Question

The escape velocity of a sphere of mass m is given by (G = Universal gravitational constant; M = Mass of the earth and R = Radius of the e earth)

• A. $\sqrt{\frac{ 2GM_e m }{ R_e}}$
• B. $\sqrt{\frac{ 2GM_e }{ R_e}}$
• C. $\sqrt{\frac{ GM_e m }{ R_e}}$
• D. $\sqrt{\frac{ 2GM_e +R_e }{ R_e}}$

Answer 1

The Correct Option is B

The gravitational potential energy of a body of mass m placed on earth's surface is given by
$ U = - \frac{GM_e m }{ R_e}$
Threfore, in order to take a body from the earth's surface to infinity, the work required is $ \frac{ GM_e m }{ R_e} $.
Hence it is evident that if we throw a body of mass m with such a velocity that its kinetic energy is
$ \frac{ GM_e m }{ R_e} $, then it will move outside the gravitational field of earth.
Hence, $ \frac{ 1}{2} m {v_e}^2 = \frac{ GM_e m }{ R_e} $ or, $v_e = \sqrt { \frac{ 2GM_e m }{ R_e}}$.