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

Answer: 99

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