Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - Problems - Page 134: 52

Answer

T = 84.5 ~minutes This result does not depend on the mass of the satellite.

Work Step by Step

$T^2 = \frac{4 \pi^2 r^3}{GM}$ $T^2 = \frac{4 \pi^2 r}{g}$ $T = \sqrt{\frac{4 \pi^2 r}{g}} = 2\pi \sqrt{\frac{r}{g}} = 2\pi \sqrt{\frac{6.38 \times 10^6 ~m}{9.80 ~m/s^2}} $ $T = 5070~s = 84.5 ~minutes$ This result does not depend on the mass of the satellite.
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