$A$ very small groove is made in the earth,and a particle of mass $m_0$ is placed at $\frac{R}{2}$ distance from the centre. Find the escape speed of the particle from that place.

If the escape velocity on Earth is $11.2 \text{ km/s}$, what is its value for a planet having double the radius and $8$ times the mass of Earth (in $\text{ km/s}$)?

$A$ planet having mass $9 M_e$ and radius $4 R_e$,where $M_e$ and $R_e$ are the mass and radius of the Earth respectively,has an escape velocity in $km/s$ given by: (Given escape velocity on Earth $V_e = 11.2 \times 10^3 \, m/s$)

The mass of the moon is $1/144$ times the mass of a planet and its diameter is $1/16$ times the diameter of a planet. If the escape velocity on the planet is $v$,the escape velocity on the moon will be:

$A$ rocket is projected in the vertically upwards direction with a velocity $kv_e$,where $v_e$ is the escape velocity and $k < 1$. The distance from the centre of the Earth up to which the rocket will reach is:

The masses and radii of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centers are at a distance $d$ apart. The minimum velocity with which a particle of mass $m$ should be projected from a point midway between their centers so that it escapes to infinity is: