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(D) By the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $r$ where velocity is zero.$-\

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$\frac{R k^2}{1-k^2}$

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(D) By the law of conservation of energy, the total energy at the surface equals the total energy at the maximum height $r$ where velocity is zero.$-\frac{GMm}{R} + \frac{1}{2} m v^2 = -\frac{GMm}{r}$Given $v = k V_{e}$ and $V_{e} = \sqrt{\frac{2GM}{R}}$, we have $v^2 = k^2 \frac{2GM}{R}$.Substituting this into the energy equation:$-\frac{GMm}{R} + \frac{1}{2} m \left( k^2 \frac{2GM}{R} \right) = -\frac{GMm}{r}$Dividing by $GMm$:$-\frac{1}{R} + \frac{k^2}{R} = -\frac{1}{r}$$\frac{1}{r} = \frac{1}{R} - \frac{k^2}{R} = \frac{1-k^2}{R}$$r = \frac{R}{1-k^2}$Since $r = R + h$, the maximum height $h$ is:$h = r - R = \frac{R}{1-k^2} - R = R \left( \frac{1 - (1-k^2)}{1-k^2} \right) = \frac{R k^2}{1-k^2}$.

$A$ particle of mass $m$ is projected with a velocity $v = k V_{e}$ $(k < 1)$ from the surface of the earth. $(V_{e} = \text{escape velocity})$. The maximum height above the surface reached by the particle is:

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