HW 27 - Morby Grant Homework 27 Due noon Inst Drummond This...

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Morby, Grant – Homework 27 – Due: Apr 14 2006, noon – Inst: Drummond 1 This print-out should have 11 questions. Multiple-choice 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 planet’s surface is exactly the same as the acceleration due to gravity on the Earth’s 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 Newton’s 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 Newton’s 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
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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 0 = 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 Δ W W = g - g 0 g = 1 - g 0 g = 1 - 9 . 74354 m / s 2 9 . 77051 m / s 2 × 100% = 0 . 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 Earth’s surface, where R is the radius of the earth. The acceleration of gravity is 9 . 8 m / s 2 . Find the speed of the object when it strikes the Earth’s surface. (Neglect any atmospheric friction. G = 9 . 8 m / s 2 . For the Earth, R = 6 . 38 × 10 6 m, M = 5 . 98 × 10 24 kg. It implies that the gravitational acceleration at the surface of the earth is g = 9 . 8 m/s 2 . ) Caution: You must take into account that the gravitational acceleration depends on dis- tance between the object and the center of the earth.
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