Bertrand's Theorem characterizes the force laws that govern stable circular orbits. It states that the only force laws permissible are the Hooke's Potential and Inverse Square Law. The proof of the theorem involves some perturbation techniques and series expansion.

The most natural things that comes to my mind when thinking about such a problem is that the effective force should be a restoring force for circular orbits to be stable.

$f_{\mathrm{eff}}(r) = \dfrac{l^2}{\mu r^3}-f(r) = 0$, for orbit to be circular.

$f'_{\mathrm{eff}}(r)<0$, for orbit to be stable. Assuming a power law, $f=Kr^n$, for the central force, solving it gives me the solution $n>-3$.

This is very weak compared to the statement of Bertrand's Theorem. Could someone explain to me the rationale behind perturbation technique used, and what is missing from my interpretation of 'stable' in my derivation?

  • $\begingroup$ what do actually want? Want us to show you the derivation of Bertrand's Theorem? $\endgroup$
    – user36790
    Commented Jun 20, 2016 at 7:23
  • $\begingroup$ I want to know why my proof is incorrect $\endgroup$
    – Jan Lynn
    Commented Jun 20, 2016 at 7:27

1 Answer 1


What you just did was to find a condition for attractive power-law forces to have stable orbits where stable means they remain bounded when perturbed around the circular orbit. You got the correct result.

The Bertrand's Theorem though says something different: the only forces whose bounded orbits imply closed orbits are the Hooke's law and the attractive inverse square force. A closed orbit is one which the particle repeats their momentum and position after some finite time - it closes in the phase space.

The idea behind the proof of Bertrand's Theorem is to consider a perturbed orbit and then calculate the periods of the angular revolution and the radial oscillations. If these periods are commensurable then the orbit is closed. You can find the proof here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.