$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

The radius of a planet is $\frac{1}{4}$ of earth's radius and its acceleration due to gravity is double that of earth's acceleration due to gravity. How many times will the escape velocity at the planet's surface be as compared to its value on earth's surface?

$A$ space station is at a height equal to the radius of the Earth. If $V_{E}$ is the escape velocity on the surface of the Earth,the escape velocity on the space station is __ times $V_{E}$.

If the escape velocity of a body of mass $1\,kg$ from the surface of the Earth is $11.2\,km/s$,then what is the escape velocity for a body of mass $10\,kg$?

$A$ 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 $R = 6400 \, km$).

The radius in kilometer to which the present radius of earth $(R = 6400 \ km)$ must be compressed so that the escape velocity is increased $10$ times is ............ $km$.