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 = 1.50 106 m/s. Relative to the center, the tangential speed is vT = 5.00 105 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 than the emitted frequency of 6.20 1014 Hz. Find the measured frequency for the light from each of the following.

(a) region A Hz

(b) region B Hz

Explanation / Answer

The difference in frequencies is because of Doppler Effect.

f' = f ( V + Vo ) / ( V + Vs ) where f' is the observed frequency , f is the actual frequency , Vo is the relative observer speed (0 here ) and Vs is the relative source speed.

For point A : Vs = Ug - Vt  

Vs = 1.50106 - 5.00105 m/s = -3.499 = -3.5 m/s ( The speed mentioned in the question may be wrong. Ug should be in the order of 106and Vt in the order of 105. I'll solve according to the values mentioned in the question.

f' = 6.201014 ( 3*108 ) / ( 3*108 - 3.5)

f' will have a negligible change in this assuming.

Similarly for point B

f' = 6.201014 ( 3*108) / ( 3*108 + ( 1.50106 + 5.00105) )

which again will have a minimal effect. I would advise you to recheck the values and solve using the same method.

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