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Ann
4725 Å
Posts: 13715
Joined: Sat May 29, 2010 5:33 am

Starsurfer just posted a picture in the Found Images thread of star cluster M6 and two adjacent nebulas:

M6 and nebulas Ciel Austral .png

Go to this page to find a lot of info on this picture (and on another version of it as well).

Anyway. I became interested in the bright star that is ionizing the large nebula RCW 132 at right. The star is HD 159176, its U magnitude is 4.88, its B magnitude is 5.74 and its V magnitude is 5.70. Its Gaia parallax is 1.1666 ±0.0707 milliarcseconds.

Yes, obviously this star is reddened, and I suppose that we could calculate by how much. Well, for now I don't care.

I also don't care about the uncertainty of the Gaia parallax, and I'm happy to say that it is, simply, 1.666 mas. According to Unit Converter, the distance to this star is 2,795.79 light-years. I'm happy to say that the distance to the star is 2,800 light-years.

All right. The apparent V luminosity is +5.70, and the distance to the star is 2,800 light-years. What is the absolute V luminosity of this star (and never mind about the reddening)?

Groan. I found this page that explained how you calculate the absolute luminosity of a star if you know its distance and apparent magnitude:
A quantity that uses the inverse square law and the logarithmic magnitude system is the ``distance modulus''. The distance modulus = the apparent magnitude - absolute magnitude. This is equal to 5 × log(distance in parsecs) - 5. The ``log()'' term is the ``logarithm base 10'' function (it is the ``log'' key on a scientific calculator). If you measure a star's apparent magnitude and its distance from its trigonometric parallax, the star's absolute magnitude = the apparent magnitude - 5 × log(distance + 5. For example, Sirius has an apparent magnitude of -1.44 and Hipparcos measured its distance at 2.6371 parsecs, so it has an an absolute magnitude of -1.44 - 5×log(2.6371) + 5 = -1.44 - (5×0.421127) + 5 = 1.45.
Yeah, right. As if I can do that.

But maybe some of you can do that? Can you help me? What is the absolute V magnitude (never mind reddening) of a V mag 5.70 star at a distance of 2,800 light-years?

Ann
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Chris Peterson
Abominable Snowman
Posts: 18495
Joined: Wed Jan 31, 2007 11:13 pm

Ann wrote: Wed Mar 02, 2022 7:38 am Starsurfer just posted a picture in the Found Images thread of star cluster M6 and two adjacent nebulas:

M6 and nebulas Ciel Austral .png

Go to this page to find a lot of info on this picture (and on another version of it as well).

Anyway. I became interested in the bright star that is ionizing the large nebula RCW 132 at right. The star is HD 159176, its U magnitude is 4.88, its B magnitude is 5.74 and its V magnitude is 5.70. Its Gaia parallax is 1.1666 ±0.0707 milliarcseconds.

Yes, obviously this star is reddened, and I suppose that we could calculate by how much. Well, for now I don't care.

I also don't care about the uncertainty of the Gaia parallax, and I'm happy to say that it is, simply, 1.666 mas. According to Unit Converter, the distance to this star is 2,795.79 light-years. I'm happy to say that the distance to the star is 2,800 light-years.

All right. The apparent V luminosity is +5.70, and the distance to the star is 2,800 light-years. What is the absolute V luminosity of this star (and never mind about the reddening)?

Groan. I found this page that explained how you calculate the absolute luminosity of a star if you know its distance and apparent magnitude:
A quantity that uses the inverse square law and the logarithmic magnitude system is the ``distance modulus''. The distance modulus = the apparent magnitude - absolute magnitude. This is equal to 5 × log(distance in parsecs) - 5. The ``log()'' term is the ``logarithm base 10'' function (it is the ``log'' key on a scientific calculator). If you measure a star's apparent magnitude and its distance from its trigonometric parallax, the star's absolute magnitude = the apparent magnitude - 5 × log(distance + 5. For example, Sirius has an apparent magnitude of -1.44 and Hipparcos measured its distance at 2.6371 parsecs, so it has an an absolute magnitude of -1.44 - 5×log(2.6371) + 5 = -1.44 - (5×0.421127) + 5 = 1.45.
Yeah, right. As if I can do that. :roll:

But maybe some of you can do that? Can you help me? What is the absolute V magnitude (never mind reddening) of a V mag 5.70 star at a distance of 2,800 light-years?

Ann
If you include "calculator" in your search for things like this, you can often find them. https://www.calctool.org/CALC/phys/astr ... _magnitude

Chris

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Chris L Peterson
Cloudbait Observatory
https://www.cloudbait.com

Ann
4725 Å
Posts: 13715
Joined: Sat May 29, 2010 5:33 am