6
$\begingroup$

Should the tags stellar-classification and spectral-type be merged? has been asked in meta two weeks ago. It requires some careful consideration but so far no response has been forthcoming, so I'm asking this question here to get a definitive answer that can then be applied there.

Wikipedia's Stellar classification begins

In astronomy, stellar classification is the classification of stars based on their spectral characteristics.

and while there is one subsection labeled simply Spectral types to my untrained eye it looks like there may be other classification schemes besides O, B, A, F, G, K, M, L, T.

Question: Do the astronomical terms "spectral type" and "stellar classification" refer to the same thing? Or at least always when applied to stars? Might "spectral types" also refer to a way to classify asteroids?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Well, the same Wikipedia article also mentions (third paragraph): “In the MK [Morgan–Keenan] system, a luminosity class is added to the spectral class using Roman numerals.” For example, the Sun is a G2 V (yellow main-sequence) star, as opposed to, say, Sadalmelik (α Aqr), which is a G2 Ib (yellow supergiant) star. So stellar classification is more than just the spectral type; it’s also the luminosity class. $\endgroup$
    – Pierre Paquette
    Dec 15, 2020 at 1:48

1 Answer 1

Reset to default
-1
$\begingroup$

Spectral type is only based on the star's color. You need additional the mass or luminosity to make a stellar classification form it. Example: Giants may belong to the same spectral class as main sequence stars.

Of course, if you don't simply look at the color but measure the spectra you see the differences. Spectroscopy is one of the most important tools for astronomers.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ Spectral type is based on a spectrum. Quoting a spectral type based on a colour is fraught with uncertainty. Neither the mass or luminosity are required to assign a spectral type. $\endgroup$
    – ProfRob
    Dec 15, 2020 at 8:01

You must log in to answer this question.

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