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solution-hw20

# solution-hw20 - lai(yl8859 HW20 Ross(15200 This print-out...

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lai (yl8859) – HW20 – Ross – (15200) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A distant star has a single planet circling it in a circular orbit of radius 7 . 11 × 10 11 m. The period of the planet’s motion about the star is 703 days. What is the mass of the star? The value of the universal gravitational constant is 6 . 67259 × 10 - 11 N · m 2 / kg 2 . Correct answer: 5 . 76416 × 10 31 kg. Explanation: Let : G = 6 . 67259 × 10 - 11 N · m 2 / kg 2 , R B = 7 . 11 × 10 11 m , and T B = 703 day . T B = (703 day) parenleftbigg 24 h 1 day parenrightbigg 3600 s 1 h = 6 . 07392 × 10 7 s . According to Newton’s explanation of Ke- pler’s third law R 3 B T 2 B = G M s 4 π 2 = const. The mass of the star is thus M s = 4 π 2 G R 3 B T 2 B = 4 π 2 6 . 67259 × 10 - 11 N · m 2 / kg 2 × (7 . 11 × 10 11 m) 3 (6 . 07392 × 10 7 s) 2 = 5 . 76416 × 10 31 kg . 002 10.0 points An object is dropped from rest from a height 6 . 9 × 10 6 m above the surface of the earth. If there is no air resistance, what is its speed when it strikes the earth? The acceleration of gravity is 9 . 81 m / s 2 and the radius of the Earth is 6 . 37 × 10 6 m. Correct answer: 8 . 06136 km / s. Explanation: Let : h = 6 . 9 × 10 6 m , g = 9 . 81 m / s 2 , and r = 6 . 37 × 10 6 m .

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solution-hw20 - lai(yl8859 HW20 Ross(15200 This print-out...

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