Essential University Physics: Volume 1 (4th Edition)

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

Chapter 3 - Section 3.6 - Uniform Circular Motion - Example - Page 47: 3.7

Answer

$ 1.52 \ h$

Work Step by Step

Using the equation for period, we find: $T = \sqrt{\frac{4\pi^2r}{a}}$ $T = \sqrt{\frac{4\pi^2(6.77\times10^6)}{8.73}}=5532\ s= 1.52 \ h$
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