The masses and radii of the earth and moon are $M_1,R_1$ and $M_2,R_2$ respectively. Their centers 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 centers so as to escape to infinity is given by 

Let P be a particle of mass m situated midway between the centers of the earth and the moon $(Fig. 6.29)$. The potential energy of particle P due to earth is

$-\frac{Gm{M}_1}{r}=-\frac{Gm{M}_1}{{\displaystyle \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\frac{2G{M}_{1}m}{d}$

and that due to moon$=-\frac{2G{M}_{2}m}{d}$

$\therefore$Total potential energy$=-\frac{2Gm}{d}(M_1+M_2)$

If the particle P is projected with a velocity v, its kinetic energy$=\frac{1}{2}m{v}^2$.

Therefore, the total initial energy of the particle is 

$E_i=-\frac{2Gm}{d}(M_1+M_2)+\frac{1}{2}m{v}^2$

If the particle is to escape to infinity, its final potential and kinetic energy will be zero. Thus the total energy $E_f= 0$. From the principle of conservation of energy,

$$\begin{gathered} E_i=E_f \\ or -\frac{2Gm}{d}(M_1+M_2)+\frac{1}{2}m{v}^2=0 \end{gathered}$$

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

Hence the correct choice is (b). Notice that v is independent of the mass m of the particle. This is the minimum value of the velocity for the particle to escape to infinity.