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Message started by wil on 09/06/11 at 09:04:12

Title: Kepler problem with unbalanced forces
Post by wil on 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.

Title: Re: Kepler problem with unbalanced forces
Post by Tony on 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.

Title: Re: Kepler problem with unbalanced forces
Post by wil on 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.

Title: Re: Kepler problem with unbalanced forces
Post by frankuitaalst on 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

Title: Re: Kepler problem with unbalanced forces
Post by wil on 09/07/11 at 13:45:06


frankuitaalst wrote:
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.


Title: Re: Kepler problem with unbalanced forces
Post by frankuitaalst on 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 .

Title: Re: Kepler problem with unbalanced forces
Post by wil on 09/07/11 at 16:43:08

Of curse, very good solution, but on one condition: we fix the Sun into space.  ;D

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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