University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 13 - Gravitation - Problems - Exercises - Page 426: 13.24

Answer

The ISS is 373 km above the surface of the earth.

Work Step by Step

We can find the period $T$ of one orbit. $T = \frac{(24~hours/day)(3600~s/hour)}{15.65~rev/day}$ $T = 5521~s$ We can find the orbital radius $R$. $T^2 = \frac{4\pi^2~R^3}{G~M}$ $R^3 = \frac{G~M~T^2}{4\pi^2}$ $R = (\frac{G~M~T^2}{4\pi^2})^{1/3}$ $R = [\frac{(6.67\times 10^{-11}~m^3/kg~s^2)(5.98\times 10^{24}~kg)(5521~s)^2}{4\pi^2}]^{1/3}$ $R = 6753~km$ The distance above the earth's surface is 6753 km - 6380 km, which is 373 km. The ISS is 373 km above the surface of the earth.
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