A remote sensing satellite of the Earth in a circular orbit at a height of `400 km` above the surface of Earth. What is the (a) orbital speed, and (b) period of revolution of satellite ? Radius of Earth `= 6 xx 10^(6) m` and acceleration due to gravity the surface of Earth is `10 m//s^(2)`.

Explanation

Here, `R = 6 xx 10^(6)m , g = 10 m//s^(2)`,
`h = 400 xx 10^(3) m = 0.4 xx 10^(6) m`
(a) Orbital speed, `upsilon = R sqrt((g)/(R + h))`
` = 6 xx 10^(6) sqrt((10)/(6 xx 10^(6) + 0.4 xx 10^(6))`
`= 7.5 xx 10^(3) m//s`
(b) Period of revolution, `T = (2 pi)/(R ) sqrt(((R + h)^(3))/(g)`
`= (2 xx (22//7))/(6 xx 10^(6)) sqrt(((6 xx 10^(6) + 0.4 xx 10^(6))^(3))/(10))`
`= 5368.5 s`