Apparent Size

I have always wondered about the following but was afraid to ask (actually not afraid, just never remembered, until now.)

When we look at stars we are obviously not seeing the surface of the star but the glare of light. Stars, even given their huge size would be too small to see at distances measured in light years.

So what is the apparent size of star for a given magnitude? Does the apparent size of say, Vega cover an area of several light hours, or more? Obviously it would be different for combinations of distance and brightness.

Just curious (but then of course that's why I love this hobby.)

Daryl

Daryl, I'm not sure I fully understand your question but If I'm right you are asking how big the image of a given star appears. From what I understand magnitude isn't indicative of actual apparent size. The star long known to display the largest apparent size was betelgeuse, I cannot remember how big this figure was but it was way below the resolving power of very large scopes, 0.05 arc seconds rings a bell. R Doradus was measured to be bigger than this in recent times but not by that great an amount. In short the image we see when we look at any star is derived by a point source, the extent of this image when viewing any bright star is the result of atmospheric effect, eye abnomalities and telescope short commings. I'm sure someone has a more technical answer for you but this is how I see it fundamently.


Ed

When speaking about apparent size of small objects, we normally express the size as an angle. The moon's apparent diameter is about 0.5 degree, corresponding to 30' (arcminutes) or 1800" (arcseconds).
Jupiter, which appears as a sizable disc in the telescope has an apparent diameter of about 40".

Generally the apparent diameter of the stars is extremely small. Due to the undulatory nature of light, they appear as discs when looking through the telescope. The size of this disc depends on the aperture of the telescope. In a 5-inch telescope the diameter of this diffraction disc is about 1" (arcsecond).

If we move the sun 10 light years away, its apparent diameter would be approximately 0.003". Most stars are much further off, and if their diameters are comparable of that of the sun, they will appear even smaller. There are a few very rare giant stars whose diameter are maybe 100 - 200x that of the sun. If such a star was 10 light years off, it would be possible to see its apparent disc in a big amateur telescope. Unfortunately (or perhaps fortunately) we have no such supergiant stars in our near neighbourhood. Betelgeuse is several light years away, therefore its diameter is about 0.05". Therefore we see the diffraction image of the stars in our telescopes, only the
greatest telescopes being able to make out the diameter (and surface features if present) of the nearest supergiant stars.

Inge S

Hello:
We're all familar with the inverse square law for the apparent brightness of stars. One thing that some are not familar with is that the surface brightness of an object (star, galaxy, etc) is constant and does not change with distance! Although stars are so far away that we cannot resolve the disk of the star (except for large, nearby stars like Betelgeuse, or by using interferometry), they do have a small but finite apparent size. The apparent brightness of the star depends on the surface brightness times the apparent surface area. The apparent size of a star is inversely proportional to the distance, and since area goes as the square of the size, the apparent surface area goes like the invserse square of the distance, which is one way to understand the inverse square law.
So, for example, we know the Sun has a magnitude of something like -26 and has a surface area of 2,544,690 square arcseconds. A G2 star with an apparent magnitude of, say, 5, should have an apparent surface area of 0.000001 square arcseconds and an apparent diameter of 0.0006 arcseconds. I hope this clears things up.

Cheers
Mike