Question

Two particles of masses $m$ & $M$ are initially at rest at an infinite distance apart. They move towards each other and gain speeds due to gravitational attraction. Find their speeds when separation between the masses becomes equal to $d$.

• A. $\sqrt{\frac{G{m}^2}{d(m+M)}},\sqrt{\frac{G{m}^2}{d(m+M)}}$
• B. $\sqrt{\frac{2G{M}^2}{d(m+M)}},\sqrt{\frac{2G{m}^2}{d(m+M)}}$
• C. $\sqrt{\frac{1}{2}\frac{G{M}^2}{d(m+M)}},\sqrt{\frac{1}{2}\frac{G{m}^2}{d(m+M)}}$
• D. $2\sqrt{\frac{G{M}^2}{d(m+M)}},2\sqrt{\frac{G{m}^2}{d(m+M)}}$
  

Answer 1

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

Let $v_1$ and $v_2$ are the speeds of two masses $m$ & $M$ respectively when they are at a separation $d$.

As initially the system of masses $m$ & $M$ are at rest & at infinite distance. So their kinetic energy & potential energy are $0$.

$\therefore$ Initial energy, $E_i=0$

Final energy of the system will be

$E_f=KE+PE$

$\Rightarrow E_f=(\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2)-\frac{GMm}{d}$

Since, there is no external and non-conservative force present. So energy of the system will be conserved.

$E_i=E_f$

$\therefore 0=\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2-\frac{GMm}{d}$

$\Rightarrow \frac{GMm}{d}=\frac{1}{2}M{v}_2^2+\frac{1}{2}m{v}_1^2........\left(1\right)$

As there is no external force on this system, its total momentum remains conserved

$\therefore p_i=p_f$

$\Rightarrow 0=m{v}_1-M{v}_2$

$\Rightarrow v_1=\frac{M{v}_2}{m}$

Putting the value of $v_1$ in eq. $\left(1\right)$

$\Rightarrow \frac{GMm}{d}=\frac{1}{2}M{v}_2^2+\frac{1}{2}m{\left(\frac{M{v}_2}{m}\right)}^2$

$\Rightarrow v_2^2(\frac{m+M}{m})=\frac{2Gm}{d}$

$\Rightarrow v_2=\sqrt{\frac{2G{m}^2}{d(m+M)}}$

and

$v_1=\sqrt{\frac{2G{M}^2}{d(m+M)}}$

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