Question

If $v_e$ is the escape velocity for earth when a projectile is fired from the surface of earth. Then the escape velocity if the same projectile is fired from its centre is

• A. $\sqrt{\frac{3}{2}}{v}_e$
• B. $\frac{3}{2}{v}_e$
• C. $\sqrt{\frac{2}{3}}{v}_e$
• D. $\frac{2}{3}{v}_e$

Answer 1

The correct option is A $\sqrt{\frac{3}{2}}{v}_e$

To calculate gravitational potential at the center of earth, take a spherical shell element at radius r.

$dV=-\frac{GM}{r}\frac{4\pi r^{2}dr}{\frac{4}{3}\pi R^3}=-\frac{3GMrdr}{R^3}$

Integrating this, we get $V=-\frac{3}{2}\frac{GM}{R}$

The potential at the surface of earth is $-\frac{GM}{R}$

Since escape velocity is proportional to the square root of potential energy and hence square root of gravitational potential, escape velocity at centre of earth will be $\sqrt{\frac{3}{2}}{v}_e$