The Solar Group is A Clock (VII)

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
The Solar Group is A Clock (VII)
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 –28th
April 2021
Abstract
Paper Question
- Can The 1 second period of time be different from a planet to another?
The question discussion
- Someone told that, the second period is a human made unit!, It's not true,
- Earth day period (24 hours) – Earth rotation period (23.9 hours) = 0.1 hour
- This fact forces us to divide the solar day into 24 hours and the 1 hour into 60
minutes – easily by using the same rate we define the second period. That means,
the time units are defined based on planets motions and cycles
- The direct question is, why we suppose (Earth second period of time) = (Mercury
Second period of time)? On what base this hypothesis is suggested?
- The idea is that, because the time unit is defined depends on planet motion
features, and these features are changed from one planet to another, that causes the
time unit (period) to be different from one planet to another…
- But why?
- in some track, many cars move by different velocities but all of them has the same
rate of time, why different velocities (or other motions features) cause different
rates of time?
- Because of the sun

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- A planet day period depends on the planets motions features, this dependency may
cause different rates of time among the planets motions – I want to say that – the
second period of time is defined based on the planet motion features and that cause
(Earth Second Period) to be different from (Mercury Second Period)
- Earth data shows that, Where, we know the solar day =86400 seconds because
Earth day period is different from Earth rotation period with 0.1hour . Mean, Earth
motion features force us to define the solar day as 86400 seconds, based on that,
the second period of time is define as a period = (1/86400) of the solar day.
- The next question should be (does planet velocity effect on its rate of time?)
- The paper discussion provides a wide range of thinking, because it asks if there's a
relationship between planet velocity and its day period and if the planet day period
effect on the planet rate of time….
- The discussion also extends its range
- Because
- If the solar group is A Great Clock, and the motion is transported from a plant to
another, in this case, does the different rates of time be found based on geometrical
necessities of the motion transportation process?
- For more flexible discussion, suppose the different rates of time are found
independently from the motion transportation, so what geometrical reason be
found behind the different rates of time creation? And how the different rates of
time be useful for the planets motions general harmony?

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1- Introduction
- (Earth velocity 29.8 km/s) / (Pluto velocity 4.7 km/s) = 6.34
- (Pluto day period 153.3 h) / (Earth day period 24 h) =6.387
- The difference is (0.8%)
- Also
- (Pluto motion distance during a solar day 406000 km) / (88000 km) = 4.61
- (the moon day period 708.7 hours) /(153.3 h) = 4.61
- Where
- 88000 km = The Earth moon displacement per a solar day
- 153.3 hours = Pluto Day Period
- The data is so interesting, because it makes Pluto velocity in proportionality with
Earth and its moon velocities by their days periods rates
- We need to know how this data is created and how it be useful in these 3 planets
motions?
- Data analysis shows that, not only the three planets velocities and their days
periods are rated to one another – but also- their rates of time are different, Where
1 hour of Earth motion = 6.387 hours of Pluto motion, and 1 hour of Pluto motion
= 4.61 hours of the moon (displacement) motion and 1 hour of Earth motion
=29.53 hours of its moon (displacement) motion – the 3 planets use 3 different
rates of time – and these rates are rated based on the planets velocities rates
- These rates of time answer 3 questions which are
o Why does Pluto day period so long in comparison with the outer planets?
o Why the moon apogee circumference is shorter than the displacements total
during its day period (29.53 days) with 2%?
o Why Earth motion distance during its day period = Pluto motion distance
during its day period = the moon displacements total during its day period?

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2- Methodology
- I use The Planets Data Analysis
- This method tries to explain how planets data is created? for example why does the
moon diameter =3475 km?
- By this method the dependency of data among the planets be shown and proved –
and many answers b provided by this method using –
- Let's discuss one example in following:
o The diameter 12430 km
o No solar planet its diameter =12430 km, This number I have concluded…
o This value is unreal, But I claim this value is a necessary value for the solar
system geometry design…
o This claim depends on the basic description argument – because – if the
solar planets are separated rigid bodies and no planet diameter =12430 km
so this value should have no effect on any planet motion – but – on the other
side – if this value has an effect on any planet motion that means the planets
diameters (and data) are created based on geometrical rules which contain
the number (12430 km) (geometrically) but not real – and that means-
Planets motions doesn't depend on their masses only (as believed now) but
depend on a whole geometrical mechanism (which supports the description
tells the solar system is one machine)
- Let's test the value 12430 km
o The value 12430 km is a middle unique rate between Venus diameter 12104
km and earth diameter 12756 km – means – no other number can do that -
o Saturn (9.7 km/s) moves during 12430 seconds a distance = 120536 km =
Saturn diameter
o Mars (24.1 km/s) moves during 12430 seconds a distance = 300000 km =
light (0.3mkm/s) motion distance during 1 second
o (Earth diameter / the moon diameter) = (Jupiter diameter/(12430 π)

