Two spherical stars A and B have densities rh | vedclass

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(A) Initial state: $R_A = R_B = R$,$M_A = M$,$M_B = 2M = 2M_A$. Density $\rho_A = \frac{M_A}{\frac{4}{3}\pi R^3}$.After interaction,star $A$ has radiu

Vedclass · Accepted Answer

$2.30$

Vedclass · Answer

(A) Initial state: $R_A = R_B = R$,$M_A = M$,$M_B = 2M = 2M_A$. Density $ho_A = \frac{M_A}{\frac{4}{3}\pi R^3}$.After interaction,star $A$ has radius $R_A' = R/2$. Since density $ho_A$ is constant,the new mass $M_A' = ho_A \cdot \frac{4}{3}\pi (R/2)^3 = M_A/8$.The escape velocity $v_A = \sqrt{\frac{2GM_A'}{R_A'}} = \sqrt{\frac{2G(M_A/8)}{R/2}} = \sqrt{\frac{GM_A}{2R}}$.Mass lost by $A$ is $\Delta M = M_A - M_A/8 = \frac{7}{8}M_A$. This mass is added to $B$ as a shell of density $ho_A$. Let the new radius of $B$ be $R_B'$.Volume of shell = $\frac{4}{3}\pi(R_B'^3 - R^3) = \frac{\Delta M}{ho_A} = \frac{7/8 M_A}{ho_A} = \frac{7}{8} \cdot \frac{4}{3}\pi R^3$.$R_B'^3 - R^3 = \frac{7}{8}R^3 \Rightarrow R_B'^3 = \frac{15}{8}R^3 \Rightarrow R_B' = R \left(\frac{15}{8}ight)^{1/3}$.The new mass of $B$ is $M_B' = M_B + \Delta M = 2M_A + \frac{7}{8}M_A = \frac{23}{8}M_A$.The escape velocity $v_B = \sqrt{\frac{2GM_B'}{R_B'}} = \sqrt{\frac{2G(23/8)M_A}{R(15/8)^{1/3}}} = \sqrt{\frac{23GM_A}{4R(15/8)^{1/3}}} = \sqrt{\frac{23GM_A}{2R(15^{1/3})}}$.Ratio $\frac{v_B}{v_A} = \sqrt{\frac{23GM_A}{2R(15^{1/3})} \cdot \frac{2R}{GM_A}} = \sqrt{\frac{23}{15^{1/3}}} = \sqrt{\frac{10 	imes 2.3}{15^{1/3}}}$.Comparing with $\sqrt{\frac{10n}{15^{1/3}}}$,we get $n = 2.30$.

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