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Sedna. In November 2003 the now-most-distant-known object

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 52E Chapter 10

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 52E

Sedna.? In November 2003 the now-most-distant-known object in the solar system was discovered by observation with a telescope on Mt. Palomar. This object known as Sedna, is approximately 1700 km in diameter, takes about 10.500 years to orbit our sun, and reaches a maximum speed of 4.64 km/s. Calculations of its complete path, based on several measurements of its position, indicate that its orbit is highly elliptical, varying from 76 AU to 942 AU in its distance from the sun, where AU is the astronomical unit, which is the average distance of the earth from the sun (1.50 × 108 km). (a) What is Sedna’s minimum speed? (b) At what points in its orbit do its maximum and minimum speeds occur? (c) What is the ratio of Sedna’s maximum kinetic energy to its minimum kinetic energy?

Step-by-Step Solution:

Solution 52E Step 1: We know that, the angular momentum of an object, L = I Where, I - moment of inertia of the object - Angular velocity of the object We know that, I = mr and = v/r Where, v - linear velocity of the object and r - distance of center of mass. Then, L = mr (v/r) L = mrv Step 2: We know that, in planetary motion, the angular momentum of the system is conserved at any point. Consider, the mass of Sedna as “m” and radius of aphelion as r and velocity at aphelion a as v . a Similarly, the radius of perihelion is r and velocity at perihelion is v . P P Therefore, we can write, L = La P Angular momentum at aphelion = angular momentum at perihelion Provided, the maximum speed of sedna is, 4.64 km/ s. This should be the velocity of sedna at perihelion because, perihelion is the maximum closest point of a planet with the sun. Provided, r = 76PU and r = 942 AU a 8 We know that, 1 AU = 1.496 × 10 km Therefore, r = 76 × 1.496 × 10 km = 1.137 × 10 km 10 P r = 942 × 1.496 × 10 km = 1.409 × 10 km 11 a Step 3: a) According to conservation of angular momentum, we can write, m v ra a r P P We cancel out the term “m” on both sides. Therefore, va a P P Substituting the values for r , v and r we get, a P P v × 1.409 × 10 km = 4.64 km/ s × 1.137 × 10 km 10 a Rearranging the equation to get v . a 10 11 va (4.64 km/ s × 1.137 × 10 km) / (1.409 × 10 km) v = 0.374 km /s a Minimum speed of sedna is, 0.374 km/s

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Chapter 10, Problem 52E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Sedna. In November 2003 the now-most-distant-known object

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