Uranus Orbital Period Analysis (II) (Revised)

IN THE ALMIGHTY GOD NAME
Through the Mother of God mediation
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Gerges Francis Tawadrous/
2nd
Course student – physics Faculty – People's Friendship University – Moscow –Russia..
mrwaheid1@yahoo.com mrwaheid@gmail.com +201022532292
1
Uranus Orbital 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 –9th
April 2021
Abstract
Paper hypothesis
- (1 hour = 1 day) this rate of time is produced by Uranus Motion Effect.
The hypothesis explanation
- The total velocities of (Mercury 47.4 + Venus 35 + Earth 29.8) = 112.2 km/s
- Pluto velocity =4.7 km /s
- means, the distance passed by the 3 planets velocities total in 1 second = the
distance passed by Pluto motion during 24 seconds…
- Uranus uses this rate to create 2 gears of motions, the first gear consisted of the 3
planets velocities total and the second gear is Pluto velocity, to create a different
rate of time
- By this geometrical mechanism, 1 hour should be = 24 hours
- But
- The real machine works by a little modifications,
- The machine input is 5040 hours but its output is 5040 Earth rotation periods (23.9
hours) –
- Means, this machine need one more modification to reach to the final rate (1 hour
is equivalent to 24 hours = 1 solar day)
Paper objective
- The paper discusses the geometrical mechanism by which this process is done.

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1- Introduction
- (Venus diameter 12104 km/ Mars diameter 6792 km)2
= π (with difference 1%)
- Where the error 1% is used frequently in the solar system motion
- Can we consider that, all (π) which is used in the solar system has this error (1%)
and that cause the error 1% in the solar system calculations
- I mean, does the solar system has its own (geometrical factors) and by that, it
doesn't use the (π) which we know (π =3.1415926) but uses its own factor (πs
=3.17586)??
- The paper discusses a significant hypothesis that, the planets motions create
interactions between each other – and these interactions create different rates of
time of the planets motions
- The basic idea in this paper is the suggestion that, there's a method to create
different rates of time depends on planets motions interaction -
- The concept depends on the machine of gears description, while one gear rotates
fast another rotates slowly, and because of that the sun rises on one planet (gear)
before another planet (gear)
- So, the day on Earth starts on (a rate of time) different from the day on Jupiter
- The paper argues, if this process can be done based on planets motions (low
velocities) and without a dependency on light motion features.
- The geometrical mechanism of this process provides significant data can help us
for better understanding for the solar system motion.

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2- Uranus Motion causes to create a different rate of time
2-1 (1 Hour =1 Day) The Geometrical Mechanism
I-Data
(1) 120536 hours = 23.91 x 5040 hours
(2) 112.2 km/s x 5040 minutes = 2 x 17 mkm
(3) 4.7 km/s x 366556 seconds = 1.72 mkm
(4) 47.7 km/s (Mercury velocity) x 366556 seconds = 17.37 mkm
II-Discussion
Equation No. (1)
120536 = 23.91 x 5040
- 120536 km = Saturn Diameter
- 5040 seconds = a period required by Mercury day period to make it =176 days
- 23.9 hours = Earth rotation period
- Equation no. (1) explains how this rate is produced, let's consider it in following:
o The equation shows input and output… the input is 5040 hours and the
output is 120536 hours and that means
o The input is 5040 hours and the output is (5040 Earth rotation periods)
o And by some modification, the period 1 hour be equivalent to 1 solar day
- Let's follow this process in details:
- Uranus (6.8 km/s) moves during (5040 seconds) a distance = 34272 km = π x
10921 km (the Earth Moon Circumference)
o But
o Pluto (4.7 km/s) moves during (5040 seconds) a distance = 23688 km = π x
7510 km (Pluto Circumference)
- Uranus (4.7 km/s) moves during (7510 seconds) a distance = 51118 km = Uranus
Diameter
o But Pluto (4.7 km/s) moves during (10921 seconds) a distance = 51118 km
- And

