Show that the orbital velocity of a satellite revolving very close to the surface of earth is `7.92 k ms^(-1)`.

Explanation

Orbital velocity of a satellite is, `v_(o)= ((gR^(2))/(R+h))^(1//2)`
If satellite revolves very close to earth then, `R+ h~~ R`
Hence, `v_(o)= ((gR^(2))/(R ))^(1//2)= sqrt(gR)`. Putting `g= 9.8 ms^(-2), R= 6.4 xx 10^(6)m`
Force `v_(o)= sqrt(9.8 xx 6.4 xx 10^(6))= 7.92 km s^(-1)`