The escape velocity for the Earth is $11 \ km/s$. The escape velocity for a planet having twice the radius and the same density as the Earth is .......... $km/s$.

If the Earth has a mass nine times and a radius twice that of a planet $P$. Then $\frac{v_e}{3} \sqrt{x} \; ms^{-1}$ will be the minimum velocity required by a rocket to escape the gravitational force of $P$,where $v_e$ is the escape velocity on Earth. The value of $x$ is

Two stars of masses $3\times10^{31} \ kg$ each,and at distance $2\times10^{11} \ m$ rotate in a plane about their common centre of mass $O$. $A$ meteorite passes through $O$ moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star,the minimum speed that the meteorite should have at $O$ is: (Take Gravitational constant $G = 6.67\times10^{-11} \ Nm^2 \ kg^{-2}$)

If the escape velocity of a body of mass $1\,kg$ from the surface of the Earth is $11.2\,km/s$,then what is the escape velocity for a body of mass $10\,kg$?

The radius and mean density of a planet are four times that of the Earth. The ratio of the escape velocity on the Earth to the escape velocity on the planet is:

If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for an orbit close to the earth's surface,then these are related by: