$A$ body of mass $m$ is taken from the Earth's surface to a height $h$ equal to twice the radius of the Earth $(h = 2R)$. The increase in potential energy will be ($g = \text{acceleration due to gravity on Earth's surface}$, $R = \text{radius of the Earth}$):

Assuming that the gravitational potential energy of an object at infinity is zero,the change in potential energy (final - initial) of an object of mass $m$,when moved to a height $h$ from the surface of the Earth (of radius $R$),is given by:

The radius of the earth is $R$. If a body is taken to a height $3R$ from the surface of the earth,then the change in potential energy will be:

An object is propelled vertically to a maximum height of $4 R$ from the surface of a planet of radius $R$ and mass $M$. The speed of the object when it returns to the surface of the planet is

$A$ body is thrown from the surface of the earth with velocity $u \ m \ s^{-1}$. The maximum height in meters above the surface of the earth up to which it will reach is ($R=$ radius of the earth,$g=$ acceleration due to gravity).

An asteroid is moving directly towards the centre of the Earth. When at a distance of $10 R$ ($R$ is the radius of the Earth) from the Earth's centre,it has a speed of $12 \; km/s$. Neglecting the effect of Earth's atmosphere,what will be the speed of the asteroid when it hits the surface of the Earth? (Escape velocity from the Earth is $11.2 \; km/s$). Give your answer to the nearest integer in $km/s$.