-2
$\begingroup$

Title pretty much says it all.

Basically, what exact variables are used, and what steps need to be taken, in the calculation of the size of a star? Preferably, some example data for how a the size of a star was recently calculated, from an official government / astronomy website, would be preferred, but not necessary.

$\endgroup$
4
  • $\begingroup$ Probably you'll need to learn about Astronomical Interferometry to get an idea for what's involved. You'll find links on that page to articles going into more detail. The first paper on this subject (about 1920) is linked to there and it's probably a good place to start from. Techniques have been continually developed and improved, which means they're also more complex. Feel free to search for modern paper on e.g. Arxiv.org (a respected preprint website fort some science papers). $\endgroup$
    – StephenG
    May 16 '20 at 0:37
  • 1
    $\begingroup$ Can you clarify whether you asking about observational measurements of the size of stars or theoretical predictions for sizes of stars ? Depending on how accurate you want it, there is a mass-radius $Radius \sim Mass^{0.7}$ but the 0.7 power does change for high and low mass stars (more details). It depends how accurate you want the answer and over what range of masses, age and metallicity you want it to apply to $\endgroup$ May 16 '20 at 0:47
  • 1
    $\begingroup$ @astrosnapper hi, preferably more observational measurement, but really both are good too $\endgroup$
    – bluejayke
    May 16 '20 at 0:58
  • 1
    $\begingroup$ Does this answer your question? Device used to measure stellar radii $\endgroup$
    – StephenG
    May 16 '20 at 13:40
2
$\begingroup$

There have been direct measurements, but they are quite delicate -- this is all the interferometry and so on.

The simplest way though is based on spectroscopy, brightness and parallax. The argument goes a follows:

  1. The spectrum (how much light of each colour it gives off) is not hard to observe -- you basically just need a prism.

  2. The spectrum of most stars basically looks like whats called a "black body spectrum" -- the light given off by a hot body with no special properties. This can be both observed on Earth and predicted from basic physics (and they agree)

  3. From the position of the peak of the black body spectrum we can tell the temperature of the star

  4. The total light emitted per unit area from a black body depends only on its temperature, so we can calculate the total light emitted per unit area by the star.

  5. For the nearer stars we can measure their distance fairly directly by observing how their apparent position moves against that of other stars as the Earth moves around the Sun

  6. Knowing the distance, the brightness per unit area, and the amount of light actually hitting our telescope, we can work out what the area (and so size) of the star must be (this is just geometry)

$\endgroup$
6
  • $\begingroup$ Ok ok very interesting, that clears up a lot of things I have been wondering regarding this. Just one more point I may need a bit more of explaining about, is point two, is a "black body spectrum" something that exists in other things on earth based on their color, and we are comparing the color given off by the stars to the colors given off by other things on earth, or is it something different? $\endgroup$
    – bluejayke
    May 17 '20 at 5:00
  • $\begingroup$ A traditional light bulb is also a black body radiator. Many things are - with more or less big deviations, including you yourself. $\endgroup$ May 17 '20 at 6:06
  • $\begingroup$ @bluejayke Most hot objects give off radiation with something close to a black body spectrum -- a piece of red hot charcoal, for instance. $\endgroup$ May 17 '20 at 6:46
  • $\begingroup$ @SteveLinton OK cool I'm just trying to understand it, so we are comparing the stars color appearance to that of other object that we observe, to get the heat, is that right? $\endgroup$
    – bluejayke
    May 17 '20 at 8:14
  • $\begingroup$ @bluejayke. essentially yes, although the process (and the physics behind it) is a lot more precise than your phrasing suggests. $\endgroup$ May 17 '20 at 8:47

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