Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm really stuck. I need to figure out the mass/light ratio of a galaxy in solar units. I know its mass is 5.7 x 1010 solar masses. I know its absolute magnitude (-17.3) and distance (29 Mpc). I'm feeling dumb here because I think this ought to be clear to me, but I'm getting hung up at the magnitude - luminosity - light part.

Here's what I think: mass/light = 100.4*(Magnitude galaxy-magnitude Sun) x (mass galaxy/mass Sun)

I get a mass to light ratio of around 90. Am I doing this right?

share|improve this question

1 Answer 1

Yes, the result is about 80 (unit: solar mass per solar intensity).

Taking "light" as meaning intensity (measured from Earth in units of power per unit area) the ratio, $R$, is:

$ R = \frac{m_g}{I_g} = \frac{5.7 \times 10^{10} m_{sun}}{I_g \times \frac{I_{sun}}{I_{sun}}} = \frac{5.7 \times 10^{10} m_{sun}}{\frac{I_g}{I_{sun}} \times I_{sun}} $

where $m_g$ is the mass of the galaxy, $I_g$ is the intensity of the galaxy and $I_{sun}$ is the intensity of the Sun.

$ \frac{I_g}{I_{sun}} $ can be computed from:

$ M_g - M_{sun} = -2.5\log_{10} ( \frac{I_g}{I_{sun}} ) $

where M are magnitudes.

Rearranging and using a value of 4.83 for the absolute magnitude of the Sun:

$ \frac{I_g}{I_{sun}} = 10^{-0.4(M_g - M_{sun})} = 10^{-0.4(-17.3 - 4.83)} = 7.1 \times 10^8$

Inserting $ \frac{I_g}{I_{sun}} $:

$ R = \frac{5.7 \times 10^{10} m_{sun}}{7.1 \times 10^8I_{sun}} = 80\ \frac{m_{sun}}{I_{sun}} $

share|improve this answer
    
Thanks! I worked it a few different ways and kept coming up with the same figure. It was the number that worried me since I believe that most galaxies are between 10 - 50 and galaxy clusters are 100-200. But thats the number (80) I'm sticking with! –  Gigi Giles Apr 4 '12 at 15:56
1  
@Gigi Giles: no problem. Now to the 'accept' phase :-) –  Peter Mortensen Apr 5 '12 at 14:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.