3
$\begingroup$

In high school we are undergoing a short course about astronomy. It is stated that you can use the apparent and absolute magnitude (m, M) to get a distance (D) of a star, with the formula

$$m - M = 5 \times \log(D) - 5$$

From which you free D:

$$D = 10^{\frac{m-M+5}{5}}$$

But how do you get both the m and M in practice, without knowing anything else? I can't imagine it.

I have asked my teacher but his explanation did not make sense to me, and as such I already forgot it.

Scenario: you have a satellite which can measure the incoming photons of a star. You now calculate the apparent magnitude using the formula:

$$m = m_\text{ref} - 2.5 \times \log(\frac{I}{I_\text{ref}})$$

Let's say you have a stable reference star and so you get $m=2$.

This is a nice start, but to determine the distance of the star you still need the absolute magnitude. How can you precisely acquire that using a HR-diagram? Because all the HR diagrams I have seen have a very fuzzy curve, and one temperature may map to multiple M.

HR diagram

This is what I'm struggling with: let's say you are somehow able to precisely measure that the star is 7500K. That means M can range from +4 to 0 (I guess we're also assuming it's a main sequence star, which we may not actually know), which gives us a possible D of about 4 to 25 (parsec?), which is an absolutely humongous range.

This can't be right, can it?

$\endgroup$

1 Answer 1

3
$\begingroup$

This is all absolutely true (pun intended). Finding absolute magnitudes is hard. For many types of star, we don't really know their absolute magnitude and so we don't really know their distance.

For close-by stars we can get the distance by measuring the parallax (how far the star appears to move over a year due to the orbit of the Earth. Nearer stars appear to move more) From this we can work out absolute magnitude.

For other stars we can get their absolute magnitude in other ways. For example, a type of large bright star called a "Cepheid variable" will have a pulse rate P that is connected to it's absolute magnitude $M_\mathrm{v}$ by the following (empirically derived) formula:

$$M_\mathrm{v} = -2.43(\log_{10}P - 1) - 4.05$$

With this you can easily measure the period in days, and use that to find the absolute magnitude. The apparent magnitude is easy to measure, it is how bright the star appears to be on Earth.

And with apparent and absolute magnitudes you can find the distance of the star.

Stars like Cepheid variables are called "standard candles", since their brightness can be determined exactly

$\endgroup$
2
  • 2
    $\begingroup$ Ah, okay. So I'm understanding that this method isn't an "end all be all" way of doing things, but rather a tool in a toolbox (and another tool would be parallax), and for some stars it might be a good way, others not so much. In my original question, was I correct about the range of the distance (0.01 parsec to 100 parsec)? $\endgroup$
    – AnnoyinC
    Oct 8, 2020 at 20:52
  • 2
    $\begingroup$ Yes, that looks correct. but the absolute magnitude method of finding distance is normally done with more distant stars (for example in globular clusters, or other galaxies) and using "standard candles". Even so, many stars have error bars of ±50% on their distance, if they are too far for parallax. $\endgroup$
    – James K
    Oct 8, 2020 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .