Solution to 1992 Problem 5

Again let m be the mass of the point mass. Then, for r \leq R,
\begin{align*}F(r) =  \frac{G m M_E \frac{4 \pi r^3}{4 \pi R^3}}{r^2} = \frac{ G m M_E r}{R^3}\end{align*}
\begin{align*}\frac{F(R)}{F(R/2)}= \frac{\frac{ G M_E R}{R^3}}{\frac{ G M_E R/2}{R^3}} = \boxed{2}\end{align*}
Therefore, answer (C) is correct.

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