Question

An earth satellite is revolving in a circular orbit of radius $a$ with a velocity $v_0$. A gun in the satellite is directly aimed toward earth. A bullet is fired from the gun with muzzle velocity $\frac{v_0}{2}$. Find the ratio of distance of farthest and closest approach of bullet from centre of earth. (Assume that mass of the satellite is very-very large with respect to the mass of the bullet.)

Answer 1

When the bullet is farthest or closest to the center of earth, the velocity of the bullet is not in radial direction to the center of the earth. Let that velocity be 'v'.

Since the satellite is orbiting the earth,
$\frac{m{v}_0^2}{a}=\frac{GMm}{a^2}$

$⟹v_0^2=\frac{GM}{a}$........ Relation 1

Now from conservation of angular momentum of bullet,

$m{v}_{0}a=mvr$

$⟹v=\frac{a}{r}{v}_0$..........Relation 2

From conservation of energy of the bullet,

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

Using relation 1 and relation 2, we get

$3r^2-8ar+4a^2=0$

$⟹r=\frac{2a}{3},2a$

Here one is the farthest distance and other is the shortest.

Ratio $=3$