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 Kepler problem with unbalanced forces (Read 3457 times)
 wil Uploader I Love YaBB 2! Posts: 41 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. Back to top IP Logged
 Tony YaBB Administrator Posts: 1058 Gender: 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. Back to top IP Logged
 wil Uploader I Love YaBB 2! Posts: 41 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. Back to top IP Logged
 frankuitaalst Ultimate Member Great site Posts: 1508 Gender: 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 Back to top IP Logged
 wil Uploader I Love YaBB 2! Posts: 41 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. Back to top IP Logged
 frankuitaalst Ultimate Member Great site Posts: 1508 Gender: 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 . Back to top IP Logged
 wil Uploader I Love YaBB 2! Posts: 41 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.     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 Back to top « Last Edit: 09/07/11 at 18:33:33 by wil »     IP Logged
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