Question

A particle is projected with velocity $kv_{e}$ in the vertically upward direction from the ground into space ( $v_{e}$ is escape velocity and $k < 1$ ). The maximum height from the centre of the earth to which it can reach is

• A. $\frac{R}{k^{2} + 1}$
• B. $\frac{k^{2} R}{1 - k^{2}}$
• C. $\frac{R}{1 - k^{2}}$
• D. $\frac{k^{2} R}{k^{2} + 1}$

Answer 1

Answer: (C)

Total energy of surface = Total energy at $h$ height
$\Rightarrow \frac{1}{2}m\left(k v_{e}\right)^{2}+\left\{\frac{- G M m}{R}\right\}=0+\left\{\frac{- G M m}{R + h}\right\}$
$\Rightarrow \mathrm{k}^2 \frac{1}{2} \mathrm{mv}_{\mathrm{e}}{ }^2+\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right)=\left(\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\right)$
$\Rightarrow k^{2}\frac{1}{2}m\left(\frac{2 G M}{R}\right)-\left(\frac{G M m}{R}\right)=-\left(\frac{G M m}{R + h}\right)$
$\Rightarrow \frac{1}{R}-\frac{k^{2}}{R}=\frac{1}{R + h}$
$\Rightarrow \frac{1 - k^{2}}{R}=\frac{1}{R + h}$
$\Rightarrow R+h=\frac{R}{1 - k^{2}}$