QTest #6-solutions - 8. 4.19525e+27 9. 1.95261e+27 10....

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Version 015/AAADD – QTest #6 – Antoniewicz – (57420) 1 This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Consider the orbit oF a typical comet around the sun, which is marked at fve di±erent po- sitions, P , X , S , U , and Z . Sun a b U X Z P S Using Kepler’s second law oF planetary mo- tion, rank those positions in order oF their rel- ative speeds, with the position For the Fastest speed frst. 1. Z P X U S 2. Z X U S P 3. U X Z P S correct 4. S U X P Z 5. P S U X Z 6. S P Z X U Explanation: The closer the comet is to the Sun, the Faster it is traveling, so ( U , X , Z , P , S ) is correct. 002 10.0 points Given: G = 6 . 672 × 10 - 11 N · m 2 / kg 2 Io, a satellite oF Jupiter, has an orbital period oF 1 . 27 days and an orbital radius oF 4 . 5 × 10 5 km. ²rom these data, determine the mass oF Jupiter. 1. 2.04922e+27 2. 1.73357e+27 3. 2.55915e+27 4. 4.47823e+27 5. 4.0417e+27 6. 3.21024e+27 7. 3.48555e+27
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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 10-11 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 10-11 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 10-11 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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QTest #6-solutions - 8. 4.19525e+27 9. 1.95261e+27 10....

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