# 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 "Use This Technique to Solve for M31's velocity"), 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.

• The speed of light is correct if the units are km/sec -- very normal for astronomy -- rather than m/sec (as you are assuming). Apr 30, 2021 at 14:51
• Okay, thanks @PeterErwin, I already found that out from the answer. Apr 30, 2021 at 14:52
• Also, if you think my other post on Aviation Stack Exchange is worthy of reopening by the community (I've just made some edits to it), please vote to undelete it; here it is, if any of you are interested: Aviation, What are its Elements? Apr 30, 2021 at 14:57

$$\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$$
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. 