$A$ body is released from a height equal to the radius $(R)$ of the earth. The velocity of the body when it strikes the surface of the earth will be: (Given $g$ = acceleration due to gravity on the earth.)

$A$ body starts from rest from a distance $R_0$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ($R=$ radius of earth,$M=$ mass of earth).

From the pole of the earth,a body of mass $m$ is imparted a velocity $v_0$ directed vertically up. If $M$ is the mass of the earth,$R$ its radius and $g$ is the free-fall acceleration on its surface,then the height $h$ to which the body will ascend is (neglect air resistance).

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:

$A$ rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V$. Assuming the radius of the earth to be $R$ and negligible air resistance,the maximum height attained by the rocket above the surface of the earth is

$A$ 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 $R = 6400 \ km$):