Two spherical planets $P$ and $Q$ have the same uniform density $\rho$,masses $M_P$ and $M_Q$,and surface areas $A$ and $4A$,respectively. $A$ spherical planet $R$ also has uniform density $\rho$ and its mass is $(M_P + M_Q)$. The escape velocities from the planets $P, Q,$ and $R$ are $V_P, V_Q,$ and $V_R$ respectively. Then:

$A$ body is projected vertically upwards from the Earth's surface with a velocity $2 v_{e}$,where $v_{e}$ is the escape velocity from the Earth's surface. The velocity of the body when it escapes the gravitational pull is

$A$ planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11 \ km/s$,the escape velocity from the surface of the planet would be ........ $km/s$.

What is the escape velocity for the surface of a black hole?

$A$ satellite is moving with a constant speed $v$ in a circular orbit around the Earth. An object of mass $m$ is ejected from the satellite such that it just escapes from the gravitational pull of the Earth. At the time of ejection,the kinetic energy of the object is

$A$ body of mass $m$ is projected with velocity $\lambda v_{e}$ in a vertically upward direction from the surface of the earth into space. It is given that $v_{e}$ is the escape velocity and $\lambda < 1$. If air resistance is considered to be negligible,then the maximum height from the center of the earth,to which the body can go,will be: ($R$: radius of earth)