Venus Rotation Period Analysis (II) (Revised)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
I do this research
Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
Venus Rotation Period Analysis (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 –20th
April 2021
Abstract
Paper hypothesis
- 1 hour of Mercury motion is equivalent to 24 hours of the moon motion
- This rate of time (1hour =24 hours) is used in the solar system motion and because
of that ( Venus & Uranus motions use this same rate
The hypothesis explanation
- The solar group is a clock, its basic task is to receive the energy of light motion for
1 second and by this energy causes a planet motion for 1 solar day – because of
that the rate of time ( 1 second is equivalent to 1 solar day)
- The question is, Why Does The Solar System Need Different Rates Of Time?
- The direct answer for this question is that (to create a configuration for the moon
orbital motion during its day period)
- The moon apogee orbital circumference = 2550973 km, where the moon
displacements total during its day period (29.53 days) be = 2598693 km
- This difference in 2 distances is found as a result of the geometrical mechanism by
which the rate of time (1 h is equivalent to 24 hours) is created.
- Shortly, because of these 2 distances difference, the moon motion has to create the
rate of time (1 hour = 24 hours) in comparison with Mercury motion – which
causes Mercury Motion to lead the moon motion. the paper discusses this process
geometrical details.

I do this research
2nd
2
- Notice
- The rate of time (1 hour = 24 hours = 1 solar day) is a rate used in the solar system
but –
- The rate of time between Mercury and the moon motions almost is (1 hour of
Mercury Motion is equivalent to 2 days of the moon motion) means (1 solar day of
the moon motion = a half of hour of Mercury Motion)
- This conclusion is added to the paper discussion after revision – in the paper
discussion we have to discuss this rate also in addition to the basic one (1 hour =24
hours).

I do this research
2nd
3
1- Introduction
- Why does the moon apogee orbital circumference (2.55mkm) is shorter than the
moon displacements total during its day period (29.53 days) (which =2.598 mkm)?
- How can the moon do that?
o The moon daily displacement = 88000 km
o During 29.53 days the total be =2598693 km
o This distance (2598693 km) should be the moon orbital circumference and
because 2598693 km = 2π x 413600 km
o That tells the moon apogee radius should be = 413600 km
o Not only that, also, the moon would have to revolve around the Earth during
all month days through this far orbit with a radius =413600 km (means, the
moon along moth can't be more near to the Earth than this distance 413600
km! because the moon daily displacement 88000 km is so long and its total
during the month (29.53 days) is a long distance = 2598693 km
o But
o The moon apogee real radius =406000 km
o And the moon revolves around Earth through more near orbits than apogee
and can reach to the perigee point (r= 363000 km) how can that be possible?
o The moon uses Pythagorean triangle in its motion
o The moon creates an angle (θ) between the moon displacement direction and
the moon orbit horizontal level. By that the real displacement (L) through
the moon orbit be not =88000 km but be = 88000 Cos (θ)
o The real displacement (L) be shorter than 88000 km and by that the
displacements total be shorter than the distance 2598693 km and because of
that the moon apogee radius be =406000 km where the moon almost never
visit the point 413600 km. but
o This intelligent techniques still need a support from Mercury motion.

I do this research
2nd
4
o The rate of time (1 hour of Mercury motion = 24 hours of the moon motion)
is a necessary geometrical tool to provide this support from Mercury motion
to the moon motion.
o The paper discusses this geometrical mechanism explains how it can create
the rate of time and provide the interaction between Mercury and the moon
motions
- The paper contents
- Point No. 2 discusses the rate of time creation
- Point No.3 discusses the moon using of Pythagorean triangle in its motion

I do this research
2nd
5
2- The Rate of Time Creation
2-1 Mercury and the moon motions interaction
2-2 Venus rotation period effect on the rate of time creation
2-3 The Moon Motion More Analysis
2-1 Mercury And The Moon Motions Interaction
I - Data
(a)
742 mkm = 88000 km (the moon daily displacement) x 8432 solar days
(b)
742 mkm x 2 = 4.095 mkm (Mercury velocity per solar day) x 365.25 solar days
(c)
Earth (29.8 km/s) moves during 6939.75 hours a distance = 742 mkm
II – Discussion
Equation No. (a)
- Equation no (a) tells that,
o The moon displacement per solar day during 8432 days be = 742 mkm
o 8432 solar days of the moon motion
o But
o Let's suppose each one day of them (8432 days) = 1 hour of Mercury motion
o 8432 hours = 2 x 175.94 days x 23.96 hours
o 175.94 solar days = Mercury Day Period
o 23.9 hours = Earth rotation period
o Where (23.96/23.9) = (361 degrees /360 degrees)
o Based on that, 1 Solar Day Of The Moon Motion Be Equivalent To 0.5
(A Half) of Hour Of Mercury Motion
o This explanation shows the rate of time (1 hour is equivalent to 24 hours)
and shows how this rate works – in our discussion – it works between the

I do this research
2nd
6
moon daily displacement (motion) and Mercury Motion – the moon period
8432 solar days needs 2 Mercury days only of Mercury motion because (1
hour of Mercury Motion = 24 hours of the moon motion).
Notice
o The rate (2) shows that one day of the moon motion = a half of hour of
Mercury motion – the next equation will explain Why?
Equation No. (b)
- Equation no (b) tells that,
- Mercury moves during 365.25 solar days a distance = 742 mkm x 2
- Let's divide the discussion into basic points as following
o (1st
Point)
o Mercury day period (175.95 solar days) x 2 = 351.9 solar days
o This value we should consider equivalent to 365.25 solar days –why?
o Because
o (365.25 days /351.9 days) the difference =3%
o The difference between 720.7 mkm and 742 mkm is 3%
o Because the cycles are designed in proportionality with each other – the
value 351.9 days can be considered as 365.25 days
o (2nd
Point)
o We have 2 Gears work interacted with Each other which are:
o The moon motion and Mercury motion
o These 2 planets motions are interacted because of the rate of time which is
created between their motions –
o The observation which we need to notice here is that –
o The moon moves a distance 742 mkm and create the different rate of time
o Based on this different rate of time Mercury moves 2 x 742 mkm

I do this research
2nd
7
o How that? Because the moon displacements period = 2 Mercury Days
=351.9 s. days - But Mercury motion period = 365.25 Solar Days
o Because we consider 351.9 days is equivalent to 365.25 days
o The input is the moon motion 742 mkm and the output is Mercury motion
2x 742 mkm
o Mercury day period (175.95 solar days) x 2 = 351.9 solar days
o For that the moon displacement daily =88000 km but the real motion should
be considered = 2 x 88000 km
o This information for which we have tried frequently to prove in the moon
orbital motion discussion – let's remember why –
o The moon moves per solar day a distance = 2.574 mkm = Earth motion
distance per a solar day otherwise they will be separated from each other
o The moon motion distance is effected by the rate (1.0725) and contracted to
be = 2.4 mkm only (where the rate 1.0725 effect on almost 35% of all the
solar system distances – I attribute it to length contraction phenomenon)
o The difficulties now be started – because the moon motion distance is less
than Earth motion distance by (176000 km) per solar day – and if the moon
doesn't move this additional distance – the moon will be separated from the
Earth through the motions course.
o For that reason – the moon has to move its daily displacement depends on
Earth gravity force (88000 km) but this distance is only (a half ) of the
required distance and the moon needs to move one more displacement to
cover the different distance.
o This is the situation – and through the moon orbital motion discussion – I
have suggested many ways to create this second displacement which is a
necessary distance the moon has to move otherwise it will be separated from
Earth through the motion

I do this research
2nd
8
o And now we get the answer of our question – how the one displacement
(88000 km) can be (=2 x displacements 88000 km)
o The rate of time (1 hour is equivalent to 24 hours) enable the moon to use
Mercury motion to cover the required distance – Now Mercury provides this
specific help for the moon to save his life
o For that, Mercury orbital period =88 days which can be 88000 km if 1000
km be equivalent to 1 solar day, this rate which we have discussed
frequently before.
o (3rd
Point)
o Mercury during 365.25 days moves a distance = 2 x 742 mkm
o The moon displacements total be 742 mkm during the period 24 x 2 x175.95
days and we supposed that the 2 periods be equal – basically because (1
hour) of Mercury motion = 24 hours of the moon motion
o But the factor (2) shows that this rate is incorrect
o The correct one is (1/2 hour of Mercury motion is equivalent to 24 hours
of the moon motion)
o Based on that – (one day of Mercury motion be equivalent to 48 days of
the moon motion)
o We still need to analyze the moon motion to see how this machine works.
So we should do that in a separated point
o In brief, the conclusion of our discussion till now tells the moon motion day
period is equivalent to a half of hour of Mercury motion.
o And the question we need to discuss later is that (Why a half of hour of
Mercury motion = 24 hours of the moon motion?) this question should be
discussed in point no. (2-3) (the moon motion more analysis)
o But why?

I do this research
2nd
9
o (4th
Point)
o Because the rate of time (1hour is equivalent to 24 hours) is a general rate
for the solar system motion – so many planets may use it – that makes the
planets cycles are interacted with each other – for that reason many cycles
uses the rate (2) which is found between the moon motion distance and
Mercury motion distance –for example – Venus day period = 2 Mercury
rotation period (approximately) – also Mercury day period = 2 Mercury
orbital period (approximately) – also Mars orbital period = 2 Nodal years
(approximately) – these cycles are members in the rate of time creation –
they use this rate (2) to enable the moon and Mercury motions interaction
o (5th
Point)
o Where (23.96/23.9) = (361 degrees /360 degrees)
o Why we need this rate
o The very small error between (23.96 h and 23.9 h) is rated to the rate
(361/360) and what's this rate
o 360 degrees is the cycle degrees
o 361 degrees is the moon orbit regression degrees – because the moon orbit
regresses during a year (365.25 days) 19 degrees which causes change for
the eclipse calendar with 19 days– so during 19 years – the complete
revolution will be 361 degrees –
o Because of that many data is rated by this rate (361/360) and that means this
small error isn't an error but it's found for a geometrical necessity because
the revolution degrees is not 360 degrees but 361 degrees and that
necessitates geometrical modifications for different data

I do this research
2nd
10
Equation No. (c)
Earth (29.8 km/s) moves during 6939.75 hours a distance = 742 mkm
- Equation no (c ) tells that, Earth moves during 6939.75 hours a distance = 742
mkm = the distance passed by the moon during (2 x 175.94 solar days x 23.96h)
- Why? Because
- 1 hour of the Earth motion = 1 day of the moon motion
- Earth uses the rate of time of Mercury and the moon uses the other rate of time
- Notice
- 6939.75 days = Metonic Cycle
- If 1 hour of Earth motion = 1 solar day of the moon motion
- That means, During Metonic Cycle (6939.75 days based on the moon rate of time)
= 6939.75 hours (based on Earth rate of time) Earth moves a distance =742 mkm

I do this research
2nd
11
Venus Diameter Analysis
I - Data
(d)
12104 x 742 = 80 x 9.9 h (Jupiter Day Period) x π
(use the rate 361/360)
(e)
78 Saturn days periods = 79 x 12104 seconds x π
(f)
2 x 30589 seconds = 5.05 x 12104 seconds
II – Discussion
Equation No. (d)
12104 x 742 = 80 x 9.9 h (Jupiter Day Period) x π
(use the rate 361/360)
- Equation no (d) tells that,
o Venus diameter =12104 km
o This value is used as 12104 seconds
o The equation tries to see this period origin and if found that the period
12104 seconds is defined based on the cycle (742) in proportionality with 80
Jupiter Days Period
o That means, the value 12104 second is a period of time (originally) because
it's a part of Jupiter day period
o Also we know that, Jupiter has a cycle called (8 days Jupiter cycle)
o Which means, the period 80 Jupiter days has a cycle also
o That tells, the period 12104 second is a period of time originally because it
originated from Jupiter day period and originated in Jupiter cycle – means –
originated in interaction between Jupiter and other planets.
o The equation uses the rate (361/360) because this period (12104 seconds)
depends on the moon orbit regression

I do this research
2nd
12
Equation No. (e)
78 Saturn days periods = 79 x 12104 seconds x π
- Equation no (e) tells that,
o The period 12104 seconds (where 12104 km = Venus diameter) is a period
defined by Jupiter day period AND by Saturn day period
o 78 Saturn Days Period = 3004560 seconds
o 79 x 12104 seconds x π = 3003975 seconds
o The equation tells that, Venus Diameter (or the value 12104 seconds) is a
point of connection between different planets – it's very important effect of
Venus diameter and motion because it works as a transportation gear
between different planets motions –
o Shortly
o Venus connects between the 3 planets (Jupiter, Saturn and Uranus) and also
connect these 3 planets with other planets motions –
o The connection is done based on the periods of time of their motions and by
that Venus support the moon rate of time which is created by Mercury
motion in comparison to the moon motion

I do this research
2nd
13
Equation No. (f)
2 x 30589 seconds = 5.05 x 12104 seconds
o 30589 solar days = Uranus Orbital Period
o 5.1deg = The Moon Orbital Inclination
o Notice 2 x 30589 seconds + 742 seconds = 17.2 h (Uranus day period)
- Equation no (f) tells that,
o The period 12104 seconds is originated (also) from Uranus orbital period, in
more clear words – the period 12104 seconds is originated basically from
Uranus motion and the connection between this period 12104 seconds with
Jupiter and Saturn motions periods are done based on the interaction
between Uranus motion on one side and the 2 planets motion on the other
side
o Venus motion is belonged basically to Uranus
o And that can be seen by analyze the provided data
Notice
o Uranus moves during 30589 hours a distance = 742 mkm (error 1%).

I do this research
2nd
14
2-2 Venus rotation period effect on the rate of time creation
I - Data
(1)
(406000 km /88000 km) = (413600 km /90560 km) (error 1%)
(2)
90560 km x 2π = 569005.25 km
(3)
569005.25 seconds = 16 Jupiter Days = (29.53/2) Saturn Days
(4)
569005.25 km = 3475 km x 327.6 = 116.75 x 4879 km
(5)
569005.25 km = 24 x 243 km x 96.7
II – Discussion
Equation no. (1)
(406000 km /88000 km) = (413600 km /90560 km) (error 1%)
Where
90560 solar days = Pluto Orbital Period
406000 km = Pluto Motion Distance During a solar day
88000 km = the moon displacement during a solar day
413600 km = (2598693 km /2π)
2598693 km = the moon displacements total during 29.53 days
- Equation no. (1) uses Pluto orbital period (90560 solar days) as a distance (=
90560 km) and we should use it in our following discussion
- This equation shows a great significance because it uses 413600 km which is the
radius of the orbital circumference 2598693 km – where this distance was the
inevitable apogee radius for the moon motion because of the moon daily
displacement 88000 km – but because the moon uses Pythagorean triangle and the
moon motion creates the rate of time (24 hours is equivalent to 1 hour of Mercury

I do this research
2nd
15
motion) because if that this apogee radius is decreased from 413600 km to 406000
km (the moon apogee radius)
- The equation still gives us benefits because the value 406000 km which is used
here doesn't refer to the moon apogee radius but refer to Pluto motion distance
during a solar day, which creates a harmony for data between Pluto motion
distance for a solar day in comparison with the moon displacement during a day.
- One more notice we need to observe here is that (The deepest interaction among
planets motions is found between Mercury and Pluto motions and velocities –
because of that – Mercury motion effect on the moon motion must create an
interaction between the moon & Pluto motions – Mercury motion doesn't effect on
any point without Pluto motion – this idea we should prove through the discussion)
- Equation no. (1) tells that, Pluto and The moon motions Proportionality creates a
rate between the value (413600 km) and Pluto orbital period which used as a
distance as 90560 km – that means
- The interaction between Pluto and the moon motions causes the moon apogee
orbit 413600 km to be decreased to 406000 km
- Shortly
- Pluto motion effect on the moon motion causes the apogee orbit to be shorter than
413600 km and be 406000 km
- (that may answer the old question, why Pluto move during a solar day a distance =
the moon apogee orbit radius)
- Now
- The distance 413600 km be rated to the value 90560 km which was originally
90560 days (Pluto Orbital Period) BUT
- 413600 km is a radius and what about its circumference? Has it a comparable
value also?
- 413600 km is a comparable to 90560 km - So
- 2598693 km is a comparable to 90560 km x 2π = 569005.25 km (Equation no. 2)

I do this research
2nd
16
Equation no. (3)
- 4 Jupiter days = 4 x 9.9x 3600 seconds =142560 seconds
- But
- Jupiter diameter =142984 km
- If 1 km = 1 second the both value will be equal (error 0.2%)
- I try to show that, it's not an invented idea to use Pluto orbital day (90560 days) as
a distance 90560 km –it's one geometrical feature of the solar system geometrical
structure – we here follow the geometrical structure design
- But
- If the solar system geometrical structure uses Jupiter diameter (142984km) with a
number very near to the value (4 Jupiter days periods) that means this value has
specific geometrical effect on the solar system
- Means,
- The 4 Jupiter days is a period mentioned geometrically
- Our value which 569005.25 km = 2π x 90560 km
- If be used as 569005.25 seconds , it will be equal 16 Jupiter days period
- 16 x 9.9 x 3600 seconds = 570240 seconds (the difference 0.2%)
- That means
- This value 569005.25 km = 4 x 142984 km (Jupiter diameter)
- Equation no. (3) tells that, the period 569005.25 seconds is a defined period in the
solar system motion – it's a required period and used for geometrical motions as
equal as any planet cycle period – why?

