Chapter 16, Problem 44Q

(a)

To determine

The ratio between the energy flux from the patch of a sunspot’s penumbra and the energy flux from an equally large patch of undisturbed photosphere. Compare both the patches.

Answer to Problem 44Q

Solution:

0.55, the photosphere will be brighter than the penumbra.

Explanation of Solution

Introduction:

State the expression for the Stefan-Boltzmann law.

$F=\sigma T^4$

Here, F is the energy flux, T is the temperature of the body, and $\sigma$ is the Stefan-Boltzmann constant.

Explanation:

Refer to the textbook section 16-8 and obtain the values of temperature for the penumbra and the photosphere, which are $5000\text{ K}$ and $5800\text{ K}$ respectively.

Recall the expression for the Stefan-Boltzmann law for a penumbra.

$F_{\text{penumbra}}=\sigma T_{\text{penumbra}}^4$

Here, $F_{\text{penumbra}}$ and $T_{\text{penumbra}}$ are the energy flux and the temperature of the penumbra respectively.

Similarly, recall the expression for the Stefan-Boltzmann law for a photosphere.

$F_{\text{photosphere}}=\sigma T_{\text{photosphere}}^4$

Here, $F_{\text{photosphere}}$ and $T_{\text{photosphere}}$ are the energy flux and the temperature of the photosphere respectively.

Now, represent the ratio of $F_{\text{penumbra}}$ and $F_{\text{photosphere}}$.

$\begin{array}{c}\frac{F_{\text{penumbra}}}{F_{\text{photosphere}}}={\left[\frac{\sigma T_{\text{penumbra}}}{\sigma T_{\text{photosphere}}}\right]}^4\\ ={\left[\frac{T_{\text{penumbra}}}{T_{\text{photosphere}}}\right]}^4\end{array}$

Substitute $5000\text{ K}$ for $T_{\text{penumbra}}$ and $5800\text{ K}$ for $T_{\text{photosphere}}$.

$\begin{array}{c}\frac{F_{\text{penumbra}}}{F_{\text{photosphere}}}={\left[\frac{5000\text{ K}}{5800\text{ K}}\right]}^4\\ ={\left[0.8620\right]}^4\\ =0.55\end{array}$

Further solve the above expression for $F_{\text{penumbra}}$.

$F_{\text{penumbra}}=0.55F_{\text{photosphere}}$

According to the above ratio, the energy flux of the penumbra is 0.55 times the energy flux of the photosphere. Thus, the photosphere will be brighter than the penumbra.

Conclusion:

Hence, the ratio between the energy flux of the penumbra and of the photosphere is 0.55, so the photosphere is brighter than the penumbra.

(b)

To determine

The ratio between the energy flux from the patch of a sunspot’s penumbra and the energy flux from an equally large patch of umbra. Also discern the brighter part.

Answer to Problem 44Q

Solution:

1.8, penumbra is brighter than the umbra.

Explanation of Solution

Introduction:

According to Stefan-Boltzmann law, energy flux is directly proportional to the fourth power of the temperature of the body. Mathematically,

$F=\sigma T^4$

Here, F is the energy flux, T is the temperature of the body, and $\sigma$ is the Stefan-Boltzmann constant.

Explanation:

Refer to the textbook section 16-8 and obtain the values of temperature for the penumbra and the photosphere, which are $5000\text{ K}$ and $5800\text{ K}$ respectively.

Recall the expression for the Stefan-Boltzmann law for a penumbra.

$F_{\text{penumbra}}=\sigma T_{\text{penumbra}}^4$

Here, $F_{\text{penumbra}}$ and $T_{\text{penumbra}}$ are the energy flux and the temperature of the penumbra respectively.

Similarly, recall the expression for the Stefan-Boltzmann law for an umbra.

$F_{\text{umbra}}=\sigma T_{\text{umbra}}^4$

Here, $F_{\text{umbra}}$ and $T_{\text{umbra}}$ are the energy flux and the temperature of the umbra respectively.

Now, represent the ratio of $F_{\text{penumbra}}$ and $F_{\text{umbra}}$.

$\begin{array}{c}\frac{F_{\text{penumbra}}}{F_{\text{umbra}}}={\left[\frac{\sigma T_{\text{penumbra}}}{\sigma T_{\text{umbra}}}\right]}^4\\ ={\left[\frac{T_{\text{penumbra}}}{T_{\text{umbra}}}\right]}^4\end{array}$

Substitute $5000\text{ K}$ for $T_{\text{penumbra}}$ and $4300\text{ K}$ for $T_{\text{umbra}}$.

$\begin{array}{c}\frac{F_{\text{penumbra}}}{F_{\text{umbra}}}={\left[\frac{5000\text{ K}}{4300\text{ K}}\right]}^4\\ ={\left[1.16\right]}^4\\ =1.8\end{array}$

Further, solve the above expression for $F_{\text{penumbra}}$.

$F_{\text{penumbra}}=1.8F_{\text{umbra}}$

According to the above ratio, the penumbra is brighter than the umbra because the energy flux of the penumbra is 1.8 times the energy flux of the umbra.

Conclusion:

Hence, the ratio of the penumbra and umbra is 1.8. So, the penumbra will be brighter than the umbra.