# Using Eddington Luminosity to calculate mass

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1. Dec 1, 2015

### Ellie Snyder

1. The problem statement, all variables and given/known data
The nuclear reactions in a star’s core are very sensitive to pressure and temperature, so high mass stars have much higher luminosities. The luminosities of massive stars have been observed to obey the following scaling relationship with mass (M):

Lstar = (34.2)*M^2.4, where M, L have units of MSun, LSun.

The Eddington limit is the maximum luminosity an object (such as a star) can achieve and still retain a balance between the outward force of energy from the center and inward pull of gravity. The following formula is for the Eddington Limit (LEdd), i.e., the luminosity which stops the inward pull of gravity:

LEdd = (3.2*10^4)*M, where M, L have units of MSun, LSun.

a. Using these two equations, calculate the maximum mass of a star in solar units.

b. A bright quasar has a luminosity of about 10^13 LSun. The source of its power is a supermassive black hole that attracts surrounding gas into a hot (~10^6 K), compact accretion disk which radiates light. If the quasar is to continue attracting gas into its central black hole, what is its minimum mass?

2. Relevant equations
Lstar = (34.2)*M^2.4
LEdd = (3.2*10^4)*M

3. The attempt at a solution
a. Solved for M in the Lstar eqn to get (Lstar/34.2)^1/2.4 and plugged that into the LEdd eqn. Got 7344.7Lstar^1/2.4, and plugged that in for LEdd to get an M of 0.2295Lstar^1/2.4, which is obviously preposterous. I guess I'm not sure how to relate the equations to solve for M.

b. Waiting to attempt this one until I have a better understanding of a.

2. Dec 1, 2015

### SteamKing

Staff Emeritus
I think the correct equation for the Eddington Luminosity Limit is

LEdd = (3.2×104) ⋅ (Mstar / Msun) ⋅ Lsun

https://en.wikipedia.org/wiki/Eddington_luminosity

3. Dec 1, 2015

### Ellie Snyder

Hmm. I believe you but since I was given that formula in the question itself, I think I should use it to answer the question, even if it is off base.

4. Dec 1, 2015

### SteamKing

Staff Emeritus
Well, the factor which was missing from your original equation was the luminosity of the sun. Perhaps it got lost when you were drafting your post.

5. Dec 1, 2015

### Nathanael

Ellie specified that M and L were in solar units (hence Msun=1 and Lsun=1) so the expression is the same.

@Ellie Snyder
You solved for M in terms of L_star, but L_star is unknown, so that isn't a useful thing to do.

I think you've realized that the maximum mass occurs when L_star = L_Edd, right? So then algebraically you can just equate those two expressions:
$L_{star}=L_{Edd}\Rightarrow 34.2M^{2.4}=32000M$
So that way the equation involves only M and not L.

6. Dec 1, 2015

### Ellie Snyder

Oh duh, ok great, so I'm getting an M of about 132.50 M_sun

7. Dec 1, 2015

### Ellie Snyder

So for part b, to find that minimum mass, would I insert the quasar's 10^13 L_sun luminosity into the first equation and solve for M?

8. Dec 1, 2015

### Nathanael

I think so, but to be honest, I don't entirely understand the physics behind that situation; it seems like some subtleties are being neglected.

9. Mar 19, 2016

### stehfahknee

I have the same questions, just put into a different scenario... I ended up taking the 1.4 root of 3.2x10^4 over 34.2 for an answer of 266, 000kg. Because i rounded to 133 Msun which makes 133x2x10^3kg...Is this completely off? My professor said I was right on... but now I am wondering if I should just leave it as 133Msun...?

10. Mar 19, 2016

### stehfahknee

WHY do I over think everything? Its the same damn thing... nevermind me I'm just arguing with myself...
As usual...