In my universe, gravity varies with the inverse of the distance (instead of the inverse square of the distance). I already know that this means that all circular orbits around the same object will have the same speed, and no true escape trajectories can exist, but there are some facts that I can't figure out about how the universe will work.

What shape will non-circular orbits take (or do they not exist)? Also, is there anything else weird about gravity in this universe?

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    $\begingroup$ Possible duplicate of worldbuilding.stackexchange.com/q/37285/627. There won't be any stable closed orbits. $\endgroup$
    – HDE 226868
    Commented Dec 14, 2016 at 21:40
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    $\begingroup$ I'd say this is not an exact duplicate since that one only talks about orbits and it omits "special cases" which I care about. $\endgroup$ Commented Dec 14, 2016 at 22:48
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    $\begingroup$ Also note that such a law would not obey the conservation of energy. $\endgroup$
    – Cort Ammon
    Commented Dec 15, 2016 at 3:38
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    $\begingroup$ How would it not obey conservation of energy? It just changes the formula for gravitational potential energy. $\endgroup$ Commented Dec 15, 2016 at 3:48
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    $\begingroup$ @CortAmmon It would. The vectorfield 1/r has no curl. $\endgroup$
    – Feyre
    Commented Dec 15, 2016 at 12:25

1 Answer 1


In order to generate an elliptical orbit, you need to have a force which is equal to the required centripetal force:

$$F=m\frac{v^2}{r}\rightarrow a=\frac{v^2}{r}$$

According to Bertrand's Theorem, this can only be solved with a potential for an inverse square force, or a radial harmonic oscillator potential.

So we cannot attain a circular orbit, is that a problem? No.

I generated a system for our sun, Earth, and moon, dependent on a linear inverse force. What we find is that we need to rescale the Gravitational constant to the negative 22nd order. (For clarity's sake I avoided using astronomical units).

So if we set $G = 6.6740831\times10^{-22}$ we find the following orbit patterns:

enter image description here enter image description here

We can further decrease the orbital eccentricity when $G \rightarrow 4\times10^{-22}$

Note however, that in the long term, the eccentricity will always increase, even for optimal $G$, take the following radial Sol-Earth distance over 500y:

enter image description here

There are more problems though, for instance, would a star even form with this Gravity configuration?

Note that in this configuration, the acceleration of gravity due to Earth on its surface would be $0.000375m/s^2$ instead of $9.8m/s^2$ As the gravity drops off more slowly, but is also significantly more massive, a habitable planet would be much more massive, but such massive planets might also more easily form under these parameters.

And here is where things get really interesting, if we suppose that our planet has a mass of $m_{earth}=5.97237\times10^{28}$, four orders higher than that of the current Earth, gravity at the same radius would be $3.75m/s^2$, and we get the following 1000 year progression:

enter image description here

My suspicion is that the collapse happens 4 orders of magnitude slower, meaning you would have at least $10^5y$ of stable orbit, possible a million (1Ma).

If you could have a planet with a mass of order $O\left(29\right)$, then you might get a near-stable orbit over evolutionary time scales, however getting such a large concentration of Earth (oxyen, quartz, aluminium, lime, iron, magnesium) might be difficult to attain, except maybe in a late-stage galaxy.

I do think the peculiar circumstances would make the formation of large planets more likely as distance is less of a factor for matter to come together. Consequently we would expect fewer planets, but of higher average mass. However, it is also possible this situation would lead to more uniformity in mass distributions. You would have to run some galaxy wide gravity calculations for that one, and recalculate the result of the background radiation. These are things beyond my scope.

  • $\begingroup$ What program did you use to generate your graphics? Can you link to code, on github maybe? $\endgroup$
    – kingledion
    Commented Dec 15, 2016 at 14:39
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    $\begingroup$ @kingledion I used my own Mathematica code. I recognize it's handier to have this in Python for sharing, but I adapted this from code I still had for a regular n-body solution. I'm no where near as good with Python yet as I am with Mathematica. $\endgroup$
    – Feyre
    Commented Dec 15, 2016 at 16:19
  • $\begingroup$ Well, good work. I was working on this in Python last night and wanted to plagiarize. :) $\endgroup$
    – kingledion
    Commented Dec 15, 2016 at 18:03
  • $\begingroup$ @kingledion I can say is I used NDSolve[], which would be roughly equivalen to odeint() for Python, except it's much more straight forward on Mathematica. If you do figure this out on Python I'd be interested to know if you can come to the same solutions. $\endgroup$
    – Feyre
    Commented Dec 15, 2016 at 18:22
  • $\begingroup$ On your second graph, is that a stable point around 1.5e10, or is that an artifact? $\endgroup$
    – Cort Ammon
    Commented Dec 15, 2016 at 18:58

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