1
$\begingroup$

For a story I am looking into the star system of Eta Cassiopeiae, which is a binary with a main sequence star called Achird (very similar to the sun) and a K7V star.

The issue I am running into is the discrepancies of what the star type is; some say it is a G0V and some say it is a F9V.

I've looked at a dozen sources so I won't link anything, but I am wondering if anyone might know the most reliable and or most current source to go off of.

$\endgroup$
2
  • $\begingroup$ Technically, they're almost the same. Wikipedia uses G0V with respect to Eta Cas A $\endgroup$ – fasterthanlight Jan 5 at 22:46
  • $\begingroup$ I've even seen it called up as a G3V. The most common information I have found puts it at slightly less massive than the sun, slightly larger and about 1.25 times the luminosity. And a very close temperature to the sun. $\endgroup$ – Markitect Jan 5 at 23:26
2
$\begingroup$

$\eta$ Cassiopeiae A has an estimated mass of 0.972 M$_\odot$, an estimated temperature of 5973 K, and a B-V color index of ~0.58[1]. In addition to the spectrum of the star, we look at these and other properties when attempting to classify main sequence stars. You can see a table [here] which shows the bulk properties of each spectral type that we can use in addition to spectrum to help classify the star. The mass estimate places it into the G5V category, while the temperature and B-V place it definitively into the G0V category[2]. Hence, I would go with G0V. The mass is a little on the light side for the G0V category, but the temperature and color index and right on for the characteristic values for G0V.

$\endgroup$
4
  • $\begingroup$ This is all over my head, but "...I would go with..." suggest that neither is conclusively wrong, and to the question "Eta Cassiopeiae star type; G0V? F9V? Both?" if you had to choose, would "Both, but with a preference" be your choice? Or do you think that G5V is wrong? What about the OP's suggestion of F9V, is that wrong? $\endgroup$ – uhoh Jan 6 at 5:14
  • 1
    $\begingroup$ Eta Cas has a lower metallicity than the Sun, which needs to be taken into account if you're going to use a mass–spectral type relationship: mostly these are calibrated for solar metallicity. In any case, spectral type refers to the observed spectrum, not the bulk properties of the star (which can in any case change as it evolves along the main sequence). $\endgroup$ – user24157 Jan 6 at 9:24
  • $\begingroup$ @antispinwards this is a great point. I updated my answer to emphasize that the bulk properties are a secondary classifier to the observed spectrum of the star, and linked to a table showing the relationship between spectral type and various properties. In addition, I think the confusion in OPs question is coming from classification ambiguities in the bulk properties, not the observed spectrum. $\endgroup$ – Joseph Farah Jan 6 at 14:41
  • $\begingroup$ @uhoh G0V and F9V are very close classifications, but I would choose G0V because F-type stars typically have masses greater than 1 M$_\odot$. antispinwards correctly points out that really the observed spectrum is the primary classifier of a star's spectral type; but we can have conversations like this because other properties of the star also slot nicely into the same categories. When you have a bunch of ways to classify stars, you can get ambiguities like in OP's question if some bulk properties point to one classification and other properties point to another. $\endgroup$ – Joseph Farah Jan 6 at 14:46
1
$\begingroup$

The spectral type of a star is determined by looking at its spectrum. Sometimes authors will use other, approximate, relationships between spectral type and colour or mass, or they will look at the spectrum compared with standard templates in different wavelength regions.

These are all possible reasons why different sources might suggest slightly different spectral types. I would doubt there is any strong reason to prefer one classification over another when they differ by just one subtype.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.