Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 8 - Exercises and Problems - Page 147: 18


$ h = 1.69\times 10^6 \ m$

Work Step by Step

We simplify the equation for the period of a satellite for r to find: $r =\sqrt[3] {\frac{GMT^2}{4\pi^2}}$ Plugging in the value of G, a constant, the mass of the earth, and 7200 seconds for the period (T), we find: $r=8.06\times 10^6$ Subtracting the radius of the earth gives: $ h = 1.69\times 10^6 \ m$
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