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 of 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 ____.

Step 1: Determine the mass and radius of star $A$ after the interaction.

The radius of star $A$ is halved while its density $\rho_A$ remains constant, meaning its volume is reduced by a factor of $8$.

$$M_A' = \rho_A \cdot \frac{4}{3}\pi\left(\frac{R}{2}\right)^3 = \frac{M_A}{8}$$

Step 2: Calculate the mass lost by star $A$.

Subtract the final mass of star $A$ from its initial mass.

$$\Delta M = M_A - M_A' = \frac{7}{8}M_A$$

Step 3: Determine the new radius of star $B$.

The lost mass is deposited as a shell of density $\rho_A$ on star $B$, which initially has radius $R$.

$$V_{\text{shell}} = \frac{\Delta M}{\rho_A} = \frac{7}{8}\cdot\frac{4}{3}\pi R^3$$ $$V_B' = V_B + V_{\text{shell}} = \frac{4}{3}\pi R^3 + \frac{7}{8}\cdot\frac{4}{3}\pi R^3 = \frac{15}{8}\cdot\frac{4}{3}\pi R^3$$ $$R_B' = R\left(\frac{15}{8}\right)^{1/3}$$

Step 4: Determine the new mass of star $B$.

Add the deposited mass to the initial mass of star $B$, where $M_B = 2M_A$.

$$M_B' = M_B + \Delta M = 2M_A + \frac{7}{8}M_A = \frac{23}{8}M_A$$

Step 5: Write the expressions for the escape velocities.

Use the formula for escape velocity $v = \sqrt{\frac{2GM}{R}}$.

$$v_A = \sqrt{\frac{2G\left(\frac{M_A}{8}\right)}{\frac{R}{2}}} = \sqrt{\frac{G M_A}{2R}}$$ $$v_B = \sqrt{\frac{2G\left(\frac{23}{8}M_A\right)}{R\left(\frac{15}{8}\right)^{1/3}}}$$

Step 6: Calculate the ratio of the escape velocities.

Divide $v_B$ by $v_A$ and simplify the expression.

$$\left(\frac{v_B}{v_A}\right)^2 = \frac{\frac{23}{4} \cdot \frac{G M_A}{R \left(\frac{15}{8}\right)^{1/3}}}{\frac{G M_A}{2R}}$$ $$\left(\frac{v_B}{v_A}\right)^2 = \frac{23}{8} \cdot 4 \cdot \left(\frac{8}{15}\right)^{1/3}$$ $$\left(\frac{v_B}{v_A}\right)^2 = \frac{23}{15^{1/3}}$$ $$\frac{v_B}{v_A} = \sqrt{\frac{23}{15^{1/3}}}$$

Step 7: Solve for $n$.

Equate the calculated ratio to the given expression.

$$\sqrt{\frac{10n}{15^{1/3}}} = \sqrt{\frac{23}{15^{1/3}}}$$ $$10n = 23$$ $$n = 2.3$$

Answer: $n = 2.3$