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- Pluto (4.7 km/s) moves during (51118 seconds) a distance = 2 x 120536 km
(2 x 120536 km = 2 Saturn Diameters)
- The previous data tells that, we have 2 motions (Uranus and Pluto) both uses the
periods 5040 seconds and pass outputted distances (10921 km and 51118 km)
which are used as periods of time to pass the required distance 120536 km x 2
- So, the motions input was 5040 seconds and the motions output was (2 x 120536
seconds)
- This result is very near to our plan, because we wanted to use (5040 hours and
produce 120536 hours) and that's what happening here, the motions use the period
(5040 seconds) and produces the distance (2 x 120536 km) which will be used as
(2 x 120536 seconds)
- Almost we perform the goal
- But nothing is understandable – What's going on here?
o Some old data may help us – let's remember them
Equation no. (I) (old)
(2 x 120536 km /2390) =(5848 mkm/57.9 mkm) = (5040/50)
Equation no. (II) (old)
47.4 km/s x 5040 sec = 2 x 119448 km (1%)
Equation no. (I) (old)
(2 x 120536 km /2390) =(5848 mkm/57.9 mkm) = (5040/50)
- 5848 mkm = Mercury Pluto distance 57.9 mkm= Mercury orb. Distance
- 120536 km= Saturn diameter 2390 km = Pluto diameter
- In our Equation no. (1) and the 2 old equations no. (I) and (II) Saturn diameter is
produced always in double value (2 x 120536km) why?
- Pluto motion (during 51118 sec) and Mercury motion during (5040 s) produce (2 x
Saturn diameter) why? and the rate between their distances also need (2 x Saturn
diameter)?! Why?

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- I think it's one motion or one geometrical reason which is produced these 3
equations – it's one geometrical effect which continues without change and
because of that the required distance (2 x Saturn diameter) isn't changed! But used
as it frequently! Why? what's the relationship between these 3 equations?
Notice (1)
(2 x 120536 km /2390) =(5848 mkm/57.9 mkm) = (5040/50)
- (5848 mkm Mercury Pluto Distance / 57.9 mkm Mercury orbital distance) =101
- (5906 mkm Pluto orbital Distance / 57.9 mkm Mercury orbital distance) =102
- But
- (Mercury velocity 47.4 km/s) / (Pluto velocity 4.7 km/s) = 10.085
- But
- (10.085)2
= 101.7
- Means
- Mercury & Pluto velocities are defined as a rate of their distances
- And because they are the greatest and smallest velocities in the solar system, that
shows the solar system motion range is defined based on the rate between these 2
distances (5848 mkm & 57.9 mkm), that tells these 2 planets control the solar
system motion range….
- Still we don't know why Saturn use double values of its diameter (2 x 120536 km)
in many different equations (among which the previous 3 equations)?
Notice (2)
- We have 3 values of Saturn diameter which are
o 120536 km (Saturn Diameter exactly)
o 119448 km (99% of Saturn Diameter) (Mercury motion during 5040/2)
o 121500 km (101 % of Saturn Diameter) (Mars motion during 5040)
o But 121500 = 243 x 500
o 243 days = Venus Rotation Period
- Light (0.3 mkm/s) need 500 seconds to pass 149.6 mkm (Earth orbital distance)

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Notice (2)
160592 km =5040 km x (97.8 deg /3.1 deg) (error 1%)
- Where
- 160592 km = Uranus Circumference
- 5040 seconds = are required for Mercury day Period to be 176 solar days
- 97.8 deg = Uranus Axial Tilt
- 3.1 deg = Jupiter Axial Tilt
- This equation shows how the value 51118 km (Uranus Diameter) depends on the
value 5040
- This equation is so important because of the following one
Notice (3)
2852 mkm = (3.1 deg/97.8 deg) x 90000 mkm
- Where
- 2872.5 mkm = Uranus Orbital Distance
- 97.8 deg = Uranus Axial Tilt
- 3.1 deg = Jupiter Axial Tilt
- 90000 mkm = C2
for 1 second
- The difference between 2852 mkm and 2872.5 mkm = 0.7% but the mathematical
series based on which the planets original supposed positions are defined tell us
Uranus Number is 75 and its series = (1+2+3+4+5+6…..+75 = 2850)
- That means, Uranus original orbital distance was 2850 mkm which makes the
equation more sufficient and has (Zero) error
- The distance 90000 mkm = (light "0.3mkm/s" motion distance during 1 second)2
- what we need in this equation is the rate (97.8/3.1) which is used here and used in
the previous one – showing a deep connection between 5040 seconds and 51118
km (Uranus diameter) or (51118 seconds).

