This is an idea for a space station on the surface of Eris (but could work on other celestial bodies), using rotating rings to simulate gravity. The idea is that these rings would sit horizontally on the surface of the dwarf planet much like you would expect in space. Instead of the floor of the rings being the inner flat edge, the rings would be slanted forming a sort of downward cone shape. You could imagine it much like how a freeway or racetrack is banked around turns.

Ideally the rotation would produce centrifugal force pulling the occupants towards the edge to simulate gravity, however, the slant of the rings would counteract the existing low gravity of Eris which is roughly 1/12 of Earth's. The closest real world comparison I could think of would be the Gravitron amusement park ride, but on a massive scale.

I imagine that first I will have to determine both the size and speed of the ring(s), but how would I calculate the slant angle? Is there a formula that I could use to plug in the measurements of the station and then calculate that value?

Please let me know if there are other values I might need for this as well. My primary goal is for station occupants to feel a force of 1/2 G pressing downward with none or minimal force pulling them towards either edge of the ring. Is this even feasible? Assume structural integrity and necessary energy are sufficient and that the ring is well balanced (or at least that some system maintains balance).

  • $\begingroup$ Let's clarify that your rings are small compared the size of Eris? $\endgroup$
    – Alexander
    Dec 14, 2018 at 0:10
  • $\begingroup$ Yes, Eris has a diameter of a bit over 2300 km. I don't imagine these rings being larger than 10 km in diameter (which makes it about 32 km around) and probably only 100 m or so tall. Those are maximum measurements for a single, large ring though. I am also considering several much smaller rings as well. Specific sizes yet to be determined. $\endgroup$ Dec 14, 2018 at 0:21

1 Answer 1


Let Eris's gravity be $g$, and the angular speed of the ring be $\omega$ and its radius be $R$.

The below assumes the ring is small compared to the size of Eris, i.e. real gravity is essentially constant direction and magnitude across the ring.

To have an effective gravity of $G$, the angular speed needs to be $\omega = \frac{(G^2 - g^2)^{1/4}}{\sqrt{R}}$ (for the given values this turns out to be $\sqrt{\frac{2.2m}{R}}$ radians per second, or $\sqrt{\frac{200.5 m}{R}}$ rpm), with an angle (between the "floor" of the ring and horizontal) of $\theta = arctan\left(\frac{\sqrt{G^2 - g^2}}{g}\right)$. For the given values this comes out to an angle of 80 degrees (or 10 degrees from vertical).

Also note that if the ring has multiple levels at significantly different radii $r$, the angle needs to be adjusted to be $arctan\left(\frac{\sqrt{G^2 - g^2}}{g}\frac{r}{R}\right)$.

In general, if $G >> g$, $\sqrt{G^2 - g^2}$ is about $G >> g$. Then (small-angle and pi ~ 3 approximations), the angle will be about $\frac{g}{G}*60$ degrees from vertical.

  • $\begingroup$ Am I assuming correctly that if my ideal force is 1/2 G instead of G I can simply plug that in it's place? $\endgroup$ Dec 14, 2018 at 0:30
  • $\begingroup$ + for real math. But I can't do that in my head and get a rough idea of the speed. It would be great to have a second section walking thru with numbers plugged in. As regards R I suspect Roche limit applies even for something as small as Eris - but maybe not? $\endgroup$
    – Willk
    Dec 15, 2018 at 15:34
  • $\begingroup$ @Willk How would Roche limit be applicable? The station isn't a satellite, it is a ground based habitat, that happens rotate to simulate gravity like space-based stations. $\endgroup$ Dec 15, 2018 at 16:33
  • $\begingroup$ Oh! So it is on a track. It is like a train girdling the asteroid. I assumed it was in orbit. Cool! $\endgroup$
    – Willk
    Dec 15, 2018 at 21:11
  • $\begingroup$ @TitaniumTurtle Here, G is 1/2 of Earth gravity when I plugged the numbers in. $\endgroup$
    – majestas32
    Dec 16, 2018 at 22:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .