Two particles of equal mass $m$ each move in a circle of radius $r$ under the action of their mutual gravitational attraction. The time period of each particle is

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R.$ If they attract each other due to gravitational force only,then the distance covered by the smaller body before collision is (in $R$)

$A$ very long (length $L$) cylindrical galaxy is made of uniformly distributed mass and has radius $R$ $(R << L)$. $A$ star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of the star is $T$ and its distance from the galaxy's axis is $r$,then:

Two particles of equal mass $(m)$ each move in a circle of radius $(r)$ under the action of their mutual gravitational attraction. Find the speed of each particle.

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

In the following four periods:
$(i)$ Time of revolution of a satellite just above the earth's surface $({T_{st}})$
$(ii)$ Period of oscillation of mass inside the tunnel bored along the diameter of the earth $({T_{ma}})$
$(iii)$ Period of simple pendulum having a length equal to the earth's radius in a uniform field of $9.8 \; N/kg \; ({T_{sp}})$
$(iv)$ Period of an infinite length simple pendulum in the earth's real gravitational field $({T_{is}})$