Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. Since mass is constant and radius becomes one-fourth, $v_e$ increases by a factor of $\sqrt{4} = 2$. Thus, $v_p = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$. Option B is correct.

Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$. When mass becomes $2M$ and radius becomes $R/2$, $v'_e = \sqrt{\frac{2G(2M)}{R/2}} = 2v_e$. Thus, $v'_e = 2 \times 11.2\text{ km/s} = 22.4\text{ km/s}$.

When a ball is dropped from a spacecraft, it possesses the same orbital velocity as the spacecraft. Therefore, it continues to move with the same speed along the original orbit.

Escape velocity is a scalar quantity and is independent of the direction or angle of projection. It depends only on the mass and radius of the planet, remaining $11.2\text{ km/s}$.

Kinetic energy of the satellite is $K = \frac{1}{2}mv^2$ and potential energy is $U = -mv^2$. Total energy $E = K + U = \frac{1}{2}mv^2 - mv^2 = -\frac{1}{2}mv^2$.

The escape velocity formula depends solely on the gravitational potential and mass/radius of the body. It is independent of the angle at which the object is launched.

Due to universal gravitational attraction between the two astronauts, they will experience a mutual attractive force and move towards each other in space.