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 $\pi = 3.14$.

The orbit of a geostationary satellite is circular. The time period of the satellite depends on $(i)$ mass of the satellite,$(ii)$ mass of the earth,$(iii)$ radius of the orbit,and $(iv)$ height of the satellite from the surface of the earth.

$A$ 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 distance between two stars of masses $3 M_S$ and $6 M_S$ is $9 R$. Here $R$ is the mean distance between the centers of the Earth and the Sun,and $M_S$ is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period $n T$,where $T$ is the period of Earth's revolution around the Sun. The value of $n$ is. . . . .

$A$ particle moves in a circular orbit of radius $r$ under a central attractive force $F = -\frac{k}{r^2}$,where $k$ is a constant. The time period of its motion shall be proportional to

$A$ satellite is revolving very close to a planet of density $\rho$. The period of revolution of the satellite is