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: 80

Answer

$7.55\times10^{-5 }\;\rm s\approx 75.5\;\mu s$

Work Step by Step

We know that the Earth is rotating around its axis and this axis starts from the north pole. In this way, the clock on north pole is stationary while the clock on the equator is rotating around the Earth's axis at a constant speed of $$v=\dfrac{2\pi R_E}{T}$$ where $T$ is the periodic time of one full rotation which complete in one day, and $R_E$ is the Earth's radius on the equator. Plugging the known; $$v=\dfrac{2\pi \times 6.38\times10^6}{24\times 60^2}=\bf 464\;\rm m/s$$ Now we need to find the time for each frame reference. $$\Delta t_{\rm equator}=\dfrac{\Delta t}{\gamma}=\dfrac{\Delta t}{\dfrac{1}{\sqrt{1-\frac{v_{\rm equator }^2}{c^2}}}}$$ $$\Delta t_{\rm equator } = \Delta t \;{\sqrt{1-\frac{v_{\rm equator }^2}{c^2}}} \tag 1$$ And the time in north pole is the same $\Delta t$ since the velocity of the clock there is zero. $$\Delta t_{\rm north\;pole}=\Delta t\tag 2$$ Therefore, the time difference between the two clocks is given by $$\Delta t_{\rm north\;pole}-\Delta t_{\rm equator } = \Delta t \;{\sqrt{1-\frac{v_{\rm equator }^2}{c^2}}}-\Delta t$$ $$\Delta t_{\rm north\;pole}-\Delta t_{\rm equator } = \Delta t \left[{\sqrt{1-\frac{v_{\rm equator }^2}{c^2}}}-1\right]$$ Using the the binomial expansion as the author told us. Since $\dfrac{v^2}{c^2}\lt \lt 1$, so $$\sqrt{1-\frac{v_{\rm equator }^2}{c^2}}=1+\frac{1}{2}\dfrac{v^2}{c^2}$$ Thus, $$\Delta t_{\rm north\;pole}-\Delta t_{\rm equator } = \Delta t \left[1+\dfrac{v_{\rm equator }^2}{2c^2}-1\right]$$ $$\Delta t_{\rm north\;pole}-\Delta t_{\rm equator } = \Delta t \left[ \dfrac{v_{\rm equator }^2}{2c^2} \right]$$ Plugging the known; $$\Delta t_{\rm north\;pole}-\Delta t_{\rm equator } = 2\times 365.25\times 24\times 60^2 \left[ \dfrac{464^2}{2(3\times10^8)^2} \right]$$ $$\Delta t_{\rm north\;pole}-\Delta t_{\rm equator } =\color{red}{\bf 7.55\times10^{-5}}\;\rm s\approx 75.5\;\mu s$$
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