$A$ body is projected vertically upwards from the Earth's surface with velocity $2 V_e$,where $V_e$ is the escape velocity from the Earth's surface. The velocity when the body escapes the gravitational pull is

How much energy is necessary for a body of $500 \, kg$ to escape from the Earth? $[g = 9.8 \, m/s^2$,radius of Earth $R = 6.4 \times 10^6 \, m]$

What is the value of escape velocity of a body lying on the surface of Earth and Moon?

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g$,where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \ km/s$,the escape speed on the surface of the planet in $km/s$ will be:

$A$ rocket is launched straight up from the surface of the earth. When its altitude is $\frac{1}{3}$ of the radius of the earth,its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth,the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is $11.2 \ km/s$.) (in $km/s$)

$A$ spaceship of mass $2 \times 10^4 \, kg$ is launched into a circular orbit close to the Earth's surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be $......$ (given $g = 10 \, m/s^2$ and radius of Earth $R = 6400 \, km$).