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:

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:

Suppose two planets (spherical in shape) of radii $R$ and $2R$,but mass $M$ and $9M$ respectively have a centre-to-centre separation $8R$ as shown in the figure. $A$ satellite of mass $m$ is projected from the surface of the planet of mass $M$ directly towards the centre of the second planet. The minimum speed $v$ required for the satellite to reach the surface of the second planet is $\sqrt{\frac{a}{7} \frac{GM}{R}}$,then the value of $a$ is $....$
[Given: The two planets are fixed in their position]

$A$ rocket is fired vertically upwards from the surface of the Earth with a velocity $V$. If $R$ is the radius of the Earth,what is the maximum height attained by the rocket?

An object of mass $1 \, kg$ is taken to a height from the surface of Earth which is equal to three times the radius of Earth. The gain in potential energy of the object will be $.... \, MJ$ [Given: $g = 10 \, m/s^2$ and radius of Earth $R = 6400 \, km$].

Two stars each of one solar mass $\left(=2 \times 10^{30} \; kg\right)$ are approaching each other for a head-on collision. When they are at a distance of $10^{9} \; km$,their speeds are negligible. What is the speed with which they collide? The radius of each star is $10^{4} \; km$. Assume the stars remain undistorted until they collide. (Use $G = 6.67 \times 10^{-11} \; N \cdot m^{2}/kg^{2}$)