Chapter 13, Problem 15P

To determine

The radii of Rigel and Barnard’s star compared to a solar radius using the Stefan-Boltzmann equation and show that the values agree with the radii in figure 13.23.

Answer to Problem 15P

The radii of Rigel is $106$ and Barnard’s star is $0.11$ compared to a solar radius using the Stefan-Boltzmann equation and the values agree with the radii in figure 13.23.

Explanation of Solution

The luminosity of the Rigel star is $L={10}^5{L}_{\Theta }$ and its temperature is $T=10000\text{ K}$.

The luminosity of the Barnard’s star is $L=0.0009L_{\Theta }$ and its temperature is $T=3000\text{ K}$. And the temperature of Sun is $T_{\Theta }=5800\text{K}$.

Write the Stefan-Boltzmann equation

    $L=4\pi R^2\sigma T^4$        (I)

Here, $L$ is the luminosity, $R$ is the radius and $T$ is the temperature.

Rearrange equation (I) to solve for $R$

$\begin{array}{c}{R}^2=\frac{L}{4\pi \sigma T^4}\\ R={\left[\frac{L}{4\pi \sigma T^4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$        (II)

Divide $\frac{R}{R_{\Theta }}$,

$\begin{array}{c}\frac{R}{R_{\Theta }}=\frac{{\left[\frac{L}{4\pi \sigma T^4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left[\frac{L_{\Theta }}{4\pi \sigma T_{\Theta }^4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ ={\left[\frac{L}{L_{\Theta }}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[\frac{T_{\Theta }^4}{T^4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ ={\left[\frac{L}{L_{\Theta }}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{T_{\Theta }^2}{T^2}\\ ={\left[\frac{L}{L_{\Theta }}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[\frac{T_{\Theta }}{T}\right]}^2\end{array}$        (III)

Substitute ${10}^5{L}_{\Theta }$ for $L_{Rigel}$, $10000\text{ K}$ for $T_{Rigel}$  and $5800\text{K}$ for $T_{\Theta }$ in equation (III) to find the value of $\frac{R_{Rigel}}{R_{\Theta }}$

$\begin{array}{c}\frac{R_{Rigel}}{R_{\Theta }}={\left[\frac{L_{Rigel}}{L_{\Theta }}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[\frac{T_{\Theta }}{T_{Rigel}}\right]}^2\\ ={\left[\frac{{10}^5{L}_{\Theta }}{L_{\Theta }}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[\frac{5800\text{K}}{10000\text{ K}}\right]}^2\\ =106\end{array}$

Substitute $0.0009L_{\Theta }$ for $L_{Barnard\text{'}sstar}$, $3000\text{ K}$ for $T_{Barnard\text{'}sstar}$  and $5800\text{K}$ for $T_{\Theta }$ in equation (III) to find the value of $\frac{R_{Barnard\text{'}sstar}}{R_{\Theta }}$

$\begin{array}{c}\frac{R_{Barnard\text{'}sstar}}{R_{\Theta }}={\left[\frac{L_{Barnard\text{'}sstar}}{L_{\Theta }}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[\frac{T_{\Theta }}{T_{Barnard\text{'}sstar}}\right]}^2\\ ={\left[\frac{0.0009L_{\Theta }}{L_{\Theta }}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[\frac{5800\text{K}}{3000\text{ K}}\right]}^2\\ =0.11\end{array}$

The radii indicated by the diagonal lines in figure 13.23 show that $\frac{R_{Barnard\text{'}sstar}}{R_{\Theta }}$ is $0.1$ which agrees with the calculated value and $\frac{R_{Rigel}}{R_{\Theta }}$ is $100$ which is closer to the calculated value $106$.

Conclusion:

Thus, the radii of Rigel is $106$ and Barnard’s star is $0.11$ compared to a solar radius using the Stefan-Boltzmann equation and the values agree with the radii in figure 13.23.