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Unformatted text preview: constants and the mass of the Sun. Therefore we can write a similar equation for any planet that orbits the Sun. T 1 2 /r 1 3 = 4 π 2 /G . M S EQ.5 T 2 2 /r 2 3 = 4 π 2 /G . M S EQ.6 And therefore can write Kepler’s third law T 1 2 /r 1 3 = T 2 2 /r 2 3 EQ.7 This was a magnificent result for Newton’s theory. Incidentally, scientists could then use these results to find the mass of the sun for the first time. Writing EQ.5 for the EarthSun system, we have T E 2 /r ES 3 = 4 π 2 /G . M S EQ.8 where T E = 3.16x10 7 sec r ES = 1.50x10 11 m G = 6.67x1011 N ⋅ m 2 /kg 2 Solving for the mass of the Sun we have M S = [4 π 2 /G] r ES 3 /T E 2 EQ.9 M S = 2.0x10 30 kg...
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 Spring '09
 JamesR.Boyd
 Force, General Relativity, Planet, circular orbit, EQ

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