Question

A particle is thrown vertically upwards from the surface of earth and it reaches to a maximum height equal to the radius of earth. The ratio of the velocity of projection to the escape velocity on the surface of earth is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

Change in kinetic energy $=$ change in potential energy
$\therefore \frac{1}{2}m{v}^2=-\frac{GMm}{R+R}-(-\frac{GMm}{R})=\frac{GMm}{2R}$

$v=\sqrt{\frac{GM}{R}}=\frac{\sqrt{\left(\frac{2GM}{R}\right)}}{\sqrt{2}}=\frac{v_e}{\sqrt{2}}$ (where $v_e=$ escape velocity)