Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems - Page 241: 79


a) The proof is below. b) The proof is below. c) $ T = 2\pi \sqrt{\frac{r_0^3}{GM}}$ d) 16 days

Work Step by Step

a) At equilibrium, the value of $\frac{dr^2}{dt^2}$ is 0, so we find: $0=4\pi r_0^2p_0-\frac{GMm}{r_0^2}$ $p_0= \frac{GMm}{4\pi r_0^4} $ b) When x is small compared to r, we see that $r_0\approx r$. We first find a value for $p$: $p =\frac{p_0r_0^5}{r^5}=\frac{GMm}{4\pi r_0^4}\frac{r_0^5}{r^5}=\frac{GMmr_0}{4\pi r^5}$ We plug this into the original equation to find: $=4\pi r^2 (\frac{GMmr_0}{4\pi r^5})-\frac{GMm}{r^2}$ $= (\frac{GMmr_0}{ r^3})-\frac{GMm}{r^2}$ We now have just r in the denominators. Using the binomial approximation, this becomes: $=-\frac{GMmx}{r_0^3}$ c) Following the steps that the book outlines, we find: $m\frac{d^2x}{dt^2}=-\frac{GMmx}{r_0^3}$ $\frac{d^2x}{dt^2}=-\frac{GMx}{r_0^3}$ Comparing this to the equations in the book that the problem asks us to look at, we find: $ T = 2\pi \sqrt{\frac{r_0^3}{GM}}$ d) Plugging in the given values, we find: $ T = 2\pi \sqrt{\frac{r_0^3}{GM}}$ $T=1,399,680 \ s \approx \fbox{16 days}$
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