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# hw8 - karna(pk4534 HW 8 coker(58245 This print-out should...

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karna (pk4534) – HW 8 – coker – (58245) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This assignment involves the last chapter that will be covered on upcoming Quiz 2. 001 10.0 points Which of the objects a. a book b. the nearest star c. the Sun d. a distant galaxy exert(s) a gravitational force on you? 1. c 2. b 3. a and b 4. b, c and d 5. Another combination 6. c and d 7. a 8. a and d 9. a, b, c and d correct 10. a, b and c Explanation: All objects exert a gravitational force on you, no matter how far away they are. 002 10.0 points Planet X has nine times the diameter and two times the mass of the earth. What is the ratio g X : g e of gravitational acceleration at the surface of planet X to the gravitational acceleration at the surface of the Earth? 1. g x g e = 8 9 2. g x g e = 3 4 3. g x g e = 4 9 4. g x g e = 9 64 5. g x g e = 1 12 6. g x g e = 2 81 correct 7. g x g e = 1 7 8. g x g e = 1 4 9. g x g e = 1 2 10. g x g e = 6 49 Explanation: Let : M x = 2 M e , and R x = 9 R e . mg e = GM e m R 2 e g e = GM e R 2 e . Similarly, g x = GM x R 2 x . The ratio is g x g e = M x R 2 e M e R 2 x = 2 M e R 2 e M e (9 R e ) 2 = 2 81 . keywords: 003 10.0 points The mass of a certain neutron star is 9 × 10 30 kg (4 . 5 solar masses) and its radius is 3500 m.

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karna (pk4534) – HW 8 – coker – (58245) 2 What is the acceleration of gravity at the surface of this condensed, burned-out star? The value of the universal gravitational con- stant is 6 . 67 × 10 11 N · m 2 / kg 2 . Correct answer: 4 . 90041 × 10 13 m / s 2 . Explanation: Let : M = 9 × 10 30 kg , r = 3500 m , and G = 6 . 67 × 10 11 N · m 2 / kg 2 . The gravitational acceleration is g = GM r 2 = ( 6 . 67 × 10 11 N · m 2 / kg 2 ) (9 × 10 30 kg) (3500 m) 2 = 4 . 90041 × 10 13 m / s 2 , several hundreds of billion times g on earth. 004 (part 1 of 2) 10.0 points You weigh 300 lb. How much would you weigh if you were standing on a mountain 200 km tall (equiva- lent to standing still at about the altitude of a space shuttle orbit)? 4.45 N = 1 lb, the grav- itational constant is 6 . 67 × 10 11 N · m 2 / kg 2 , the radius of the Earth is 6 . 4 × 10 6 m, and the mass of the Earth is 6 × 10 24 kg. Correct answer: 1251 . 54 N. Explanation: Let : W = 300 lb , h orbit = 200 km = 2 × 10 5 m , r E = 6 . 4 × 10 6 m , and G = 6 . 67 × 10 11 N · m 2 / kg 2 . The gravitational acceleration at an alti- tude of 200 km is g = GM E r 2 = GM E ( r E + h orbit ) 2 = (6 . 67 × 10 11 N · m 2 / kg 2 ) (6 × 10 24 kg) (6 . 4 × 10 6 m + 2 × 10 5 m) 2 = 9 . 18733 m / s 2 . Your weight is W = 300 lb × 4 . 45 N lb = 1335 N , so your mass is m = W g = 1335 N 9 . 8 m / s 2 = 136 . 224 kg . At an altitude of 200 km, your weight would be W = mg = (136 . 224 kg) (9 . 18733 m / s 2 ) = 1251 . 54 N . 005 (part 2 of 2) 10.0 points How much does this differ from your weight on the surface of the Earth? Correct answer: 18 . 7553 lb. Explanation: W - W = 1335 N - 1251 . 54 N = 83 . 461 N × 1 lb 4 . 45 N = 18 . 7553 lb , which would be easily detected on a bathroom scale.
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hw8 - karna(pk4534 HW 8 coker(58245 This print-out should...

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