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 $2v_e$. 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

Answer: 3

Applying conservation of energy,
$-\frac{G{M}_{e}m}{R_e}+\frac{1}{2}m{\left(2v_e\right)}^2=\frac{1}{2}m{v}^{{}^{\prime }2}$
$\Rightarrow \frac{-G{M}_{e}m}{R_e}+\frac{1}{2}m{\left(2\sqrt{\frac{2G{M}_e}{R_e}}\right)}^2=\frac{1}{2}m{v}^{{}^{\prime }2}$
$\Rightarrow \frac{3G{M}_{e}m}{R_e}=\frac{1}{2}m{v}^{{}^{\prime }2}$
$v^{{}^{\prime }2}=\frac{6G{M}_e}{R_e}=3\times \frac{2G{M}_e}{R_e}$
$v^{\prime }=\sqrt{3}\sqrt{\frac{2G{M}_e}{R_e}}=\sqrt{3}{v}_e$