Q106 / 200 · NEET 2021 · Physics

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

Solution

$v = k V_{e} (k < 1)$

$\frac{1}{2} m v^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{R + h}$

$\frac{1}{2} m k^{2} \times \frac{2 G M}{R} - \frac{G M m}{R} = - \frac{G M m}{R + h} (∵ V_{e} = \sqrt{\frac{2 G M}{R}})$

$G M m \frac{k^{2}}{R} = - G M m (\frac{1}{R + h} - \frac{1}{R})$

$\frac{k^{2}}{R} = \frac{h}{R (R + h)} \Rightarrow 1 + \frac{R}{h} = \frac{1}{k^{2}}$

$\Rightarrow \frac{R}{h} = \frac{1}{k^{2}} - 1$

$\Rightarrow h = \frac{R K^{2}}{1 - k^{2}}$

Hence, the correct answer is option: 'C'.