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 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 ______.

For a uniform solid sphere of mass $$M$$ and radius $$R$$ the escape speed from its surface is
$$v = \sqrt{\dfrac{2GM}{R}} \; \; -(1)$$

Initial data
Both stars have the same initial radius $$R$$.
Masses are related by $$M_B = 2M_A$$.
Since mass $$M = \dfrac{4}{3}\pi R^{3}\rho$$ for a sphere of density $$\rho$$, we have
$$\dfrac{\rho_B}{\rho_A} = \dfrac{M_B}{M_A} = 2 \;\;\; \Rightarrow \;\;\; \rho_B = 2\rho_A$$.

1. Final parameters of star A
Its radius is halved: $$R_A' = \dfrac{R}{2}$$.
Its density is still $$\rho_A$$, hence its final mass is
$$M_A' = \dfrac{4}{3}\pi \left(\dfrac{R}{2}\right)^3 \rho_A = \dfrac{1}{8}\left(\dfrac{4}{3}\pi R^{3}\rho_A\right)=\dfrac{M_A}{8}$$.
Mass lost by A
$$\Delta M = M_A - M_A' = M_A\left(1-\dfrac{1}{8}\right)=\dfrac{7M_A}{8}$$.

2. Final parameters of star B
The whole mass $$\Delta M$$ forms a spherical shell of density $$\rho_A$$ around B.
Let the outer radius after coating be $$R_B'$$.
Mass of the shell:
$$\dfrac{4}{3}\pi\bigl(R_B'^{3}-R^{3}\bigr)\rho_A = \dfrac{7M_A}{8} = \dfrac{7}{8}\left(\dfrac{4}{3}\pi R^{3}\rho_A\right).$$
Cancelling the common factor $$\dfrac{4}{3}\pi\rho_A$$ gives
$$R_B'^{3}-R^{3}= \dfrac{7}{8}R^{3}\;\;\;\Rightarrow\;\;\;R_B'^{3} =\left(1+\dfrac{7}{8}\right)R^{3}= \dfrac{15}{8}R^{3}.$$

Therefore
$$R_B' = R\left(\dfrac{15}{8}\right)^{1/3} \;\;\; -(2)$$
Final mass of B:
$$M_B' = M_B + \Delta M = 2M_A + \dfrac{7M_A}{8}= \dfrac{23M_A}{8} \;\;\; -(3)$$

3. Escape speeds after interaction
From (1):
Escape speed from A:
$$v_A = \sqrt{\dfrac{2G M_A'}{R_A'}} = \sqrt{\dfrac{2G\left(\dfrac{M_A}{8}\right)}{R/2}} = \sqrt{\dfrac{G M_A}{2R}} \;\;\; -(4)$$
Escape speed from B:
$$v_B = \sqrt{\dfrac{2G M_B'}{R_B'}} = \sqrt{\dfrac{2G\left(\dfrac{23M_A}{8}\right)} {R\left( \dfrac{15}{8}\right)^{1/3}}} \;\;\; -(5)$$

4. Required ratio
From (4) and (5):
$$\left(\dfrac{v_B}{v_A}\right)^2 = \dfrac{\,\dfrac{2G(23M_A/8)}{R(15/8)^{1/3}}\,}{\,\dfrac{G M_A}{2R}\,} = \left(\dfrac{2\cdot23}{8}\right)\cdot2\cdot\left(\dfrac{8}{15}\right)^{1/3} = \dfrac{23}{2}\left(\dfrac{8}{15}\right)^{1/3}.$$

Note that $$\left(\dfrac{8}{15}\right)^{1/3} = \dfrac{2^{3/3}}{15^{1/3}}= \dfrac{2}{15^{1/3}}.$$ Hence
$$\left(\dfrac{v_B}{v_A}\right)^2 = \dfrac{23}{2}\cdot\dfrac{2}{15^{1/3}} = \dfrac{23}{15^{1/3}}.$$

Taking the square root,
$$\dfrac{v_B}{v_A}= \dfrac{\sqrt{23}}{15^{1/6}} = \dfrac{\sqrt{10\,n}}{15^{1/6}},$$ which matches the given form $$\sqrt{\dfrac{10n}{15^{1/3}}}$$ because $$1/15^{1/6} = \sqrt{1/15^{1/3}}.$$

Equating the numerators:
$$\sqrt{10\,n} = \sqrt{23}\;\;\;\Rightarrow\;\;\;10n = 23 \;\;\;\Rightarrow\;\;\;n = 2.3.$$

The required value is 2.30.