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Unformatted text preview: Bautista, Aldo – Homework 14 – Due: Dec 7 2005, 4:00 am – Inst: Maxim Tsoi 1 This printout should have 18 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 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 = 4 R . 3. R p = 1 2 R . 4. R p = √ 2 2 R . 5. R p = 2 R . Explanation: From Newton’s second law and the law of universal gravitation, the gravitational force near the surface is F g = mg = G M m r 2 g = GM r 2 . Now M p = 2 M e and g p = g e , so GM e R 2 = GM p R 2 p = 2 GM 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 = 2 GM 2 D 2. v 2 = 2 GM D 3. v 2 = 4 GM 2 D 4. v 2 = M GD 5. v 2 = 4 GM D 6. v 2 = GM 2 D correct 7. v 2 = GM D 2 8. v 2 = GM 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 = GM D 2 = ⇒ v 2 = GM 2 D . 003 (part 1 of 2) 10 points Bautista, Aldo – Homework 14 – Due: Dec 7 2005, 4:00 am – Inst: Maxim Tsoi 2 When it orbited the Moon, the Apollo 11 spacecraft’s mass was 15500 kg, and its mean distance from the Moon’s center was 2 . 44807 × 10 6 m. Assume its orbit was circu lar and the Moon to be a uniform sphere of mass 7 . 36 × 10 22 kg. Given the gravitational constant G is 6 . 67259 × 10 11 Nm 2 / kg 2 , calculate the or bital speed of the spacecraft. Correct answer: 1416 . 36 m / s. Explanation: The Gravitational force is going to be the centripetal force, which is, GMm d 2 = m v 2 d (where d is the mean distance from the space craft to the Moon’s center), we find v = r GM d = q 6 . 67259 × 10 11 Nm 2 / kg 2 × s 7 . 36 × 10 22 kg 2 . 44807 × 10 6 m = 1416 . 36 m / s . 004 (part 2 of 2) 10 points What is the minimum energy required for the craft to leave the orbit and escape the Moon’s gravitational field? Correct answer: 1 . 55471 × 10 10 J. Explanation: The total energy is E = K + U = 1 2 mv 2 GmM d ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
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