Question

A body is thrown with a velocity equal to $n$ times the escape velocity $\left(v_e\right)$. Velocity of the body at a large distance away will be

• A. $v_e\sqrt{n^2-1}$
• B. $v_e\sqrt{n^2+1}$
• C. $v_e\sqrt{1-n^2}$
• D. $noneofthese$

Answer 1

The correct option is A $v_e\sqrt{n^2-1}$

Kinetic energy at a larger distance = PE + KE at the surface of the earth

$\frac{1}{2}m{v}^2=-\frac{GMm}{R}+\frac{1}{2}m(n{v}_e{)}^2$

But, the escape velocity,

$v_e^2=\frac{2GM}{R}$

So,$\frac{GMm}{R^2}=\frac{1}{2}m{v}_e^2$

Hence,

$\frac{1}{2}m{v}^2=-\frac{1}{2}m{v}_e^2+\frac{1}{2}m{n}^2{v}_e^2$

$v^2=-v_e^2+n^2{v}_e^2$

$v=\sqrt{(n^2-1)}{v}_e$