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 = 2.00 × 106 m/s. Relative to the center, the tangential speed is vT = 0.380 × 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 5.077 × 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

Given that
uG  = 2x106 m/s

vT = 0.38x 106 m/s

fs = 5.077 x 1014 Hz

Speed at location A is

          VA = uG - vT

                = (2x106 m/s)- ( 0.38x 106 m/s)

                 = 1.62 x 106 m/s
Speed at location B is

        VB = uG + vT

             = (2x106 m/s)+ ( 0.38x 106 m/s)

            = 2.38 x106 m/s

a) The measured frequency at A is
                   fA = fs ( 1 - VA /c)

                      = (5.077 x 1014 Hz) ( 1 -   (1.62x 106 m/s) /( 2.998 x 108 m/s) )

                        = 5.0495 x 1014Hz
b) frequency at B is
                   fB = fs ( 1 - VB /c)

                       = (5.077 x 1014 Hz) ( 1 -   (2.38 x 106 m/s) /( 2.998 x 108 m/s) )

                        = 5.0366x 1014Hz

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