Question

The gravitational force between two particles with masses $m$ and $M$, initially at rest at great separation, pulls them together. When their separation becomes $d$, then the speed of either particle relative to the other will be

• A. $\sqrt{\frac{G(m+M)}{2d}}$
• B. $\sqrt{\frac{G(m+M)}{d}}$
• C. $\sqrt{\frac{4G(m+M)}{d}}$
• D. $\sqrt{\frac{2G(m+M)}{d}}$

Answer 1

The correct option is D $\sqrt{\frac{2G(m+M)}{d}}$


Let the speed of mass $M$ and $m$ be $v$ and $u$ respectively when the distance between them reduces to $d$.

Since external forces are absent, the linear momentum of the system will be conserved.

$\therefore P_i=P_f$

$\Rightarrow 0=Mv-mu$

$\Rightarrow u=\frac{Mv}{m}....\left(1\right)$

Also, since external and non-conservative forces are absent, so the mechanical energy of the system will be conserved.

$\therefore U_i+K{E}_i=U_f+K{E}_f$

$\Rightarrow 0+0=\frac{-GMm}{d}+\frac{1}{2}M{v}^2+\frac{1}{2}m{u}^2$

$[\because$ at a large distance $U_i=0$ and initial the particles are at rest, hence $K{E}_i=0]$

from eq. $\left(1\right)$,

$\Rightarrow 0=\frac{-GMm}{d}+\frac{1}{2}M{v}^2+\frac{1}{2}m\frac{M^2}{m^2}{v}^2$

$\Rightarrow \frac{2Gm}{d}=v^2+\frac{M}{m}{v}^2=v^2\frac{(m+M)}{m}$

$\Rightarrow v=\sqrt{\frac{2G{m}^2}{(m+M)d}}=m\sqrt{\frac{2G}{(m+M)d}}$

putting the value in equation $\left(1\right)$,

$u=\frac{M}{m}\times m\sqrt{\frac{2G}{(m+M)d}}$


$\Rightarrow u=M\sqrt{\frac{2G}{(m+M)d}}$

Since, both particles are moving in opposote direction, so relative speed will be

$v-(-u)=v+u=(m+M)\sqrt{\frac{2G}{(m+M)d}}$

$\Rightarrow v+u=\sqrt{\frac{2G(m+M)}{d}}$

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

$$\begin{array}{c}\text{Why this question ?}\\ \text{Key Concept - When the external force}\\ \text{on a system is zero, linear momentum}\\ \text{of the system is conserved.}\end{array}$$