Consider a particle moving in the potential $U (r)= -A/r^n$, where $A>0$. What are the values of $n$ which admit stable circular orbits?

I tried to solve by putting $dr/dt=0$ in the total energy equation $E= T + U_\mathrm{eff}$, but it didn't work. Then I came across a solution which said that for the orbit to be circular, $U_\mathrm{eff}(r)$ needs to have a minimum when plotted against $r$, where $U_\mathrm{eff}$ is the effective potential $(L^2/2mr^2+ U (r))$. But I don't understand why it has to, because when $n=1$, where circular orbits are possible, $U_\mathrm{eff}$ does not have a minimum since it varies with $1/r$.

  • $\begingroup$ It is tempting to mention Bertrand's theorem: One may show that if a slightly deformed circular orbit should remain closed, then the central force has to be the inverse square law or Hooke's law. $\endgroup$
    – Qmechanic
    Oct 3, 2017 at 19:21

1 Answer 1


I'm not sure where you got this idea:

when $n=1$, where circular orbits are possible, $U_\mathrm{eff}=L^2/2mr^2+ U (r)$ does not have a minimum since it varies with $1/r$.

Here, have a look at that function:

Mathematica graphics

At small $r$, the $+1/r^2$ dominates and the function is positive and monotonously decreasing. At large $r$, the $-1/r$ term dominates and the function is negative but monotonously increasing. The only way to reconcile those two behaviours is to have a minimum in the middle, which can easily be found by setting $\frac{\mathrm dU_\mathrm{eff}}{\mathrm dr}=0$.

  • $\begingroup$ I'm sorry about that. But is there any explanation behind why $Ueff $ should have a minima for circular orbits? $\endgroup$ Oct 3, 2017 at 13:34
  • $\begingroup$ @PrashantGovind Have you ever seen a particle sit still, stably, in anything other than a minimum? $\endgroup$ Oct 3, 2017 at 13:37
  • $\begingroup$ Also, some formatting notes: 'minima' is plural, 'minimum' is singular. And you typeset $U_\mathrm{eff}$ as $U_\mathrm{eff}$. $\endgroup$ Oct 3, 2017 at 13:38

Your Answer

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

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