$A$ satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth,the orbital speed of the satellite is

$A$ satellite of mass $\frac{M}{2}$ is revolving around the Earth in a circular orbit at a height of $\frac{R}{3}$ from the Earth's surface. The angular momentum of the satellite is $M \sqrt{\frac{GMR}{x}}$. The value of $x$ is . . . . . . ,where $M$ and $R$ are the mass and radius of the Earth,respectively. ($G$ is the gravitational constant)

Two particles of equal mass $m$ move in a circle of radius $r$ under the action of their mutual gravitational attraction. The speed of each particle will be ($G=$ Universal gravitational constant).

What is the period of revolution of an Earth satellite? Ignore the height of the satellite above the surface of the Earth.
Given:
$(1)$ The value of gravitational acceleration $g = 10 \ ms^{-2}$.
$(2)$ Radius of Earth $R_E = 6400 \ km$. Take $\pi = 3.14$.

$A$ satellite can be in a geostationary orbit around a planet at a distance $r$ from the centre of the planet. If the angular velocity of the planet about its axis doubles,a satellite can now be in a geostationary orbit around the planet if its distance from the centre of the planet is

Define the periodic time (orbital period) of a satellite.