Derive the time period of satellite orbiting the Earth.

Explanation

Time period of the satellite:
The distance covered by teh satellite during one rotation in its orbit is equal to `2 pi (R_E + h)` and time taken for it, is the time period, T. Then speed,
`v = ("Distance travelled")/("Time taken") = (2pi(R_E + h))/(T)`
From equation
`sqrt((GM_E)/((R_E + H))) = (2pi(R_E + h))/(T)" "......(1)`
`T = (2pi)/(sqrt(GM_E)) (R_(E) + h)^(3//2)" "......(2)`
Squaring both sides of the equation (2), we get
`T^(2) = (4pi^2)/(GM_E) = (R_(E) + h)^(3)`
`(4pi^2)/(GM_E)` = constant say c
`T^(2) = c(R_E + h)^(3)" ".....(3)`
Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler's law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth `R_(E)`. Then,
`T^(2) = (4pi^2)/(GM_E) R_(E)^(3) = (4pi^2)/(GM_E//R_E^2) cdot R_(E)`
`T_(2) = (4pi^2)/(g) R_E`
(Since `GM_E//R_E^2 = g)`
`T = 2pi sqrt((R_E)/(g))`.