If gravity, for instance, dropped off with the cube instead of the square of distance from the Sun, would the planets still follow elliptical paths?
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5Check out Bertrand's theorem. Related: http://physics.stackexchange.com/q/137768/2451 , http://physics.stackexchange.com/q/191950/2451 and links therein. – Qmechanic Jan 04 '16 at 16:53
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I don't think so, we can try solving the orbit differential equation, taking $V=-\alpha/(r^3) $, but I don't think it can be easily solved. – Courage Jan 04 '16 at 16:54
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2@TheGhostOfPerdition You probably mean $V = -a/r^2$, as the OP seems to be talking about the force law... – dmckee --- ex-moderator kitten Jan 04 '16 at 21:51
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No you cannot get a stable orbit. With animations, it's very interesting what you get with a combination of 1/r² + 1/r³. You get an elliptical orbit that precesses like you would expect with relativity (ie. the ellipse rotates around the central object). You can control the speed of precession by adjusting the amount of the cubed term that you use. I am not sure why this works (I have my guesses), but its a great trick if you want to do a quick animation of an elliptical orbit that is precessing. Using only 1/r² produces an ellipse that never precesses no matter the starting positions and momentum of a 2-body orbit.
AnimatedPhysics
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The short answer is no, if you require that the orbits be stable and closed.
Whereas any central potential can produce "circular" orbits given the correct initial conditions, it has been shown (by Bertrand as mentioned in the comments) that there are only two potentials that can possibly produce stable and perfectly closed orbits. The first is the gravitational potential and the second is a harmonic oscillator (which is probably not what you were thinking).
dannuic
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