Question

The frequency of light reaching Earth from a particular galaxy is 18% lower than the frequency the light had when it was emitted. a) explain whether the galaxy is moving toward or away from earth. b) What is the speed of this galaxy relative to the Earth? Give your answer as a fraction of the speed of light.

Answer 1

(a) So the first equation you should remember is c = LF ; where c = speed of light, F = frequency, and L = Wavelength. Now that you know that, you can relate wavelength to frequency since c is constant.

If F decreases, so does c/L, therefore L must increase! Okay so we now know that we are observing a longer wavelength than the wavelength that was emitted. So now what? Well, now we refer to the doppler effect. We know that things moving away are redshifted, and things moving towards you are blueshifted. We found that the galaxy produces a longer wavelength, which is the same as saying it's being redshifted. Therefore the galaxy is moving away from the Earth.

(b) 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 0.18 * F. This will tell us what we need to multiply L by to cancel out the 0.18. From pure inspection you can tell that C = 1/0.18 or -> C = 100/18 = 50/9

Now you know that the wavelength observed is 50/9 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 and delta L = change in wavelength

The right side equals (50/9 - 1) / 1 = 41/9.
So V_r = 41/9 * 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 = 41/9 * c

Now we square everything

(V_r ^ 2)/(1 - v^2/c^2) = (41/9 * c)^2
V_r^2 + (41/9)^2 V_r^2 = (41/9 * c)^2

So V_r = 0.977 * c