The escape velocity of a sphere of mass $m$ is given by ($G =$ Universal gravitational constant; $M_e =$ Mass of the earth and $R_e =$ Radius of the earth).

$A$ particle of mass $m$ is thrown upwards from the surface of the earth with a velocity $u$. The mass and the radius of the earth are $M$ and $R$, respectively. $G$ is the gravitational constant and $g$ is the acceleration due to gravity on the surface of the earth. The minimum value of $u$ so that the particle does not return back to earth is:

The mass of the Earth is $81$ times that of the Moon and the radius of the Earth is $3.5$ times that of the Moon. The ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon will be:

$A$ body is projected from the earth's surface with thrice the escape velocity. What will be its velocity when it escapes the gravitational pull?

Maximum height reached by an object projected perpendicular to the surface of the earth with a speed equal to $50\%$ of the escape velocity from the earth's surface is - ($R =$ Radius of Earth)

Escape velocity of a body of $1\, kg$ mass on a planet is $100\, m/s$. Gravitational potential energy of the body on the surface of the planet is ......... $J$.