For a planet having mass equal to the mass of the Earth but a radius that is one-fourth of the radius of the Earth,the escape velocity for this planet will be (in $km/s$): (in $km/s$)

The escape velocity for a planet whose radius is $1.7 \times 10^6 \ m$ and acceleration due to gravity is $1.7 \ m s^{-2}$ is

An object is thrown vertically upwards from the surface of the earth with a velocity $x$ times the escape velocity on the earth $(x < 1)$. The maximum height to which it rises from the center of the earth is (radius of earth is $R$):

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

$A$ body of mass $m$ is situated at a distance equal to $2R$ ($R$ is the radius of the Earth) from the Earth's surface. The minimum energy required to be given to the body so that it may escape out of the Earth's gravitational field is:

The angular velocity of rotation of a star (of mass $M$ and radius $R$) at which the matter starts to escape from its equator will be: