Rotation of Earth, Sun and
Planets
If a small body of mass (m) orbits a large body of mass (M) then, may be we can calculate the sidereal rotation period of the large body (M) by the rotation equation
The rotation equation can calculate the rotation period for a Star of mass (M) which circled by another mass (m).
The nine Planets orbit our sun so, we can calculate the rotation period of our sun.
From the rotation Equation we can calculate the rotation period of Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto because all these planets are circled by another moons.
The Earth rotation period is going to be faster as a result of increasing Earth- Moon distance.
The Earth rotation period increases when the moon orbit inclination plane is perpendicular to the Earth equator
The Earth rotation period decreases when the moon orbit inclination plane reaches to it's max. inclination
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| Rotation equation (1) |
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| Rotation equation (2) |
=The angular velocity of the large body ;
=The density of the large body;
=The density of the small body
T=Sidereal rotation period of the large body (sec)
K=Moment of inertia factor of the large body.
G=gravitational constant = 6.67300 ? 10-11 m3 kg-1 s-2.
M=Large body mass (kg)
Re= Equatorial radius for the large body (meter)
a=Semi major axis of the mass center of all satellites that orbit large body (meter)
m=Total mass of all satellites that orbit the large body (kg)
re =Equatorial radius of all satellites that orbit large body (meter)
rp =Polar radius of all satellites that orbit large body (meter)
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| The observed and calculated values from equation (2) |
Larg body
The observed period
The calculated period
Earth
23.9
24
Mars
24.6
24.4
Jupiter
9.25
9.95
Saturn
10.6
9.4
Uranus
17.2
16.9
Neptune
16.1
16.9
Pluto
153.3
116.2
Sun
609.1
609.1 2
3-Determination of Earth rotation period
M=5.9736E+24 (kg)
Re= 6.3781E+6 (meter)
The Earth has one satellite (the Moon)
a=3.844E+8 (meter)
m=7.35E+22 (kg)
re =1738100 (meter)
rp =1736000 (meter)
From equation 2 (T) of Earth is 24.007
The Earth rotation period is going to be faster as a result of increasing Earth- Moon distance.
From equa.2, the Earth is fastening it's rotation period by -0.43 milliseconds/century (3.82 cm/year)
M=6.4185E+23 (kg)
Re= 3397000 (meter)
There are two satellites orbit Mars
m= Total mass of all satellites that orbit the large body Mars (kg)
= m1 (Phobos)+m2 (Deimos) =1.3E+16
a=semi major axis of the center of mass of all satellites that orbit Mars (meter)
=(m1 a1 +m2 a2 )/(m1 +m2 )=1.1977E+7 (meter)
re= Equatorial radius of the total satellites (meter)
re=[(r1)^3+(r2)^3]^0.3333=[(13400)^3+(7500)^3)]^0.3333=14128 (meter)
rp= Polar radius of the total satellites (meter)
rp=[(r1)^3+(r2)^3]^0.3333=[(9200)^3+(5200)^3)]^0.3333=9714 (meter)
From equation 2 (T) of Mars is 24.36
Mars rotation period is going to be slower as a result of decreasing the distance of it's moons' mass center.
From the rotation equation, Mars is slowing it's rotation period by 0.67 second/century (1.8 meter/year)
By the same way, we can calculate the rotation period (T) from equation 2
All moons of small masses are negligible according to the equation results.
There is no effect for the small moons on the rotation period
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| MJD from 10/Jul/2006 to 23/Nov./2006 (lunar declination is 29) |
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| MJD from 10/Jul/2006 to 23/Nov./2006 (lunar declination is 29) |
UTC is the time given by broadcast time since 1972 = Coordinated Universal Time.
UT1=Universal time. Defined by the Earth's rotation, and determined by astronomical observations.
(UT1 - UTC) is a standard Earth Orientation product.
((UT1-UTC)today)-((UT1-UTC)yesterday)=The difference between any two successive days
From UT1-UtC, the Earth rotation period increases when the moon orbit inclination plane is perpendicular to the Earth equator
From UT1-UTC, the Earth rotation period decreases when the moon orbit inclination plane reaches to it's max. inclination
The Low points in Fig.(9) represents the highest speed of the earth rotation. (close to the earth equator)
The Low points in Fig.(10) represents the Min. Lunar declination.
The Peak points in Fig.(11) represents the Max.earth radius in the lunar orbital plane.(the equator radius)
The Earth rotation rate reaches to it's Max. when lunar declination is zero. (the highest Earth diameter)
There is a perfect correlation in the three curves.
From Fig.(9), (10) and (11), the Earth speed increased when the lunar declination decreased.
The Earth diameter is in the moon inclination plane
Thanks to Eng. Franz J. Heeke; (Germany) for his useful internet web.
Thanks to Dr. Jerald Lee ; (USA) for his useful internet discussion and advice.
If a small body of mass (m) orbits a large body of mass (M) then, may be we can calculate the sidereal rotation period of the large body (M).
The moons are "driving" the rotation of their parent planet in the same way, as sun's rotation is being controlled by the planets.
The Earth has one satellite (the Moon)
The Earth rotation period is going to be faster as a result of increasing Earth- Moon distance.)
From the rotation equation, the Earth is fastening it's rotation period by -0.43 milliseconds/century.(if the moon orbits the Earth in a circular pass)
If a small body of mass (m) orbits a large body of mass (M) then, the rotation period of (M) increased by decreasing the moment of inertia factor (K) for (M).
The Moment of inertia factor of Haumea may be expected to be 0.37 from the rotation equation.
A perfect correlation between lunar declination and the changes of leap seconds.(UT1-UTC)-(UT1-UTC)
From UT1-UTC, the Earth rotation period increases when the moon orbit inclination plane is perpendicular to the Earth equator
From UT1-UTC, the Earth rotation period decreases when the moon orbit inclination plane reaches to it's max. inclination