Question

A particle is projected with velocity $k{v}_e$ 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
$\Rightarrow \frac{1}{2}m{\left(k{v}_e\right)}^2+\left\{\frac{-GMm}{R}\right\}=0+\left\{\frac{-GMm}{R+h}\right\}$
$\Rightarrow {\mathrm{k}}^2\frac{1}{2}{\mathrm{mv}}_{\mathrm{e}}{}^2+\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right)=\left(\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}\right)$
$\Rightarrow k^2\frac{1}{2}m\left(\frac{2GM}{R}\right)-\left(\frac{GMm}{R}\right)=-\left(\frac{GMm}{R+h}\right)$
$\Rightarrow \frac{1}{R}-\frac{k^2}{R}=\frac{1}{R+h}$
$\Rightarrow \frac{1-k^2}{R}=\frac{1}{R+h}$
$\Rightarrow R+h=\frac{R}{1-k^2}$