Two point masses of mass $4m$ and $m$ respectively,separated by a distance $d$,are revolving under their mutual force of attraction. The ratio of their kinetic energies is:

Four particles,each of mass $M$,move along a circle of radius $R$ under the action of their mutual gravitational attraction as shown in the figure. The speed of each particle is:

$A$ spherical body of radius $R$ consists of a fluid of constant density $\rho$ and is in equilibrium under its own gravity. If $P(r)$ is the pressure at a distance $r$ from the center $(r < R)$,then the correct option$(s)$ is(are):
$(A) P(r=0) = P_c$ (maximum pressure at center)
$(B) \frac{P(r=3R/4)}{P(r=2R/3)} = \frac{63}{80}$
$(C) \frac{P(r=3R/5)}{P(r=2R/5)} = \frac{16}{21}$
$(D) \frac{P(r=R/2)}{P(r=R/3)} = \frac{20}{27}$

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 $R$)

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:

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?