Planetary motion: with a different nature of potential. Condition for circular orbit

Question

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 $1/r$.

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:

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