Two spherical stars A and B have densities ρ A and ρ 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 ρ A . The entire mass lost by A is deposited as a thick spherical shell on B with the densi

Step-by-step solution

Given here: R A = R B = R and M B = 2 M A . Now, after interaction process, radius of remaining star A is R A ' = R 2 and its mass is M A ' = ρ A 4 3 π R A 2 3 = M A 8 Applying conservation of energy, - G M A ' m R A ' + 1 2 m v A 2 = 0 Escape velocity of star A is v A = 2 G M A 8 × R 2 = v 0 2 Now, for B , mass collected over B is M B ' = M A - M A 8 = 7 8 M A . Let the radius of star B after interaction becomes r . Applying mass conservation, 4 3 π r 3 - R 3 ρ A = 4 3