A star has an element in its atmosphere that normally emits a line of frequency fs = 7.5 x 10^14 vib/s. If astronomers measure the frequency of this line to be fo = 7.7 x 10^14 vib/s, then how fast are the Earth and this star traveling relative to each other? Remember that the correct equation for the speed v is given by v = [(fo^2 - fs^2) / (fo^2 + fs^2)] c Remember fo^2 means "fo squared."
A star has an element in its atmosphere that normally emits a line of frequency fs = 7.5 x 10^14 vib/s. If astronomers measure the frequency of this line to be fo = 7.7 x 10^14 vib/s, then how fast are the Earth and this star traveling relative to each other? Remember that the correct equation for the speed v is given by v = [(fo^2 - fs^2) / (fo^2 + fs^2)] c Remember fo^2 means "fo squared."
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter39: Relativity
Section: Chapter Questions
Problem 39.85AP
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A star has an element in its atmosphere that normally emits a line of frequency fs = 7.5 x 10^14 vib/s. If astronomers measure the frequency of this line to be fo = 7.7 x 10^14 vib/s, then how fast are the Earth and this star traveling relative to each other? Remember that the correct equation for the speed v is given by v = [(fo^2 - fs^2) / (fo^2 + fs^2)] c Remember fo^2 means "fo squared."
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