Spectra: What is that Supposed to Mean? I've found this website (click here for the link), the link leads you to a problem about the velocity Andromeda is going towards the Milky way galaxy, and you can try three approaches. I like to challenge myself, so I click on the second one, about spectra. I already know what the Electromagnetic Spectrum is, but this one is a tad different. It tells you about the redshift equation, which I already know about, and I've used it a few times in examples I made for myself, but I am confused about the next page (to get there, click the button at the bottom that says &quot;Use This Technique to Solve for M31's velocity&quot;), and the next page is worded in a way that I can't really understand it, and using the clues I get from the Spectrum Graph, I can't figure out what the answer is. What is Relative intensity of light, and what does the Calcium have to do with it? Also, isn't their value for the speed of light inaccurate, as far as I know, it's $3\times 10^8$, not $3\times 10^5$. Even further on, they want you to calculate the radial velocity from the graph, which I have no idea how to do.
If an answer about this here is acceptable, please give me a feasibly worded explanation of all this madness, and thanks in advance!
Okay, well, calcium is just an easily recognizable spectral feature, so I guess that's why they used it. The y-axis shows the relative intensity. I am assuming this is a spectrum that is normalized or scaled down so that the highest value of the intensity is 1, but this is not so important. The laboratory wavelengths of the absorption lines are shown in red in the graph. If you look carefully, you can see that the minimum of each spectral line does not exactly line up with these red tick marks. That is because the lines are Doppler shifted. M31 is moving and thus the light that we see has a different color (or frequency) from what we would observe in the lab. Let's focus on the left absorption line. The minimum is a bit to the left of the red mark. I conceive this to be approximately 4 angstrom off. The wavelength that you measure here on Earth will thus be 3630 angstrom and not 3934. If I take the formula from the previous page and rewrite I get
$$\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$$
where $\Delta \lambda$ is the difference between the observed and the lab wavelength of the spectral line. In this case $\Delta \lambda = -4$ angstrom and $\lambda_0 = 3934$ angstrom. The speed of light is given in km/s, not in m/s, so the value is indeed correct. Now you can get the speed of M31 from
$$v = c\frac{\Delta \lambda}{\lambda_0} = 3 \times 10^5 \frac{4}{3934} = -305km/s$$
I entered this value and it was accepted.