Question

The masses of a planet and its satellite are 3m and m. Their centres are separated by a distance d. The minimum speed with which a body should be projected from the mid point of line joining their centres so that body escapes to infinity is

• A. $V_e=\sqrt{(8G\times 4m/d)}$
• B. $V_e=\sqrt{(4G\times 4m/d)}$
• C. $V_e=\sqrt{(4G\times 8m/d)}$
• D. $V_e=\sqrt{(4G\times 2m/d)}$

Answer 1

The correct option is B $V_e=\sqrt{(4G\times 4m/d)}$

Let the mass of particle is $m_1$ and escape velocity is $V_e$.

Gravitational potential energy of the particle $PE=-\frac{G3m{m}_1}{d/2}-\frac{Gm{m}_1}{d/2}=-\frac{8Gm{m}_1}{d}$,

Kinetic energy of the particle $KE=\frac{1}{2}{m}_1{V}_e^2$

$\therefore$ Total Energy $=PE+KE=-\frac{8Gm{m}_1}{d}+\frac{1}{2}{m}_1{V}_e^2$

The total energy of the particle should be zero, if it had to escape from the surface of the earth. Hence, Total Energy $=-\frac{8Gm{m}_1}{d}+\frac{1}{2}{m}_1{V}_e^2=0$ $\Rightarrow V_e=\sqrt{\frac{16Gm}{d}}=\sqrt{(4G\times 4m/d)}$

The correct option is (b)