How can I calculate the eccentricity of the orbit of a planet (with mass equal to that of the Earth) around a star (with mass equal to that of the Sun) assuming that the force of gravity is given by the formula: $$ F= -\frac{GMm}{R^{5/2}} \hat{u}_r $$

Actually it would be enough for me to demonstrate that the eccentricity is between 0 and 1 to say that the planet follows a closed orbit.

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    $\begingroup$ You can't calculate eccentricity from just a law of gravitation, an orbital trajectory depends on velocity. An earth-sized rock could be put into a circular orbit around the sun, or an eccentric orbit, or it could just fall straight into the sun, depending only on its initial velocity and regardless of what power (>1) you put into the gravity formula. Are you asking if orbits would still be stable/closed with different powers in the denominator? $\endgroup$ Mar 30 at 13:35
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    $\begingroup$ Only potentials with $\phi \propto 1/r$ or $\phi \propto r^2$ allow for closed trajectories. If you can define the eccentricity of a non-closed trajectory, then maybe you'd want to ask that first. $\endgroup$ Mar 30 at 13:41
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    $\begingroup$ A circular orbit is always mathematically possible on any center-seeking attractive force whose strength is based on distance from the central body, because the circular orbit is a balanced-on-knife-point special case that doesn't actually occur in practice. $\endgroup$
    – notovny
    Mar 30 at 15:55
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    $\begingroup$ I’m voting to close this question because the question doesn't make sense. As @AtmosphericPrisonEscape noted, the only potential functions that yield elliptical orbits are inversely proportional to distance (e.g., gravity, static electricity) and proportional to the square of distance (e.g., springs). $\endgroup$ Mar 31 at 9:07
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    $\begingroup$ I'm voting to leave the question open because I don't make up custom close reasons all the time and none of the actual Stack Exchange reasons apply. The effect of closing is answer prevention and I think the best way for the OP to understand the flawed premise is for good answers to be posted. JamesK's answer is a great start, I'll try to add something soon as well. $\endgroup$
    – uhoh
    Apr 2 at 6:45

1 Answer 1


A planet following this law would either:

  1. Follow a circular orbit. In this case the value of R would be constant and so the magnitude of the central force would also be constant, only the direction would change. Any central force permits circular orbits, and circular orbits have eccentricity zero. This kind of orbit has a zero probably of occurring if the initial velocity and separation are chosen randomly

  2. Not have a closed orbit. Here is a picture of a bit of such an orbit:

enter image description here (generated by code adapted by Zack Fizell

As such an orbit isn't an ellipse, the eccentricity is not defined. Such an orbit has probably 1 of occurring if the initial conditions are random.

Your question cannot be answered. The eccentricity or more generally the shape of an orbit is not completely determined by the central force. Even with Newtonian gravity, you can't determine the eccentricity only from the law of gravitation.

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    $\begingroup$ Note, an orbit doesn’t need to be an ellipse for eccentricity to be defined, this is standard practice when discussing orbits in general relativity. $\endgroup$
    – TimRias
    Apr 1 at 22:07
  • $\begingroup$ -1 "Not have a closed orbit" is an unsupported assertion and would be okay for Quora but not Stack Exchange. The -5/2 power law central force problem does have an exact solution. This suggests that whether it is closed or not is going to be worked out and easy to find a supporting link. Just saying stuff without supporting it is not how Stack Exchange works. It is however exactly how ChatGPT works, and we've gone to great lengths to differentiate Stack Exchange answers from chatterbot answers. $\endgroup$
    – uhoh
    Apr 2 at 4:38
  • $\begingroup$ @TimRias do you mean likke arxiv.org/pdf/2206.13532.pdf "The concept of eccentricity is uniquely defined in Newtonian gravity. An extension to General Relativity is not strictly uniquely or even well defined, " $\endgroup$
    – James K
    Apr 2 at 5:59
  • $\begingroup$ I was more thinking of the limit of a test particle orbiting a black hole, where there is a canonical widely excepted definition of eccentricity. Full blown GR is more tricky, but even then you can find a well defined coordinate invariant definition of eccentricity: arxiv.org/abs/2302.11257 $\endgroup$
    – TimRias
    Apr 2 at 23:05

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