Question

A satellite is launched to orbit the Earth at an altitude of 2.20 * 10^7 m for use in the Global Positioning system (GPS). Take the mass of the Earth to be 5.97 * 10^24 kg and its radius 6.38 * 10^6 m. (a) What is the orbital period of this GPS satellite? (b) With what speed does it orbit the Earth?

Answer 1

a)
Period of the satellite is given by,
t^2 = (4π^2* r^3)/(G * Me)
where
G = 6.67 * 10^-11 m^3/(kg-s^2),
Me = 5.97 * 10^24 kg and
r = 6.38 * 10^6 m + 2.20 * 10^7 m = 2.838 * 10^7 m

Substitung Values,
t^2 = (4*π^2* (2.838 * 10^7)^3)/(6.67 * 10^-11 * 5.97 * 10^24)
t =  47604.5 s
Converting to days,
t = 0.55 days

b)
velocity of the satellite is given by,
v^2 = G(Me)/r
v^2 = (6.67 * 10^-11 * 5.97 * 10^24)/(2.838 * 10^7)
v = 3745.8 m/s
v = 3.746 km/s