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.)

Answer 1

Given that
u_G  = 2x10⁶ m/s

v_T = 0.38x 10⁶ m/s

f_s = 5.077 x 10¹⁴ Hz

Speed at location A is

          V_A = u_G - v_T

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

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

        V_B = u_G + v_T

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

            = 2.38 x10⁶ m/s

a) The measured frequency at A is
                   f_A = f_s ( 1 - V_A /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
                   f_B = f_s ( 1 - V_B /c)

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

                        = 5.0366x 10¹⁴Hz