Two uniform solid spheres of equal radii $R$,but mass $M$ and $4M$ have a centre-to-centre separation $6R$,as shown in the figure. The two spheres are held fixed. $A$ projectile of mass $m$ is projected from the surface of the sphere of mass $M$ directly towards the centre of the second sphere. Obtain an expression for the minimum speed $v$ of the projectile so that it reaches the surface of the second sphere.

(N/A) The projectile is acted upon by two mutually opposing gravitational forces from the two spheres. The neutral point $N$ is defined as the position where the two forces cancel each other exactly. If $ON = r$,we have:
$\frac{GMm}{r^2} = \frac{4GMm}{(6R - r)^2}$
$(6R - r)^2 = 4r^2$
$6R - r = \pm 2r$
$r = 2R$ or $r = -6R$ (ignoring the negative value as it lies outside the region between the spheres).
Thus,the neutral point is at $r = 2R$ from the centre $O$.
It is sufficient to project the particle with a speed that enables it to reach $N$. Thereafter,the greater gravitational pull of the sphere of mass $4M$ will pull it towards the second sphere.
The mechanical energy at the surface of the sphere of mass $M$ is:
$E_i = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{4GMm}{5R}$
At the neutral point $N$,the speed approaches zero. The mechanical energy at $N$ is purely potential:
$E_N = -\frac{GMm}{2R} - \frac{4GMm}{4R} = -\frac{GMm}{2R} - \frac{GMm}{R} = -\frac{3GMm}{2R}$
From the principle of conservation of mechanical energy $(E_i = E_N)$:
$\frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{4GMm}{5R} = -\frac{3GMm}{2R}$
$\frac{1}{2}v^2 = \frac{GM}{R} + \frac{4GM}{5R} - \frac{3GM}{2R} = \frac{GM}{R} \left( 1 + 0.8 - 1.5 \right) = \frac{GM}{R} (0.3) = \frac{3GM}{10R}$
$v^2 = \frac{6GM}{10R} = \frac{3GM}{5R}$
$v = \sqrt{\frac{3GM}{5R}}$