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 - General Problems - Page 770: 72

Answer

$1.8 \times10^{22 }\;\rm J$

Work Step by Step

We know that the work needed is given by $$W=\Delta K=\left(\gamma-1\right)mc^2$$ And hence, $$W= \left(\sqrt{\dfrac{1}{1-\frac{v^2}{c^2}}}-1\right)mc^2\tag 1$$ Now we need to find the speed of the spaceship $v$ that will give this work. According to the onboard astronauts, the distance traveled is given by $$L= \dfrac{ L_{1}}{\gamma}$$ where $L_1$ is the distance traveled according to an observer on Earth. And w eknow that the time the journey would take relative to the onboard astronauts is given by $$t=\gamma t_1=1\;\rm year$$ where $t_1$ is the time interval the journey would take according to an observer on Earth. Therefore, the velocity of the spaceship is given by $$v=\dfrac{L}{t}$$ Plugging from the previous two formulas; $$v=\dfrac{ \overbrace{L_{1}}^{6.6c}}{\gamma t}=\dfrac{ 6.6c}{ \sqrt{\dfrac{1}{1-\frac{v^2}{c^2}}} \;t}$$ $$v= \dfrac{6.6c\;\times \;\sqrt{ 1-\frac{v^2}{c^2 }}}{ t}$$ $$v^2= \dfrac{6.6^2c^2\;\times \; \left[1-\frac{v^2}{c^2 }\right] }{ t^2}$$ $$v^2= \dfrac{6.6^2c^2\;\times \; \left[ \frac{c^2\;-\;v^2}{c^2 }\right] }{ t^2}$$ $$44.56 v^2= \dfrac{6.6^2 c^2 }{ t^2}$$ Thus, $$\dfrac{v^2}{c^2}= \dfrac{6.6^2 }{ 44.56 t^2}$$ Plugging into (1); $$W= \left(\sqrt{\dfrac{1}{1-\dfrac{6.6^2 }{ 44.56 t^2}}}-1\right)mc^2 $$ Plugging the known; $$W= \left(\sqrt{\dfrac{1}{1-\dfrac{6.6^2 }{ 44.56 \times 1^2}}}-1\right)\times 36000\times(3\times10^8)^2 $$ $$W=\color{red}{\bf1.8 \times10^{22}}\;\rm J$$
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