# Finding the distance modulus and the absolute magnitude

1. Sep 29, 2007

### Benzoate

1. The problem statement, all variables and given/known data
If a star has an apparent magnitude of -.4 and a parallax of .3'' what is :
a) the distance modulus
b) the absolute magnitude

2. Relevant equations
m is the apparent magnitude and M is the absolute magnitude
m-M =5 log d - 5
M= m + 5+5 log(pi''), pi'' is the parallex angle
3. The attempt at a solution

In order to find the Absolute magnitude, M, I apply the equation M= m + 5 + 5log(pi'')

I can easily find M since m and pi'' are already given in the problem. The only trouble I'm having is I don't know what units of measurement I'm supposed to convert pi'' to or if I'm suppose to leave pi'' the way it is.

Last edited: Sep 29, 2007
2. Sep 29, 2007

### lightgrav

the parallax angle, in arcsec, is the reciprocal distance in parsec.
So, (recall abs.Mag.definition) 10 pc means .1" , which has
M = m + 5 + 5*log(.1) , = m like it should.

3. Sep 29, 2007

### dynamicsolo

The parallax angle, $$\pi$$, is customarily given in arcseconds, but it is the (narrow) triangle involved that makes it clear how to use it. A parallax angle of 1" is the angle subtended by the mean radius of the Earth's orbit (more accurately, the semi-major axis), which is 1 AU, at a distance of 1 parsec. This automatically defines the parsec in terms of astronomical units (it also explains the name of the unit...).

What you'd want to think about it how that angle changes for other distances. You then have a simple relation between the stellar distance, d, in parsecs, and the parallax angle in arcseconds. The angle $$\pi$$ is often used interchangeably with the distance. The distance d in parsecs is what goes into your distance modulus equation (the modulus (M-m) is also used by some astronomers interchangeably with distance).

Noting many of the threads you've started lately, you wouldn't happen to be in an introductory astrophysics course, would you?