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)

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 $N$,$\frac{GMm}{r^2}=\frac{G\left(4M\right)m}{{\left(6R-r\right)}^2}$
${\left(6R-r\right)}^2=4r^2$
$6R-r=\pm 2r$ 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 of $m$ 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 $N$, the 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,
$E_i=E_N$
$\frac{1}{2}m{v}^2-\frac{GMm}{R}-\frac{4GMm}{5R}$
$=-\frac{GMm}{2R}-\frac{GMm}{R}$
$v^2=\frac{2GM}{R}\left[\frac{4}{5}-\frac{1}{2}\right]$
$=\frac{3}{5}\frac{GM}{R}$ or $v={\left(\frac{3}{5}\frac{GM}{R}\right)}^{1/2}$