Chapter 5, Problem 45QAP

(a)

To determine

The flux from the each square meter of the planet’s surface if its temperature is $400\text{K}$.

Answer to Problem 45QAP

The flux per square meter of the planet is $2.72\times {10}^{-13}\text{K}/{\text{m}}^{\text{2}}$.

Explanation of Solution

The radius of the Earth is $6371\text{km}$ whereas the radius of the planet is $1.7\text{times}$ the radius of the Earth i.e. $10830.7\text{km}$.

Write the expression for the surface area of the planet.

$A=4\pi R^2$        (I)

Here, $A$ is the total surface area of the planet and $R$ is the radius of the planet.

Write the expression for the flux from a square meter of the planet’s surface.

$\varphi =\frac{T}{A}$        (II)

Here, $\varphi$ is the flux from the surface and $T$ is the temperature of the star.

Conclusion:

Substitute $10830.7\text{km}$ for $R$ in equation (I).

$\begin{array}{c}A=4\pi {\left(10830.7\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^2\\ =12.566\left(1.17\times {10}^{14}\right){\text{m}}^{\text{2}}\\ \approx 1.47\times {10}^{15}{\text{m}}^{\text{2}}\end{array}$

Substitute $400\text{K}$ for $T$ and $1.47\times {10}^{15}{\text{m}}^{\text{2}}$ for $A$ in equation (II).

$\begin{array}{c}\varphi =\frac{400\text{K}}{1.47\times {10}^{15}{\text{m}}^{\text{2}}}\\ =2.72\times {10}^{-13}\text{K}/{\text{m}}^{\text{2}}\end{array}$

Thus, the flux per square meter of the planet is $2.72\times {10}^{-13}\text{K}/{\text{m}}^{\text{2}}$.

(b)

To determine

The luminosity of the planet that has the temperature of $400\text{K}$.

Answer to Problem 45QAP

The luminosity of the planet will be $2.133\times {10}^{18}\text{W}$.

Explanation of Solution

Write the expression for the Luminosity of the planet.

$L=\sigma A{T}^4$        (III)

Here, $L$ is the luminosity of the planet and $\sigma$ is the Stefan-Boltzmann constant.

Conclusion:

Substitute $5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}$ for $\sigma$, $1.47\times {10}^{15}{\text{m}}^{\text{2}}$ for $A$ and $400\text{K}$ for $T$ in equation (III).

$\begin{array}{c}L=\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}\right)\left(1.47\times {10}^{15}{\text{m}}^{\text{2}}\right){\left(400\text{K}\right)}^4\\ =2.133\times {10}^{18}\text{W}\end{array}$

Thus, the luminosity of the planet will be $2.133\times {10}^{18}\text{W}$.

(c)

To determine

The peak wavelength of the radiations emitted from the planet.

Answer to Problem 45QAP

The peak wavelength of emission of the planet is $7242.5\text{nm}$.

Explanation of Solution

Write the expression for the peak wavelength of emission from the planet.

${\lambda }_{peak}=\frac{2.897\times {10}^{-3}\text{m}\cdot \text{K}}{T}$        (IV)

Here, ${\lambda }_{peak}$ is the peak wavelength of emission.

Conclusion:

Substitute $400\text{K}$ for $T$ in equation (IV).

$\begin{array}{c}{\lambda }_{peak}=\frac{2.897\times {10}^{-3}\text{m}\cdot \text{K}}{400\text{K}}\\ =7.24\times {10}^{-6}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ \approx \text{7242}\text{.5}\text{nm}\end{array}$

Thus, the peak wavelength of emission of the planet is $7242.5\text{nm}$.