An earth satellite is revolving in a circular orbit of radius ‘a’ with velocity ${\upsilon }_0.$ A gun is in the satellite and is aimed directly towards the earth. A bullet is fired from the gun with muzzle velocity $\frac{v_0}{2}.$ Neglecting resistance offered by cosmic dust and recoil of gun, calculate maximum distance of bullet from the centre of earth during its subsequent motion.

Orbital speed of satellite is $v_0=\sqrt{\frac{GM}{a}}\mathrm{.................}\left(i\right)$

From conservation of angular momentum at P and Q, we have $ma{v}_0=mvr$ or $v=\frac{a{v}_0}{r}\mathrm{...............}\left(ii\right)$

From conservation off mechanical energy at P and Q, we have

$\frac{1}{2}m(v_0^2+\frac{v_0^2}{4})-\frac{GMm}{a}=\frac{1}{2}m{v}^2-\frac{GMm}{r}$ or $\frac{5}{8}{v}_0^2-\frac{GM}{a}=\frac{v^2}{2}-\frac{GM}{r}$ 

Substituting values of $v$ and $v_0$ from equation(i) and (ii), we get 

$\frac{5}{8}\frac{GM}{a}-\frac{GM}{a}=\frac{a^2}{r^2}\left(\frac{GM}{2a}\right)-\frac{GM}{r}$or $-\frac{3}{8a}=\frac{a}{2r^2}-\frac{1}{r}$ or $-3r^2=4a^2-8ar$

or $3r^2-8ar+4a^2=0$ or $r=\frac{8a\pm \sqrt{64a^2-48a^2}}{6}$ or $r=\frac{8a\pm 4a}{6}$ or $r=2aand\frac{2a}{3}$

Hence, the maximum and minimum distance are 2a and $\frac{2a}{3}$ respectively.