Solution to 1992 Problem 87

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.

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