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.

Match the column $-I$ with column $-II$ for a satellite in circular orbit:

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

(where $M_E$ is the mass of the earth,$m$ is the mass of the satellite,and $r$ is the radius of the orbit)

Four identical particles of equal masses $m = 1 \, kg$ are made to move along the circumference of a circle of radius $R = 1 \, m$ under the action of their own mutual gravitational attraction. The speed of each particle will be

Particles of masses $m, 2m, 3m$ and $4m$ are placed at the corners of a square of side $a$. What is the net gravitational force on a particle of mass $m$ placed at the center?

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 $R$)

Take the mean distance of the moon and the sun from the earth to be $0.4 \times 10^6 \, km$ and $150 \times 10^6 \, km$ respectively. Their masses are $8 \times 10^{22} \, kg$ and $2 \times 10^{30} \, kg$ respectively. The radius of the earth is $6400 \, km$. Let $\Delta F_1$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $\Delta F_2$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then,the number closest to $\frac{\Delta F_1}{\Delta F_2}$ is