Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 354: 30


The orbital period is 4.2 hours.

Work Step by Step

Let $M$ be the mass of the earth. We can use the satellite's orbital speed to find the radius of the satellite's orbit. $v = \sqrt{\frac{G~M}{R}}$ $R = \frac{G~M}{v^2}$ $R = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)}{(5500~m/s)^2}$ $R = 1.32\times 10^7~m$ We can find the time $T$ of one orbit. $T = \frac{distance}{speed}$ $T = \frac{2\pi~R}{v}$ $T = \frac{(2\pi)(1.32\times 10^7~m)}{5500~m/s}$ $T = 15,080~s = 4.2~hours$ The orbital period is 4.2 hours.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.