Question

A satellite orbits the earth at an altitude of 700 km on the equatorial plane
of the earth and it revolves in the same direction as the direction of rotation
of the earth. Considering the radius of a spherical earth as 6300 km and the

acceleration due to gravity as 10 m/s', the tangential velocity of the satellite
In the orbit is
ki/s. [round off to 2 decimal places]

Answer 1

m g = m v^2 / r ... v^2 = g r ... v = √(g r) = √[(10 m/s^2) (7E6 m)]