The radius in kilometer to which the present radius of earth $(R = 6400 \ km)$ must be compressed so that the escape velocity is increased $10$ times is ............ $km$.

$A$ 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 $km/s$)

$A$ small asteroid is orbiting around the sun in a circular orbit of radius $r_0$ with speed $v_0$. $A$ rocket is launched from the asteroid with speed $v = \alpha v_0$,where $v$ is the speed relative to the sun. The highest value of $\alpha$ for which the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effects of other planets).

The mass of a planet is half that of the Earth and the radius of the planet is one-fourth that of the Earth. If we plan to send an artificial satellite from the planet,the escape velocity will be (escape velocity on Earth $v_e = 11 \ km \ s^{-1}$): (in $km \ s^{-1}$)

If a planet of mass $6.4 \times 10^{23} \ kg$ can be compressed into a sphere such that the escape velocity from its surface is $8 \times 10^4 \ m/s$,then what should be the radius of the sphere (in $km$)? (Gravitational constant,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius four times that of Earth and the same mass density is: