hw12 - patel(kp9583 HW12 Mackie(20211 This print-out should...

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patel (kp9583) – HW12 – Mackie – (20211) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points The period oF the earth around the sun is 1 year and its distance is 150 million km From the sun. An asteroid in a circular orbit around the sun is at a distance 229 million km From the sun. What is the period oF the asteroid’s orbit? Correct answer: 1 . 88632 year. Explanation: Let : T e = 1 year , r e = 150 million km , and r a = 229 million km . ±rom Kepler’s laws, T 2 e r 3 e = T 2 a r 3 a T a = p r a r e P 3 / 2 T e = p 229 million km 150 million km P 3 / 2 (1 year) = 1 . 88632 year . 002 (part 2 oF 2) 10.0 points What is the orbital velocity oF the asteroid? Assume there are 365 days in one year. Correct answer: 24187 . 6 m / s. Explanation: v a = 2 π r a T a = 2 π (2 . 29 × 10 11 m) 1 . 88632 year 1 y 365 d × 1 d 24 h 1 h 3600 s = 24187 . 6 m / s . 003 (part 1 oF 2) 10.0 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 . A 744 kg geosynchronous satellite or- bits a planet similar to Earth at a radius 1 . 95 × 10 5 km From the planet’s center. Its angular speed at this radius is the same as the rotational speed oF the Earth, and so they ap- pear stationary in the sky. That is, the period oF the satellite is 24 h . What is the Force acting on this satellite? Correct answer: 767 . 255 N. Explanation: Let : G = 6 . 67259 × 10 11 N m 2 / kg 2 , M satellite = 744 kg , and T = 86400 s . Solution: v = 2 π r T F = M satellite v 2 r = M satellite 4 π 2 r T 2 (1) = (744 kg) 4 π 2 (1 . 95 × 10 8 m) (86400 s) 2 = 767 . 255 N . 004 (part 2 oF 2) 10.0 points What is the mass oF this planet? Correct answer: 5 . 87681 × 10 26 kg. Explanation: Using the general gravitation law and Eq. 1, we have F = G M satellite M planet r 2 = M satellite 4 π 2 r T 2 G M satellite M planet r 2 = M satellite 4 π 2 r T 2 , so
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patel (kp9583) – HW12 – Mackie – (20211) 2 M planet = 4 π 2 GT 2 r 3 = 4 π 2 (6 . 67259 × 10 11
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hw12 - patel(kp9583 HW12 Mackie(20211 This print-out should...

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