The potential energy is
$\begin{align*}U(r) = -\int F dr = \frac{K}{2 r^2} + C\end{align*}$
We are told that the potential energy is $0$ at infinity, therefore we take the constant $C$ to be equal to $0$ .
The kinetic energy can be found by using the condition $F = m v^2/r$ which always holds for circular motion.

$\begin{align*}m v^2/r = K/r^3 \Rightarrow K = \frac{1}{2} m v^2 = \frac{K}{2 r^2}\end{align*}$
Therefore, the total energy is
$\begin{align*}T = K + U = \frac{K}{2 r^2} + \frac{K}{2 r^2} = \boxed{\frac{K}{r^2}}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1992 Problem 87