Orbital velocity
`V = sqrt((GM)/(R+h))`
Suppose the orbit of a satellite is exactly 35780 km above the earth's surface , how much time the satellite will take to complete one revolution around the earth ?

Explanation

Given : `V = 3.08 km//s`
`h = 35780 km`
r = 6400 km
To find : t = ?
Formula : `V = (2pir)/(T)`
solution `V = (2pir)/(T)`
`T = (2 pi r)/(V)`
`T = (2pi)/(V) (R +h)`
` T = 2 xx (3.14)/(3.08)xx (6400 + 35780)`
` = 2 xx (3.1)/(3.1) xx 42180`
`=2 xx 42.12 xx 10^(3)`
` = 2 xx 43 xx 10^(3)`
` = 86 xx 10^(3)s or 8600 s`
Also `T = (86000)/(3600)` hrs = 23.98 hrs = 23 hrs 54 mins
`therefore` Satellite will take 23 Hrs. 54 mins to complete one revlution.