Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $\text{CoD-33}°\text{7795 B}$ star whose surface temperature is $2550\text{ K}$.

Answer to Problem 54Q

Solution:

L

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550\text{ K}$.

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

(b)

To determine

The radius (in kilometers) of $\text{CoD-33}°\text{7795 B}$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L_{\odot }$ and its surface temperature is $2550\text{ K}$.

Answer to Problem 54Q

Solution:

$0.26R_{\odot }$, $1.8096\times {10}^5\text{ km}$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L_{\odot }$ and its surface temperature is $2550\text{ K}$.

Formula used:

The relation between luminosity of different stars is:

$\frac{L}{L_{\odot }}={\left(\frac{R}{R_{\odot }}\right)}^2{\left(\frac{T}{T_{\odot }}\right)}^4$

Here, $L$ is the luminosity of the star, $L_{\odot }$ is the luminosity of the Sun, $R$ is the radius of the star, $R_{\odot }$ the radius of the Sun, $T$ is the temperature of the star and $b_{\odot }$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T_{\odot }=5800\text{ K}$.

The relation between luminosity of different stars is:

$\frac{L}{L_{\odot }}={\left(\frac{R}{R_{\odot }}\right)}^2{\left(\frac{T}{T_{\odot }}\right)}^4$

Substitute $2550\text{ K}$ for $T$, $5800\text{ K}$ for $T_{\odot }$ and $0.0025L_{\odot }$ for $L$.

$\begin{array}{c}\frac{0.0025L_{\odot }}{L_{\odot }}={\left(\frac{R}{R_{\odot }}\right)}^2{\left(\frac{2550\text{ K}}{5800\text{ K}}\right)}^4\\ {\left(\frac{R}{R_{\odot }}\right)}^2={\left(\frac{5800\text{ K}}{2550\text{ K}}\right)}^4\left(\frac{0.0025L_{\odot }}{L_{\odot }}\right)\\ \frac{R}{R_{\odot }}=\sqrt{{\left(\frac{5800\text{ K}}{2550\text{ K}}\right)}^4\left(0.0025\right)}\\ \frac{R}{R_{\odot }}=0.26\end{array}$

Further solve,

$R=0.26R_{\odot }$

Refer to Appendix 6, the value of Solar radius is $1R_{\odot }=6.9599\times {10}^8\text{ m}$.

Therefore,

$\begin{array}{c}R=0.26R_{\odot }\\ =0.26\left(6.9599\times {10}^8\text{ m}\right)\\ =0.26\left(6.9599\times {10}^8\text{ m}\right)\left(\frac{1\text{ km}}{{10}^3\text{ m}}\right)\\ =1.8096\times {10}^5\text{ km}\end{array}$

Therefore, the value of the radius in kilometer is $1.8096\times {10}^5\text{ km}$.

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984\text{ km}$.

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=\frac{d}{2}$

Use this expression to determine the radius of Jupiter.

$\begin{array}{c}{R}_{\text{Jupiter}}=\frac{142984\text{ km}}{2}\\ =71492\text{ km}\end{array}$

The radius of $\text{CoD-33}°\text{7795 B}$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R_{\odot }$ in term of solar radius and $1.8096\times {10}^5\text{ km}$ in terms of kilometer.