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 = 0.350 × 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 7.489 × 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

Frequency relation because of Doppler effect in light can be given by:

fo = f / [1 + (vs /c)]

fo - observed frequency, f - actual frequency, vs - velocity of source, c - velocity of light. (here, we take velocity of observer as zero).

vs is +ve, if the source recedes away from the observer. Then equation can be rewritten as:

fo = f [1 + (vs /c)]–1 = f [ 1 – (vs/c)]         (if vs << c)

(The above result is obtained using binomial expansion of (1 + x)–1 and neglecting higher powers of x as x << 1)

Here, the velocity of recedence is of the order of 106 and velocity of light is of the order of 108. Therefore, the above approximaiton can be used.

The galaxy is receding at a speed uG with respect to earth. But as it rotates, the point A has a relative motion with respect to the centre of galaxy towards the earth. No doubt, A is also moving away from earth, but not with the speed uG, but with a speed vA = uG – vT with respect to earth. This is definitely away from earth.

On the otherhand, the point B has an additional velocity vT in the direction of uG. Therefore, point B recedes with a speed vB = uG + vT with respect to the earth. As the velocities of recedence are different for A and B, the measured frequencies from A and B will be different.

We take the equation fo = f [ 1 – (vs/c)] and apply separately for points A and B.

a) For region A,      measured frequency: foA = f [1 –(vA/c)]

f = 7.489 × 1014 Hz, vA = uG – vT = 1.50 ×106 – 0.350 ×106 = 1.15 ×106 m/s, c = 2.998 ×108 m/s

Substituting; foA = 7.489 × 1014 [ 1 – (1.15 ×106 /2.998 ×108 ) ] = 7.460 × 1014 Hz

b) For region B,      measured frequency: foB = f [1 –(vB/c)]

f = 7.489 × 1014 Hz, vB = uG + vT = 1.50 ×106 + 0.350 ×106 = 1.85 ×106 m/s, c = 2.998 ×108 m/s

Substituting; foB = 7.489 × 1014 [ 1 – (1.85 ×106 /2.998 ×108 ) ] = 7.443 × 1014 Hz

Hope the concepts and steps are clear..

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