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Please remember
o Jupiter (13.1 km/s) moves during (10921 s) a distance = Jupiter diameter
o Uranus (6.8 km/s) moves during (7510 s) a distance = Uranus diameter
o Where
o 10921 km = The Moon Circumference
o 7510 km = Pluto Circumference
- I try to show that, the value 12430 km is used as any other planet diameter in
different calculations...
- We have a reason to consider that the value 12430 km is a value mentioned by the
geometrical design, But Why? we should know that through the paper discussion
- The previous example is a method to show how the planets data analysis can direct
us for some conclusions- based on the previous data – we have to consider the
value 12430 km is one of the planets data spite it's a created number but the other
planets depend on it in different motions features – if we consider this value 12430
km is unreal we have to explain why the other planets use it as equal any other
planet diameter?!
- Planets data analysis provides chances for similar conclusions – for example –
Uranus orbital circumference =19 Earth orbital circumferences but the Earth moon
rotates Metonic Cycle which is a cycle continues for 19 years – the data tells if
Uranus motion effect on the moon motion so the number 19 should be found in
this effect – and based on that – we conclude – Metonic Cycle is a result of Uranus
motion effect on the moon motion…
- The planets data analysis is a method similar to the living creature genes analysis
to know how this planet creation and motion data is created.

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3- Can Different Rates Of Time Be Found In The Solar Group?
3-1 The hypothesis Discussion
3-2 The hypothesis Proves
3-3 The Proves Difficulties
3-4 The Solar Group as A Clock.
3-1 The hypothesis Discussion
- The hypothesis tells, (V1/V2) = (P2/P1)
- V = Planet velocity
- P = Planet Rate Of Time
- The hypothesis tells, The Planet Rate Of Time Is Defined As The 2 Planets
Velocities Rate
- So, because (Earth velocity / Pluto velocity) = 6.378,
- 1 hour of Earth motion = 6.378 hours of Pluto motion
- This is the supposed rule to define the different rates of time between the 2 planets.
- We should use this rule and test it if it woks correctly
- How to prove that, this rule works sufficiently?
o It's an interesting and strong method of proof
o For example
o 1 hour of Earth motion = 6.378 hours of Pluto motion, and
o 1 hour of Pluto motion = 4.61 hours of the Earth moon motion, based on that
o 1 hour of Earth motion must be = 29.53 hours of the Earth moon motion
o The three planets have different motions data, if these data follow the rule
based on the previous rates, that means, the rule is correct
o This network can contain more than 3 planets, means, all solar planets can
be used in this form and by that if the planets data follow their rates of time
that can prove the rule sufficiency
o Also, that proves the solar planets aren't separated rigid bodies from each
other – which is the new vision discussed in point no. (3-4)

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3-2 The hypothesis Proves
(1) –The rate of time between Mercury motion and the moon displacement
I- Data
(1)
742 mkm = 88000 km (the moon daily displacement) x 8432 solar days
(2)
8432 hours = 2 x 175.94 days x 23.96 hours
(3)
742 mkm x 2 = 4.095 mkm (Mercury velocity per solar day) x 365.25 solar days
II- Discussion
Equations No. (1 and 2) tell that
o The moon displacements total during 8432 solar days be = 742 mkm
o 8432 solar days of the moon motion
o But
o Let's suppose each 2 days of this (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, 48 Hours Of The Moon Motion Be Equivalent To One
Hour Of Mercury Motion
o That shows the rate of time is (1 hour is equivalent to 48 hours)
Equation No. (3)
- Equation no (3) 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

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o (1st
Point)
o We have to suppose that the periods 365.25 days and (2 x 175.94 days) are
equivalent. Based on that, 1 hour of Mercury motion be equivalent to 48
hours of the moon motion
o (Why the periods 365.25 days and (2 x 175.94 days) are equivalent? The
question is answered in the point (3-4)
o (2nd
Point)
o The rule which controls this process tells (Equal Distances Create
Different Rates Of Time) – that means – If Mercury moves 742 mkm and
not 2 x 742 mkm – that will cause the rate of time to be 1 hour of Mercury
motion is equivalent to 24 hours of the moon motion – but because Mercury
moves 2 x 742 mkm that causes the rate to be 1 hour of Mercury motion be
equivalent to 48 hours of the moon motion.
o (3rd
Point)
o The question is
o Why these 2 planets (mercury and the moon) are required to be
interacted in their motions?
o Because the moon daily displacement = 88000 km where it's necessary to
make this displacement equal (176000 km = 2 x 88000 km) – (in the moon
orbital motion description (Point No. 4) of the current paper we review why
this distance 176000 km is required)
o So, The 2 Gears (Mercury and the moon use the moon displacement 88000
km daily to produce the required distance 176000 km)
o So, the moon motion distance = 742 mkm and Mercury motion distance = 2
x 742 mkm – similar to that – the moon daily displacement be 88000 km
and be = 176000 km by The Interaction Between Mercury And The Moon
Motion –

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o (Because of that, Mercury orbital period =88 days and its day period
=175.95 days, If this data is used through the rate 1000 km = 1 solar day,
these 2 values will be equal the 2 distances of the moon motion)
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

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(2) –The rate of time between Mars motion and the moon displacement
I- Data
(4)
24.1 (km/s) (Mars Velocity) x 29.53 hours = 2562074.856
(5)
The moon displacement total during 29.53 solar days = 2598693km
The difference between both = 1.4%
II- Discussion
- Equations no. (4 and 5) tell that
- Mars moves during 29.53 hours a distance equal (approximately) the moon
displacements total during 29.53 solar days
- From this data we conclude that
- 1 hour of Mars Motion be equivalent to 24 hours of the moon motion
- We use the same rule (Equal Distances Create Different Rates Of Time)
- But
- There's error 1.4% between the 2 distances, that tells the rate (1 hour = 24 hours) is
correct but in the detailed discussion the accuracy needs to take into consideration
this error 1.4%
- The rule is still solid and correct.
Notice
- 1 hour of Mercury Motion be equivalent to 48 hours of the moon motion
- And
- 1 hour of Mars Motion be equivalent to 24 hours of the moon motion
- So
- 2 hours of Mars Motion must be equivalent to 1 hour of Mercury Motion, Is it
true? Let's answer in following

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I-Data
(6)
Mercury (47.4 km/s) moves during its rotation period (1407.6 hours) = 240.2 mkm
Mars (24.1 km/s) moves during Venus day period (2802 hours) = 243.1 mkm
II-Discussion
Equation no. (6)
Mercury (47.4 km/s) moves during its rotation period (1407.6 hours) = 240.2 mkm
- The difference between the distance 240.2 mkm and 243.1 mkm is around (1%)
- Equation no. (6) tells that both distances are equal (approximately)
- Let's remember the rule (Equal distances create different rates of time) means, the
periods of time are equal, so the mercury period (1407.6 hours) is equal Mars
period (2802 hours)
- By that 2 hours of Mars Motion is equivalent to 1 hour of Mercury Motion
- This same rate we have concluded from the moon motion (supposed) interaction
with these 2 planets – that means – we have 2 different data prove this same rate of
time which may show that it's a fact
Notice
- Mercury velocity (47.4 km/s) = 2 x Mars velocity (24.1 km/s) (error 1.6%)
- The rate of time (2: 1) depends on the rate of velocities
- But why Mars moves 243 mkm during Venus day period? This question should be
answer in point no. (3-4) of this current paper.