I do this research
2nd
17
Equation no. (3) (continued)
- Saturn day period =10.7 h x 3600 s = 38520 seconds
- 29.53 x Saturn days periods = 29.53 x 38520 s =1137496 seconds
- 1137496 seconds =2 x 568748 seconds (= 569005.25 se error 0.04%)
- That means
- 29.53 Saturn days = 2 x 569005.25 seconds
- Where
- 29.53 solar days = the moon day period
- Equation no.(3) tells that, the value 569005.25 seconds is a mentioned value
geometrically and defined by effect of Jupiter and Saturn motions interaction
- Please remember
- (708.7 h/655.7 h) = (10.7 h/9.9h)
- Where
- 708.7 h = the moon day period 29.53 solar days
- 655.7 h = the moon rotation period 27.32 solar days
- 10.7 h = Saturn day period
- 9.9 h = Jupiter day period

I do this research
2nd
18
Equation no. (4)
569005.25 km = 3475 km x 327.6 = 116.75 x 4879 km
Where
3475 km = The Moon Diameter
4879 km = Mercury Diameter
327.6 solar days = The Moon Sidereal Year
116.75 solar days = Venus Day Period
- Equation no .(4) shows that, the value 569005.25 km is a value used geometrically
as a reference for different values of the Moon and Mercury motions data
- The geometrical mechanism behind these interactions can be discussed more easy
through the following equation o. (5) but we use this equation to show that the
value 569005.25 km is used for different motions data concerning mercury & the
moon motions

I do this research
2nd
19
Equation no. (5)
569005.25 km = 24 x 243 km x 96.7
Where
243 solar days = Venus rotation period
96.7 degrees = 90 degrees + 6.7 degrees (= The Moon Axial Tilt)
- Equation no. (5) tells that, each 1 solar day of Venus rotation period (243 days) =
24 hours of this period 569005.25 hours with an effect of the moon axial tilt
- Let's see that as deep as possible
o 24 x 243 x 96.7 = 563954 seconds
o 569005.25 -563945 = 5050 Seconds
o 5050 Seconds = π x 10 x 161
o But
o 161 = 24 x 6.7
Notice
- Mercury Day period needs 5040 seconds to be 176 solar days
- Almost this value (5040) is our discussion value (5050 Seconds) and it's produced
by the interaction between Mercury and the moon motions which causes Mercury
day period to be less than 176 solar day with 5040 seconds
- The rate (24) which is used frequently in the data shows that the rate of time (1
hour is equivalent to 24 hours) is produced with this process – where this rate
effects on Mercury motion data
Notice
- Uranus motion almost uses the moon rate of time – means- 1 hour of Mercury
motion is equivalent to 24 hours of Uranus motion
- Uranus uses the moon motion rate of time because Uranus motion effect on the
moon motion causes to create Metonic Cycle and by that Uranus be a partner with
the moon motion and be effected by its rate of time.

I do this research
2nd
20
Uranus Motion Rate Of Time
- (1 hour of Mercury Motion is Equivalent to 24 hours of Uranus motion)
o Mercury rotation period = 1407.6 hours = 58.66 solar days
o 2 Mercury rotation periods = 2 x 1407.6 h = 2815.2 hours
o If 1 mkm = 1 hour
o 2815.2 hours will be = 2815.2 mkm = Mercury Uranus Distance
o The input is
o 2815.2 mkm is used as 2815.2 hours by Mercury motion (1 mkm =1 hour)
o Mercury moves during (1407.6 h x 2) a distance =2 x 243 mkm
o 243 solar days = Venus rotation period
o The output is
o 2 x 243 mkm where the rate is (1 mkm = 1 solar day)
o Mercury motion data shows the rate (1 h =24 h)
Notice
Why Uranus Distance is used by the rate (1 mkm = 1 hour)
17687 mkm = 24 x 735 mkm
17687 mkm = 2π x 2815 mkm (Mercury Uranus Distance)
- Where
- 735 mkm = Venus Motion distance during 243 solar days
- The previous data tells that, each 1 mkm of Venus motion is equivalent to 24 mkm
of the distance (17687 mkm) where this distance is the circumference of Mercury
Uranus Distance
- Simply it tells, Venus uses the rate (1: 24) and this rate is created by Mercury
Motion interaction with the moon (and may with Uranus) motions.
- For that reason the distance (742 mkm) which is passed by Uranus during the
period 30589 hours be equivalent to Venus motion distance during 243 days (1%)

I do this research
2nd
21
2-3 The Moon Motion More Analysis
Let's remember the question which we need to discuss here
The Question
Why a half of hour of Mercury motion = 24 hours of the moon motion?
This question depended on the following two equations
Equation No. (a)
- And
- 8432 hours = 2 x 175.94 days x 23.96 hours
- (175.94 solar days = Mercury Day)
Equation No. (b)
- Mercury ( 1/2 of hour) of its motion = 24 hours of the moon motion because the
motions equal distances causes different rates of time for these motions
- Mercury moves double value of the distance (2 x 742 mkm) but the moon moves
only (742 mkm) because of that, the rate of time is not (1 hour is equivalent to 1
solar day) but (A half of hour is equivalent to 1 solar day)
- this point is important because it's a method to discover the different rates of time
between different planets motions and then to test this rate if it's really works based
on the equal distances
- the concluded rule can be (Equal Distances Create Different Rates Of Time)
- let's take a look on the moon motion details in following

I do this research
2nd
22
(1st
Point)
- The moon orbital motion details are explained clearly in point no. (3) of this paper
(3- The Moon Orbital Motion Description).
- The moon has 3 basic distances per a solar day of its motion which are:
- The moon motion distance per a solar day = 2.574 mkm = Earth motion distance
per a solar day – because – Earth and The moon motion distance should be equal
otherwise they will be separated from each through their motions course
- The moon motion distance per a solar day be effected by the rate 1.0725 and
contracted to be 2.4 mkm –
- Because the moon motion distance per a solar day became 2.4 mkm less than Earth
motion distance per a solar day 2.574 mkm by a different distance = 176000 km,
the moon has to move this additional distance per a solar day
- The moon moves its daily displacement 88000 km depends on Earth gravity but
this distance is not enough and the moon needs to pass another displacement
(88000 km) during the same solar day
- So, the three distances are
o 2.574 mkm
o 2.4 mkm
o 88000 km
- These 3 distances are defined distances of the moon motion per a solar day
Notice
- 2.4 mkm = 88000 km x 27.27
- Where
- 27.32 solar days = the moon orbital period
- That means, the moon orbital period (or the nodal month) is defined based on these
3 distances (motions) interaction with each other – that shows the machine is so
complex and causes different effects on the moon motion data

I do this research
2nd
23
(2nd
Point)
- The moon displacement per a solar day =88000 km
- Pluto motion distance per a solar day = 406000 km
- 406000 km = 88000 km x 4.61
- Equal Distance Create Different Rates Of Time
- That means,
- Each 1 hour of Pluto motion should be equal 5.2 hours of the moon motion
- Because
- The moon needs 4.61 folds of the period Pluto uses to pass equal distances
- That creates a rate of time between Pluto and the moon motions = (1: 5.2)
- But
- We know that, the moon orbital inclination (5.1 degrees) is created by some effect
of Pluto orbital inclination – let's remember that in following
- 5.1deg (the moon orbital inclination) x3.4 deg (Venus orbital inclination)=17.4 deg
- Where
- 17.2 deg = Pluto orbital inclination
- But
- 17.4 deg = The Inner Planets Orbital Inclinations Total (The difference (1%))
- This discussion shows that, there are 2 used data by the rate (5.1 and 5.2) between
Pluto and the moon motions –
- I claim these 2 rates are created by the interaction between Pluto and the moon
motion by which the rate of time (1 : 5.2) is created
- I want to say that, the rate of time is not a secret word has no a real effect – on the
contrary – it effects on the planet motion data clearly and because of that we
should find this rate frequently in the planet motion data
- So, the hypothesis which tells that, 1 hour of Pluto motion is equivalent to 5.2
hours of the moon motion is not an imaginary idea – it's try to explain the motion
data which is found really.

I do this research
2nd
24
Notice (1)
- The moon displacement during (5.2 hours) =19066.6 km = 7510 km x 5.08
- Where
- 7510 km = Pluto Circumference
- 5.1 deg = the moon orbital inclination
- (Also 19066.6 km = 2390 km (Pluto Diameter) x 8)
Notice (2)
- 7510 km = 5.14 x 1461 km
- Where
- 1461 days = The Earth Cycle ( 365 + 365+ 365 +366 = 1461 days)
Notice (3)
- Uranus day period effects on Pluto and the Earth motions – for that – Uranus day
period 17.2 hours can be used also as 17.4 hours
- (similar to Pluto orbital inclination 17.2 deg which be compared with the inner
planets orbital inclinations total 17.4 degrees)
- That show the planets motions interaction
- For that
- 17.4 hours = (30589 seconds x 2 + 1462 seconds)
Notice (4)
- An interaction is found between Pluto and Venus motions (as seen from the
equation 17.4 deg = 5.1 deg x 3.4 deg)
- This interaction effects on both planets motions and because of that
- (Venus velocity 35/ 4.7 Pluto velocity) = (90560 /12104) (error 0.4%)
- Where
- 12104 km = Venus diameter
- 90560 solar days = Pluto Orbital Period

I do this research
2nd
25
3- The Moon Orbital Motion Description
3-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
3-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
3-3 The Moon Orbital Motion Analysis
3-4 The Moon Orbital Motion Equation

I do this research
2nd
26
3-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
- Let's summarize this question answer in following:
o The moon uses Pythagorean triangle basically to decrease its displacement
daily through its orbit
o The moon daily displacement = 88000 km and the moon has to move this
distance every day without any decreasing (later we will know why!)
o But
o If the moon moves by this displacement as its orbital displacement the moon
would revolve around Earth through its apogee orbit only (r=0.406 mkm)
o For that reason
o The moon creates an angle between its motion direction and its orbit
horizontal level to create a displacement through its orbit less than (88000
km)
o As a result of this technique, the moon can revolve around Earth through
more near orbits than apogee orbit (r=0.406 mkm)
o Simply, because the moon uses this technique the moon can revolve around
Earth through perigee orbit (r=0.363 mkm)
o Let's explain this intelligent technique with some details to show the useful
result of using Pythagorean triangle by the moon orbital motion….

I do this research
2nd
27
3-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
- The moon moves daily (88000 km) on the right triangle hypotenuse (AC), but the
moon creates an angle (θ) between its motion direction and its orbit horizontal
level, by that the real displacement through the moon orbit will be (L= 88000 km
cos (θ)), and by that, spite the moon moves 88000 km, but the real orbital
horizontal displacement be less than (88000 km) and this is the objective for which
the moon uses Pythagorean triangle –
As an example,
- If (θ) =28.63 degrees, the real displacement (L== 88000 km cos (θ)) = 77237 km,
So, if the moon real displacement daily be (77237 km), during 29.53 days the
moon will pass a distance = 2.28 million km and this will be the moon orbital
circumference, where 2.28 mkm = 2π x (0.363 mkm)
- The Moon Orbital Perigee Radius =0.363 mkm
- That means, the moon by a real displacement =77237 km can move around Earth
through the perigee orbit (radius =0.363 mkm), this is the useful result the moon
performs by using Pythagorean triangle,
- Now let's suppose the moon doesn't use Pythagorean triangle, what would happen?
- The moon daily displacement = 88000 km, during 29.53 days the moon moves a
distance = 2.598 mkm where 2.598 mkm = 2π x (0.413 mkm)
- The Moon Orbital Apogee Radius =0.406 mkm
- So the moon will move along month revolving around Earth through its apogee
orbit (or even far from apogee orbit) because the total distance can't be passed
through any more near orbit around Earth…
- The data shows how Pythagorean triangle is so useful for the moon orbital motion.

