This is the practical form of Kepler's 3rd law for Earth satellites. Our imagined satellite skimming the surface of the Earth (r' = 1) has a period
T = 5063.5 sec = (5063.5/60) minutes = 84.4 minutes
The space shuttle must clear the atmosphere and goes a bit higher. Say it orbits at r' = 1. 05, with SQRT(r') = 1. 0247. Then
T = (5063.5) (1.05) (1.0247) = 5448 seconds = 90.8 minutes
International communication satellites are in the equatorial plane of the Earth and have orbits with a 24-hour period. As the Earth rotates, they keep pace with it and always stay above the same spot. What is their distance?
Here T is known and we need to find r':
T = 24 hours = 86 400 sec = 5063.5 √(r')3
√(r')3 = 86 400/5063.5 = 17.0632
If all numbers on the last line are equal, their squares are equal, too
(r')3 = (17.0632)2 = 291.156
Now you need a calculator able to derive cubic roots (or else, the 1/3 = 0.333. . . power). This gives
r' = 6.628 earth radii
as the distance of "synchronous" satellites. The satellites of the global positioning system (GPS), by which a small, handheld instument can tell one's location on the globe with amazing accuracy, are in 12-hour orbits. Can you calculate their distance?