The luminosity $L$ (total power output in watts, W) of a star is given by the formula $$L=L( R,T) =4\pi R^{2}\sigma T^{4}$$

where $R$ is the radius of the star (in meters), $T$ is its effective surface temperature (in kelvin, K), and $\sigma$ is the Stefan–Boltzmann constant. For the sun, $L_{s}=(3.90\times\;10^{26})$ $\rm{W}$, $R_{s}=(6.94\times\;10^{8})\;\rm{m}$ , and $T_{s}=4800\;\rm{K}$. Suppose in a billion years, the changes in the Sun will be $\Delta R_{s}=(0.08\;\times\;10^{8})\;\rm{m}$ and $\Delta T_{s}=100\;\rm{K}$. What will be the resulting percent increase in luminosity?

Solution We begin with $L=4\pi\;R^{2}\sigma\;T^{4}.$ Then $$dL=4\pi \sigma (2R) T^{4}dR+4\pi \sigma R^{2}( 4T^{3}) dT=8\pi \sigma RT^{3}( T dR+2R\ dT)$$

The relative error in luminosity is $$\begin{eqnarray*} \frac{\Delta L}{L}&\approx& \frac{dL}{L}=\dfrac{8\pi \sigma RT^{3}}{4\pi R^{2}\sigma T^{4}}\left( TdR+2RdT\right) =2\left( \frac{dR}{R}+2\frac{dT}{T} \right) =2\frac{\Delta R}{R}+4\frac{\Delta T}{T}\\[4pt] &=&\frac{2(0.08\times 10^{8})}{6.94\times 10^{8}}+\frac{4(100)}{4800}\approx 0.106 \end{eqnarray*}$$

The percent increase in luminosity will be approximately $10.6\%$.