Question

There is a crater of depth $\frac{ R }{100}$ on the surface of the moon (radius $R$). A projectile is fired vertically upward from the crater with velocity, which is equal to the escape velocity ' $v$ ' from the surface of moon. Find the maximum height attained by the projectile (in terms of $R$ ).

Answer 1

Speed of particle at $A$,
$v _{ A }= v =\sqrt{\frac{2 GM }{ R }}$
At point $B , v _{ B }=0$
By conservation of energy,
decrease in $KE =$ increase in GPE
$\Rightarrow \frac{1}{2} Mv _{ A }^{2}= U _{ B }- U _{ A }$
$\therefore \frac{1}{2} M \cdot\left(\frac{2 GM }{ R }\right)= M \left( V _{ B }- V _{ A }\right)$
$\therefore \frac{G M}{R}=-\frac{G M}{R +h}-\left[\frac{-G M}{2 R}\left[3-\frac{\left(R-\frac{R}{100}\right)^{2}}{R^{2}}\right]\right]$
$\therefore \frac{1}{R}=-\frac{1}{R +h}+\frac{3}{2 R}-\left(\frac{1}{2}\right)\left(\frac{99}{100}\right)^{2} \cdot \frac{1}{R}$
$\therefore \frac{1}{ R }=-\frac{1}{ R + h }+\frac{3}{2 R }-\left(\frac{1}{2}\right)\left(\frac{0.98}{ R }\right)$
$\therefore \frac{1}{ R + h }=\frac{1}{2 R }[1-0.98]$
$\therefore 2 R =0.02( R + h )$
$\therefore h =\frac{1.98}{0.02}$
$\therefore h =99\, R$