Question

A satellite is orbiting around the earth in a circular orbit. Its orbital speed is $V_0$. A rocket on board is fired from the satellite which imparts a thrust to the satellite directed radially away from the centre of the earth. The duration of the engine burn is negligible so that it can be considered instantaneous. Due to this thrust, a velocity variation $\Delta V$ is imparted to the satellite. The minimum value of the ratio $\frac{\Delta V}{V_0}$ for which the satellite will escape out of the gravitational field of the earth is

Answer 1

Orbital velocity of a satellite,
$V_0=\sqrt{\frac{GM}{r}}\to \left(1\right)$
This orbital velocity is tangential in nature.
As thrust is imparted radially, $\Delta V$ is imparted radially.
So, speed after firing of the rocket is given by,
$V=\sqrt{V_0^2+(\Delta V{)}^2}$
For satellite has to escape,
$V=V_e=\sqrt{\frac{2GM}{r}}$
$\frac{V_0^2+(\Delta V{)}^2}{2}=\frac{GM}{r}$
$V_0^2+\Delta V^2=2V_0^2$
$\Rightarrow \Delta V=V_0\Rightarrow \frac{\Delta V}{V_0}=1$