Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B,$ then the ratio of the escape velocity $\frac{V_A}{V_B}$ is

$A$ black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass $= 5.98 \times 10^{24} \, kg$) have to be compressed to be a black hole?

$A$ body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity of that planet. Then,the maximum height attained by the body is

Escape velocity at the surface of the Earth is $11.2 \, km/s$. If the radius of a planet is double that of the Earth but its mean density is the same as that of the Earth,then the escape velocity will be ........ $km/s$.

$A$ mass of $6 \times 10^{24} \ kg$ is to be compressed into a sphere in such a way that the escape velocity from the sphere is $3 \times 10^8 \ m/s$. What should be the radius of the sphere? (Given: $G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

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