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(3) –The rate of time between Mars motion and Pluto motion
I-Data
(7)
Mars velocity (24.1 km/s) = Pluto velocity (4.7 km/s) x 5.12
(8)
(112.2 km/s) = 24 x (4.7 km/s)
II-Discussion
Equation no. (7)
Mars velocity (24.1 km/s) = Pluto velocity (4.7 km/s) x 5.12
- We remember that
- The moon day period 708.7 hours = 4.61 x 153.3 h (Pluto day period)
- And
- 1 hour of Mars motion = 24 hours of the moon motion
- Based on that,
- 1 hour of Mars motion should be = 5.12 hours of Pluto motion
- But, it's not true!!
- 1 hour of Mars motion should be = 4.61 hours of Pluto motion,
- Why?
Equation no. (8)
(112.2 km/s) = 24 x (4.7 km/s) = 4.61 x (24.1 km/s) Where 24 = 5.2 x 4.61
- The velocity 112.2 km/s = the velocities total of (Mercury + Venus + Earth)
- The equation tells, we have 2 gears, the 3 planets motions together form one gear
and Pluto motion alone form the other gear between both gears the rate of time (1
hour =24 hours) is created
- The 3 planets motions total effect on Pluto (4.7 km/s) and Mars (24.1 km/s)
motions and because of that the rate 4.61 is used with the 2 planets.

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- 1 hour of the 3 planets motions total = 4.61 hours of Mars motion
- And
- 1 hour of Mars motion = 4.61 hours of Pluto motion
- And
- 1 hour of Pluto motion = 4.61 hours of the Earth moon motion
- The data tells that there's a geometrical necessity to use the same with 3 planets
- It's a transportation of motion is done from point to another point through the solar
group.
- The secret is in the equation ((24 hours/ 4.61) = 5.2), also because Mars velocity
=24.1 km/s, this all (similar data) doesn't tell about pure coincidence of numbers
but tells about a motion transportation mentioned to create the data values similar
to one another for geometrical necessities
- What's This Geometrical Necessity?
The Velocities Total 112.2 km/s
- How to create this velocity 112.2 km/s ?
- It's the total of (Mercury 47.4 + Venus 35+Earth) velocities =112.2 km/s
- But
- How to create this velocity geometrically? Relative to what point this velocity
(112.2 km/s) can be defined?
- This velocity (112.2 km/s) is defined relative to the sun point, it's the only possible
method, because the 3 planets revolve around the sun in the same direction and
theoretically their velocities can be added to each other relative to the sun point.
- Imagine these 3 planets as 3 gears, how to create their total velocities? Their
velocities total can be defined relative to their center
- Let's see the following equation

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Equation no. (9)
1392000 km = 112.2 km/s x 12430 seconds
- Where
- 1392000 km = The Sun Diameter
- 12430 seconds =we have discussed deeply in the current paper Methodology
- Even if the geometrical mechanism is still absent but the data shows that our
expectation is correct, the velocity 112.2 km/s is used for the sun data
- That tells there's a geometrical necessity for this using – and this velocity (112.2
km/s) is a real one
Notice
- Earth motion distance during its day period = 2574720 km
- The velocity (112.2 km/s) passes this same distance in 6.374 hours
- This is the rate between Earth and Pluto velocities (and also their days periods)
- That tells this velocity (112.2 km/s) is a basic player in the solar system geometry,
how this velocity (112.2 km/s) effect on the other planets motions data? In point
no. 3-3 we should discuss this question
Notice
- 113.4 km x 0.99 = 112.2 km
- But
- 113.4 degrees = 90 degrees + 23.4 degrees (Earth Axial Tilt)
- That tells this velocities total (112.2 km/s) effect on Earth Axial Tilt creation

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(4) –The general rate of time between the 4 planets
I-Data
(10)
48 hours of the moon motion =1 hour of Mercury Motion = 2 hours of Mars
Motion = 9.23 hours of Pluto Motion
(11)
9.23 x 153.3 hours = 1415 hours
Equation no. (11)
9.23 x 153.3 hours = 1415 hours
- Where
- 153.3 hours = Pluto Rotation Period
- 1407.6 hours = Mercury Rotation Period (the error 0.5% with 1415)
- So, both 2 planets rotation periods depends on this same rate 9.23
- We have a reason to consider that, this data is created based on a geometrical
machine and that tells the different rates of time is almost a fact tool used in the
solar system geometry.
Notice
- But the rate 4.61 doesn't define Mars motion with Mercury motion
- Where
- 4.61 x 24.6 hours (Mars rotation period) = 113.4 hours
- And
- Mercury rotation period =1407.6 hours?

