Chapter 5, Problem 26Q

(a)

To determine

The energy flux at the surface of Sirius, if the bright star Sirius in the constellation of Canis Major has a radius of $1.67{\text{ R}}_{\odot }$.

Answer to Problem 26Q

Solution:

$5.75\times {10}^8{\text{ W/m}}^2$

Explanation of Solution

Given data:

The bright star Sirius in the constellation of Canis Major has a radius of $1.67{\text{ R}}_{\odot }$.

The luminosity is $25{\text{ L}}_{\odot }$.

Formula used:

Write the expression for energy flux of the Sun’s luminosity $\left(F_{\odot }\right)$.

$F_{\odot }=\frac{L_{\odot }}{4\pi r^2}$

Here, $F_{\odot }$ is energy flux of the Sun’s luminosity or the solar constant or the flux of sunlight, $L_{\odot }$ is the luminosity of the Sun and $r$ is the distance from the Sun to the Earth.

Explanation:

Recall the expression for energy flux of the Sun’s luminosity $\left(F_{\odot }\right)$.

$F_{\odot }=\frac{L_{\odot }}{4\pi r^2}$

The bright star Sirius in the constellation of Canis Major has a radius of $1.67{\text{ R}}_{\odot }$. Therefore,

$\begin{array}{c}R=1.67{\text{ R}}_{\odot }\\ \frac{R_{\odot }}{R}=\frac{1}{1.67}\end{array}$

The luminosity is about $25{\text{ L}}_{\odot }$. Therefore,

$\begin{array}{c}L=25{\text{ L}}_{\odot }\\ \frac{L}{L_{\odot }}=25\end{array}$

The relation between flux and luminosity of the different stars is:

$\frac{F}{F_{\odot }}=\frac{L}{L_{\odot }}\frac{R_{\odot }^2}{R^2}$

Substitute $\frac{1}{1.67}$ for $\frac{R_{\odot }}{R}$ and $25$ for $\frac{L}{L_{\odot }}$.

$\begin{array}{c}\frac{F}{F_{\odot }}=25{\left(\frac{1}{1.67}\right)}^2\\ =8.96\\ F=8.96F_{\odot }\end{array}$

Refer to Box 5-2 for the value of the energy flux of the Sun’s luminosity.

$F_{\odot }=6.41\times {10}^7{\text{ W/m}}^2$

Substitute $6.41\times {10}^7{\text{ W/m}}^2$ for $F_{\odot }$.

$\begin{array}{c}F=8.96\left(6.41\times {10}^7{\text{ W/m}}^2\right)\\ =5.75\times {10}^8{\text{ W/m}}^2\end{array}$

Conclusion:

The energy flux at the surface of Sirius is $5.75\times {10}^8{\text{ W/m}}^2$.

(b)

To determine

The surface temperature of Sirius. Also compare the value with that given in Box 5-2, if the bright star Sirius in the constellation of Canis Major has a radius of $1.67{\text{ R}}_{\odot }$.

Answer to Problem 26Q

Solution:

$100035\text{ K}$; it is very high as compared to the surface temperature of the Sun.

Explanation of Solution

Given data:

The bright star Sirius in the constellation of Canis Major has a radius of $1.67{\text{ R}}_{\odot }$.

The luminosity is $25{\text{ L}}_{\odot }$.

Formula used:

The expression for Stefan Boltzmann law is:

$F=\sigma T^4$

Here, $T$ is the temperature of the object, $F$ is the energy flux per square unit per second and $\sigma$ is the Stefan constant $\left(5.67\times {10}^{-8}{\text{ Wm}}^{-2}{\text{K}}^{-4}\right)$.

Explanation:

Recall the result of part (a).

$5.75\times {10}^8{\text{ W/m}}^2$

The expression for Stefan Boltzmann law is:

$F=\sigma T^4$

Substitute $5.75\times {10}^8{\text{ W/m}}^2$ for $F$ and $5.67\times {10}^{-8}{\text{ Wm}}^{-2}{\text{K}}^{-4}$ for $\sigma$.

$\begin{array}{c}5.75\times {10}^8{\text{ W/m}}^2=\left(5.67\times {10}^{-8}{\text{ Wm}}^{-2}{\text{K}}^{-4}\right)T^4\\ T={\left(\frac{5.75\times {10}^8{\text{ W/m}}^2}{5.67\times {10}^{-8}{\text{ Wm}}^{-2}{\text{K}}^{-4}}\right)}^{\frac{1}{4}}\\ T=100035\text{ K}\end{array}$

Refer to Box 5-2 for the value of the surface temperature of the Sun.

$T_{\odot }=5800\text{ K}$

The surface temperature of Sirius is very high as compared to the surface temperature of the Sun.

Conclusion:

The surface temperature of Sirius is $100035\text{ K}$ and it is very high as compared to the surface temperature of the Sun.