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Q.

A ball of mass m is fired vertically upward from the surface of the earth with velocity ${nv}_{e}$ where v_e is the escape velocity and n < 1. Neglecting air resistance, to what height will the ball rise ? Take radius of the earth as R

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a

$R / n^{2}$

b

$R / (1 - n^{2})$

c

${Rn}^{2} / (1 - n^{2})$

d

${Rn}^{2}$

answer is C.

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Detailed Solution

According to this problem
$\begin{array}{l} \frac{1}{2} {mv}^{2} = \frac{GMm}{R} - \frac{GMm}{(R + h)} \\ = \frac{GMmh}{R (R + h)} = \frac{mgh}{(R + h)} (∵ g = \frac{GM}{R}) \end{array}$
Further $\frac{1}{2} {mv}^{2} = \frac{1}{2} {mn}^{2} v_{e}^{2}$
$= \frac{1}{2} {mn}^{2} (2 gR) (∵ v_{e} = \sqrt{2 gR})$
$∴ \frac{1}{2} {mn}^{2} (2 gR) = \frac{mgh}{(R + h)}$
Solving, we get $h = {Rn}^{2} / (1 - n^{2})$

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