Question

(a) Find the speed of a satellite moving around the earth in a circular orbit that has a radius equal to six times the earth's radius of 6.38 106 m.
(b) Find the satellite's orbital period.

Answer 1

Becasue of the inverse swquare law of gravity, the value of g at that orbital altitude will be 1/36 of the value at the earth's surface, which make is g/36. Thus

Centripetal acceleration =
V^2/(6*Re) = g/36
where Re is the radius of the Earth.
V^2 = Re*g/6 = 1.043*10^7
V = 3.32*10^3 m/s

For the orbit period T:
2 pi Re = V*T
T = 1.241*10^4 s = 207 minutes