Question

F = $-\frac{\partial u}{\partial r}=\frac{K}{r^3}$
Since it is performing circular motion
F = $\frac{m{v}^2}{r}=\frac{K}{r^3}$
mv² = $\frac{K}{r^2}$
$\Rightarrow K.E. =\frac{1}{2}m{v}^2=\frac{K}{2r^2}$
Total energy = P.E. + K.E.
$=-\frac{K}{2r^2}+\frac{K}{2r^2}$ = Zero