Question

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

• 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)

$\sqrt{\frac{3}{5}\frac{GM}{R}}$

Le the projectile of mass m is fired with minimum velocity, v from the surface of the sphere of mass M to reach the surface of the sphere of mass 4M. Let N be a neutral point at a distance r from the center of the sphere of mass M. At neutral point N, $\frac{GM}{r^2}=\frac{G\left(4M\right)m}{{(6R-r)}^2}$

${(6R-r)}^2=4r^2$

$6R-r=\pm 2r$ 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 that would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice.

The mechanical energy at the surface of M is $E_i=\frac{1}{2}m{v}^2-\frac{GMm}{R}-\frac{G\left(4M\right)m}{5R}$

At the neutral point N, speed approaches zero.

$\therefore$ The mechanical energy at N is $E_N=-\frac{GMm}{2R}\frac{G\left(4M\right)m}{4R}=-\frac{GMm}{2R}-\frac{GMm}{R}$

According to law of conservation of mechanical energy, $E_i-E_N$

$\frac{1}{2}m{v}^2-\frac{GMm}{R}\frac{G\left(4M\right)m}{5R}=-\frac{GMm}{2R}-\frac{GMm}{R}$

$v^2=\frac{2GM}{R}[\frac{4}{5}-\frac{1}{2}]=\frac{3}{5}\frac{GM}{R}$ or $v={\left(\frac{3}{5}\frac{GM}{R}\right)}^{1/2}$