I do this research
2nd
28
The Angle θ
- The angle (θ) should get our attention for its specific effect…let's summarize the
idea in following
o The angle (θ) changes the real displacement (L = 88000 cos (θ)), through the
moon orbit..
o We know that, when the real displacement (L) be shorter the moon can
move through near orbits to Earth and by that the moon can be near or at
Perigee radius (0.363 mkm)
o When the real displacement (L) be greater the moon has to move through
orbits far from Earth and by that the moon can be near or at apogee orbit
(r=0.406 mkm)
o That means, the angle (θ) changes the real displacement (L) and also
changes the distance between the moon to perigee or to apogee, shortly, the
angle (θ) defines the moon position (as a ship) between 2 river banks….
- The angle (θ) defines the moon orbital motion basic features and we have to
discuss is deeply with the moon orbital motion equation (θ1= θ0 + 1.7 degrees),
but before we need to analyze the moon orbital motion
Notice
o We know that (363000)2
+ (86000)2
= (373000)2
o In Pythagoras triangle with dimensions (363000 km, 373000km, 86000 km),
what's the angle (θ)? The angle (θ) = 13.33 degrees
o Also (396800)2
+ (86000)2
= (406000)2
the angle (θ) = 12.229 degrees
o I have used (363000 km and 406000 km) because they are the perigee and
apogee radiuses between which the moon moves.
o The difference between angles = 1.1 degrees
i.e.,
The angle (1.1 deg.) controls the moon motion from perigee to apogee, we will need
this notice later in our discussion

I do this research
2nd
29
3-3 The Moon Orbital Motion
- The moon moves per a solar day a motion typical to the Earth motion to avoid the
separation from Earth through their motions, based on this rule, the moon moves
per a solar day 2.573 million km with an angle declines on the horizontal level
0.98562 degrees as typical to Earth motion
- If there's no Lorentz Length Contraction Phenomenon effect on the moon motion,
the moon motion trajectory would to be a parallel line to Earth Motion Trajectory,
But Lorentz Length Contraction effects on the moon motion daily distance (2.573
mkm) with a rate 1.0725 and causes this distance to be contracted (2.399 mkm)
- The moon difficulties are started here, because the difference between both
distances (0.17 mkm) will cause the moon to be separated from Earth motion
inevitably
- We should notice that, these motions are done far from our observation, means, we
see nothing of this motion distance, because the moon moves on the Earth orbital
circumference revolving around the sun, but, even if we can't observe this motion
distance the motion is still fact and proved by its power, because the Earth moves
per a solar day 2.573 mkm and if the moon doesn't move this same distance every
solar day that necessities the moon to be separated from the Earth through their
motions course – based on that- the facts prove this motion regardless our
observation ability for it.
- Now the moon has an additional distance to be passed (0.17 mkm) and the moon
has to pass this distance on the same solar day to avoid the separation from the
Earth during their motions.
- Because of that, the moon moves its daily displacement (88000 km) depends on
Earth gravity force (by which we see the moon in the Earth sky), but the different
distance (0.17 mkm) to be covered still needs the moon to move one more
displacement (= 88000 km)

I do this research
2nd
30
- The previous explanation tells that, the moon has to move 2 displacements each =
88000 km, while we see one displacement only because it's done through the
moon orbital motion around Earth but the other displacement should be done also
because this total distance (0.17 mkm) is required to cover the different distance
and create the total (2.573 mkm) which saves the moon and Earth motions
accompanying.
- Now we have 2 basic information about the moon orbital motion
o (1st
information) the moon uses Pythagorean triangle in its orbital motion
o (2nd
information) the moon has to move 2 displacements each =88000 km
and their total distance =0.17 mkm which is a required distance necessary to
cover the difference between the moon and Earth motions distances.
- This explanation helps us to understand why the moon uses Pythagorean triangle
in its motion, because the moon can't decrease its daily displacement (88000 km)
because the moon needs this distance to cover the different distance between its
contracted motion distance (2.399 mkm) and Earth motion distance (2.573 mkm),
So the moon needs to move this displacement perfectly, but if it's used as a
displacement through the moon orbit, the moon would be always a prisoner in the
apogee orbit (r=0.406 mkm) as we have discussed before, because of that, the
moon creates Pythagorean triangle technique by which the moon moves actually
88000 km daily but the real displacement through the moon orbit became less (L =
88000 Cos θ) and by that the moon can achieve 2 objectives, First to pass the
required distance (88000 km) and Second to move in near orbits to Earth, that
shows the intelligent moon motion technique…
- (Notice, Lorentz Length Contraction Effect Discussion is in Appendix No. 1)

I do this research
2nd
31
The Moon Orbital Motion Needs One More Orbit
- The previous explanation tells that, the moon moves 2 displacements each =88000
km, we see one of these 2 displacements but where's the other displacement?!
- We know that, the moon original motion (2.573 mkm) which is contracted to be
(2.399 mkm) isn't seen by us because the moon moves this distance revolving with
Earth around the sun along the Earth Orbital Circumference
- We may accept that, the 2nd
displacement the moon does on this same trajectory
and isn't seen by us.
- So,
- There must be one more orbit for the moon to move through this 2nd
displacement.
means,
- There's 2nd
Orbit For The Moon Motion
- But
- How can we discover this second orbit if we can't observe the 2nd
displacement
motion?
- We can discover this 2nd
orbit by the moon orbit data analysis. So we should
depend on the moon orbital triangle data analysis to define this 2nd
orbit position.
- For that we have to discuss the moon 2nd
orbit in our deep analysis of The Moon
Orbital Triangle Geometrical Structure.

