Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 36 - Relativity - Exercises and Problems - Page 1059: 16

Answer

(a) $v = 0.9965~c$ (b) $d = 119.6~ly$

Work Step by Step

(a) Let $t' = 120~y$ and let $t_0 = 10~y$. We can find the required speed in the moving reference frame: $t' = \gamma~t_0$ $t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~t_0$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{t_0}{t'}$ $1-\frac{v^2}{c^2} = (\frac{t_0}{t'})^2$ $\frac{v^2}{c^2} = 1-(\frac{t_0}{t'})^2$ $v^2 = [1-(\frac{t_0}{t'})^2] \times c^2$ $v = \sqrt{1-(\frac{t_0}{t'})^2}~\times c$ $v = \sqrt{1-(\frac{10~y}{120~y})^2}~\times c$ $v = 0.9965~c$ (b) We can find the distance to the distant star according to Mission Control: $d = v~t$ $d = (0.9965~c)(120~y)$ $d = 119.6~ly$
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