The kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$) to infinity is:

$A$ body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity. The time taken by it to reach a height of three times the radius of the earth is (acceleration due to gravity $g = 9.8 \ m/s^2$ and radius of the earth $R = 6400 \ km$). (in $min$)

When the speed of a satellite is increased by $x$ percentage,it will escape from its orbit,where the value of $x$ is ....... $\%$

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 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

$A$ particle of mass $m_0$ is projected from the midpoint of the line joining two fixed particles,each of mass $m$. If the separation between the fixed particles is $l$,the minimum velocity of projection of the particle so as to escape is equal to: