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 up = 1.6 x 10^6 m/s. Relative to the center, the tangential speed is vr = 0.4 x 10^6 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 emitted frequency of 6.200 x 10^14 Hz. Find the measured frequency for the light from (a) region A and (b) region B.

Answer 1

(a)

At location A, the relative speed of the galaxy and the tangential velocity are opposite in direction to each other.

Therefore, the velocity of the galaxy at location A is,

V_A = u_G - v_r = 1.6*10⁶ - 0.4*10⁶ = 1.2*10⁶ m/s

The expression for the observed frequency when the source is moving away is,

f_o = f_s*(1 - V_A/c)

f_o = 6.2*10¹⁴*(1 - 1.2*10⁶/3*10⁸)

f_o = 6.17*10¹⁴ Hz

(b)

V_B = u_G + v_r = 1.6*10⁶ + 0.4*10⁶ = 2*10⁶ m/s

The expression for the observed frequency when the source is moving away is,

f_o = f_s*(1 - V_B/c)

f_o = 6.2*10¹⁴*(1 - 2*10⁶/3*10⁸)

f_o = 6.15*10¹⁴ Hz