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(A) Let $m$ be the mass of the object and $v$ be its speed when it returns to the surface of the planet.At the maximum height,the distance from the ce

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$2 \sqrt{\frac{2 G M}{5 R}}$

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(A) Let $m$ be the mass of the object and $v$ be its speed when it returns to the surface of the planet.At the maximum height,the distance from the center of the planet is $r_{max} = R + 4R = 5R$. At this point,the velocity of the object is $0$.Using the principle of conservation of mechanical energy between the maximum height and the surface of the planet:$E_{initial} = E_{final}$$U_{max} + K_{max} = U_{surface} + K_{surface}$$-\frac{G M m}{5R} + 0 = -\frac{G M m}{R} + \frac{1}{2} m v^2$$\frac{1}{2} v^2 = \frac{G M}{R} - \frac{G M}{5R}$$\frac{1}{2} v^2 = \frac{G M}{R} (1 - \frac{1}{5}) = \frac{G M}{R} (\frac{4}{5})$$v^2 = \frac{8 G M}{5 R}$$v = \sqrt{\frac{8 G M}{5 R}} = 2 \sqrt{\frac{2 G M}{5 R}}$Thus,the speed of the object when it returns to the surface is $2 \sqrt{\frac{2 G M}{5 R}}$.

An object is propelled vertically to a maximum height of $4 R$ from the surface of a planet of radius $R$ and mass $M$. The speed of the object when it returns to the surface of the planet is

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