If a GPS satellite (circular orbit) has an orbital period of 12 hours, what is its distance from the Earth?

Question

If a GPS satellite (circular orbit) has an orbital period of 12 hours, what is its distance from the Earth? And what is its velocity? (Hint convert time to seconds)

Answer 1

At that altitude R, the centripetal acceleration must be equal to earth's gravity

Ca = w^2 R

w = 2Pi rad/(12*3600) s = 1.45E-4 rad/s

g = G M/R^2 = 6.67 × 10−11 N.m^2/kg^2 x 5.9742×10^24 kg/R^2

w^2R = GM/R^2
R^3 = GM/w^2
R = (GM/w^2)^1/3

R = cubicroot(6.67 × 10−11 N.m^2/kg^2 x 5.9742×10^24 kg/(1.45E-4 rad/s)^2) = 26 661 824m

Earth radius is 6 372 797 m

So the satellite is at altitude

26 661 824m - 6 372 797m = 20 289 027m = 20 289 km