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 $M_B=2 M_A$. 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 $\frac{v_B}{v_A}=\sqrt{\frac{10 n}{15^{1 / 3}}}$. The value of $n$ is ___

Answer 1

Given $R _{ A }= R _{ B }= R$
$M _{ B }=2 M _{ A }$
Calculation of escape velocity for $A$:
Radius of remaining star $=\frac{ R _{ A }}{2}$.
Mass of remaining star $=\rho_{ A } \frac{4}{3} \pi \frac{ R _{ A }^3}{8}=\frac{ M _{ A }}{8}$
$\frac{- GM _{ A B B }}{ R _{ A / 2}}+\frac{1}{2} mv _{ A }^2=0$
$ \Rightarrow v _{ A }=\sqrt{\frac{2 GM _{ A / B }}{ R _{ A / 2}}}=\sqrt{\frac{ GM _{ A }}{2 R }}$
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{23 GM _{ A }}{8 \times 15^{1 / 3} \frac{ R }{2}}=\frac{23 GM _{ A }}{4 \times 15^{1 / 3} R } $
$ \therefore V _{ B }=\sqrt{\frac{23 GM _{ 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$