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 16 - Traveling Waves - Exercises and Problems - Page 454: 79


The galaxy is receding from the earth with a speed of $1.5\times 10^6~m/s$

Work Step by Step

Since the wavelength of the light from the galaxy is longer than the wavelength measured in a laboratory, the galaxy is receding from the earth. We can find the speed of the galaxy; $\lambda' = \lambda_0~\frac{\sqrt{1+v/c}}{\sqrt{1-v/c}}$ $1.005~\lambda_0 = \lambda_0~\frac{\sqrt{1+v/c}}{\sqrt{1-v/c}}$ $1.005 = \frac{\sqrt{1+v/c}}{\sqrt{1-v/c}}$ $(1.005)^2 = \frac{1+v/c}{1-v/c}$ $(1.005)^2(c-v) = c+v$ $(1.005)^2~v + v = (1.005)^2~c - c$ $v = \frac{(1.005)^2 - 1}{(1.005)^2 + 1}~c$ $v = \frac{(1.005)^2 - 1}{(1.005)^2 + 1}~(3.0\times 10^8~m/s)$ $v = 1.5\times 10^6~m/s$ The speed of the galaxy is $1.5\times 10^6~m/s$.
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