Chandra Education - Velocity of the X-Ray Source

HINT:#

The question asks us to determine the velocity of the x-ray source based on the Doppler shift of the observed signal's frequency.

HINT:#

The problem tells us the maximum and minimum observed frequencies of observed x-rays.

HINT:#

Δf/f_av = v/c
Where,
Δf = (f_max - f_min)/2
Note: Δf is divided by two as the frequencies that we are using for calculation were the maximum and minum values observed, corresponding to the maximum doppler shift. These maxima occur when an emitter moves as close as it can to directly towards the observer and when it moves away from the observer in the same fashion. So, the instantaneous doppler shift due to the objects movement is only half of the change.
f_av = (f_max + f_min)/2
substituting the above, v = (f_max - f_min)/2)/(f_max + f_min)/2)*c

HINT:#

f_min = 0.20783/s
f_max = 0.2083/s
c = 3.0 x 10⁸ m/s

HINT:#

Let's use unit analysis:
v(m/s) = (f_max - f_min s^-1)s^-1/2)/(f_max + f_min)/2)s^-1*c (m/s)
The only units that will not cancel are those of c (m/s), those desired for our final answer!
No units need to be converted!

HINT:#

v(m/s) = (f_max - f_min s^-1)s^-1/2)/(f_max + f_min)/2)s^-1*c (m/s)
= ((0.2083 - 0.20783) s^-1/2)/((0.2083 + 0.20783) s^-1/2) * 3.0 x 10⁸ m/s
= 339 x 10³m/s = 339 km/s

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