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 149: 70


The proof is below.

Work Step by Step

We know that their velocity is equal to $v=\frac{T}{2\pi r}$. We know there is a centripetal force due to gravity, so it follows: $\frac{Mv^2}{r}=\frac {GM^2}{r^2}$ Using these relationships, we find: $\frac{r}{GM^2}=\frac{T^2}{m4\pi^2 r^2}$ This simplifies to: $T^2 = \frac{4\pi^2r^3}{GM}$
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