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(5) –The rates among other planets
- 24 hours of Uranus Motion = 1 hour of Mercury Motion = 1.35 hour of Venus
Motion = 1.6 hours of Earth Motion= 2 hours of Mars Motion = 48 hours of
the moon motion = 4.61 hours of Pluto Motion
I- Data
(12)
Earth (29.8 km/s) moves during its day period (24 h) a distance = 2574720 km
(13)
Pluto (4.7 km/s) moves during its day period (153.3 h) a distance = 2593836 km
(14)
The Earth moon displacements total during 29.53 days be = 2598693 km
(15)
Mars (24.1 km/s) moves during (29.53 h) a distance = 2562075 km
(16)
Uranus (6.8 km/s) moves during 6 Uranus days (103.2h) a distance =2526336 km
II- Discussion
- let's remember the rule
- Equal Distances Cause To Create A Different Rate Of Time
o (Point No. A)
o The moon displacement daily =88000 km, and its total during 29.53 days =
Earth motion distance during its day period (the difference =1%)
o Based on this data
o 29.53 solar days of the moon motion =1 solar day of Earth motion
o i.e.
o 29.53 hours of the moon motion =1 hours of Earth motion.

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o (Point No. B)
o Earth (29.8 km/s) moves during its day period (24h) a distance= the distance
passed by Pluto (4.7 km/s) during its day period (153.3 h) (error 1%)
o So
o 153.3 hours of Pluto Motion = 24 hours of Earth Motion
o i.e.
o 1 hour of Earth motion = 6.3875 hours of Pluto Motion
o Also
o The moon displacements total during 29.53 days = Pluto motion distance
during its day period (153.3 h)
o So
o 153.3 hours of Pluto motion = 29.53 days of the moon motion
o i.e.
o 1 hour of Pluto motion = 4.61 hours of the Earth moon motion
o (Point No. C)
o Mars (24.1 km/s) moves during (29.53 h) a distance = the distance passed
by Earth during its day period (24 h) = the moon displacements total
distance during 29.53 solar days = the distance passed by Pluto during its
day period (153.3 h)
o Based on that
o 29.53 hours of Mars motion = 29.53 solar days of the moon displacement
motion = 1 solar day of Earth motion = 153.3 hours of Pluto motion
o i.e.
o 1 hour of Mars motion = 24 hours of the moon motion
o 1.23 hour of Mars motion = 1 hour of the Earth motion
o 1 hour of Mars motion should be equal 5.2 hours of Pluto motion (But this
rate isn't correct, instead 1 hour of Mars motion =4.62 hours of Pluto
motion) (Why? we have discussed that)

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o (Point No. D)
o Earth (29.8 km/s) moves during its day period (24 h) a distance = the
distance passed by Pluto during its day period (153.3 h) = The moon
displacements total during 29.53 days be = the distance passed by Mars
during (29.53 h) = the distance passed by Uranus during 6 Uranus days
(103.2h)
o So
o 24 hours of Earth motion = 153.3 hours of Pluto motion = 29.53 solar days
of the moon motion = 29.53 hours of Mars motion = 103.2 hours of Uranus
Motion
o i.e.
o 1 hour of Earth motion = 6.3875 hours of Pluto motion = 1.23 solar days of
the moon motion = 1.23 hours of Mars motion = 4.3 hours of Uranus
Motion
o But, not
o (1 hour of Uranus Motion = 2 hours of the moon motion) why? we need
to answer this question later
Shortly
Pluto & Uranus rates of time are different from the defined rates based on the used
rule
Let's discuss why both rates should be different in following

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Pluto and Uranus Motions Rates Of Time
- (1st
Pluto Rate of time) (Revision)
- Equations no. (13 and 15) tell that, Pluto motion distance during its day period
(153.3 hours) = Mars motion distance during (29.53 hours)
- Based on that
- (153.3 hours /29.53 hours) = 5.2
- Means
- 1 hour of Mars Motion should be = 5.2 hours of Pluto motion
- But
- The fact is that
- 1 hour of Mars Motion should be = 4.61 hours of Pluto motion
- And
- 1 hour of Mercury Motion = 2 hours of Mars motion ,
- So
- 1 hour of Mercury Motion = 9.23 hours of Pluto motion.
- Why Pluto rate is 4.62 hours and not 5.2 hours?
- Because
- The total velocities of (Mercury + Venus +Earth)=112.2 km/s
- These velocities total work together as one gear in opposite to Pluto (4.7km/s) as
another gear
- 112.2 km/s = 24 x 4.7 km/s = 24.1 x 4.61
- And
- 24 = 5.2 x 4.62
- That means, Pluto works as a common gears between 2 sides, on the first side the
three planets velocities total and on the second side Mars velocity
- For that, the rate (5.2) changed to be (4.62) because of the rate 24 effect which is
found between the 3 planets velocities total on one side and Pluto on the other side.

