The ratio of the kinetic energy required to be given to a satellite to escape from the Earth's surface to the kinetic energy required to be given to the same satellite to revolve around the Earth in an orbit just above the Earth's surface is ...........

$A$ particle released at a large distance from a planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre. If $v_e$ is the escape velocity of a body at the planet,then the particle's speed at the centre of the planet 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 $min$)

$A$ body of mass $m$ is thrown vertically upward with speed $\sqrt{3} v_e$,where $v_e$ is the escape velocity of a body from the Earth's surface. The final velocity of the body is

$A$ body of mass $m$ is projected with velocity $\lambda v_{e}$ in a vertically upward direction from the surface of the earth into space. It is given that $v_{e}$ is the escape velocity and $\lambda < 1$. If air resistance is considered to be negligible,then the maximum height from the center of the earth,to which the body can go,will be: ($R$: radius of earth)

The magnitude of potential energy per unit mass of an object at the surface of the Earth is $E$. Then,the escape velocity of the object is ..........