$A$ particle is dropped on Earth from a height $R$ (radius of Earth) and it bounces back to a height $R/2$. The coefficient of restitution for the collision is (ignore air resistance and rotation of Earth).

$A$ particle of mass $m$ starts from rest at a distance $R$ from the centre and along the axis of a fixed ring of radius $R$ and mass $M$. Its velocity at the centre of the ring is:

Four identical particles of mass $M$ are located at the corners of a square of side $a$. What should be their speed if each of them revolves under the influence of the others' gravitational field in a circular orbit circumscribing the square?

Imagine a narrow tunnel between two diametrically opposite points of the Earth. $A$ particle of mass $m$ is released in this tunnel. The time period of oscillation is ..........

$A$ solid sphere of radius $R$ gravitationally attracts a particle of mass $m$ placed at a distance of $3R$ from its centre with a force $F_{1}$. Now,a spherical cavity of radius $\frac{R}{2}$ is made in the sphere (as shown in the figure) such that the cavity is tangent to the centre of the sphere and the outer surface. The force on the particle becomes $F_{2}$. The value of $F_{1} : F_{2}$ is:

Four similar particles of mass $m$ are orbiting in a circle of radius $r$ in the same direction because of their mutual gravitational attractive force. The velocity of a particle is given by