Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - Problems - Page 767: 10

Answer

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Work Step by Step

a. An Earth observer measures 21.6 ly as the (rest) length. The time is the distance divided by the speed. $$t_E=\frac{21.6\;ly}{0.950c}=22.74y\approx 22.7y$$ b. The time measured by the spacecraft is the proper time. Use equation 26–1a. $$\Delta t_0=\Delta t\sqrt{1-v^2/c^2}=(22.74y) \sqrt{1-(0.950)^2}=7.10y$$ c. The spacecraft observers measure the contracted distance. Use equation 26–3a. $$\mathcal{l} = \mathcal{l}_{o} \sqrt{1-\frac{ v^{2} } { c^{2} }} = 21.6ly\sqrt{1-0.950^2}=6.74ly$$ d. Spacecraft observers calculate their speed as distance divided by time. $$v=\frac{6.7446ly}{7.10y}=0.950c$$ This is the expected answer.
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