(21a) Applying Kepler's Third Law

For circular orbits around Earth, we found

T² = (4π²/g R_E²) r³

with T in seconds and r in meters. The distance of a satellite from the center of Earth in meters is an inconveniently big number, even before we raise it to the 3rd power. We may however multiply the expression on the right by (R_E³/R_E³) = 1 and then rearrange the terms:

T² = (4π²/g R_E²) (R_E³/R_E³) r³ = (4π²R_E/g) (r/R_E)³

The ratio r' = (r/R_E) is the orbital distance measured in units of the Earth's radius. That number is usually between 1 (at the Earth's surface, r = R_E) and 60 (at the Moon's orbit, r ~ 60 R_E).

Also, that ratio is always the same, whether r and R_E are in meters, yards or nautical miles, as long as both r and R_E are measured in the same units. The other term is calculated below, with multiplication denoted by blank spaces between parentheses; you may check it with your calculator.

(4π²R_E/g) = (4) (9.87) (6 371 000)/9.81 = 25 638 838

Deriving a square root

√(25 638 838) = 5063.5 From this

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 √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?