Question

Two spherical stars $A$ and $B$ have densities ${\rho }_A$ and ${\rho }_B$, respectively. $A$ and $B$ have the same radius, and their masses $M_A$ and $M_B$ are related by $M_B=2M_A$. Due to an interaction process, star $A$ loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains ${\rho }_A$. The entire mass lost by $A$ is deposited as a thick spherical shell on $B$ with the density of the shell being ${\rho }_A$. If $v_A$ and $v_B$ are the escape velocities from $A$ and $B$ after the interaction process, the ratio $\frac{v_B}{v_A}=\sqrt{\frac{10n}{15^{1/3}}}$. The value of $n$ is ___

Answer 1

Answer: 2.30

Given $R_A=R_B=R$
$M_B=2M_A$
Calculation of escape velocity for $A$:
Radius of remaining star $=\frac{R_A}{2}$.
Mass of remaining star $={\rho }_A\frac{4}{3}\pi \frac{R_A^3}{8}=\frac{M_A}{8}$
$\frac{-G{M}_{ABB}}{R_{A/2}}+\frac{1}{2}m{v}_A^2=0$
$\Rightarrow v_A=\sqrt{\frac{2G{M}_{A/B}}{R_{A/2}}}=\sqrt{\frac{G{M}_A}{2R}}$
Calculation of escape velocity for $B$
Mass collected over $B=\frac{7}{8}{M}_A$
Let the radius of B becomes $r$.
$\therefore \frac{4}{3}\pi \left(r^3-R_B^3\right){\rho }_A=\frac{7}{8}{\rho }_A\frac{4}{3}\pi R_A^3$
$\Rightarrow {\pi }^3=\frac{7}{8}{R}_A^3+R_B^3=\frac{(15{)}^{1/3}R}{2}$
$\therefore \frac{V_B^2}{2}=\frac{23G{M}_A}{8\times 15^{1/3}\frac{R}{2}}=\frac{23G{M}_A}{4\times 15^{1/3}R}$
$\therefore V_B=\sqrt{\frac{23G{M}_A}{2\times 15^{1/3}R}}$
$\therefore \frac{V_B}{V_A}=\sqrt{\frac{23}{15^{1/3}}}=\sqrt{\frac{10\times 2.30}{15^{1/3}}}$
$n=2.30$