HW 28 - Morby, Grant Homework 28 Due: Apr 14 2006, noon...

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Morby, Grant – Homework 28 – Due: Apr 14 2006, noon – Inst: Drummond 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Consider the orbit oF a typical comet around the sun, which is marked at fve di±erent po- sitions, X , S , Z , P , and U . Sun a b P S U X Z Using Kepler’s second law oF planetary mo- tion, rank those positions in order oF thier rel- ative speeds, with the position For the Fastest speed frst. 1. X Z P S U 2. Z P S X U 3. U X S P Z 4. Z X U S P 5. U S P Z X 6. P S U X Z correct Explanation: Note : ( f, 0) = p a 2 - b 2 , 0 · e = a 2 - b 2 a . The closer the comet is the the Sun, the Faster it is traveling, thereFore ( P , S , U , X , Z ) is correct. 002 (part 1 oF 1) 10 points A “geosynchronous” Earth satellite can re- main directly overhead in which oF the Follow- ing cities? 1. Sidney 2. Singapore correct 3. San ²rancisco 4. London 5. Moscow Explanation: An Earth satellite is geosynchronous when its period and the direction oF motion are the same as Earth’s spin motion. This occurs only as the satellite is overhead in a point which lies on the Earth’s equator. 003 (part 1 oF 1) 10 points The radius oF the satellite’s orbit around the center oF the Earth is r = 9 2 R . Hint: You may fnd it useFul to take into account that the gravitational Force is a con- servative Force. Hint: The universal gravitational Force law is ~ F = G M m r 2 ˆ r . Earth Satellite R 9 2 Caution: Neglect the rotational kinetic en- ergy due to the Earth’s rotation.
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Morby, Grant – Homework 28 – Due: Apr 14 2006, noon – Inst: Drummond 2 Find the energy required to launch a satel- lite from Earth into the circular orbit at the speci±ed radius r = 9 2 R . 1. E = 10 9 GM m R 2. E = 7 8 GM m R 3. E = 2 3 GM m R 4. E = 7 9 GM m R 5. E = 8 9 GM m R correct 6. E = 7 6 GM m R 7. None of these 8. E = 11 9 GM m R 9. E = 9 8 GM m R 10. E = 3 2 GM m R Explanation: Basic Concepts: Force of gravity between two masses m 1 and m 2 at a distance r F g = G m 1 m 2 r 2 and U g = G m 1 m 2 r , where G is the gravitational constant. Centripetal acceleration for circular motion is a r = v 2 r , for the radial acceleration, where v is the tangential speed. Solution: Since the gravitational force sup- plies the centripetal force, ma r = m v 2 r = G mM r 2 so that mv 2 = G mM r = G mM 9 2 R . Then the kinetic energy of the satellite in orbit is found to be K r = 1 2 mv 2 = G mM 2 µ 9 2 R . Conservation of energy gives us U R + Δ E = U r + K r Δ E = U r - U R + K r = - GM m 9 2 R + GM m R + GM m 9 R = GM m R µ - 2 9 + 9 9 + 1 9 = 8 9 GM m R . 004
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HW 28 - Morby, Grant Homework 28 Due: Apr 14 2006, noon...

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