A distant galaxy is simultaneously rotating and receding from the earth. As the
ID: 1786646 • Letter: A
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 wr . 5.00 x 105 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 than the emitted frequency of 6.20 1014 Hz. Find the measured frequency for the light from each of the following (a) region A Hz (b) region B Hz Additional Materials Unpolarized light whose intensity is 1.90 W/m2 is incident on the polarizer in the figure (a) What is the intensity of the light leaving the polarizer? W/m2 (b) If the analyzer is set at an angle of -76" with respect to the polarizer, what is the intensity of the light that reaches the photocell? W/m2Explanation / Answer
Given that,
Ug = 1.8 x 10^6 m/s ; Vt = 5 x 10^5 m/s; Fs = 6.2 x 10^14 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 - 5 x 10^5 = 1.3 x 106 m/s
Putting this V|(rel) into the formula we get
fo(A) = 6.2 x 10^14 ( 1 - (1.3 x 10^6 / 3 x 10^8 )) = 6.2 x 1014 x (0.996) = 6.18 x 10^14 Hz
Hence, at region A, fo(A) = 6.18 x 10^14 Hz
For region B, v(rel) will be given by:
v(rel) = 1.8 x 10^6 + 5 x 10^5 = 2.3 x 10^6 m/s
fo(B) = 6.2 x 10^14 ( 1 - (2.3 x 10^6 / 3 x 10^8 ) = 6.2 x 1014 x 0.992 = 6.15 x 10^14 Hz
Hence, at region A, fo(B) = 6.15 x 10^14 Hz.
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