Gravitational potential difference between the surface of a planet and a point situated at a height of $20\,m$ above its surface is $2\,J/kg$. If the gravitational field is uniform,then the work done in taking a $5\,kg$ body to a height of $4\,m$ above the surface will be ........ $J$.

$A$ 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 $G = 6.67 \times 10^{-11}\, Nm^2 / kg^2$).

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)

$A$ mass of $50 \, \text{kg}$ is placed at the centre of a uniform spherical shell of mass $100 \, \text{kg}$ and radius $50 \, \text{m}$. If the gravitational potential at a point $25 \, \text{m}$ from the centre is $V \, \text{J/kg}$,find the value of $V$.

$A$ hollow spherical shell is compressed to half its radius. The gravitational potential at the centre

An infinite number of objects,each of mass $1 \ kg$,are placed on the $x$-axis on both sides of $x=0$ at positions $x = \pm 1 \ m, \pm 2 \ m, \pm 4 \ m, \pm 8 \ m, \ldots$ and so on. The magnitude of the resultant gravitational potential (in $SI$ units) at $x=0$ is ($G$ is the universal gravitational constant).