Escape velocity on Earth is $11.2 \, km/s$. What would be the escape velocity on a planet whose mass is $1000$ times and radius is $10$ times that of Earth?

The escape velocity of an atmospheric particle which is $1000 \, km$ above the Earth's surface is .......... $km/s$ (radius of Earth is $6400 \, km$ and $g = 9.8 \, m/s^2$).

If the acceleration due to gravity $g$ doubles and the radius of earth becomes half that of the present value,then the value of escape velocity is (Assume,$g=10 \ m/s^2$ and radius of earth,$R=6400 \ km$)

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:

If an object is projected vertically upwards with a speed equal to half the escape speed of Earth,then the maximum height attained by it is ($R$ is the radius of Earth).

$A$ rocket is launched straight up from the surface of the earth. When its altitude is $\frac{1}{3}$ of the radius of the earth,its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth,the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is $11.2 \ km/s$.) (in $km/s$)