Two heavenly bodies $S_1$ and $S_2$,not far from each other,are seen to revolve in orbits:

Given below are two statements:
Statement $I$: $A$ satellite is moving around the Earth in an orbit very close to the Earth's surface. The time period of revolution of the satellite depends upon the density of the Earth.
Statement $II$: The time period of revolution of the satellite is $T = 2\pi\sqrt{\frac{R_e}{g}}$ (for a satellite very close to the Earth's surface), where $R_e$ is the radius of the Earth and $g$ is the acceleration due to gravity.
In the light of the above statements, choose the correct answer from the options given below:

Three identical bodies of equal mass $M$ each are moving along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each body is

Two satellites,$A$ and $B,$ have masses $m$ and $2m$ respectively. $A$ is in a circular orbit of radius $R,$ and $B$ is in a circular orbit of radius $2R$ around the earth. The ratio of their kinetic energies,$K.E._A / K.E._B ,$ is

$A$ satellite $S$ is moving in an elliptical orbit around the Earth. The mass of the satellite is very small compared to the mass of the Earth.

$A$ satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius $R_e$. By firing rockets attached to it,its speed is instantaneously increased in the direction of its motion so that it becomes $\sqrt{\frac{3}{2}}$ times larger. Due to this,the farthest distance from the centre of the earth that the satellite reaches is $R$. The value of $R$ is $....R_e$.