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 \(\frac{v_B}{v_A}=\sqrt{\frac{10 n}{15^{\frac 13}}}\). The value of $n$ is ___ .

Answer 1

Correct Answer: 2.3

Star Interaction and Escape Velocity Calculation

Initially:

Star A has mass \( M_A \), and star B has mass \( M_B = 2M_A \), both having the same radius \( R \).

Due to an interaction process, star A loses some of its mass and radius becomes smaller. Let the new mass of star A be \( M_A' \). The density of star A remains the same in both cases.

Initially, the density of star A is: \[ \rho_A = \frac{M_A}{\frac{4}{3} \pi R^3} \]

Finally, the density of star A is: \[ \rho_A = \frac{M_A'}{\frac{4}{3} \pi (R')^3} \]

Since density remains the same: \[ \frac{M_A}{\frac{4}{3} \pi R^3} = \frac{M_A'}{\frac{4}{3} \pi (R')^3} \Rightarrow 8M_A' = M_A \]

Therefore, the lost mass by star A is: \[ A = M_A - M_A' = M_A - \frac{M_A}{8} = \frac{7M_A}{8} \] This lost mass is attached to star B, and the radius of star B changes to \( R_2 \).

Finally:

The density of the removed part from star A is: \[ \rho_A = \frac{\frac{7M_A}{8}}{\frac{4}{3} \pi (R^3 - (R_2)^3)} \]

The density of the added part in star B remains the same as \( \rho_A \), so we get: \[ \frac{\frac{7M_A}{8}}{\frac{4}{3} \pi (R^3 - (R_2)^3)} = \frac{\frac{7M_A}{8}}{\frac{4}{3} \pi (R_2^3 - R^3)} \]

Solving for \( R_2 \): \[ R_2 = 15^{1/3} R \]

Escape Velocity:

Escape velocity from star A after interaction is: \[ V_A = \sqrt{\frac{2GM_A}{R/2}} \]

Escape velocity from star B after interaction is: \[ V_B = \sqrt{\frac{2G(23M_A/8)}{15^{1/3} R}} \]

The ratio of escape velocities is: \[ \frac{V_B}{V_A} = \sqrt{\frac{23}{15}} \times \frac{M_A}{8} \Rightarrow \frac{V_B}{V_A} = 10n \quad \text{(where n = 2.3)} \]

Final Answer:

The value of \( n \) is \( \boxed{2.3} \).