Chapter 24, Problem 37 A distant galaxy is simultaneously rotating and receding
ID: 2007581 • Letter: C
Question
Chapter 24, Problem 37 A distant galaxy is simultaneously rotating and receding from the earth. As the drawing shows, the galactic center is receding from the earth at a relative speed of uG = 1.2 x 106 m/s. Relative to the center, the tangential speed is vT = 0.31 x 106 m/s for locations A and B, which are equidistant from the center. When the frequencies of the light coming from regions A and B are measured on earth, they are not the same and each is different from the emitted frequency of 6.841 x 1014 Hz. Find the measured frequency for the light from (a) region A and (b) region B. (Give your answer to 4 significant digits. Use 2.998 x 108 m/s as the speed of light.)Explanation / Answer
tangential speed vT = 0.31 x 106 m/s relative speed o uG = 1.2 x 106 m/s frequency fs = speed of the light c = 2.998 x 108 m/s when electromagnetic waves and source and the observer of the waves all travel along the same line in a vacuum , the single equation that specifies the Doppler effect is f0 = fs[1 ± (vrel/c)] ........... (1) for region A : relative speed (galaxy travels away from the earth) : relative speed (galaxy travels away from the earth) : vrel = [1.2 x 106 m/s] -[ 0.31 x 106 m/s] = 0.89*106 m/s from eq (1), we have observed frequency f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (0.89*106/2.998 x 108)] = 6.82*1014 Hz ..................................................... for region B : relative speed (galaxy travels away from the earth) : vrel = [1.2 x 106 m/s] + [ 0.31 x 106 m/s] = 1.51*106 m/s from eq (1), we have f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (1.51*106/2.998 x 108)] = 6.80*1014 Hz vrel = [1.2 x 106 m/s] -[ 0.31 x 106 m/s] = 0.89*106 m/s from eq (1), we have observed frequency f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (0.89*106/2.998 x 108)] = 6.82*1014 Hz ..................................................... for region B : relative speed (galaxy travels away from the earth) : vrel = [1.2 x 106 m/s] + [ 0.31 x 106 m/s] = 1.51*106 m/s from eq (1), we have f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (1.51*106/2.998 x 108)] = 6.80*1014 Hz = 0.89*106 m/s from eq (1), we have observed frequency f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (0.89*106/2.998 x 108)] = 6.82*1014 Hz ..................................................... for region B : relative speed (galaxy travels away from the earth) : vrel = [1.2 x 106 m/s] + [ 0.31 x 106 m/s] = 1.51*106 m/s from eq (1), we have f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (1.51*106/2.998 x 108)] = 6.80*1014 Hz relative speed (galaxy travels away from the earth) : vrel = [1.2 x 106 m/s] + [ 0.31 x 106 m/s] = 1.51*106 m/s from eq (1), we have f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (1.51*106/2.998 x 108)] = 6.80*1014 Hz vrel = [1.2 x 106 m/s] + [ 0.31 x 106 m/s] = 1.51*106 m/s from eq (1), we have f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (1.51*106/2.998 x 108)] = 6.80*1014 Hz = 1.51*106 m/s from eq (1), we have f0 = fs[1 - (vrel/c)] f0 = (6.841*1014)[1 - (1.51*106/2.998 x 108)] = 6.80*1014 HzRelated Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.