The time period of a satellite orbiting Earth in a circular orbit is independent of

Explanation

The distance covered by the satellite during one rotation in its orbit is equal to `2pi(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 the above 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)`
`T^(2)=(4pi^(2))/(GM_E//R_(E)^(2))R_(E)`
`T^(2)=(4pi^(2))/(g)R_(E)`
Since `(GM_E)/(R_(E)^(2))=g`
`T=2pisqrt((R_E)/(g))`