$A$ 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 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)

Explain the escape energy of a body of mass $m$ lying on the surface of the Earth.

If the acceleration due to gravity on the surface of a planet is two times that on the surface of the Earth and its radius is double that of the Earth,then the escape velocity from the surface of that planet in comparison to the Earth will be:

The escape velocity of a planet having mass $6$ times and radius $2$ times as that of Earth is

The escape velocity for a body projected vertically upwards from the surface of the Earth is $11 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ........... $km/s$.