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: 53

Answer

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Work Step by Step

a) We know the speed of our solar system (which is 230 km/s) and we know the radius of its circular path around the center of the galaxy (which is 25,000 light-year). Thus, the period is given by $$v=\dfrac{2\pi R}{T}$$ $$T=\dfrac{2\pi R}{v}$$ Plugging the known; where $$R=25,000\;\rm ly(\frac{3\times 10^8\;m}{1\;s})(\frac{365.25\;day}{1\;yr})(\frac{24\;h}{1\;day})(\frac{60^2\;s}{1\;h})\\ =\bf2.367\times10^{20}\;\rm m\tag 1$$ $$T=\dfrac{2\pi (2.367\times10^{20})}{230\times 1000}=\bf 6.47\times 10^{15}\;\rm s$$ $$T=\color{red}{\bf2.05\times10^8}\;\rm yr$$ ________________________________________________ b) The number of full rotations is given by $$N=\dfrac{\Delta t}{T}=\dfrac{5\times 10^{9}}{2.05\times10^8}=24.4\rm \;rev\approx \color{red}{\bf24}\;rev$$ ________________________________________________ c) According to Newton's gravitational law, the force exerted by the center of the Milky Way galaxy on our solar system is given by $$F=\dfrac{GM_{galaxy} \color{red}{\bf\not}M_{solar}}{R^2}= \color{red}{\bf\not}M_{solar}a_r$$ Hence, $$ \dfrac{GM_{galaxy} }{R^{ \color{red}{\bf\not}2}} =\dfrac{v^2}{ \color{red}{\bf\not}R}$$ So $$M_{galaxy}=\dfrac{Rv^2}{G}$$ Plugging the known and plug from (1); $$M_{galaxy}=\dfrac{(2.367\times10^{20})(230\times 10^3)^2}{6.67\times 10^{-11}}$$ $$M_{galaxy}=\color{red}{\bf 1.88\times10^{41}}\;\rm kg$$ ________________________________________________ d) The number of stars is given by $$N_{star}=\dfrac{M_{galaxy}}{M_{Sun}}=\dfrac{1.88\times10^{41}}{1.99\times 10^{30}}$$ $$N_{star}=\color{red}{\bf 9.43\times 10^{10}}\;\rm Star$$
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