$A$ satellite revolves in the geostationary orbit but in a direction east to west. The time interval between its successive passing above a point on the equator is ........ $hrs$.

The radius of the orbit of a geostationary satellite is (mean radius of earth is $R$,angular velocity about own axis is $\omega$,and acceleration due to gravity on earth's surface is $g$).

$A$ satellite of mass $m$ is revolving close to the surface of a planet of density $d$ with time period $T$. The value of the universal gravitational constant $G$ in terms of $d$ and $T$ is given by:

$A$ 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:

The period of a satellite in a circular orbit around a planet is independent of

$A$ spaceship orbits around a planet at a height of $20 \, km$ from its surface. Assuming that only the gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in $24 \, hours$ around the planet? [Given: Mass of planet $= 8 \times 10^{22} \, kg$, Radius of planet $= 2 \times 10^6 \, m$, Gravitational constant $G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2$]