Re: astronomy lesson

[ammagal@mpcnet.com.br]

Michael,

> 3. How can I start to get an estimate of the interstellar extinction A?

One way you may estimate the distance is by assuming an average extinction per unit distance. On or near the Galactic plane, for instance, an average value is about 2 mag/kpc = .002 mag/pc.

The equation for the distance modulus then becomes

m - M = 5 log(d/10) + 0.002d,

which can be solved, for instance, by iteration. For that purpose, using your m-M=10.90, you may write it as

d = 10 x 10^(2.18 - 0.0004d)

Using your first estimate (1510 pc, for A=0), I end up with

d = 755 pc

after 20 or so iterations (i.e., plugging 755 into the right side gives me 755 back). I've double checked the calculation with Maple, which gave me the same result (w/o the fun, of course...).

Cheers,

Antonio Mario

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Dr. Antonio Mario Magalhães    http://astroweb.iag.usp.br/~mario/
Instituto Astronomico e Geofisico  mario@astro.iag.usp.br
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----- Original Message -----
From: "Michael Koppelman" <lolife@bitstream.net>
To: "Tass" <tass@listserv.wwa.com>
Sent: Sunday, February 23, 2003 9:22 PM
Subject: astronomy lesson


> So GSC 748-1618 may well be a RR Lyrae star. Just for fun, I wanted to
> assume it was and see if I could figure out it's distance. I'm making
a
> lot of assumptions at first.
>
> According to Santis et al 2002 a fairly accurate relation is:
>
> Mv = -1.842 log P - 0.137Ab + 0.26
>
> where Mv = the absolute visual magnitude and Ab = the amplitude in the
> color B.
>
> For my star P=0.7774 days. I do not know the blue amplitude but the V
> amplitude is 0.29m. I could probably assume that the blue amplitude is
> somewhat bigger than this. Just using the V amplitude for now I get:
>
> Mv= -1.842 log(0.7774) - 0.137(0.29) + 0.26 = 0.42
>
> This gives a distance modulus of 10.90. To convert this to distance:
>
> d = 10^(mv-Mv+5-A)
>
> If we assume that A is zero (incorrectly) we get a distance of 1.51
kpc
> or almost 5000 ly. If I bump the blue amplitude up to 0.4m I get 1.52
> kpc -- not a huge difference. A, however, makes a huge difference. I
> don't really have a way to estimate that at this point. A=1 brings the
> star closer to 3100 ly and A=5 all the way to 500 ly. Clearly, as Arne
> warned me, extinction is going to be a big factor. If I try to solve
> for A using nearby stars where the distance is known, I get very small
> values -- like 0.04 or less.
>
> My questions to you are:
>
> 1. Does an absolute visual magnitude of 0.42 seem at all reasonable?
> According to Astrophysical Quantities, these things should be around
> Mv=0.7 but it mentions that this depends on the metallicity.
> 2. Is there a relation between metallicity (Fe/H) and BVRI photometry?
> Any other way (for me) to determine metallicity?
> 3. How can I start to get an estimate of the interstellar extinction
A?
> 4. Where else am I off base?
>
> It seems safe to say at this point that this star is less than 5000 ly
> away.
>
> Thanks!
> Michael Koppelman
>
>
>



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