Question

Escape velocity for a projectile at the earth's surface is $v_{e} .$ A body is projected from the earth's surface with velocity $2\, v_{e}$. The velocity of the body when it is at infinite distance from the centre of the earth is $\sqrt{x} v_{e}$. Find $x$.

Answer 1

Applying conservation of energy,
$-\frac{G M_{e} m}{R_{e}}+\frac{1}{2} m\left(2 v_{e}\right)^{2}=\frac{1}{2} m v^{'2} $
$\Rightarrow \frac{-G M_{e} m}{R_{e}}+\frac{1}{2} m\left(2 \sqrt{\frac{2 G M_{e}}{R_{e}}}\right)^{2}=\frac{1}{2} m v^{'2} $
$\Rightarrow \frac{3 G M_{e} m}{R_{e}}=\frac{1}{2} m v^{'2} $
$v^{'2}=\frac{6 G M_{e}}{R_{e}}=3 \times \frac{2 G M_{e}}{R_{e}} $
$v'=\sqrt{3} \sqrt{\frac{2 G M_{e}}{R_{e}}}=\sqrt{3} v_{e}$