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Equation No. (2)
112.2 km/s x 5040 Minutes = 2 x 17 mkm
- 5040 seconds = a period required by Mercury day period to make it =176 days
- 112.2 km/s =Mercury, Venus and Earth velocities total (47.4+35+29.8 =112.2)
- The total velocities of the three planets pass 17 mkm x 2 during the period 5040
seconds x 60 m = 5040 minutes
- 17.6 mkm = the solar planets motions distances total during 1 solar day
- Between 17.6 mkm and 17 mkm the difference around 3%
- Equation no. (2) tells the 3 planets move during 5040 minutes a distance = the
planets motions distances total during 2 solar days (error 3%) (Why 2? )
- The equation shows that the passed distance = 2 x 17 mkm –
- We know that, Pluto orbital inclination =17.2 degrees
- But, 86400 seconds = 5040 seconds x 17.2
- The difference between 17.2 and 17 around 1% and because of that we don’t use
the value (34 mkm) because it's unknown but we use the value (2 x 17 mkm) and
by that we discover the equation uses the rate (2) for the produced distance!
Notice
- The equation uses 5040 minutes and not 5040 hours
- Because we need to test our hypothesis we have to use the value 5040 hours, so we
have to modify this value later

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Equation No. (3)
4.7 km/s x 366556 seconds = 1.72 mkm
- Where
- 366556 km = The Outer Planets Diameters Total
- Equation No. (3) tells that, Pluto moves during 366556 seconds a distance =1.72
mom
Notice (1)
(366556 km /Pluto diameter 2390 km) =153.3
But
Pluto Day Period =153.3 hours.
Notice (2)
(366556 km /The moon diameter 3475 km) =105.5
And
105.5 x 2π = 663 km
But
The Moon Rotation Period =655.7 hours. (error 1%)
Notice (3)
Mercury, Venus & Earth velocities total = 112.2 km/s
Pluto velocity =4.7 km /s
But
112.2 km/s = 24 x 4.7 km/s
Based on these 4 planets motions the rate of time (1 day = 1 hour) is produced.
Notice (4)
- 5040 seconds x 0.99 = 4989.6 minutes
- 4989.6 minutes x 60 = 299376 seconds
- Light (0.3 mkm/s) travels during 299376 seconds a distance = 89812.8 mkm but
we minus from this value a distance = 86400 mkm and so the result will be 3413
mkm (= The 4 Inner Planets Orbital Circumferences Total)

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Let's summarize the whole idea in following :
o We have one machine
o The input is -
o A motion done by 3 planets velocities total which are (Mercury + Venus +
Earth velocities total) (112.2 km/s) during a period = 5040 minutes
o The output is
o Pluto (4.7 km/s) motion distance during a period (366556 seconds) a
distance = 1.72 mkm - Let's remember the paper hypothesis
o The input should be 1 hour and output should 1 Earth rotation period or
o The input should be 5040 hours and output be 5040 Earth rotation periods
o Now the input is 5040 minutes
o The output should be 5040 minutes x 23.9 = 120536 minutes
o But 120536 km = Saturn diameter
- The input is 5040 minutes - But the output is 1.72 mkm!!
- What's the relationship between the (366556) and Saturn 120536 km?
- 366556 km = The Outer Planets Diameters Total
- 366556 km = π x 116724 km
- Saturn diameter 120536 km = 116724 km +3812 km
- 3812 km x 9 = 34308 km
o 34308 km = π x 10921 km (the moon circumference) and
o 34308 km = the distance Uranus 6.8 (km/s) moves during 5040 s
o That shows geometrical mechanism behind the previous data and supports
the paper hypothesis
Notice
- We need the input to be 5040 hours and Not Minutes
- And the output should be 5040 (Earth rotation period 23.9 hours)
- So, the previous data doesn't perform the goal correctly and we need to create
some modification for this data to perform our goal

