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: 43

Answer

$3.71\times 10^5\;\rm m/s$

Work Step by Step

Since the three stars are identical and since they are released from rest from equal distances, their crash speeds are identical as well. Let's assume that the system [the three stars] is isolated which means that energy and momentum are conserved. $$E_i=E_f$$ $$K_i+U_{gi}=K_{f}+U_{gf}$$ Recall that $K_i=0$ since the three stars are released from rest. $$0+ \dfrac{-GM^2}{R_{i,12} }+ \dfrac{-GM^2}{R_{i,13} }+ \dfrac{-GM^2}{R_{i,23} }=\\ \frac{1}{2}Mv_{1f}^2+\frac{1}{2}Mv_{2f}^2+\frac{1}{2}Mv_{3f}^2 +\dfrac{-GM^2}{R_{f,12} }+ \dfrac{-GM^2}{R_{f,13} }+ \dfrac{-GM^2}{R_{f,23} } $$ From the figure below, we can see that $R_{i,12}=R_{i,13}=R_{i,23}=d$, and that $R_{f,12}=R_{f,13}=R_{f,23}=2R_s$ where $R_s$ is the radius of the star. Also, their final speeds are equal, so $v_{1f}=v_{2f}=v_{3f}=v_f$ $$ \dfrac{- \color{red}{\bf\not}3G M^{\color{red}{\bf\not}2}}{d }=\frac{ \color{red}{\bf\not}3}{2} \color{red}{\bf\not}Mv_f^2 + \dfrac{- \color{red}{\bf\not}3G M^{\color{red}{\bf\not}2} }{2R_s } $$ $$ \dfrac{- 2G M}{d }=v_f^2 + \dfrac{- GM }{R_s } $$ Solving for $v_f$; $$v_f=\sqrt{ GM \left(\dfrac{-2}{d }+\dfrac{ 1}{R_s }\right)}$$ Plugging the known; $$v_f=\sqrt{ (6.67\times 10^{-11})(1.99\times 10^{30})\left(\dfrac{-2}{(5\times 10^9) }+\dfrac{ 1}{(6.96\times 10^8) }\right)}$$ $$v_f=\color{red}{\bf 3.71\times 10^5}\;\rm m/s$$
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