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 setellite ` 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_2 + 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 rel `T62 = (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 = 2pi sqrt( (R_E)/(g) ) `
By substituting the values of `R_E= 6.4 xx 10^6 m` and g` = 9.8 ms^(-2)` , the orbital time period is obtained as T = 85 minutes.