Question 55: Gravitation

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 __________ .

Accepted Answer

Due to an interaction process, star A losses some of it's mass and radius becomes $${R \over 2}$$. Let new mass of star A is M'
A
. Here in both cases density of star A remains same $${ho _A}$$.
$$	herefore$$ Initially $$({ho _A}) = {{{M_A}} \over {{4 \over 3}\pi {R^3}}}$$
Finally $$({ho _A}) = {{M{'_A}} \over {{4 \over 3}\pi {{\left( {{R \over 2}} ight)}^3}}}$$
Density remains same,
$$	herefore$$ $${{{M_A}} \over {{4 \over 3}\pi {R^3}}} = {{M{'_A}} \over {{4 \over 3}\pi {{\left( {{R \over 2}} ight)}^3}}}$$
$$ \Rightarrow 8M{'_A} = {M_A}$$
$$ \Rightarrow M{'_A} = {{{M_A}} \over 8}$$
$$	herefore$$ Lost mass by $$A = {M_A} - M{'_A} = {M_A} - {{{M_A}} \over 8} = {{7{M_A}} \over 8}$$
This lost mass $${{7{M_A}} \over 8}$$ is attached on the star B and density of the attached mass stay $${ho _A}$$. So new radius of star B is $${R_2}$$.
Density of the removed part from star A is,
$${ho _A} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {{R^3} - {{\left( {{R \over 2}} ight)}^3}} ight)}}$$
Density of the added part in star B stay's same as $${ho _A}$$,
$$	herefore$$ $${ho _A} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {R_2^3 - {R^3}} ight)}}$$
$$	herefore$$ $${{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {{R^3} - {{\left( {{R \over 2}} ight)}^3}} ight)}} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {R_2^3 - {R^3}} ight)}}$$
$$ \Rightarrow {R^3} - {{{R^3}} \over 8} = R_2^3 - {R^3}$$
$$ \Rightarrow R_2^3 = 2{R^3} - {{{R^3}} \over 8}$$
$$ \Rightarrow R_2^3 = {{15{R^3}} \over 8}$$
$$ \Rightarrow {R_2} = {(15)^{{1 \over 3}}} 	imes {R \over 2}$$
Escape velocity from star A after interaction process,
$${V_A} = \sqrt {{{2G\left( {{{{M_A}} \over 8}} ight)} \over {{R \over 2}}}} $$
And escape velocity from star B after interaction process,
$${V_B} = \sqrt {{{2G\left( {{{23{M_A}} \over 8}} ight)} \over {{{(15)}^{{1 \over 3}}} 	imes {R \over 2}}}} $$
$$	herefore$$ $${{{V_B}} \over {{V_A}}} = \sqrt {{{{{23{M_A}} \over 8}} \over {{{(15)}^{{1 \over 3}}} 	imes {R \over 2}}} 	imes {{{R \over 2}} \over {{{{M_A}} \over 8}}}} $$
$$ = \sqrt {{{23} \over {{{(15)}^{{1 \over 3}}}}}} $$ ..... (1)
Given,
$${{{V_B}} \over {{V_A}}} = \sqrt {{{10n} \over {{{(15)}^{{1 \over 3}}}}}} $$ ..... (2)
Comparing equation (1) and (2), we get,
$$10n = 23$$
$$ \Rightarrow n = 2.3$$

Gravitation

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