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Unformatted text preview: practice 07 ALIBHAI, ZAHID Due: Mar 4 2007, 4:00 am 1 Question 1 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 = GM D 2 2. v 2 = 2 GM D 3. v 2 = 4 GM 2 D 4. v 2 = M GD 5. v 2 = 2 GM 2 D 6. v 2 = GM 2 D correct 7. v 2 = 4 GM D 8. v 2 = GM 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 = GM D 2 = v 2 = GM 2 D . Question 2 part 1 of 2 10 points Compare the gravitational force on a 45 . 5 kg mass at the surface of the Earth ( R E = 6 . 4 10 6 m, M E = 6 10 24 kg) with that on the surface of the Moon ( M M = 1 81 . 3 M E , R M = 0 . 27 R E ). What is it on the Earth? Correct answer: 444 . 558 N (tolerance 1 %). Explanation: Let : m = 45 . 5 kg , and R E = 6 . 4 10 6 m . F = GmM E R 2 E = (6 . 67 10 11 N m 2 / kg 2 ) (45 . 5 kg) (6 . 4 10 6 m) 2 (6 10 24 kg) = 444 . 558 N . Question 3 part 2 of 2 10 points What is it on the Moon? Correct answer: 75 . 0085 N (tolerance 1 %). Explanation: F = GmM M R 2 M = (6 . 67 10 11 N m 2 / kg 2 ) (45 . 5 kg) (0 . 27 6 . 4 10 6 m) 2 (6 10 24 kg / 81 . 3) = 75 . 0085 N . practice 07 ALIBHAI, ZAHID Due: Mar 4 2007, 4:00 am 2 Question 4 part 1 of 1 10 points If a book is sitting on a table, the force of gravity is pulling it down. Why doesnt it fall? 1. The book has a pushing force on the table. 2. The book is not heavy enough. 3. The table is pushing up on it with a force equal to the weight of the book. correct Explanation: The table exerts an equal but opposite force on the book. Question 5 part 1 of 2 10 points Given: G = 6 . 6726 10 11 N m 2 / kg 2 . Three masses are arranged in the ( x,y ) plane (as shown in the figure below, where the scale is in meters). y ( m ) 5 3 1 0 1 2 3 4 5 x ( m ) 5 4 3 2 1 1 2 3 4 5 6 kg 3 kg 4 kg What is the magnitude of the resulting force on the 6 kg mass at the origin? Correct answer: 9 . 30337 10 11 N (tolerance 1 %). Explanation: Let : m o = 6 kg , ( x o ,y o ) = (0 m , 0 m) , m a = 3 kg , ( x a ,y a ) = (3 m , 4 m) , m b = 4 kg , ( x b ,y b ) = (5 m , 1 m) . Newtons Universal Gravitational Law for m o and m a is F ao = G m o m a radicalbig ( x a x o ) 2 + ( y a y o ) 2 = (6 . 6726 10 11 N m 2 / kg 2 ) (6 kg) (3 kg) radicalbig (3 m) 2 + (4 m) 2 = 4 . 80427 10 11 N ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Center Of Mass, Mass, Work

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