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I am hoping that someone can clarify this for me.

With these equations. (Boltzmann Law) and radius of Radius of star how does surface temp scale with mass

$R _x \approx R_\bigodot (\frac{M_x}{M_\bigodot})^.5$

$L _x \approx L_\bigodot (\frac{M_x}{M_\bigodot})^3.5$

$\frac{L}{\pi R^2} = \sigma T^4$

$T_\bigodot = 5800 K$ = Surface Temp of Sun

What I have done is that I have substituted $R_x$ and $L_x$ in the 3rd equation and simplified it in terms of $T$ where my final expression is

$T = (\frac{L_\bigodot M_x^2.5}{\pi\sigma M_\bigodot^2.5 R_\bigodot})^0.25$

Am i done? What am I suppose to do next? Thanks

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I think it would look nicer if you defined a $T_{\bigodot}$. – Raskolnikov Mar 13 '11 at 0:00
Hi, I have added the $T_\bigodot$.. – Kartik Mar 13 '11 at 0:01
Yeah, but that's not really what I meant. The point is you can find a direct relationship between $T/T_{\bigodot}$ and $M/M_{\bigodot}$. – Raskolnikov Mar 13 '11 at 0:04
Well.. That is what I hoping to understand. I am not sure how to tackle the question. What do I do to get the ratio? – Kartik Mar 13 '11 at 0:06
Never mind, I'll post an answer. – Raskolnikov Mar 13 '11 at 0:07
up vote 4 down vote accepted

Notice that

$$\frac{L_{\bigodot}}{\pi R_{\bigodot}^2} = \sigma T_{\bigodot}^4$$

So that, dividing through the relation for an arbitrary star and that for the Sun gives:

$$\frac{L/L_{\bigodot}}{R^2/R_{\bigodot}^2} = T^4/T_{\bigodot}^4$$

Using the other relations

$$\frac{(M/M_{\bigodot})^{3.5}}{M/M_{\bigodot}} = (T/T_{\bigodot})^4$$


$$\left(\frac{M}{M_{\bigodot}} \right)^{2.5} = \left(\frac{T}{T_{\bigodot}}\right)^4$$

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