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Unformatted text preview: 8. 4.19525e+27 9. 1.95261e+27 10. 2.36104e+27 Correct answer: 4 . 47823 10 27 kg. Explanation: Given : T = 1 . 27 days = 1 . 09728 10 5 s , R = 4 . 5 10 5 km = 4 . 5 10 8 m , and G = 6 . 672 1011 N m 2 / kg 2 . The orbital speed oF Io is v = 2 R T . The gravitational Force supplies the required centripetal acceleration, so GM m R 2 = mv 2 R M = Rv 2 G = R G p 2 R T P 2 = 4 2 R 3 GT 2 = 4 2 6 . 672 1011 N m 2 / kg 2 (4 . 5 10 8 m) 3 (1 . 09728 10 5 s) 2 = 4 . 47823 10 27 kg . Direct Solution: rom Keplers third law, we have M = 4 2 G R 3 T 2 Version 015/AAADD QTest #6 Antoniewicz (57420) 2 = 4 2 6 . 672 1011 N m 2 / kg 2 (4 . 5 10 8 m) 3 (1 . 09728 10 5 s) 2 = 4 . 47823 10 27 kg ....
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 Spring '08
 Turner
 Physics

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