Question

Two uniform solid spheres held fixed of equal radii $R$, but mass $M$ and $4M$ have centre to centre separation $6R$. 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.

• A. $(\frac{3GM}{5R}{)}^{1/2}$
• B. $(\frac{5GM}{3R}{)}^{1/2}$
• C. $(\frac{2GM}{5R}{)}^{1/2}$
• D. $(\frac{5GM}{2R}{)}^{1/2}$

Answer 1

The correct option is A $(\frac{3GM}{5R}{)}^{1/2}$


Let $\text{N}$ be the null point between sphere, which is at distance $x$ from the centre of $M$.

At null point, field due to mass $M$ will be equal to mass $4M$,

$\therefore E_M=E_{4M}$

$\Rightarrow \frac{GM}{x^2}=\frac{G\times 4M}{(6R-x{)}^2}$

$\Rightarrow \frac{1}{x}=\frac{2}{6R-x}$

$\Rightarrow x=2R$ (Neglecting $-$ve value)

If $m$ just crosses null point $\text{N}$ with zero velocity, it will be pulled towards $4M$ due to its gravitational attraction.

Thus, the minimum speed will be such that $m$ just reaches $\text{N}$ with zero speed.

Since external and non-conservative force are absent in this scenario, mechanical energy of the system $(M+4M+m)$ will be conserved.

Now, conserving mechanical energy between the point when $m$ is at the surface of $M$ and when it is at the null point.

$U_i+K_i=U_f+K_f$

$\Rightarrow -\frac{G\left(M\right)\left(m\right)}{R}-\frac{G\left(4M\right)\left(m\right)}{5R}+\frac{1}{2}m{v}^2=-\frac{G\left(M\right)\left(m\right)}{2R}-\frac{G\left(4M\right)\left(m\right)}{4R}+0$

$\Rightarrow -\frac{9GMm}{5R}+\frac{1}{2}m{v}^2=-\frac{3GMm}{2R}$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{9}{5}\frac{GMm}{R}-\frac{3GMm}{2R}=\frac{3}{10}\frac{GMm}{R}$

$\therefore v=\sqrt{\frac{3GM}{5R}}$

Hence, option $\left(a\right)$ is correct.

Why this question ? Key Concept - The particle just has to cross the null point in order to reach the other particle. This is a frequently asked concept.