If the gravitational potential on the surface of the earth is $V_0$,then the potential at a point at a height equal to half of the radius of the earth is ..........

An infinite number of bodies,each of mass $2 \, kg$,are situated on the $x-$axis at distances $1 \, m, 2 \, m, 4 \, m, 8 \, m, \dots$ respectively,from the origin. The resulting gravitational potential due to this system at the origin will be:

$A$ particle of mass $10\, g$ is kept on the surface of a uniform sphere of mass $100\, kg$ and radius $10\, cm$. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $G = 6.67 \times 10^{-11}\, Nm^2 / kg^2$).

$A$ particle of mass $m$ is placed at the centre of a uniform spherical shell of mass $3m$ and radius $R$. The gravitational potential on the surface of the shell is

If one wants to remove all the mass of the earth to infinity in order to break it up completely,the amount of energy that needs to be supplied will be $\frac{x}{5} \frac{GM^2}{R}$,where $x$ is .......... (Round off to the Nearest Integer). ($M$ is the mass of earth,$R$ is the radius of earth,$G$ is the gravitational constant)

From a solid sphere of mass $M$ and radius $R$,a spherical portion of radius $R/2$ is removed,as shown in the figure. Taking gravitational potential $V = 0$ at $r = \infty$,the potential at the centre of the cavity thus formed is
($G =$ gravitational constant)