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 = 1.80 × 106 m/s. Relative to the center, the tangential speed is vT = 0.530 × 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 5.699 × 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 = 1.8 x 10^6 m/s ; Vt = 0.53 x 10^6 m/s; Fs = 5.699 x 1014 Hz

For region A :

We know that, from Doppler effect that

fo = fs ( 1 +/- V(rel)/c) ; Where fo is the observed frequency; Fs is the frequency of source and V(rel) is the relative velocity (relative to the observer) and c is the speed of light.

We notice that from region A, the galaxy is moving away from Earth with relative velocity of

V(rel) = 1.8 x 10^6 - 0.53 x 10^6 = 1.27 x 106 m/s

Putting this V|(rel) into the formula we get

fo(A) = 5.699 x 10^14 ( 1 - 1.27 x 10^6 / 3 x 10^8 ) = 5.675 x 10^14 Hz

Hence, at region A, fo(A) = 5.675 x 10^14 Hz.

For region B, v(rel) will be given by:

v(rel) = 1.8 x 106 + 0.53 x 10^6 = 2.33 x 106 m/s

fo(B) = 5.699 x 10^14 ( 1 - 2.33 x 10^6 / 3 x 10^8 ) = 5.655 x 10^14

Hence, at region A, fo(B) = 5.655 x 10^14 Hz.

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