Hi
Check this.
David
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_*A simple and approximate method to determine the gravitational
torque between two astronomical bodies.*_
_Introduction_
Bodies in space rotate slower the farther away from each other they are.
Look at
Sun Mercury Venus
almost still fast not that fast
Also look at
Sun planet's left side planet's right side
almost still wants to move fast wants to move not that fast
The left side of the planet want to move faster than the right side.
The reason is the same as in the case with several planets. If the
left side experiences a higher speed than the right we get an angular
momentum or a torque on the planet. The torque contributes to the
planets rotation around its own axis. The planets' rotation around
their axises is a mystery since this motion causes a lot of viscous
heat production in the planet slowing it down. The tidal force on the
planet would also cause it to slow down. Despite this the planets
continue to rotate. Lets see if the gravitational torque can explain
why.
_Calculation of the torque_
One way is to view how the position of the sun changes as gravity is
experienced in the front versus the back of the planet. Assume that
gravity moves at the speed of light, as many say. (There are evidence
against this showing it moves instantaneously. This is currently
ignored.)
Say the Sun is in horizontal position when viewed from the central of
Earth. This means the Sun will be a little above the horizon when
viewed from the equator (say it is midsummer's day for simplicity). On
midnight on the other side of earth the gravity of the Sun will be
experienced an equal distance below the horizon.
This little difference in experiencing the Sun will cause a little
tangential force on earth.
Earth at t1
Sun Earth at t0
Earth at t-1
The torque on earth is symmetrical in a plane through its axis. Lets
calculate the effect from that plane to the further end and multiply
by two.
Some values
R = distance between Sun and Earth
r = distance from Earths center
ft = tangential force on Earth
fs = Suns force on Earth
d = displacement of Earth seen from non rotating frame on earth.
vs = Earths speed relative the Sun
te = time elapsed since gravity passed Earths center = r/c
re = Earths radius
ft is proportional to d/R so
ft=fs*d/R (1)
The displacement d is determined by Earths speed and the time in takes
gravity to pass Earth so
d=vs*te=vs/c*r (2)
(2) in (1) makes
ft=fs*vs*te/R=fs*vs/c*r/R
Suns force fs on a an element dm of Earth mass
df=G Ms dm / R²
and the tangential component
dft=G Ms dm / R² *vs/c*r/R = vs/c r G Ms dm / R³
Earths mass per unit length
dm/dr = rho pi (re²-r²)
The torque is
T= r f
dT = r dft = r vs/c r G Ms dm / R³ = r² vs/c G Ms rho pi (re² - r²) dr / R³
T=2 vs/c G Ms rho pi / R³ integral (r=0 to re) r² (re² - r²) dr =
T=2 vs/c G Ms rho pi / R³ (r=0 to re of) (re² r³/3 - r^5/5) =
T=4 vs/c G Ms rho pi / R³ re^5/15
This formula is general for any two bodies. The power is P = T*omega
where omega is the angular velocity of Earth around its axis. Lets
calculate it to see if it corresponds with the observed value of
Earths heat production 500 W/m² equal to 2.5*10^17 watts.
Omega for Earth in radians per seconds is 2 pi / 86400
For earth P = T omega =
4*10^8/365/86400/3/10^8*6.7*10^-11*2*10^30*5515*3.14/(1.5*10^11)^3*(6.3*10^6)^5/15*2*3.14/86400
= 1.4*10^12 watt = 1 400 gigawatt
Far too little by a factor 10 000. Maybe I did something wrong? Is
gravitational torque really this small? Please correct me in that
case.
David
