Homework 14

# Homework 14 - Bautista Aldo – Homework 14 – Due Dec 7...

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Unformatted text preview: Bautista, Aldo – Homework 14 – Due: Dec 7 2005, 4:00 am – Inst: Maxim Tsoi 1 This print-out should have 18 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 = 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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Homework 14 - Bautista Aldo – Homework 14 – Due Dec 7...

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