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Unformatted text preview: Morby, Grant Homework 27 Due: Apr 14 2006, noon Inst: Drummond 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A newly discovered planet has twice the mass of the Earth, but the acceleration due to grav ity on the new planets surface is exactly the same as the acceleration due to gravity on the Earths surface. The radius R p of the new planet in terms of the radius R of Earth is 1. R p = 2 R . correct 2. R p = 2 R . 3. R p = 4 R . 4. R p = 1 2 R . 5. R p = 2 2 R . Explanation: From Newtons second law and the law of universal gravitation, the gravitational force near the surface is F g = m g = G M m r 2 g = G M r 2 . Now M p = 2 M e and g p = g e , so G M e R 2 = G M p R 2 p = 2 G M e R 2 p 1 R 2 = 2 R 2 p R p = 2 R . 002 (part 1 of 1) 10 points Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass, as shown below. Each star has mass M and speed v . G is the universal gravitational constant. D v v M M Which of the following is a correct relation ship among these quantities? 1. v 2 = G M 2 D correct 2. v 2 = G M D 2 3. v 2 = 4 G M 2 D 4. v 2 = G M D 5. v 2 = M G D 6. v 2 = 2 G M D 7. v 2 = 2 G M 2 D 8. v 2 = 4 G M D Explanation: From circular orbital movement, the cen tripetal acceleration is a = v 2 D 2 . Using the Newtons second law of motion, we know the acceleration is a = F M , where F is the force between two stars and is totally supplied by the universal force. So we obtain 2 v 2 D = a = F M = G M D 2 = v 2 = G M 2 D . 003 (part 1 of 1) 10 points Morby, Grant Homework 27 Due: Apr 14 2006, noon Inst: Drummond 2 How much less would you weigh on the top of Mount Everest (elevation 8850 m) than at sea level? 1. 0.5% 2. 5% 3. 0.3% correct 4. 2% 5. 1.5% 6. 3% 7. 0.2% 8. 0.4% 9. 0.1% Explanation: Let : M E = 6 10 24 kg , R E = 6 . 4 10 6 m , h = 8850 m , and G = 6 . 67 10 11 N m 2 / kg 2 . At sea level, g = G M E R 2 E = (6 . 67 10 11 N m 2 / kg 2 ) (6 10 24 kg) (6 . 4 10 6 m) 2 = 9 . 77051 m / s 2 . At the top of Mount Everest, g = G M E ( R E + h ) 2 = (6 . 67 10 11 N m 2 / kg 2 ) (6 10 24 kg) (6 . 4 10 6 m + 8850 m) 2 = 9 . 74354 m / s 2 . Since W = m g , then W W = g g g = 1 g g = 1 9 . 74354 m / s 2 9 . 77051 m / s 2 100% = . 27599% . 004 (part 1 of 1) 10 points A m = 78 . 7 kg object is released from rest at a distance h = 0 . 624272 R above the Earths surface, where R is the radius of the earth. The acceleration of gravity is 9 . 8 m / s 2 ....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.
 Spring '09
 KLEINMAN
 Physics, Work

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