$A$ particle is kept at rest at a distance $R$ from the surface of the Earth (of radius $R$). The minimum speed with which it should be projected so that it does not return is

The Earth is assumed to be a sphere of radius $R$. $A$ platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$,where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is

$A$ spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $V$. Due to the rotation of the planet about its axis,the acceleration due to gravity $g$ at the equator is $1/2$ of $g$ at the poles. Find the escape velocity of a particle on the planet in terms of $V$.

$A$ rocket is launched normal to the surface of the Earth,away from the Sun,along the line joining the Sun and the Earth. The Sun is $3 \times 10^5$ times heavier than the Earth and is at a distance $2.5 \times 10^4$ times larger than the radius of the Earth. The escape velocity from the Earth's gravitational field is $v_e = 11.2 \text{ km s}^{-1}$. The minimum initial velocity $(v_s)$ required for the rocket to be able to leave the Sun-Earth system is closest to:
(Ignore the rotation and revolution of the Earth and the presence of any other planet)

$A$ 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 is projected vertically upward from the surface of the earth of radius $R$ into space with velocity $n V_e$ (where $n < 1$ and $V_e$ is the escape velocity). The maximum height from the surface of the earth to which the body can reach is