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Unformatted text preview: reyes (mr29667) Homework 7 chelikowsky (59005) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points If the earth were of uniform density, what would be the value of g inside the earth at half its radius? (The value of g at the surface of earth is 9.8 m/s 2 .) 1. 19.6 m/s 2 2. 9.8 m/s 2 3. 4.9 m/s 2 correct 4. 39.2 m/s 2 Explanation: Consider the spherical shell idea: only the mass of the earth in the inner shell con tributes to g ; the mass of the sphere of ra dius r 2 . To determine this smaller mass, con sider the effect on the volume: V r 3 , so V parenleftbigg 1 2 r parenrightbigg 3 = 1 8 r 3 ; 1 8 of the original volume means 1 8 of the original mass. Now G m r 2 , so G 1 8 m parenleftbigg 1 2 r parenrightbigg 2 = 1 8 m 1 4 r 2 = 1 2 m r 2 , and the accleration due to gravity at r 2 is g 2 . 002 10.0 points The planet Mars has a mass of 6 . 1 10 23 kg and radius of 3 . 8 10 6 m. What is the acceleration of an object in free fall near the surface of Mars? The value of the gravitational constant is 6 . 67259 10 11 N m 2 / kg 2 . Correct answer: 2 . 81875 m / s 2 . Explanation: Let : M = 6 . 1 10 23 kg , R = 3 . 8 10 6 m , and G = 6 . 67259 10 11 N m 2 / kg 2 . Near the surface of Mars, the gravitation force on an object of mass m is F = G M m R 2 , so the acceleration of an object in free fall is a = F m = G M R 2 = (6 . 67259 10 11 N m 2 / kg 2 ) 6 . 1 10 23 kg (3 . 8 10 6 m) 2 = 2 . 81875 m / s 2 . 003 10.0 points An apparatus like the one Cavendish used to find G has large lead balls that are 8 . 4 kg in mass and small ones that are 0 . 067 kg. The center of a large ball is separated by 0 . 058 m from the center of a small ball.from the center of a small ball....
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This note was uploaded on 07/24/2010 for the course PHY 59005 taught by Professor Chelikowsky during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Chelikowsky

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