A satellite orbiting above the equator of the earth is geosynchronous if the period is T = 24 hours (in this case, the satellite stays over a fixed point on the equator). Use Kepler’s Third Law to show that in a circular geosynchronous orbit, the distance from the center of the earth is R ≈ 42,246 km. Then compute the altitude h of the orbit above the earth’s surface. The earth has mass M ≈ 5.974 × 10²⁴ kg and radius R ≈ 6371 km.