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

At what altitude will the acceleration due to gravity be $25\% $ of that at the earth’s surface (given radius of earth is $R$) ?

$Assertion$ : The escape speed does not depend on the direction in which the projectile is fired.
$Reason$ : Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the earth rotates about its axis.

A body of mass  $m$  falls from a height  $R$  above the surface of the earth, where $R$  is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$ )

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 particle of mass $M$ is situated at the centre of a spherical shell of same mass and radius $a$. The gravitational potential at a point situated at $\frac{a}{2}$ distance from the centre, will be