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(B) The gravitational force $F$ between two masses $m$ and $(M-m)$ separated by a distance $r$ is given by $F = \frac{G m (M - m)}{r^2}$.To maximize t

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$\frac{1}{2}$

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(B) The gravitational force $F$ between two masses $m$ and $(M-m)$ separated by a distance $r$ is given by $F = \frac{G m (M - m)}{r^2}$.To maximize the force,we differentiate $F$ with respect to $m$ and set it to zero: $\frac{dF}{dm} = 0$.$\frac{d}{dm} \left( \frac{G M m - G m^2}{r^2} \right) = 0$.Since $G$ and $r^2$ are constants,we have $\frac{d}{dm} (M m - m^2) = 0$.$M - 2m = 0$.$M = 2m$.Therefore,the ratio $\frac{m}{M} = \frac{1}{2}$.

$A$ mass $M$ is split into two parts,$m$ and $(M-m)$,which are then separated by a certain distance. What ratio of $\frac{m}{M}$ maximizes the gravitational force between the two parts?

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