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(2nd
Uranus Rate of time)
- Let's suppose Uranus rate of time as following
- 1 hour of Mercury Motion = 24 hours of Uranus Motion =9.23 of Pluto
Motion
- Based on that
- 1 hour Pluto motion =2.6 h Uranus Motion
- Uranus rate of time is changed (from the rule suppose result) because of Pluto rate
change from 5.2 to 4.62
- The rate 2.6 = (5.2/2)
- And the rate 5.2 connects Uranus and Pluto motion with the moon motion (because
5.1 degrees = the moon orbital inclination)
- Also
- 97.8 deg (Uranus axial tilt) = 19 x 5.14 deg (the moon orbital inclination)
- 17.2 deg (Pluto orbital inclination) = 5.1 deg x 3.37 deg (Venus orbital inclination
3.4 deg)
Notice
- Because 1 hour Mercury Motion = 24 hours of Uranus Motion
- 2815 mkm (Mercury Uranus Distance), this distance is used as a period of time
with the rate 1 mkm =1 hour
- And by that
- 2815 mkm = 2815 hours = 2 x 1407.6 hours (Mercury Rotation period)
Notice
- Mercury (47.4 km/s) moves during its rotation period (1407.6 hours) a distance =
240.2 mkm
- It different 1% from the distance 243 mkm
- If this distance is used as a period of time with the rate 1 mkm = 1 solar day
- It will be =243 solar days = Venus Rotation Period

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- Also
- Venus (35 km/s) moves during 365.25 solar days a distance = 1106 mkm = 2π x
175.94 mkm
- If this distance is used as a period of time with the rate 1 mkm =11 solar day
- It will be =175.94 solar days (Mercury Day period)
Notice
- 153.3 hours/ 29.53 (hours) = 5.2 =2 x 2.6
- That shows Uranus motion effects on the earth moon and Pluto motions to create
their days periods
- That tells this geometrical mechanism is effected basically by Uranus motion
which cause a unification for the whole process
Notice (please remember)
- 1 hour Mercury Motion = 9.23 hours Pluto Motion
- Because of that
- 153.3 h (Pluto rotation period) x 9.23 = 1415 hours
- (where 1407.6 hours = Mercury rotation period ) (error 0.5%)
- Based On Uranus Rate Of Time
- 1 hour of Mercury Motion = 24 hours of Uranus Motion =9.23 of Pluto
Motion
- Let's define the rest planets rates of time in following
- 1 hour of Earth Motion = 1.235 hours of Mars Motion
- But
- 1 hour of Mercury Motion =2 hours of Mars Motion
- Based on that
- 1 hour of Mercury Motion =1.6 hours of Earth Motion
- This is true because (Mercury velocity 47.4) / (Earth velocity 29.8) =1.6

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- 1 hour of Mercury Motion = 1.354 hours of Venus Motion
- Based on that
- 1 hour of Venus Motion = 17.7 hours of Uranus Motion
Notice
17.2 hours = Uranus day period
And
17.7 hours = 17.2hours = 0.5 hours = 1800 seconds
Uranus (6.8 km/s) moves during 1800 seconds a distance = 12104 km (error 1%)
Where
12104 km = Venus Diameter.
Notice
2 x 2815 hours (2 Mercury rotation periods) = 17.2 hours (Uranus day) x 327.6
Where
327.6 solar days = the moon sidereal year
- But the rate 4.61 doesn't define Mars motion with Mercury motion
- Where
- 4.61 x 24.6 hours (Mars rotation period) = 113.4 hours
- And
- 113.4 deg = 90 deg +23.4 deg (Earth Axial Tilt)

