$A$ geostationary satellite is orbiting the earth at a height of $6R$ from the earth's surface ($R$ is the earth's radius). What is the period of rotation of another satellite at a height of $2.5R$ from the earth's surface?

$A$ satellite is to be placed in an equatorial geostationary orbit around the Earth for communication.
$(a)$ Calculate the height of such a satellite.
$(b)$ Find out the minimum number of satellites that are needed to cover the entire Earth,so that at least one satellite is visible from any point on the equator.
Given: $M = 6 \times 10^{24} \ kg$,$R = 6400 \ km$,$T = 24 \ h$,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$.

The earth (mass $M = 6 \times 10^{24} \ kg$) revolves around the sun with an angular velocity $\omega = 2 \times 10^{-7} \ rad/s$ in a circular orbit of radius $R = 1.5 \times 10^8 \ km$. The force exerted by the sun on the earth in newtons is:

An earth satellite of mass $m$ revolves in a circular orbit at a height $h$ from the surface of the earth. $R$ is the radius of the earth and $g$ is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

Two satellites $A$ and $B$ go round a planet in circular orbits having radii $4R$ and $R$,respectively. If the speed of satellite $A$ is $3v$,then the speed of satellite $B$ is:

$A$ satellite of mass $m$ is moving in a circular orbit of radius $r$ around the earth. The angular momentum of the satellite about the center of orbit is ($M=$ mass of earth,$G=$ gravitational constant).