In his 1970 science fiction novel Ringworld, author Larry Niven describes the eponymous Ringworld, a gigantic structure shaped as a ring with a radius of around 1 AU, rotating around a star in the center of the ring. This system is described as physically stable; however many readers have complained that it is actually unstable and the structure will drift away in time.

My first thought reading this is that the Ringworld's center of mass is identical to the star's center of mass, so the system should be stable. Why is it unstable?

Some additional details of the structure:

  • Radius: ~1 AU
  • Mass: ~1 Solar Mass
  • Year duration: ~220 hours

The Ringworld is also made of material strong enough to withstand the stresses affecting it in such a system.


3 Answers 3


It is only for a spherically symmetric shape that you can treat an extended body as if it were a point mass at the CoM.

The Ringworld is

  • stable against axial displacements after which it will gently bob back and forth around the star.
  • unstable against transverse ones because the gravitational attraction of the near-side is greater than that of the far-side.
  • 1
    $\begingroup$ Why is the gravitational attraction of the near-side larger than that of the far-side? After all, there's more far side, isn't there? $\endgroup$
    – Oak
    Commented Oct 19, 2012 at 20:10
  • 13
    $\begingroup$ @Oak "After all, there's more far side, isn't there?" Yes, but there is more in linear proportion to the distance, while the force per unit mass drops by the square of the distance. That is also where spherical shells just balance out: there is more in quadratic proportion to the distance. $\endgroup$ Commented Oct 19, 2012 at 20:16
  • 3
    $\begingroup$ If your math is up to it, it is best just to work the integrals for a few cases, otherwise it is hard to be really convincing rather than hand-wavy convincing. $\endgroup$ Commented Oct 19, 2012 at 21:31
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    $\begingroup$ The first sentence of the answer may be a bit misleading because it implies that a spherical shell such as a Dyson sphere would behave as a point mass at its own center, which would make it stable. Actually the shell theorem says that the sphere's field vanishes on its interior, so the shell's force on the sun is zero. By Newton's third law, the sun's force on the shell is also zero, so this is not a stable equilibrium but a neutral one. $\endgroup$
    – user4552
    Commented Apr 7, 2013 at 21:02
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    $\begingroup$ I don't think it's necessary to do grotty integrals in order to get a rigorous result. The stability against axial displacements is trivial. Instability against transverse displacements then follows by applying Gauss's law to the field of the ring at its own center. $\endgroup$
    – user4552
    Commented Apr 7, 2013 at 21:24


The short version: a Dyson sphere is stable because even if the sphere gets off-center, there is no net attraction or repulsion (the extra mass of the parts further away help offset any attraction from parts closer to the star).

The ring doesn't have this advantage because, being confined to a plane (mostly), it doesn't have sufficient mass further away to counteract the runaway attraction that would happen if the ring moved off-center.

  • 5
    $\begingroup$ The situation with a solid spherical shell is sometimes described as "neutral stability"...if you give it a push it will keep going, but steadily such that to stop it only requires an equivalent push. $\endgroup$ Commented Oct 19, 2012 at 19:56
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    $\begingroup$ Actually,isn't light pressure from the star going to be a restoring force? $\endgroup$ Commented Aug 12, 2013 at 17:29
  • 3
    $\begingroup$ @JamesBowery No. For the same reason that gravity isn't a restoring force. Both fall by $1/r^2$, so both are subject to the same math. $\endgroup$ Commented Dec 18, 2014 at 23:53

Both are unstable (ring and sphere) for the same reason: the potential everywhere inside either one is zero.

This is true for gravitational as well as electrical and magnetic forces, which are all inverse square law / central force situations, and it is that pattern which causes the result.

The proof requires calculus, but is considered an elementary derivation easily done by first-year students in physics or math.

  • $\begingroup$ The potential inside a sphere is everywhere zero, but the potential inside a ring is only zero at the center. The sphere has an exponent of 2 (surface area) canceling an exponent of -2 (gravity), but the ring only has an exponent of 1 on the surface area term. $\endgroup$
    – Mark
    Commented Jun 6, 2023 at 21:25

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