= Solution
The radial <force> is $F_r=-dU/dr=-3Ar^2$. A nontrivial <circular orbit> therefore requires $A>0$, so that the force is inward. Balancing its magnitude against the required <centripetal acceleration> gives
$$
\frac{mv^2}{r}=3Ar^2,\qquad K=\frac12mv^2=\frac32Ar^3.
$$
Thus
$$
\boxed{\frac{K}{U}=\frac32.}
$$
This is the <circular-orbit energy ratio in a power-law potential>, $2K=rU'(r)=sU$ for $U=Ar^s$, with $s=3$. It uses the zero of <potential energy> specified by $U=Ar^3$; adding a constant would change the ratio without changing the motion. For $A<0$, the force is outward and no such free circular orbit exists.
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