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PeterSurma
member
Reged: 08/24/06
Posts: 87
Loc: Heidelberg, Germany
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Re: Any resolvable stars outside Milky Way?
07/10/08 05:20 AM
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Hi,
it's petty easy to calculate this on your own. You have to know how luminous the object is M (absolute magnitude), that you wanna detect and you have to know the distance, even better know the so-called distance modulus (m-M) (which is just another measure for distance than pure miles and lyrs).
Your object has M = -8mag of absolute luminosity. Your galaxy has a distance modulus of (m-M)=+25mag (corresponding to 1 million pc = 3.2 million ly). So this object looks to us like a faint thing of apparent magnitude m = M + (m-M) = -8 + 25 = +17mag. This is the example calculated for M31 now. 17mag are reachable with a 20" scope under good conditions. So the VERY brightest M31 stars can be seen with a 20".
The most luminous stars typically have visual absolute magnitude of M=-8. So at a given distance they shine with apparent magnitude m, that you can calculate now.
Of, course, you can also turn it around and calculate the maximum distance modulus (--> distance) where you can still see some object of absolute magnitude M, if you know your limiting magnitude m.
As another example of such calculations you may checkout an observing report of mine on M13. [Edit: shortened the hyperlink]
There are some arguments around seeing faint stars in there.
Best wishes,
Peter
Some links, but they are partly hard to understand:
http://en.wikipedia.org/wiki/List_of_most_luminous_stars
- note here that M_v = M_bol +4mag for hot stars!!!
http://www.peripatus.gen.nz/Astronomy/SteMag.html
http://en.wikipedia.org/wiki/Distance_modulus
Edited by Olivier Biot (07/13/08 07:40 AM)
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