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Unformatted text preview: Alvarado, Patrick – Homework 9 – Due: Nov 9 2006, 10:00 am – Inst: Andrei Sirenko 1 This printout should have 17 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 2) 10 points When it orbited the Moon, the Apollo 11 spacecraft’s mass was 8350 kg, and its mean distance from the Moon’s center was 2 . 39367 × 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: 1432 . 37 m / s. Explanation: The gravitational force is 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 G M d = q 6 . 67259 × 10 11 Nm 2 / kg 2 × s 7 . 36 × 10 22 kg 2 . 39367 × 10 6 m = 1432 . 37 m / s . 002 (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: 8 . 56574 × 10 9 J. Explanation: The total energy is E = K + U = 1 2 m v 2 G m M d . In the first part of the problem we found an expression for v in terms of G , M and d . Substituting this expression in the formula for the total energy, we obtain E = 1 2 m G M d G m M d = G m M 2 d = 6 . 67259 × 10 11 Nm 2 / kg 2 × 8350 kg × 7 . 36 × 10 22 kg 2 × 2 . 39367 × 10 6 m = 8 . 56574 × 10 9 J . The minimum energy required for the craft to leave the orbit and escape the Moon’s gravi tational field is equal to the absolute value of the total energy: E min = k ~ E k = 8 . 56574 × 10 9 J . keywords: 003 (part 1 of 1) 10 points Earth’s gravitational field is 7 . 61 N / kg at the altitude of the space shuttle. What is the size of the force of attraction betweenastudentofmass51 . 3kg andEarth? Correct answer: 390 . 393 N. Explanation: The force of attraction between the student and the Earth is his weight, W = m g keywords: 004 (part 1 of 1) 10 points Given: G = 6 . 67259 × 10 11 Nm 2 / kg 2 . An apparatus like the one Cavendish used to find G has large lead balls that are 5 . 7 kg in mass and small ones that are 0 . 0733 kg. The center of a large ball is separated by 0 . 054 m from the center of a small ball. Mirror m Light source M r κ Alvarado, Patrick – Homework 9 – Due: Nov 9 2006, 10:00 am – Inst: Andrei Sirenko 2 Schematic diagram of the Cavendish ap paratus for measuring G . As the small spheres of mass m are attracted to the large spheres of mass M , the rod between the two small spheres rotates through a small angle. A light beam reflects from a mirror, fixed to the axis of the small masses. A torsion ribbon rotates (shown with torsion constant κ ) holding the mir ror and the small masses. The skewed dashed line represents the position of the rod when the masses are at a large dis tance....
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This note was uploaded on 09/04/2008 for the course PHYS 106 taught by Professor Opyt during the Spring '08 term at NJIT.
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