For a planet having mass equal to the mass of the Earth but a radius that is one-fourth of the radius of the Earth,the escape velocity for this planet will be (in $km/s$): (in $km/s$)

The value of escape velocity on a certain planet is $2 \, km/s$. Then the value of orbital speed for a satellite orbiting close to its surface is

The escape velocity from the moon's surface is less than that on the earth's surface,because:

$A$ satellite is orbiting close to the Earth and has a kinetic energy $K$. The minimum extra kinetic energy required by it to just overcome the gravitational pull of the Earth is

$v_e$ and $v_p$ denote the escape velocity from the Earth and another planet having twice the radius and the same mean density as the Earth. Then:

The mass and radius of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a body of mass $m$ should be projected from a distance $\frac{2d}{3}$ from the centre of $M_1$ so as to escape to infinity is: