Two particles of equal mass $m$ move in a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their center of mass is

The distance between the Sun and the Earth is $1.6 \times 10^{11} \,m$ and the radius of the Earth is $6.4 \times 10^6 \,m$. The ratio of the angular momentum of the Earth around the Sun to the angular momentum around its own axis is approximately (Assume the Earth as a solid sphere with uniform mass density and that it rotates around the Sun in a circular path.)

Two bodies of masses $m_1$ and $m_2$ initially at rest at infinite distance apart move towards each other under gravitational force of attraction. Their relative velocity of approach when they are separated by a distance $r$ is ($G=$ universal gravitational constant.)

Three particles of mass $1 \, kg$ each are placed at $(0, 0)$,$(0, 0.2 \, m)$,and $(0.2 \, m, 0)$. What is the net gravitational force on the particle placed at the origin?

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:

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]