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2-2 Jupiter Uranus Distance 2094 mkm
I-Data
(1)
- The 3 planets (112.2 km/s) moves during 5040 minutes a distance = 2 x17 mkm
- The 3 planets (112.2 km/s) moves during 5040 hours a distance = 2040 mkm
- Jupiter Uranus Distance = 2094 mkm (2.5%)
(2)
- Pluto (4.7 km/s) moves during 366556 seconds a distance =1.72 mkm
- Pluto (4.7 km/s) moves during 120536 hours a distance = 2040 mkm
II-Discussion
- The machine input is 5040 hours (as the hypothesis tells)
- The machine output is 120536 hours (as the hypothesis tells)
- So
- The output 120536 hours = (5040 x Earth Rotation Periods)
- Based on that,
- The input is 1 hour and the output is 1 Earth rotation period – and the machine
produces a different rate of time where (1 hour is equivalent for 23.9 hours)
- The data performs the task
But
- The 3 planets velocities total (112.2 km/s) = 23.9 x Pluto velocity (4.7 km/s), that
means, any data can perform the previous explanation because the rate 23.9 is
found between the 2 velocities – to prove this machine is a correct idea – we need
to prove that – the previous data (specifically) is mentioned- that means we need to
prove that – Jupiter Uranus Distance specifically is required for this process
- So, we need to analyze this distance as deep as possible in following..

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Jupiter Uranus Distance More Analysis
(1st
Point)
- The solar system geometry always shows 3 steps of stairs – let's see that here
(1)
o Jupiter Uranus Distance = 2094 mkm
(2)
o Light (0.3 mkm/s) travels during 6939.75 s a distance = 2082 mkm
o But
o Light supposed velocity (1.16 mk/s) travels during 3600 s = 2 x 2088 mkm
(3)
o The distance 2073 mkm = 99% of 2094 mkm
- The data always gives us 3 values for specific number – the geometrical reason is
still absent but the data forces us to accept this rule – for example
o 17.4 degrees = the inner planets orbital inclinations total
o 17.2 degrees = Pluto orbital inclinational
Also
o 23.6 degrees = the outer planets orbital inclinations total
o 23.4 degrees = Earth Axial Tilt
o Also
o 28.66 degrees = (180 deg /2π)
o 28.3 degrees = Neptune Axial Tilt
Also
o The moon apogee orbital circumference =2550973 km (100%)
o Earth motion distance per a solar day =2576000 km (101%)
o The moon displacements total during 29.53 days = 2598640 km (102%)

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(2nd
Point)
- The 3 planets moves during 5040 minutes a distance = 17 mkm (100%)
- But
- Pluto orbital inclination =17.2 deg = 17.2 mkm (101%)
- The inner planets orbital inclinational total =17.4 deg = 17.4 mkm (102%)
(Mercury motion during 366556 seconds = 17.37 mkm)
- By that
- 2x 17 x 60 = 2040 mkm But
- 2x 17.2 x 60 = 2064 mkm
- 2x 17.4 x 60 = 2088 mkm
- The data tells we deal with 2 groups, each group is consisted of 3 stairs – the first
group is (17, 17.2 and 17.4) where the second group is (2040 mkm, 2064 mkm and
2088 mkm)
- Both are defined relative to each other based on the geometrical rule which creates
these 3 stairs through the solar system data
- The data analysis shows that, both groups are mentioned to one another.
Please Note
- Light (Light (0.3 mkm/s) travels during 6939.75 s a distance = 2082 mkm
- Where
o Metonic Cycle Period = 6939.75 days
o In this equation, light uses 1 day period of the moon (Planet) motion as 1
second period for light motion – that shows the general rate of time – but in
this discussion we deal with small part of this general rate of time – we deal
with a rate of time creates 1 solar day for a planet in comparison to 1 hour of
another planet.

