Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 374: 50

Answer

$ T =\sqrt{\dfrac{4\pi^2r^3}{G\left(M+\dfrac{ m }{4 }\right)}} $

Work Step by Step

To find the period of the two planets, we need to find the net force exerted on one of them. $$\sum F_1=F_{G,\rm star\;on\;1}+F_{G,\rm 1\;on\;2}=ma_r=m\dfrac{v^2}{r}$$ where $r$ is the radius of the circular path of the planet. Hence, $$\dfrac{GM \color{red}{\bf\not}m}{r^2}+\dfrac{Gm \color{red}{\bf\not}m}{(2r)^2}=\dfrac{ \color{red}{\bf\not}mv^2}{r}$$ $$\dfrac{GM }{r^2}+\dfrac{Gm }{4r^2}=\dfrac{ v^2}{r}$$ recalling that $v=2\pi r/T$ $$\dfrac{GM }{r^2}+\dfrac{Gm }{4r^2}=\dfrac{ \left[\dfrac{2\pi r}{T}\right]^2}{r}=\dfrac{4\pi^2r^{ \color{red}{\bf\not}2}}{T^2 \color{red}{\bf\not}r}$$ $$\dfrac{G }{r^2}\left(M+\dfrac{ m }{4 }\right)= \dfrac{4\pi^2r }{T^2 }$$ Thus, $$T^2=\dfrac{4\pi^2r^3}{G\left(M+\dfrac{ m }{4 }\right)}$$ $$\boxed{T =\sqrt{\dfrac{4\pi^2r^3}{G\left(M+\dfrac{ m }{4 }\right)}}}$$
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.