Can we apply Earnshaw theorem in circular/elliptical/hyperbolic orbits? I was wondering how did the moon stop at its current stationary orbit. It seems to me the fragments that gave origin to the moon didn't have the minimum kinetic energy to escape Earth's gravitational field and they had a velocity component that got conserved and another one that decreased as moon's got distance. Still, this result seems to kind of contradict Earnshaw's theorem as this orbit should not be stable. Any thoughts?


1 Answer 1


No, Earnshaw's theorem applies to statics. It doesn't apply to a dynamical equilibrium.

  • $\begingroup$ are you saying the circular orbit is stable? $\endgroup$
    – mranon
    Oct 20, 2019 at 14:39
  • $\begingroup$ Any orbit of two gravitational objects is stable. Earnshaw's theorem is about multiple fixed charges where only one of the charges is free to move. The problem you describe is dynamic: both objects can move. $\endgroup$ Oct 20, 2019 at 19:16
  • $\begingroup$ I disagree, in order for an orbit to be stable the gravitational aceleration must be contrary to the displacment from equilibrium position. A body can lose its orbit forever if it moves one centimeter closer to the attracting body. The moon should fall back on Earth after some time $\endgroup$
    – mranon
    Oct 21, 2019 at 13:19

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