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Index 18b. Momentum 18c. Work 18d. Work against Electric Forces 19.Motion in a Circle 20. Newton's Gravity 21. Kepler's 3rd Law 21a.Applying 3rd Law 21b. Fly to Mars! (1) 21c. Fly to Mars! (2) 21d. Fly to Mars! (3) 22.Reference Frames 22a.Starlight Aberration 22b. Relativity 22c. Flight (1) 22d. Flight (2) 
For circular orbits around Earth, we found T^{2} = (4π^{2}/g R_{E}^{2}) r^{3} 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_{E}^{3}/R_{E}^{3}) = 1 and then rearrange the terms: T^{2} = (4π^{2}/g R_{E}^{2}) (R_{E}^{3}/R_{E}^{3}) r^{3} = (4π^{2}R_{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. (4π^{2}R_{E}/g) = (4) (9.87) (6 371 000)/9.81 = 25 638 838 Deriving a square root √(25 638 838) = 5063.5
From this T= 5063.5 seconds √(r')^{3} = 5063.5 sec r' √(r') 
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 24hour 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} 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 12hour orbits. Can you calculate their distance? 
Optional next stop (1st of 3): #21b Flight to Mars: How Long? Along what Path?
Next Regular Stop: #22 Frames of Reference: The Basics
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Author and Curator: Dr. David P. Stern
Greenbelt, Maryland
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 9222004
Reformatted 24 March 2006