If the gravitational acceleration at the surface of the Earth is $g$,then the increase in potential energy in lifting an object of mass $m$ to a height equal to half of the radius of the Earth from the surface will be:

$A$ body is projected vertically upwards from the earth's surface. If its kinetic energy of projection is equal to half of its minimum value required to escape from the gravitational influence,then the height up to which it rises is ($R =$ radius of the earth).

$A$ projectile is thrown straight upward from the Earth's surface with an initial speed $v = \alpha v_E$,where $\alpha$ is a constant and $v_E$ is the escape speed. The projectile travels up to a height $h = 800 \ km$ from the Earth's surface before it comes to rest. The value of the constant $\alpha$ is (Radius of the Earth $R = 6400 \ km$):

Two planets have masses $M$ and $16\, M$ and their radii are $a$ and $2\, a$ respectively. The separation between the centres of the planets is $10\, a.$ $A$ body of mass $m$ is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach the surface of the smaller planet,the minimum firing speed needed is

$A$ body of mass $m$ starts falling from a point $2R$ above the Earth's surface. What is its kinetic energy when it has fallen to a point $R$ above the Earth's surface? [$R$ = Radius of Earth,$M$ = Mass of Earth,$G$ = Gravitational Constant]

$A$ body which is initially at rest at a height $R$ above the surface of the earth of radius $R$,falls freely towards the earth. Its velocity on reaching the surface of the earth is: