moon orbital motion and orbital geometrical structure
and the moon day period all this data are created based on this concept where the distance value is used as period of time.
- The moon tells us that, the conclusion is correct
- But
- The question is still behind …. The conclusion tells us… the moon orbital circumference =2.55 mkm because the moon day
period =2.55 million seconds….! Why? by what mechanism these 2 values are created based on each other? !
We need to discuss this question with the following data (may later we will do)
2.58 mkm = 29.8 km/s x 86400 seconds
2872.5 mkm = 1153 mkm x 2.5
2872.5 mkm x 0.406 mkm = 1166 mkm

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3- The Moon Day Period Discussion
3-1 The Moon Day Period Analysis 3-2 How can Planet Circumference be used as a period of time?
3-1 The Moon Day Period Analysis
I-Data
(h)
The moon orbital circumference at apogee radius (r=0.406 mkm) = 2.55 mkm But the moon day period 29.53 solar days
=2.55 million seconds (Zero Error) (Note Please (6.8 km/s x 378675 seconds = 2.55 mkm))
(i)
Earth moves per a solar day a distance = 2.573 mkm (with difference 1% with the value 2.55 mkm)
(j)
The moon moves daily a displacement =88000 km and during 29.53 days the total distance = 2.5977 mkm
(with difference 1% with the value 2.573 mkm)
(k)
25920 mkm = 2.5977 mkm x 9978 (But 9978 = 365.25 x 27.3)
(L)
25920 mkm = 2.573 mkm x 10072 (But 0.3 mkm/ s = 29.8 km/sec x 10072)
(m)
2.55 mkm = 116750 km x 21.8 ( BUT if 1000 = 1 day, So 116750 km =116.75 days = Venus Day)
(n)
327.6 mkm = 2.55 mkm x 128.4 (BUT 128.4 km = 27.32 seconds x 4.7 km /s).

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II-Discussion
- I use the chance of this paper discussion to face one old puzzled data, the equations no. (h, I and J) tells about this dilemma…
Equation (h)
The moon orbital circumference at apogee radius (r=0.406 mkm) = 2.55 mkm But the moon day period 29.53 solar days
=2.55 million seconds (Zero Error) (Note Please (6.8 km/s x 378675 seconds = 2.55 mkm))
Equation (i)
Earth moves per a solar day a distance = 2.573 mkm (with difference 1% with the value 2.55 mkm)
Equation (j)
The moon moves daily a displacement =88000 km and during 29.53 days the total distance = 2.5977 mkm
(with difference 1% with the value 2.573 mkm)
- let's try to form the question as clear as possible in following:
o The moon day period =29.53 solar days =2.55 million seconds, but the moon orbital circumference =2.55 mkm, here
we will suppose that 1 second is used as 1 km (although we don't know how!) but this supposition will give us a way to
explain the previous puzzled data…. But
o Earth moves per solar day a distance = 2.573 mkm AND the moon has to move this same distance per solar day
otherwise the moon will be separated from Earth through their motions, that means, The moon motion distance per a
solar day is greater than the moon orbital circumference at apogee orbit (r=0.406 mkm) with difference 1% why?
o Also the moon displacement = 88000 km and during 29.53 days the total will be 2.5977 mkm with difference 1% (with
2.573 mkm) and difference 2% (with 2.55 mkm) – no way to consider these differences as errors!

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- The question is, what's happening here?!
- I claim that we deal with geometrical structure- means- There's a geometrical necessity to create a difference = 1% between
these 3 values! BUT what's this geometrical necessity?!
Equation (k)
25920 mkm = 2.5977 mkm x 9978 (But 9978 = 365.25 x 27.3)
Equation (L)
25920 mkm = 2.573 mkm x 10072 (But 0.3 mkm/ s = 29.8 km/sec x 10072)
- 25920 km = 0.3 mkm/ s x 86400 seconds, means, Light known velocity (0.3 mkm/s) passes a distance = 25920 km during 1
solar day (86400 seconds)
- The distance 25920 km tells that the values (2.5977 mkm and 2.573 mkm) are created by geometrical interaction with Earth
velocity (29.8 km/s) and Earth orbital period (365.25 days) with the moon orbital period.
- The basic importance of this data is that, the data shows the geometrical mechanism inside the planets motions interaction…
We have a rate (1%) is used here for a geometrical necessity, we try to know Why, How And Where this rate is used….
- 23.6 degrees x 0.99= 23.45 degrees (Earth Axial Tilt) And (23.6 degrees= The Outer Planets Orbital Inclinations Total),
this previous equation may tell us how this (1%) effects on Earth Data Creation…
- 5.4 km/s x 7511 seconds = 40560 km (Earth Circumference = 40080 km the difference = (1%)), this Equation tells that,
Neptune (Whose velocity 5.4 km/s m0oves during 7511 seconds a distance = Earth Circumference with difference (1%)).
- The basic data may be that, the moon orbital inclination =5.1 degrees and can be extending to 5.2 degrees with range (2%)

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- The difference (1%) is significant for the moon data because
o 300000 km =10921 km x 27.47, We know this equation, it tells that, if the moon rotates around its axis once per solar
day as Earth does, so in this case the moon would move during its orbital period (27.3 solar days) a distance =300000
km = light known velocity motion distance during 1 second………. But
o 27.47 days doesn't equal the moon orbital period (27.3 days), BUT (27.47 days x 0.99 = 27.2 days) where the perigee
and apogee month = 27.2 solar days.
o By this same rule, 27.57 days x 0.99 =27.32 days (the moon orbital period)
o 25920 mkm = 27.57 x 940 mkm (Earth orbital circumference around the sun).
Clearly
- I can't catch the point! Why? because the difference (1%) is spreading in the moon and Earth data (and all other planets) we
don't know from where this difference is created or for what geometrical reason it's used, simply there are 2 values of the data
one =100% and the other =99% and both are used as real data (as used in the moon data).
Equation (m)
2.55 mkm = 116750 km x 21.8 ( BUT if 1000 = 1 day, So 116750 km =116.75 days = Venus Day)
- 21.8 = (Jupiter Mass / Uranus Mass)
- The moon orbital circumference (2.55 mkm) = (Jupiter Mass / Uranus Mass) x 116750 km
- The planets masses rate effect on the planets data regardless the distances! This notable observation we have caught frequently
in different data…

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- If 1000 km = 1 day so 116750 km will = 116.75 days = Venus day period, why this supposition is significant?! Because if
1000 km = 1 day so the moon displacement 88000 km will cause Mercury orbital period 88 days,
- That means, this data is supported even by Mercury data also… but we don't know by what geometrical mechanism 1000 km
= 1 day, but this rate (1000 km =1 solar day) is an important rate because it connects Mercury and Venus Motions with the
Moon Motion and because these 3 planets (Mercury, Venus and The earth moon) have the longest days periods in the solar
system because of that we consider this connection between them is so strong which makes this rate a specific one.
Equation (n)
327.6 mkm = 2.55 mkm x 128.4 (BUT 128.4 km = 27.32 seconds x 4.7 km /s).
- If 1 mkm = 1 day (the usual rate) So 327.6 mkm will =327.6 days = The Moon Sidereal Year.
- The Moon Sidereal Year (327.6 days) depends on the moon orbital circumference (2.55 mkm) or on the moon day period
(2.55 million seconds)
- We may remember (51118 km = 4.7 km/s x 10921 seconds, this equation told us that, Pluto (4.7 km/s) moves during 10921
seconds a distance = 51118 km = Uranus diameter BUT this equation tells also that the moon (whose circumference = 10921
km, needs to rotate around its axis 4.7 times to move by rotation a distance = 51118 km = Uranus diameter but the moon
rotates around its axis once each 27.32 days and that means this distance (51118 km) needs (4.7 x 27.32 = 128.4 days),
- The moon sidereal year (327.6 days) depends on these 2 values (2.55) and (128.4)

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- Let's remember one more important data in following:
- 2094 mkm = Jupiter Uranus Distance
- 2082 mkm = a distance light known velocity (0.3mkm/s) passed during 6939.75 seconds (Where Metonic Cycle period
=6939.75 solar days), and also the value 2072 mkm = 2094 mkm x 0.99 is used geometrically…
- This distance is very similar to the moon data (2.55 mkm, 2.573 mkm and 2.5977 mkm) where the differences are so small
(1%) but are real differences
- These differences are created by geometrical necessities in this distance (Jupiter Uranus Distance) and in the moon motion
data.
- Shortly
- Still the moon data needs more analysis… but I have to leave it here
Because
- We need to answer how planet circumference can be used as a period of time- we should do that in the next point.

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3-2 How can Planet Circumference be used as a period of time?
(1st
) The Main Idea
- Let's summarize the main idea in following….
- We try to explain how planet circumference or diameter can be used as a period of time!
- Light motion creates the equation (x = ct) if (c= 1), So (x =t) and by that the distance value can be used as a period of time.
- When a planet rotates around its axis, the planet moves a distance = its circumference
- As seen in the figure, the planet (the blue circle) moves on the trajectory a distance = The Planet Circumference
- This distance can be used by light motion as a period of time, so when we use some equation we find this same distance
consider that this is the planet circumference and it's not true, it's the distance on the trajectory passed by the planet rotation
around its axis which it = the planet circumference…. So
- 7511 seconds x 6.8 km/sec = 51118 km, where 7511 km = Pluto Circumference, this equation doesn't show the real steps
because Pluto circumference = a distance on the trajectory passed by Pluto rotation around its axis (as the figure shows) and
the equation uses this distance (7511 km) as a period of time and not Pluto circumference itself ….But
- What about the planet diameter? for example 12104 seconds x 9.7 km/sec = 120536 km where 12104 km = Venus Diameter,
this equation uses the same previous explanation but for geometrical necessities the equation uses (π) and by that the planet
circumference is reduced to be its diameter (what do we try to do? We try here to decrease the gab between the physics book

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and the planets data, because the data show that the planets diameters or circumferences are used as periods of time and I try
to create a physical explanation how that's happened…)
- The previous explanation tells that, (A Light Beam Motion Is Found In Parallel To A Planet Motion), this is the
cornerstone in the answer why Planet circumference is used as a period of time? But, is there any other method to use planet
circumference as a period of time without using light motion?! I don't think so, but let's test one more suggestion in following:
o Suppose the Earth become a gas planet and no crust of the earth be longer, and we have to fly into the Earth sky because
no crust we can stand on…
o In The Total Solar Eclipse, suppose that, the area which is covered by the total solar eclipse umbra became a solid land
and be (A Part Of The Crust Of Earth), so we can stand on this part of land and don't need to fly more
o But this part of Earth Crust will be available in change value with a period of time because of the total solar eclipse
mechanism, that means, the land be available as a function of time because the total solar eclipse will be finished and
this land will be dissolved again.
o That makes the distance (part of Earth crust) is created with a period of time depending on each other, by that, I can
measure the distance by a time unit, for example, 1 km is provided each 2 minutes, so this area (1 km2
) can be called an
area of (2 minutes).
o This simple suggestion try to show that, the distance and time can be created depending on each other and by that they
can be measured by exchangeable units
- Regardless this suggestion, the distance and time can be measured by exchangeable units as advantage of light motion features
because of that we should discuss how to discover a light motion which is accompanying the planet motion.

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(2nd
) Light Motion
- How a light motion can be accompanying with a planet motion? Let's consider the following suggestion:
- Suppose the light known velocity (0.3 mkm/sec) is a total energy, from which planet matter is created and its orbital distance
to enable the planet motion and by the rest of energy the planet velocity is defined, so we see planet (as Pluto for example)
moves by a velocity 4.7 km/s but Pluto total data causes this velocity to be = (0.3 mkm/sec = light known velocity),
- I want to say that, the light known velocity (0.3mkm/sec) is similar to a great gear in a machine of gears, planet velocity is a
small gear in this same machine, when planet moves 4.7 km/second, in this same second, the light moves 0.3 mkm/sec as a
result of the planet motion- the great gear (light) moves by the small gear (planet) motion because of the machine gears
between them.
- So, Uranus move 6.8 km per second, but each 6.8 km of Uranus motion causes 0.3mkm of light motion in this same second.
- Means, light motion is accompanying all planets motions but we see planet velocity because the gears are hidden behind the
planets data … for example
o 365.25 days x 27.32 days = 9978, and light velocity / Earth velocity =10072, Why, because light beam motion with
known velocity (0.3mkm/sec) is accompanying with Earth motion where the gears (365.25 days = Earth orbital period)
and (27.3 days = the moon orbital period), causes each 29.8 km passed by Earth to cause the light to move 0.3 mkm in
the same second.
o 9.7 km/s x 6378 seconds = 61920 km (But Uranus day period =61920 seconds) and Earth Radius =6378 km, and earth
radius is used as a period of time because of light motion accompanying Earth motion but the explanation is compl4eex

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and difficult because we don't completely understand how the gears works where the gears are (365.25 days and 27.32
days).
Shortly
o We try here to discover the light motion accompanying the planets motions and by this light motion we can't discover
the geometrical mechanism by which the planet circumference or diameter can be used as periods of time, Let's see the
data in following
I-Data
(o)
0.3 mkm = 9.7 km x 30589 seconds (1%) (BUT142984 km = 4.7 km/s x 30589s) (AND π x 466884 km= 47.4 km/s x 30589 s)
(p)
0.3 mkm = 27.78 km x 10921 seconds (1%)
(q)
0.3 mkm = 47.4 km x 6378 seconds (1%)
(s)
4x 0.3 mkm = 35 km x 10921 seconds x π
(t)
0.3 mkm = 24.1 km x 12448 seconds x π (BUT 12448 second x 9.7 km/s = 120536 km).

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II-Discussion
Equation no. (o)
0.3 mkm= 9.7 km x 30589 seconds (1%) (142984 km = 4.7 km/s x 30589s) (AND π x 466884 km= 47.4 km/s x 30589 s)
- This equation tells us that, the period 30589 seconds is used with 3 planets velocity, (30589 days = Uranus orbital period),
- Saturn (9.7 km/s) moves during 30589 seconds a distance= light known velocity (0.3mkm/s) motion distance during 1 second,
this equation has error 1%,
Also
- Pluto (4.7 km/s) moves during (30589s) a distance = 142984 km = Jupiter Diameter, BUT we know 142984 km = 13.1 km/s x
10921 seconds, that shows more interaction between Pluto and the moon data
Also
- Mercury (47.4 km/s) moves during 30589 s a distance = Jupiter motion distance during its day period (466884 km) x π, this
equation also has error 1%
- There are 3 motions done based on the value (30589 seconds), and by these motions 3 interactions are created,
- Saturn and Uranus data equation (0.3 mkm= 9.7 km x 30589 seconds) produces light known velocity, and that shows specific
effect of this equation especially because Uranus orbital distance = 2 Saturn orbital distance means this interaction is
extending into a great effect in both planets data.
- The previous discussion shows specific effect of 30589 seconds (where Uranus orbital period 30589 days), we may remember
o 30589 days =346.6 x 88 days +88 days (88 days Mercury Orbital Period)
o 1120 = 30589 days / 27.3 days BUT (2872.5 mkm Uranus orbital distance = 1120 x 2.573 mkm)

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Equation (p)
0.3 mkm = 27.78 km x 10921 seconds (1%)
- 27.78 km /s = the moon velocity
- The moon moves during 10921 km a distance = 300000 km (light motion distance for 1 second) but Jupiter (13.1 km/s) moves
during 10921 seconds a distance = 142984 km = Jupiter diameter
- Note Please
- 27.78 km x 10921 seconds = 303386 km (1%) means, the distance is greater than light motion for 1 second with 1% but
- 27.47 day x 10921 km = 300000 km, this equation tells, if the moon rotates around its axis 1 time per solar day so it will
pass during its orbital period (27.3 days) a distance = light motion distance for 1 second (0.3mkm/sec)
- 27.47 days x 0.99 =27.2 days (Perigee Month)
- Equation no. (p) shows that a light beam travels in parallel with the moon motion, this light motion is hidden behind the planer
motion data but the light motion features are seen simply and clearly effective on planet motion
Equation (q)
0.3 mkm = 47.4 km x 6378 seconds (1%)
- Mercury (47.4 km/s) moves during 6378 seconds a distance = 300000 km but 6378 km = Earth Radius.
- The equation shows the light beam motion accompanying planet motion …
- The next equation is similar to this one, let's discuss the last equation

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Equation (t)
0.3 mkm = 24.1 km x 12448 seconds x π (BUT 12448 second x 9.7 km/s = 120536 km).
- This equation is important because…
o 9.7 km/s x 12104 seconds = 117409 km
o 9.7 km/s x 12756 seconds = 123733 km
o Both planets diameters (Venus and Earth) can't produce Saturn diameter without great error, because the correct answer
is 12448 seconds which is a value in average between Venus diameter (12104 km) and Earth diameter (12756 km), this
value 12448 is defined by Mars motion , for that reason Mars Orbital Circumference = Saturn Orbital Distance.
- The basic equation in this discussion is no. (o) because it uses Saturn and Uranus data to produce light motion, let's see it here
again
Equation no. (o)
0.3 mkm= 9.7 km x 30589 seconds (1%) (142984 km = 4.7 km/s x 30589s) (AND π x 466884 km= 47.4 km/s x 30589 s)
We may remember also
- 6.8 km/s x 153.3 hours = 9.7 km /s x 10.7 hours x 10
- This equation also tells the interaction between Saturn and Uranus is so deep and effective greatly on the planets motions.

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References
Light Motion Features Are Discovered in Planet Motion
https://www.academia.edu/44286772/Light_Motion_Features_Are_Discovered_in_Planet_Motion
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
Curriculum Vitae http://vixra.org/abs/1902.0044
E-mail mrwaheid@gmail.com
Linkedln https://eg.linkedin.com/in/gerges-francis-86a351a1
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List of publications http://vixra.org/author/gerges_francis_tawdrous

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