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3-3 The Proves Difficulties
- The paper discussed the hypothesis proves with strong data – we have 2 basic
points need to refer
- (1st
Point)
- The rate of time depends on the planets velocities rates
- But
- Because of the 3 planets velocities total (112.2 km/s) interacts with Pluto and Mars
motions that causes 1 hour of Mars motion to be =4.61 hours of Pluto motion
instead of the value 5.2 which defined by the using rule
- Because of that we have 2 series of rates
- The rates are defined by the rule
- And
- The rates effected by the change in Pluto rate relative to Mars
- This fact effect on Uranus rate of time as we have seen
- This interaction of Pluto motion must be effected by Uranus and by this interaction
Pluto day period became 153.3 hours (so long in comparison with any outer
planets)
- That shows this interaction has a great effect on Pluto motion (and in the Earth
moon motion also)
- Now the situation can be expected
- Uranus motion effects on Pluto motion and cause to change the rate from 5.2 to
4.61 (notice 2.6 x2 =5.2) – in this process – Pluto day period became 153.3 hours –
and the moon apogee radius became 406000 km in place of the value 413600 km
which should be
- Where 406000 km = Pluto motion distance during a solar day

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- (2nd
Point)
- We can't the geometrical mechanism
- How does the velocity (112.2 km/s) work?
- The planets aren't gears, but what rules the motion is transported among them?
Because the data shows clearly that, the motion is transported, but how? Based on
geometrical rules that's done
- For example,
- Uranus day period =17.2 hours but 1 hour of Venus =17.7 hours
- The difference 0.5 hour Uranus needs to move a distance = Venus diameter,
- The data tells some interaction is found between Venus and Uranus motions, but
how this interaction is done? How this data is created?

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3-4 The Solar Group as A Clock.
- The solar planets aren't rigid bodies separated from each other in creation and
motion data, revolve round the sun by the masses gravity force- this description
fails frequently to explain the planets motions data
- The solar group is one trajectory of Energy, or one light beam, and each planet is a
point on this same trajectory of Energy
- This description makes the solar group as one cable and each planet is a knot on it.
- Also the solar group can be similar to a puppets theater, all puppets are binding
with the same one rope
- Based on this description, each planet data is created depends on the other planets
data and no planet can be found alone - let's explain this meaning-
- The double production experiment is the best example to explain how the planets
data is created complementary with one another
- From gamma ray, 2 particles are created and electron and an positron.
- Because gamma ray is (Zero) Charge, the produced electron forces a positron to be
created otherwise both can't be created – and so 2 electrons can't be created from
gamma spite their masses are equal but the charge reservation law forces the
process to produce an electron and a positron.
- Similar to that, the planets data is created complementary with each other
- Because the data is complementary with each other – the motion is transported
from one planet to another as among gears
- But the planets aren't gears, how the motion can be transported from one planet to
another?
- The discussion tries to discover the features of the motion transportation among
the planets
- The first discovered geometrical feature of this process is that, the motion
transportation among planets may create different rates of time for these planets
motions

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- And because of that, 1 hour of Mercury motion = 24 hours of Uranus motion and
that means, while a clock moves 1 second in Mercury the other clock moves 24
seconds in Uranus –
- The planets different rates of time depend on the planets velocities –
- Let's answer the lest questions in this discussion in following
The Answers Of The Basic Questions
- (1st
Question) why the rate of time (4.61) is used between Pluto and Mars
Motions?
- 1 hour of Mars Motion = 24 hours of The Moon Motion (no. 1)
- 1 hour of Pluto Motion = 4.62 hours of the moon motion (no.2)
- From 1 and 2, 1 hour of Mars Motion should be = 5.2 hours of Pluto motion
- But it's not true – on the contrary – 1 hour of Mars Motion = 4.62 hours of Pluto
Motion- Why?
- Because
- 24 hours = 4.62 x 5.2
- 112.2 km/s = 24 x 4.7 km /s (Pluto Velocity)
- The 3 planets velocities total (Mercury 47.4 km/s + Venus 35 km/s + Earth 29.8
km/s) = 112.2 km/s, this total velocity creates the rate 24 with Pluto motion
- That means 1 hour of this total velocity motion = 24 hours of Pluto motion
- This (24 hours) is the value which changes between the rates 4.62 and 5.2 –
- By that the rate 5.2 (which the original one between Mars and Pluto motions is
changed to 4.62 which we have used because of these (24 hours)
- Means
- The interaction between the three planets velocities total on one side and Pluto
velocity on the other side – this interaction causes to change the rate from 5.2 to
4.62

