Assume that a tunnel is dug along a chord of the Earth,at a perpendicular distance $(R / 2)$ from the Earth's centre,where $R$ is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel,it will execute a simple harmonic motion with a time period:

$A$ body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest

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?

Consider a spherical gaseous cloud of mass density $\rho(r)$ in free space,where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common center with the same kinetic energy $K$. The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant in time,the particle number density $n(r) = \rho(r) / m$ is:
[$G$ is the universal gravitational constant]

Two planets $A$ and $B$ having masses $m_1$ and $m_2$ move around the sun in circular orbits of $r_1$ and $r_2$ radii respectively. If angular momentum of $A$ is $L$ and that of $B$ is $3L$,the ratio of time period $\left(\frac{T_A}{T_B}\right)$ is:

$A$ 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).