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Kepler problem with unbalanced forces (Read 2810 times)
wil
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Kepler problem with unbalanced forces
09/06/11 at 09:04:12
 
Kepler problem with unbalanced forces.
 
m - planet, M - the Sun.
 
F_m = mGM / r ^ 2 + Solor Wind = mG'M / r ^ 2;
F_M = MGm / r ^ 2
 
F_M > F_m; Force acting on the Sun is greater!
 
G '- effective gravitational constant of the Sun;
dG / G = (G-G') / G =~ 1 / 33500.
 
What will be the movement of both bodies?
Or three bodies - Jupiter, Earth and the Sun.
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Tony
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Re: Kepler problem with unbalanced forces
Reply #1 - 09/06/11 at 20:31:17
 
IIRC, light pressure is greater than solar wind.  The effect on the planets is that it's as if they orbited a slightly less-massive star.  So their periods are ever so slightly lengthened.
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wil
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Re: Kepler problem with unbalanced forces
Reply #2 - 09/07/11 at 09:57:52
 
You are talking about the case of balanced forces, but reduced:
 
F_m = mG'M/r^2;
F_M = MG'm/r^2;
F_m = F_M.
 
G == G'.
This is normal Kepler - the shape of the orbits does not depend on the symbol of G constant.
 
In the case of unbalanced forces, center of mass is not static, but revolves.
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Re: Kepler problem with unbalanced forces
Reply #3 - 09/07/11 at 11:30:43
 
I think you can see the Solar wind as an extra force acting on the planet .  
Imagine a rocket which pushes the planet in radial direction .  
The sun doesn't experience this force while the solar wind is radially balanced .  
So in my opinion the planet is slowly driven outwards
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Re: Kepler problem with unbalanced forces
Reply #4 - 09/07/11 at 13:45:06
 
Quote from frankuitaalst on 09/07/11 at 11:30:43:
So in my opinion the planet is slowly driven outwards

 
I do not agree with that.
Angular momentum is conserved (only radial forces), so the system must be stable.
 
Probably masses revolve only around another point:
M ------c---o----------------- m
 
around the 'o' instead of c - the center of mass;
 
And in the case of three bodies, the situation will be much more interesting.
 
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Re: Kepler problem with unbalanced forces
Reply #5 - 09/07/11 at 14:54:45
 
Tony 's answer gives a very good approximation of what happens .  
You obatin the same answer if you change the solar mass in the formulas you provide .  
Murray and Dermott "Solar Systems Dynamics " , section 2.9 , further describe the orbital perturbation which results  in a da/dt and de/dt , which corresponds to what Tony mentionned .
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Re: Kepler problem with unbalanced forces
Reply #6 - 09/07/11 at 16:43:08
 
Of curse, very good solution, but on one condition: we fix the Sun into space.  Grin
 
Force is lower, so you can reduce the mass of Sun, but only the gravitational mass; inertial mass must be preserved!
 
Besides, in the case of two bodies, this problem can be easily solved analytically.
 
Jupiter - Sun's barycenter: 740000 km.
The baycenter revolves around circle of radius: 740000 / 33500 = 22 km;
 
Earth: 13.56m
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« Last Edit: 09/07/11 at 18:33:33 by wil »  
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