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

Answer: (B)

$\sqrt{\frac{ 2GM_e }{ R_e}}$

Answer 2

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, $v_e = \sqrt { \frac{ 2GM_e m }{ R_e}}$.