Question

The masses and radii of the earth and moon are$M_1$, $R_1$ and $M_2$, $R_2$ respectively. Their centres are a distance d apart. The minimum speed with which a particle of mass m should be projected from a point midway between the two centres so as to escape to infinity is given by

• A. $2{\left[\frac{G(M_1+M_2)}{md}\right]}^{\frac{1}{2}}$
• B. $2{\left[\frac{G(M_1+M_2)}{d}\right]}^{\frac{1}{2}}$
• C. $2{\left[\frac{G(M_1-M_2)}{md}\right]}^{\frac{1}{2}}$
• D. $2{\left[\frac{G(M_1-M_2)}{d}\right]}^{\frac{1}{2}}$

Answer 1

The correct option is B

$2{\left[\frac{G(M_1+M_2)}{d}\right]}^{\frac{1}{2}}$

Let P be a particle of mass m situated midway between the centres of the earth and the moon as shown in the figure. The potential energy of particle P due to earth is
$U_1=-\frac{Gm{M}_1}{r}=-\frac{Gm{M}_1}{\frac{d}{2}}=-\frac{2G{M}_{1}m}{d}$
and that due to moon is $U_2=-\frac{2G{M}_{2}m}{d}$

If the particle P is projected with a velocity v, its kinetic energy is $K=\frac{1}{2}m{v}^2$. Therefore, the total initial energy of the particle is
$E_i=U_1+U_2+K=-\frac{2Gm}{d}(M_1+M_2)+\frac{1}{2}m{v}^2$
If the particle is to escape to infinity, its total energy should be greater than or equal to zero. So, the minimum velocity required is given by
$-\frac{2Gm}{d}(M_1+M_2)+\frac{1}{2}m{v}^2=0$
$\Rightarrow v=2{\left[\frac{G(M_1+M_2)}{d}\right]}^{\frac{1}{2}}$
Hence, the correct choice is (b).