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o The other part of the discussion should interest for the rate of time which
uses 1 hours of a planet motion as 1 second of light motion –and these 2
rates of time creates the general rate of time in the solar system.
Notice
- 2094 mkm (Jupiter Uranus Distance) = 9.688 mkm x 216.4 solar days
- 9.688 mkm = Mercury velocity per solar day 4.095 mkm + Venus velocity per
solar day 3.02 mkm + Earth velocity per solar day 2.57 mkm
- Venus orbital distance =108.2 mkm and its orbital distance =216.4 mkm
- The equation uses this value (216.4 mkm) as a period of time (216.4 days)
- Why? Because
- Light known velocity (0.3 mkm/s) travels during 216.4 seconds a distance = 720.7
mkm = Mercury Jupiter Distance
- Mercury Jupiter distance is related to Jupiter Uranus distance because
- 778.6 mkm (Jupiter orbital distance) = 1.0725 x 720.7 mkm
- The rate 1.0725 we have discussed before
- This discussion tries to show that a deep connection is found between Jupiter
Uranus distance and Mercury Jupiter Distance…
Notice
- Mercury (4.095 mkm per solar day) moves during 1433.5 days a distance = 5848
mkm (Mercury Pluto Distance)
- Where 1433.5 mkm = Saturn orbital distance used here as a period of time
showing Saturn data effect on Pluto and Mercury motions
- Mercury (4.095 mkm per solar day) moves during 84 days a distance = 346.6
mkm (notice 5040 seconds =84 minutes)
- Mercury (4.095 mkm per solar day) moves during 346.6 days (the nodal year) a
distance = 1433.5 mkm

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2-3 Uranus Day Period More Analysis
I- Data
(1)
Uranus day period 61920 seconds = (2 x 30589 seconds +742 seconds)
Where
Uranus orbital period = 30589 days
(2)
742 seconds x 112.2 km/s (Mercury, Venus & Earth velocities total) = 83252.4km
(3)
742 seconds x 4.7 km/s (Pluto Velocity) = 3475 km (The Earth moon diameter)
II- Discussion
- We try to answer the question – is this data mentioned data or depend on the rate
23.9 which is found between the 2 velocities (112.2 km and 4.7 km)?
- The data tells us that,
- The moon diameter 3475 km is defined as a distance passed by Pluto (4.7 km/s)
motion during a period 742 seconds
- During the same period 742 seconds the 3 planets velocities total (112.2 km/s)
moves a distance = 83252.4 km
- Where the moon orbital circumference = 2581346.8 km = 83252.4 km x π3
- That means,
- The 4 planets motions effect also on the moon orbital circumference creation in
addition to the moon diameter definition – I try to show some great geometrical
mechanism is found under this data.
Notice
- The moon displacements total during 29.53 days = 2598693 km But
- 2598693 km - 2581346.8 km = 17346 km
- The 3 Planets Velocities Total (112.2 km/s) moves the distance 17346 km in a
period = 153.3 seconds (where Pluto day period =153.3 hours).