I do this research
2nd
32
3-4 The Moon Orbital Motion Equation
3-4-1 The Equation Concept
3-4-2 The Equation Test and Accuracy
3-4-1 The Equation Concept
The Moon Orbital Motion Equation
(θ1= θ0 + 1.7 degrees)
- The moon orbital motion equation is created depending on the concept we have
discussed, which is (the moon uses Pythagorean triangle in its orbital motion)
- The moon uses Pythagorean triangle and by this intelligent technique the moon be
under control of the angle (θ) change
- The angle (θ) defines almost all the moon motion features.…
- The moon uses this technique, aiming to create a real displacement shorter than its
actual displacement (88000 km) based on the equation (L =88000 cos (θ)) and by
that while the moon moves a displacement =88000 km but the real displacement
(L) through its orbit be shorter than 88000 km and by that the moon can revolve
around Earth through more near orbits than its apogee orbit (r=0.406 mkm).
- The moon orbital motion equation depends on this concept and, the equation
uses (the constant) 1.7 degrees as the moon daily motion degrees, and the equation
uses the previous day angle (θ0) to produce the today angle (θ1)
- We have 3 questions in this equation study which are:
o How does this equation work?
o Is this equation trustee and correct?
o Why does the equation use the angle 1.7 degrees for the moon daily motion?
Let's try to answer….

I do this research
2nd
33
How to use this equation?
- Perigee Radius =0.363 mkm, so Its Orbital Circumference =2.28 mkm
- Suppose the moon will revolve around Earth through perigee orbit only during
29.53 days, so
- (2.28 mkm /29.53 days) = 77237 km
- This is (the real displacement = L = 88000 km Cos θ = 77237 km),
- What's the angle θ value? the angle θ = 28.63 degrees
- Suppose the moon stand on this point yesterday with the angle (θ) =28.63 degrees,
where the moon will move today?
- From Perigee (the most near point to Earth) the moon will move in Ascending
motion because it moves from perigee (0.363 mkm) to apogee (0.406 mkm)
- In Ascending motion we use (-1.7 degrees) because the angle (θ) is decreased
where the real displacement (L) is increased, So let's do that in following
o (θ1= θ0 - 1.7 degrees)
o (θ1= 28.63 degrees - 1.7 degrees) = 26.93 degrees
o L = 88000 Cos (26.93 degrees) = 78454 km
o During 29.53 days so (78454 km x 29.53 days = 2.316 mkm)
o 2.316 mkm = 2π x 368722 km
That means
o The moon was (before motion) on Perigee radius (r=0.363 mkm) and starts
its motion displacement 88000 km. For day motion the equation uses 1.7
degrees, that means, the moon on perigee uses Pythagorean triangle with
angle (28.63 degrees) and during one solar day the moon uses - 1.7 degrees
and by that the angle will be (26.93 degrees)…... The angle 1.7 degrees
expresses The Moon Daily Motion
o By using Pythagorean triangle its angle (θ) = 26.93 deg, the displacement
(88000 km) will create a real displacement through the moon orbit = 78454
km and the moon will finish its motion today at a distance 368722 km

I do this research
2nd
34
means the moon is far from perigee radius with (368722 km-363000 km
=5722 km )
o So, the moon after 1 day motion will be at the point 368722 km and will
have the Pythagorean triangle its angle 26.93 degrees.
The Descending Motion
o When the moon moves from apogee (0.406 mkm) to perigee (0.363 mkm),
so the angle (1.7 degrees) will be positive (+1.7 degrees) because the angle
(θ) is increased and the real displacement (L = 88000 Cos (θ)) be shorter.
So
o If the moon in apogee radius (r=0.406 mkm), what's the angle (θ)?
o The apogee orbital circumference = 0.406 mkm x2π =2.55 mkm = 29.53
days x 86400 km, the angle (θ) = 10.96 degrees (=11 deg approx.)
o The moon moves from apogee to perigee (descending motion)
o (θ1= θ0 + 1.7 degrees) means (θ1= 11 degrees + 1.7 degrees) = 12.7 deg.
o L = 88000 Cos (12.7 degrees) = 85847 km
o During 29.53 days so (85847 km x 29.53 days = 2.535 mkm)
o 2.535 mkm = 2π x 403467 km
So
o After one day the moon will be on 403467 km far from apogee (406000 km)
with 2540 km
Now let's see this equation test and efficiency in following

I do this research
2nd
35
3-4-2 The Equation Test and Accuracy
- I have tested the Equation with real data for 2 months June 2020 and October 2020
- The results are very good and I provide the results here for better vision
concerning the equation efficiency
1st
Test June 2020
Day Registered Data The Results (1.7) Difference
6-6-2020 369418 km
7-6-2020 373729 km 374772.5 - 1044
8-6-2020 378917 km 378821.5 96
9-6-2020 384534 km 383667.7 867
10-6-2020 390096 km 388890 1206
11-6-2020 395156 km 394000 1156
12-6-2020 399345 km 398604.2 741
13-6-2020 402395 km 402361.3 34
14-6-2020 404153 km 405052.8 -900
15-6-2020 404574 km ---- ---
16-6-2020 403718 km 401848.5 1870
17-6-2020 401733 km 400876.1 857
18-6-2020 398840 km 398640.7 200
19-6-2020 395303 km 395417.4 115
20-6-2020 391409 km 391521.2 -113
21-6-2020 387432 km 387273.4 159
22-6-2020 383607 km 382968.4 639
23-6-2020 380110 km 378852 1258
24-6-2020 377044 km 375107 1937
25-6-2020 374451 km 371836.5 2615
26-6-2020 372338 km 369077 3262
27-6-2020 370703 km 366855.6 3847
[

