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(a) Suppose you are at the earth’s equator and observe a

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

Solution for problem 55P Chapter 13

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 55P

(a) Suppose you are at the earth’s equator and observe a satellite passing directly overhead and moving from west to east in the sky. Exactly 12.0 hours later, you again observe this satellite to be directly overhead. How far above the earth’s surface is the satellite’s orbit? (b) You observe another satellite directly overhead and traveling east to west. This satellite is again overhead in 12.0 hours. How far is this satellite’s orbit above the surface of the earth?

Step-by-Step Solution:

Solution 55P Step 1 a) For a satellite to orbit around the earth, the gravitational force should be equal to the centripetal force. That is, GMm/R = mv /R 2 2 2 Or we can write, GMm/R = m R Where, M - Mass of the earth m - Mass of the satellite R - Distance of the satellite from the centre of the earth - Angular velocity of the satellite We can write, = 2 / T Where T is the time period of the satellite. Therefore, GM/R = 4 / T 2 2 2 2 3 Or, T = 4 R /GM Step 2 Mass of the earth, M = 5.792 ×10 24kg Rearranging the previous equation to get R 3 3 2 2 R = GMT / 4 Provided, period of the satellite, T = 12 hours. Converting it into seconds we get, T = 12 × 60 × 60 seconds = 43200 s 3 -11 2 2 24 2 2 2 Then, R = (6.67×10 Nm /kg ×5.792×10 kg×43200 s ) / 4×3.14 R = (1.83×10 22m ) Taking cube root on both sides, R = 2.63 ×10 m 7

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Chapter 13, Problem 55P is Solved
Step 4 of 4

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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