Escape velocity ratio of stars after mass transfer

To find the value of $n$, we analyze the parameters of stars $A$ and $B$ before and after the interaction process.

1. Initial Parameters of Stars $A$ and $B$

Let $R$ be the initial radius of both spherical stars $A$ and $B$.

• The initial mass of star $A$ is $M_A$, given by: $M_A = \frac{4}{3} \pi R^3 \rho_A$
• The initial mass of star $B$ is $M_B = 2M_A$.

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2. Parameters of Star $A$ After Mass Transfer

After the interaction:

• The radius of star $A$ becomes $R_A' = \frac{R}{2}$.
• The density remains $\rho_A$.

The new mass of star $A$, $M_A'$, is: $M_A' = \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 \rho_A = \frac{1}{8} \left(\frac{4}{3} \pi R^3 \rho_A\right) = \frac{M_A}{8}$

The mass lost by star $A$ is: $\Delta M = M_A - M_A' = M_A - \frac{M_A}{8} = \frac{7}{8} M_A$

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3. Parameters of Star $B$ After Mass Transfer

The mass lost by $A$, $\Delta M$, is deposited as a thick spherical shell on $B$ with density $\rho_A$.

• Inner radius of the shell = $R$
• Let $R_B'$ be the outer radius of star $B$ with the shell.

The volume of the shell is: $V_{\text{shell}} = \frac{4}{3} \pi \left((R_B')^3 - R^3\right)$

Since the mass of the shell is $\Delta M$: $\frac{4}{3} \pi \left((R_B')^3 - R^3\right) \rho_A = \frac{7}{8} M_A = \frac{7}{8} \left(\frac{4}{3} \pi R^3 \rho_A\right)$

Canceling common terms: $(R_B')^3 - R^3 = \frac{7}{8} R^3$ $(R_B')^3 = \frac{15}{8} R^3 \implies R_B' = \frac{15^{1/3}}{2} R$

The total mass of star $B$ after the interaction is: $M_B' = M_B + \Delta M = 2M_A + \frac{7}{8} M_A = \frac{23}{8} M_A$

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4. Ratio of Escape Velocities

The escape velocity from a spherical body of mass $M$ and radius $r$ is given by: $v = \sqrt{\frac{2GM}{r}}$

• Escape velocity from $A$ ($v_A$): $v_A = \sqrt{\frac{2 G M_A'}{R_A'}} = \sqrt{\frac{2 G \left(\frac{M_A}{8}\right)}{\frac{R}{2}}} = \sqrt{\frac{G M_A}{2 R}}$

• Escape velocity from $B$ ($v_B$): $v_B = \sqrt{\frac{2 G M_B'}{R_B'}} = \sqrt{\frac{2 G \left(\frac{23}{8} M_A\right)}{\frac{15^{1/3}}{2} R}} = \sqrt{\frac{23 G M_A}{2 \cdot 15^{1/3} R}}$

• Ratio $\frac{v_B}{v_A}$: $\frac{v_B}{v_A} = \frac{\sqrt{\frac{23 G M_A}{2 \cdot 15^{1/3} R}}}{\sqrt{\frac{G M_A}{2 R}}} = \sqrt{\frac{23}{15^{1/3}}}$

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5. Determination of $n$

Comparing the derived ratio with the given expression: $\sqrt{\frac{23}{15^{1/3}}} = \sqrt{\frac{10n}{15^{1/3}}}$

Equating the numerators inside the square root: $10n = 23 \implies n = 2.3$