Question

The frequency of light reaching Earth from a particular galaxy is 16% lower than the frequency the light had when it was emitted.

(b) What is the speed of this galaxy relative to the Earth? Give your answer as a fraction of the speed of light.
c

If you could show your work as well that'd be awesome! Thank you!

Answer 1

As far as the speed of the galaxy is concerned, we can find this by using the first equation again.
c = LF

Now we replace L by C*L and F by 0.16 * F.
This will tell us what we need to multiply L by to cancel out the 0.16.
From pure inspection you can tell that
C = 1/0.16
or
C = 100/16 = 6.25

Now you know that the wavelength observed is 6.25 times longer than the wavelength emitted.
Now simply use the doppler formula:

(V_r) / c = (delta L) / L

where V_r = radial velocity,
c = speed of light,
L = wavelength
delta L = change in wavelength

The right side equals
(6.25 - 1) / 1 = 5.25.

So
V_r = 5.25 * c .

But this clearly isn't right. And that's because we haven't taken relativity into account. To do so, all you do is change the formula to:

[(V_r) * (gamma)] / c = (delta L) / L

where
gamma = 1 / sqrt(1 - v^2/c^2)

Repeat the math with this formula and get

V_r * gamma = 5.25 * c

Now we square everything

(V_r ^ 2)/(1 - v^2/c^2) = (5.25 * c)^2

V_r^2 + (5.25)^2 V_r^2 = (5.25 * c)^2

So

V_r = 0.982 * c