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?

Select the correct choice$(s)$:

Three equal masses of $m \; kg$ each are fixed at the vertices of an equilateral triangle $ABC.$
$(a)$ What is the force acting on a mass $2 \; m$ placed at the centroid $G$ of the triangle?
$(b)$ What is the force if the mass at the vertex $A$ is doubled?
Take $AG = BG = CG = 1 \; m$.

Let $\omega$ be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth $d$ below the earth's surface at a pole $(d << R)$. The value of $d$ is

$A$ particle of mass $M$ is situated at the centre of a spherical shell of same mass $M$ and radius $a$. The gravitational potential at a point situated at a distance of $\frac{a}{2}$ from the centre will be:

$A$ 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)
$(B) \frac{P(r=3R/4)}{P(r=2R/3)} = \frac{63}{80}$
$(C) \frac{P(r=3R/5)}{P(r=2R/5)} = \frac{16}{21}$
$(D) \frac{P(r=R/2)}{P(r=R/3)} = \frac{20}{27}$