Physics - Gravitation Question with Solution | TestHub

Given here: $R_A=R_B=R$ and $M_B=2M_A$.

Now, after interaction process, radius of remaining star $A$ is ${R_A}^{\text{'}}=\frac{R}{2}$ and its mass is ${M_A}^{\text{'}}={\rho }_A\frac{4}{3}\pi {\left(\frac{R_A}{2}\right)}^3=\frac{M_A}{8}$

Applying conservation of energy,

 $\frac{-G{M_A}^{\text{'}}m}{{R_A}^{\text{'}}}+\frac{1}{2}m{v}_A^2=0$

Escape velocity of star $A$ is $v_A=\sqrt{\frac{2G{M}_A}{8\times \left({\displaystyle \frac{R}{2}}\right)}}=\frac{v_0}{2}$

Now, for $B,$ mass collected over $B$ is ${M_B}^{\text{'}}=M_A-\frac{M_A}{8}=\frac{7}{8}{M}_A$.

Let the radius of star $B$ after interaction becomes $r$.

Applying mass conservation, 

$\frac{4}{3}\pi \left(r^3-R^3\right){\rho }_A=\frac{4}{3}\pi R^3\times \frac{7}{8}{\rho }_A$

$\Rightarrow r={\left(\frac{15}{8}\right)}^{\frac{1}{3}}R$

Escape velocity of star $B$ is 

$\therefore v_B=\sqrt{\frac{2G\times \left(2M_A+{\displaystyle \frac{7}{8}}{M}_A\right)}{{\left(15\right)}^{1/3}\left({\displaystyle \frac{R}{2}}\right)}}$

$=\sqrt{\frac{2G{M}_A}{R}}\sqrt{\frac{2\left({\displaystyle \frac{16+7}{8}}\right)}{{\left(15\right)}^{{\displaystyle \frac{1}{3}}}}}$

$=v_0\times \sqrt{\frac{23\times 2}{8\times {\left(15\right)}^{\frac{1}{3}}}}=\frac{v_0}{2}\times \sqrt{\frac{23}{{\left(15\right)}^{\frac{1}{3}}}}$

Now, the ratio $\frac{v_B}{v_A}=\sqrt{\frac{23}{{\left(15\right)}^{\frac{1}{3}}}}=\sqrt{\frac{2.30\times 10}{{\left(15\right)}^{\frac{1}{3}}}}$

$\therefore n=2.30$