Question

Two uniform solid spheres of masses $M$ and $4M$ but of equal radii $R$ , have a centre to centre separation $6R$ , as depicted in the figure. A projectile of mass $m$ is projected from the surface of the sphere of mass $M$ directly towards the center of the second sphere. What should be the minimum speed of projection so that it reaches the surface of the second sphere?

• A. $\sqrt{\frac{4}{5}\frac{GM}{R}}$
• B. $\sqrt{\frac{5}{4}\frac{GM}{R}}$
• C. $\sqrt{\frac{3}{5}\frac{GM}{R}}$
• D. $\sqrt{\frac{5}{3}\frac{GM}{R}}$

Answer 1

Answer: (C)

Let the projectile of mass $m$ be fired with minimum velocity, $v$ from the surface of sphere of mass $M$ to reach the surface of sphere of mass $4M$ . Let $N$ be neutral point at a distance $r$ from the centre of the sphere of mass $M$ .
At neutral point $N$ ,
$\frac{GMm}{r^2}=\frac{G(4M)m}{(6R-r{)}^2}$
$(6R-r{)}^2=4r^2$
$6R-r=\pm 2r$ or $r=2R$ or $-6R$
The point $r=-6R$ does not concern us.
Thus, $ON=r=2R.$
It is sufficient to project the projectile with a speed which would enable it to reach $N$ . Thereafter, the greater gravitational pull of $4M$ would suffice.
The mechanical energy at the surface of $M$ is
$E_i=\frac{1}{2}m{v}^2-\frac{GMm}{R}-\frac{G(4M)m}{5R}$
At the neutral point $N$ , the speed approaches zero.
The mechanical energy at $N$ is
$E_N=-\frac{GMm}{2R}-\frac{G(4M)m}{4R}=-\frac{GMm}{2R}-\frac{GMm}{R}$
According to law of conservation of mechanical energy,
$E_i=E_N$
$\frac{1}{2}m{v}^2-\frac{GMm}{R}-\frac{4GMm}{5R}=-\frac{GMm}{2R}-\frac{GMm}{R}$