Two identical spherically symmetric planets, each of mass M, are somehow held at rest with respect to each other. Each planet has radius R, and the distance between the centers of the planets is 4R. If a rocket is launched from the surface of one planet with speed v, what is the minimum speed v so that the rocket can reach the other planet?

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$\sqrt{\frac{2 G M}{3 R}}$

$\sqrt{\frac{2 G M}{R}}$

$\sqrt{\frac{3 G M}{4 R}}$

$\sqrt{\frac{G M}{R}}$

answer is D.

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Detailed Solution

The gravitational force vanishes at the midway point between
the planets, so the rocket only needs to
have enough energy to get there. The initial and final gravitational potential energies are
$U_{i} = - \frac{G M m}{R} - \frac{G M m}{3 R} = - \frac{4 G M m}{3 R}$ an $U_{f} = - \frac{2 G M m}{2 R} = - \frac{G M m}{R}$

Thus, the initial kinetic energy needed is
$\frac{1}{2} m v^{2} = U_{f} - U_{i} = \frac{G M m}{3 R}$
which implies $v = \sqrt{\frac{2 G M}{3 R}}$

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