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hw18 - khounvivongsy(sk27799 Homework 18 Weathers(17104...

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khounvivongsy (sk27799) – Homework 18 – Weathers – (17104) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Halley’s comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0 . 714 AU and its greatest dis- tance being 32 . 2 AU (1 AU=the Earth-Sun distance). If the comet’s speed at closest approach is 31 . 3 km / s, what is its speed when it is farthest from the Sun? You may assume that its angular momen- tum about the Sun is conserved. Correct answer: 0 . 694043 km / s. Explanation: Using conservation of angular momentum, we have L apogee = L perihelion , or ( m r a 2 ) ω a = ( m r p 2 ) ω p , thus m r a 2 v a r a = m r p 2 v p r p , giving r a v a = r p v p , or v a = r p r a v p = (0 . 714 AU) (32 . 2 AU) (31 . 3 km / s) = 0 . 694043 km / s . 002 10.0 points A distant star has a single planet circling it in a circular orbit of radius 5 . 75 × 10 11 m. The period of the planet’s motion about the star is 781 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: 2 . 47024 × 10 31 kg. Explanation: Let : G = 6 . 67259 × 10 - 11 N · m 2 / kg 2 , R B = 5 . 75 × 10 11 m , and T B = 781 day . T B = (781 day) parenleftbigg 24 h 1 day parenrightbigg 3600 s 1 h = 6 . 74784 × 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 × (5 . 75 × 10 11 m) 3 (6 . 74784 × 10 7 s) 2 = 2 . 47024 × 10 31 kg . 003 10.0 points Given: G = 6 . 67259 × 10 - 11 N m 2 / kg 2 The acceleration of gravity on the surface of a planet of radius R = 6850 km is 7 . 24 m / s 2 . What is the period T of a satellite in circu- lar orbit h = 25345 km above the surface? Correct answer: 62273 . 4 s. Explanation: Since the gravity on the surface of a planet is given by g = GM/R 2 , we can solve for the mass of the planet: M = gR 2 G = (7 . 24 m / s 2 )(6 . 85 × 10 6 m) 2 6 . 67259 × 10 - 11 N m 2 / kg 2 = 5 . 09126 × 10 24 kg .
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