Luminosity-Radius-Temperature Relation

$L=4\pi R^2 T^4 \Rightarrow L \propto R^2 T^4$

$T\propto\left(\frac{L}{R^2}\right)^{1/4}=\frac{L^{\frac{1}{4}}}{R^\frac{1}{2}}= \frac{L^{.25}}{R^{.5}}$

$R\propto\sqrt{\frac{L}{T^4}}=\frac{\sqrt{L}}{T^2}$

$L_{Sun}=4\times 10^{26}~ W = 1 L_\odot$
$R_{Sun}=7\times 10^5~ km= 1 R_\odot$
$T_{Sun}= 5800 K=T_\odot$

Example:

Betelgeuse: L=60,000$L_\odot$, R=630 $R_\odot$
What is the temperature of Betelgeuse compared to the Sun?

$T_{Sun}=1 T_\odot \propto \frac{L^{.25}}{R^{.5}}=\frac{(1 L_\odot)^{.25}}{(1 R_\odot)^{.5}}$

$\frac{T_{B}}{T_{Sun}}=\frac{\frac{(60,000 L_\odot)^{.25}}{(630 R_\odot)^{.5}}}{\frac{(1 L_\odot)^{.25}}{(1 R_\odot)^{.5}}}= \frac{\frac{15.65}{25.10}}{1}=0.6$

So: $T_{B}=.6\times T_{Sun}=.6 T_\odot= .6 \times 5800~ K = 3600 K$