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Notice
- 2550973 km (the moon apogee circumference) = 116724 km x 21.8
- Where
- 21.8 = (Jupiter Mass / Uranus Mass)
- That shows the moon apogee orbital radius is defined based on the value 116724
km by support the rate of mass between Jupiter and Uranus, the most effective 2
planets on the moon motion –
Notice
- If 1000 km = 1 day (the rate we have used before)
- 116724 km = 116.724 days (Venus day period 116.75 days)
- That shows more deep interaction between the planets data.
Where's the rate of time is shown?
- The 3 planets (Mercury, Venus and Earth) moves during 5040 minutes a distance =
2 x 17 mkm
- And
- Pluto moves during 366556 seconds a distance = 17.2 mkm (error 1%)
- The rate of time is
- 1 hour is equivalent to (1 solar day or 1 Earth rotation period)
- So, the rate of time must be effected on the value 17.2
- That means, Uranus day period 17.2 hours
- This value must be effected by the rate of time (1 hour is equivalent to 1 solar day)

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2-4 Mercury Motion Effect
I- Data
(A)
During 5201 hours
- Pluto (4.7 km/s) moves a distance = 88 mkm
- The 3 planets velocities total pass a distance = 2094 mkm
- Where
- 88 days = Mercury Orbital Period
- 2094 mkm = Jupiter Uranus Distance
(B)
- The period 5201 hours = 216.4 solar days
(C )
- Mercury (47.4 km/s) moves during 366553 seconds a distance =17.37 mkm
- But
- The 3 planets velocities total pass during 5040 minutes a distance = 17 mkm
II- Discussion
- The previous data forms the general idea as clear as possible –
- The machine has 2 gears, the (3) planets is one gear and Pluto is another gear
- Both moves during 5201 hours 2 distances (88 mkm and 2094 mkm) between both
the rate around 24, so (88) works as (88 days) (Mercury orbital period) and the
other value works as (2094 hours) (where 2094 mkm = Jupiter Uranus distance)
- The period 5201 hours = 216.4 solar days , where Venus orbital distance (108.2
mkm) is used as a period of time (108.2 days x 2 = 216.4 solar days)
- The machine works perfectly and the 2 produced values are mentioned as results
of the process
- Why mercury moves during 366556 sec a distance = the 3 planets motions during
5040 minutes? We should answer this question once - but we need to review the
rate of time (1 second of light motion = 1 hour of a planet motion).

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1 second of light motion is equivalent to 1 hour of planet motion
- Let's remember this discussion in following…
o Light supposed velocity (1.16 mkm/s) during 4222.6 seconds travels a
distance =4900 km = Jupiter Orbital Circumference
o Where 4900 km = 2π x 778.6 mkm (Jupiter Orbital Distance)
o Mercury Day Period = 4222.6 hours
o Mercury moves during its day period (4222.6 h) a distance = 720.7 mkm =
Mercury Jupiter Distance
o 778.6 mkm (Jupiter Orbital Distance) = 1.0725 x 720.7 mkm Mercury
Jupiter Distance
o The rate 1.0725 is used frequently in Jupiter distances, whatsoever its
geometrical reason – we can use it as a reason of the following data
o 778.6 mkm = 1.0725 x 720.7 mkm
o 720.7 mkm = 1.0725 x 671 mkm (Venus Jupiter Distance)
o 671 mkm = 1.0725 x 627 mkm (Earth Jupiter Distance)
o 627 mkm = (1.0725)2
x 550.7 mkm (Mars Jupiter Distance)
o The previous data creates a geometrical mechanism between Jupiter and the
5 inner planets – and can tell us these distances are created based on Jupiter
orbital distance (or Jupiter orbital circumference) which is defined by light
motion
Shortly
o Mercury moves during 4222.6 hours a distance related to light motion
distance during 4222.6 seconds and that create the different rates of time
between both – where 1 second of light motion is equivalent to 1 hour of
mercury motion.

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Notice
- The value 4222.6 controls the inner planets –
- 4222.6 hours = Mercury Day Period
- 4222.6 mkm = 2π x671 mkm (Jupiter Venus Distance)
- (4222.6 s/5040s) = (778.6 mkm/928 mkm)
o Where
o 778.6 mkm = Jupiter Orbital Distance
o 928 mkm = Earth Jupiter Distance (but both should be at 2 different
sides from the sun and so 928 mkm =149.6 mkm + 778.6 mkm)

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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

Uranus Orbital Period Analysis (II) (Revised)

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Uranus Orbital Period Analysis (II) (Revised)