Two bodies of equal masses are at some distance apart. If $20 \%$ of the mass is transferred from the first body to the second body,then the gravitational force between them

$A$ particle of mass $m$ moves in a circular orbit under the central potential field,$U(r) = -\frac{C}{r}$,where $C$ is a positive constant. The correct radius-velocity graph of the particle's motion is:

$A$ particle of mass $m$ is dropped from a height $R$ equal to the radius of the earth above the tunnel dug through the earth as shown in the figure. Hence the correct statement is

Two particles of equal mass $m$ go round a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

The solar constant on the surface of the earth is $S$. What will be its value on the surface of another planet which is about $5.3 \, AU$ away from the sun?

$A$ uniform sphere $A$ with radius $R$ exerts a force $F$ on a small particle $B$ situated at a distance $2R$ from the centre of the sphere. $A$ spherical portion of diameter $R$ is cut from the sphere $A$ as shown in the figure. If $F^{\prime}$ is the new gravitational force between the remaining part of the sphere $A$ and the particle $B$,then find the correct relation between $F$ and $F^{\prime}$.