I do this research
2nd
36
The 1st
Test Results Analysis:
- The Total Results Are 20 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 2 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 3 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 20) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
Category of results defines the moon position in error range from (1300
km to 2000 km) = error (4.5%), that means (2 values of 20) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (3 values of 20) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

I do this research
2nd
37
2nd
Test October 2020
Day Registered Data Results (1.7) Difference
5-10-2020 405,690 km --- ---
6-10-2020 404,171 km 403125.3 km 1046 km
7-10-2020 401,649 km 401390 km 259 km
8-10-2020 398,073 km 398545.6 Km - 473 km
9-10-2020 393,464 km 394568.8 km -1105 km
10-10-2020 387,944 km 389510 km -1567 km
11-10-2020 381,763 km 383520 km -1758 km
12-10-2020 375,302 km 376875.3km -1574 km
13-10-2020 369,063 km 369981km -919 km
14-10-2020 363,617 km 363363.4km 254 km
15-10-2020 359,530 km 357612 km 1918 km
16-10-2020 357,269 km 353307 km 3962 km
17-10-2020 357,105 km ---- --
18-10-2020 359,048 km --- --
19-10-2020 362,851 km 364979.7 km - 2129 km
20-10-2020 368,058 km 368579.3 km -522 km
21-10-2020 374,101 km 373492.4 km 609 km
22-10-2020 380,412 km 379168.3 Km 1244 Km
23-10-2020 386,497 km 385059.3Km 1438 km
24-10-2020 391,989 km 390694.3 km 1295 km
25-10-2020 396,659 km 395729.5 km 930 km
26-10-2020 400,395 km 399958.7 km 437 km
27-10-2020 403,181 km 403299 km 112 km
28-10-2020 405,059 km 405738.5 km -680 km
29-10-2020 406,104 km 407359.4 km -1256 km
[

I do this research
2nd
38
The Test Results Analysis:
- The Total Results Are 22 Values
(1st
Category)
o 15 values, defines the moon position in range 1300 km (Error 3%)
(2nd
Category)
o 5 values, defines the moon position in range 1300-2000 km (Error 4.6 %)
(3rd
Category)
o 2 values, defines the moon position in range 2000-3500 km (Error 8 %)
- The Results Explanation
- The distance from perigee to apogee =43000 km…
o 1st
Category of results defines the moon position in error range (1300 km) =
error (3%), that means, (15 values of 22) defines the moon position with
error (3%) only (Small Error Range)
o 2nd
km to 2000 km) = error (4.5%), that means (5 values of 22) defines the
moon position with error (4.5%) (Average Error Range)
o 3rd
km to 3500 km) = error (8%), that means (2 values of 22) defines the moon
position with error (8%) (Great Error Range)
- The Equation Accuracy
o The previous explanation shows that, the equation has a good range of
accuracy and its error is in the acceptable error range
The Conclusion
The Equation Is correct and trustee
And
It's a useful tool to define the moon position daily

I do this research
2nd
39
3-4-3 The Value 1.7 degrees
- The 3rd
question was, why the equation uses 1.7 degrees?
Because
1.7 degrees = 0.98562 degrees + 0.712 degrees
Where
- 0.98562 degrees = Earth motion daily degrees, and it equals the moon daily
motion degrees because the moon has to move an equal distance to Earth motion
daily distance to save their motions accompanying
- This question and the angle 0.712 degrees is discussed deeply before.

I do this research
2nd
40
The Moon Motion Difficulties
- There are 2 basic difficulties are observed in the moon orbital motions, let's refer
to them in following:
o (1st
Difficulty) The moon moves per day different distances from perigee to
apogee…..
o We know the moon moves from perigee to apogee (go and back) during
Anomalistic month (27.55 solar days)
o (43000 km x 2) / 27.55 days = 3122 km
o The moon doesn't use this rate (3122 km) in its motion, instead the moon
can move (6000 km) on one day only and on another day may move only
2500 km (or even less)!
o The moon orbital equation tries to solve this difficulty by using the rate 1.7
degrees in the equation (θ1 = θ0 + 1.7 degrees), the value 1.7 degrees is a
great number and enables the moon to move around (5000 km) per solar day
and by that if the moon moves per solar day 4000 km the different distance
will be 1000 km and if the moon moves 6000 km the different will be
– 1000 km, it’s the same difference, and by that, the error be minimized as
possible enabling the equation to be more efficient..
o (2nd
Difficulty) The moon stays in perigee and apogee points long time….
o That means, while the moon be on perigee or apogee, the moon doesn't use
the equation and doesn't change its distance to perigee or apogee for long
days…we may notice that in the equation tests, when the moon reach to
perigee or apogee the equation stops its work and stays 2 or 3 days to return
to its work… because the moon consumes long time to leave the points
(perigee and apogee)…

I do this research
2nd
41
References
The Moon Orbital Motion Geometry (II)
https://www.academia.edu/45181646/The_Moon_Orbital_Motion_Geometry_II_
or
https://www.slideshare.net/Gergesfrancis/the-moon-orbital-motion-geometry-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
The Moon Motion Trajectory Analysis (II)
https://www.academia.edu/44368860/The_Moon_Motion_Trajectory_Analysis_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
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

Venus Rotation Period Analysis (II) (Revised)

More Related Content

What's hot

Similar to Venus Rotation Period Analysis (II) (Revised)

More from Gerges francis

Recently uploaded

Venus Rotation Period Analysis (II) (Revised)