## (21a) Applying Kepler's Third LawFor circular orbits around Earth, we found T 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 T The ratio Also, that ratio is always the same, whether (4p Using computer notation for the square root SQRT (25 638 838) = 5063.5
From this ^{2}= (5063.5)^{2} (r')^{3}T= 5063.5 seconds SQRT(r') ^{3} = 5063 sec r'SQRT=(r')This is the 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's 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 T = 24 hours = 86 400 sec = 5063.5 SQRT(r') If all numbers on the last line are equal, their squares are equal, too (r') 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? |

**Next Regular Stop: #22 Frames of Reference: The Basics**

**
**

*Author and curator: David P. Stern, u5dps@lepvax.gsfc.nasa.gov*