Derive the time period of satellite orbiting the Earth.

Explanation

Time period of the satellite: 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 " "...(1)`
Speed of the satellite, `V = sqrt((GM_(E))/((R_(E)+h)))`
`sqrt((GM_(E))/((R_(E)+h))) = (2pi(R_(E)+h))/T`
`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)`...(2)
`(4pi^(2))/(GM_(E)) = ` constant say c
`T^(2) =c(R_(E)+h)^(3)`
Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that 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)rArr T^(2) = (4pi^(2))/g R_(E)`
Since ` (GM_(E))/(R_(E)^(2)) = g `
`T = 2pisqrt(R_(E)/g)`
By substituting the values of`R_(E) = 6.4 xx10^(6) ` m and g = `9.8 " m s"^(-2)` the orbital time period is obtained `T cong 85` minutes .