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 ............ $km$.

What is the escape energy of an object of mass $m$ at a height of $4R_e$ from the surface of the Earth? ($R_e$ = radius of the Earth)

$A$ mass of $6 \times 10^{24} \,kg$ is to be compressed in the form of a solid sphere such that the escape velocity from its surface is $3 \times 10^4 \,ms^{-1}$. The radius of the sphere is (Universal gravitational constant $G = 6.66 \times 10^{-11} \,N \,m^2 \,kg^{-2}$) (in $\,km$)

The escape velocity of a body on an imaginary planet,which has thrice the radius of the Earth and double the mass of the Earth,is (where $v_e$ is the escape velocity of Earth):

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,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

When the total energy of a satellite-planet system is positive,the satellite will .......