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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 $1/r$.

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  • $\begingroup$ 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
    Commented Oct 3, 2017 at 19:21

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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 $\frac{\mathrm dU_\mathrm{eff}}{\mathrm dr}=0$.

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

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