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 = 1.60 × 106 m/s. Relative to the center, the tangential speed is vT = 0.580 × 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 8.645 × 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 × 108 m/s as the speed of light.)

Explanation / Answer

The speed of region A relative to the earth is

1.60*10^6 +(-0.580*10^6) = 1.02*10^6 m/s

the speed of region B relative to the earth is

1.60*10^6 +(0.580*10^6) = 2.18*10^6 m/s

formula for doppler shift is

fo = fs(1- v_rel/c)

(a) the light from region A is observed with a frequency

foA = (8.645*10^14)(1 - 1.02*10^6 / 2.998*10^8)
      = 8.615*10^14 Hz

(b) the light from region B is observed with a frequency

foB = (8.645*10^14)(1 - 2.18*10^6 / 2.998*10^8)

      = 8.5821*10^14 Hz

rounding to 4 significant figures

foB = 8.582*10^14 Hz

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