Question

Two particles $\text{A}$ and $\text{B}$ have masses $m$ and $2m$ respectively. They are held at separation $r_0$ in space. $\text{A}$ is given a velocity $v_0$ along the line joining the two masses (away from $\text{B}$) and $\text{B}$ is released simultaneously. Find the range of velocity $v_0$ for which the two particles would remain bound under their mutual gravitation.

• A. $v_0>\sqrt{\frac{6Gm}{r_0}}$
  
• B. $v_0<\sqrt{\frac{6Gm}{r_0}}$
  
• C. $v_0<\sqrt{\frac{2Gm}{r_0}}$
  
• D. $v_0<\sqrt{\frac{12Gm}{r_0}}$
  

Answer 1

The correct option is B $v_0<\sqrt{\frac{6Gm}{r_0}}$


Since, there is no net external force involved, the linear momentum of the system remains conserved.


Due to gravitational attraction, $\text{A}$ will slow down and $\text{B}$ will speed up.

At a point the two particles will have same speed, after which they will start getting closer. Conserving the linear momentum of the system

$\Rightarrow P_i=P_f$

$\Rightarrow m{v}_o+2m\times 0=(m+2m)v$

$\Rightarrow v=\frac{v_0}{3}$

where, $v$ is the common speed attained.

Applying energy conservation principle $U_i+K_i=U_f+K_f$

$\Rightarrow \frac{-Gm\left(2m\right)}{r_0}+\frac{1}{2}mv{{}_0}^2=\frac{1}{2}\left(3m\right){\left(\frac{v_o}{3}\right)}^2-\frac{Gm\left(2m\right)}{r}$

$\Rightarrow \frac{Gm\left(2m\right)}{r}=\frac{2G{m}^2}{r_0}-\frac{1}{3}mv{{}_0}^2$

For $r<\infty$ [bounded system]

$\Rightarrow \frac{2G{m}^2}{r_0}>\frac{1}{3}mv{{}_0}^2$

$\Rightarrow v_0<\sqrt{\frac{6Gm}{r_0}}$

Hence, option (b) is correct answer.