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(2nd
Question) Why Mars moves the distance 243 mkm in 116.75 solar days
(Venus Day Period) –
Where
- The answer is simple
- That because there's an interaction between Mars and Venus Motions, they move
together and use common data
- Let's try to prove that
- (a) 1433 mkm = 2.13429 mkm x 670.5 days
- (b) (734.8 mkm /353.05) = (1433 mkm/ 687 mkm)
- (c) 734.8 mkm = 2.142279 mkm x 342.88 days
- (d) 2 x 734.8 mkm = 2.13429 mkm x 687 rotation
- Let's explain this data in following
- Equation no. (a)
- 1433 mkm = 2.13429 mkm x 670.5 rotation
- Where
- 1433 mkm = Mars Orbital Circumference
- 2.13429 mkm = Mars Motion distance during Mars rotation period
- 670.5 mkm = Venus Jupiter Distance
- 24.6 hours = Mars rotation period
- Equation no (a) tells that, Venus Jupiter distance (670.5 mkm) is used as a period
of time (670.5 Mars Rotation)
- So, there's a great interaction between Mars and Venus Motions data, because
many motions of both planets depend on the other planet data
- Notice/ to use the distance value as a period of time it's usual behavior between
Mars and Venus (also between Mercury and Venus).

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Equation no. (b)
(734.8 mkm /353.05) = (1433 mkm/ 687 mkm)
- Where
- 734.8 mkm = the distance passed by Venus during 243 days
- 353.05 mkm = The distance passed by Venus during Venus day period (2802
hours)
- 1433 mkm = Mars Orbital Circumference
- 687 days = Mars orbital period (but used as 687 mkm)
- Equation no. (b) shows more interactions
- I wish I can explain the idea as clear as possible – the idea is that – both planets
Mars and Venus Motions data are rated to one another in different motions data,
that shows an interaction behind because the contact points between the 2 planets
are several and found in different motions data
Equation no. (c)
734.8 mkm = 2.142279 mkm x 342.88 days
- Where
- 243 solar days = Venus rotation period
- 2.142279 mkm = Mars Motion Distance during Mars Day period (24.7 h)
- 346.6 solar days = the nodal year (but the value 342.88 days is different by 1%)
- Equation (c) show more interaction proves of both planets motions
Equation no. (d)
2 x 734.8 mkm = 2.13429 mkm x 687 rotation
- Where
- 2.13429 mkm = Mars Motion Distance during Mars Rotation period (24.6 h)
- 687 solar days = Mars orbital period (but the equation uses Mars rotation period)

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- (3rd
Question) Why the period 351.88 solar days is equivalent to 365.25 solar
days?
- Let's remember this question (we have this question with the 3 first equations
discussion – which are)
(1)
742 mkm = 88000 km (the moon daily displacement) x 8432 solar days
(2)
8432 hours = 2 x 175.94 days x 23.96 hours
(3)
- The equations discussion depended on that the 2 values (351.88 and 365.25 days)
are equivalent. Let's explain why in following
o As in the moon orbit regression (361 degrees), the moon regresses per year
19 degrees and during 19 years the total revolution be =361 degrees – so-
the real value is Not 360 degrees but 361 degrees – because of that any data
is defined based on the value 360 degrees needs a modification to be defined
on 361 degrees – it's the effect on the moon regression cycle on other
planets motions
o Also
o (243 days /224.7 days) = (29.53 days /27.3 days)
o Where
o 243 solar days = Venus rotation period
o 224.7 solar days = Venus orbital period
o 29.53 solar days = the moon day period
o 27.3 solar days = The Moon Orbital Period
o These are examples try to explain the meaning of 2 values configuration
based motions interactions- they are 2 different values but they behave as

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equal values why? because of the configuration …. The moon revolves
around Earth in 27.3 days but the configuration forces the moon to compete
its day by 29.53 days – the 2 values are different from each other – but both
are found from one motion and both are found depending on each other –
and both are found for configuration –
o It's the gears effect on each other.

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4- The Moon Orbital Motion Description
4-1 Why Does The Moon Use Pythagorean Triangle In Its Motion?
4-2 How Does The Moon Use Pythagorean Triangle In Its Motion?
4-3 The Moon Orbital Motion Analysis
4-4 The Moon Orbital Motion Equation

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4-1 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….

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4-2 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.

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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

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4-3 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)

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- 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)

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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.

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4-4 The Moon Orbital Motion Equation
4-4-1 The Equation Concept
4-4-2 The Equation Test and Accuracy
4-4-1 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….

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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

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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

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4-4-2 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
[

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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

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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
[

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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

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4-4-3 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.

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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)…

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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

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