A distant galaxy is simultaneously rotating and receding from the earth. As the

Question

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 = 1.9 x 106 m/s. Relative to the center, the tangential speed is vT = 0.56 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 7.481 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

angential speed vT = 0.56 x 106 m/s relative speed o uG = 1.9 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) :         vrel = [1.9 x 106 m/s] -[ 0.56 x 106 m/s]               = 1.34*106 m/s from eq (1), we have observed frequency f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (1.34*106/2.998 x 108)]                   = 7.44*1014 Hz ..................................................... for region B : relative speed (galaxy travels away from the earth) :         vrel = [1.9 x 106 m/s] + [ 0.56 x 106 m/s]               = 2.46*106 m/s from eq (1), we have           f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (2.46*106/2.998 x 108)]                   = 7.35*1014 Hz angential speed vT = 0.56 x 106 m/s relative speed o uG = 1.9 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 = [1.9 x 106 m/s] -[ 0.56 x 106 m/s]               = 1.34*106 m/s from eq (1), we have observed frequency f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (1.34*106/2.998 x 108)]                   = 7.44*1014 Hz ..................................................... for region B : relative speed (galaxy travels away from the earth) :         vrel = [1.9 x 106 m/s] + [ 0.56 x 106 m/s]               = 2.46*106 m/s from eq (1), we have           f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (2.46*106/2.998 x 108)]                   = 7.35*1014 Hz         vrel = [1.9 x 106 m/s] -[ 0.56 x 106 m/s]               = 1.34*106 m/s from eq (1), we have observed frequency f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (1.34*106/2.998 x 108)]                   = 7.44*1014 Hz ..................................................... for region B : relative speed (galaxy travels away from the earth) :         vrel = [1.9 x 106 m/s] + [ 0.56 x 106 m/s]               = 2.46*106 m/s from eq (1), we have           f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (2.46*106/2.998 x 108)]                   = 7.35*1014 Hz               = 1.34*106 m/s from eq (1), we have observed frequency f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (1.34*106/2.998 x 108)]                   = 7.44*1014 Hz ..................................................... for region B : relative speed (galaxy travels away from the earth) :         vrel = [1.9 x 106 m/s] + [ 0.56 x 106 m/s]               = 2.46*106 m/s from eq (1), we have           f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (2.46*106/2.998 x 108)]                   = 7.35*1014 Hz relative speed (galaxy travels away from the earth) :         vrel = [1.9 x 106 m/s] + [ 0.56 x 106 m/s]               = 2.46*106 m/s from eq (1), we have           f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (2.46*106/2.998 x 108)]                   = 7.35*1014 Hz         vrel = [1.9 x 106 m/s] + [ 0.56 x 106 m/s]               = 2.46*106 m/s from eq (1), we have           f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (2.46*106/2.998 x 108)]                   = 7.35*1014 Hz               = 2.46*106 m/s from eq (1), we have           f0 = fs[1 - (vrel/c)]            f0 = (7.481*1014)[1 - (2.46*106/2.998 x 108)]                   = 7.35*1014 Hz

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