Chapter 10, Problem 10.22P

To determine

The radii of a $0.072M_{\odot }$ star and $85M_{\odot }$ star also the ratio of radii of both the star.

Answer to Problem 10.22P

The radius of star one is $\text{0}{\text{.082R}}_{\odot }$ the radius of star second is $\text{13}{\text{.1R}}_{\odot }$ and ratio of radius of both stars is $0.0063$.

Explanation of Solution

Write the expression Stefan-Boltzmann equation for radius of star is given as,

    $R={\left(\frac{L}{4\pi \sigma T^4}\right)}^{1/2}$        (1)

Here, $L$ is luminosity, $T$ is surface temperature and $\sigma$ is numerical constant.

Conclusion:

Substitute ${10}^{6.006}{L}_{\odot }$ for $L$ and $5.76\times {10}^{-8}$ for $\sigma$ and $50506.87$ for $T$ in equation (1)

    $\begin{array}{c}{R}_1={\left(\frac{{10}^{6.066}{L}_{\odot }}{4\pi \left(5.76\times {10}^{-8}\right){\left(50506.87\right)}^4}\right)}^{1/2}\\ ={\left(\frac{3.9\times {10}^{32}}{4\pi \left(5.76\times {10}^{-8}\right){\left(50506.87\right)}^4}\right)}^{1/2}\\ ={\left(\frac{1.92\times {10}^{22}}{4\pi \left(5.76\times {10}^{-8}\right)\left(6.50731\times {10}^{18}\right)}\right)}^{1/2}\\ ={\left(\frac{1.92\times {10}^{22}}{\left(4.7101\times {10}^{12}\right)}\right)}^{1/2}\end{array}$

Thus, radius of small star is $\text{0}{\text{.082R}}_{\odot }$.

Substitute ${10}^{-4.3}{L}_{\odot }$ for $L$ and $5.76\times {10}^{-8}$ for $\sigma$ and $1661.46$ for $T$ in equation (1)

    $\begin{array}{c}{R}_2={\left(\frac{{10}^{-4.3}{L}_{\odot }}{4\pi \left(5.76\times {10}^{-8}\right){\left(1661.46\right)}^4}\right)}^{1/2}\\ ={\left(\frac{1.92\times {10}^{22}}{4\pi \left(5.76\times {10}^{-8}\right){\left(1661.46\right)}^4}\right)}^{1/2}\\ ={\left(\frac{1.92\times {10}^{22}}{4\pi \left(5.76\times {10}^{-8}\right)\left(7.62017347\times {10}^{12}\right)}\right)}^{1/2}\\ ={\left(\frac{1.92\times {10}^{22}}{\left(5515656.421\right)}\right)}^{1/2}\end{array}$

On further simplifying the above expression.

    $\begin{array}{c}{R}_2={\left(\frac{1.92\times {10}^{22}}{\left(3.254\times {10}^{14}\right)}\right)}^{1/2}\\ ={\left(3.481\times {10}^{15}\right)}^{1/2}\\ =9.1\times {10}^9\text{m}\\ \text{=13}{\text{.1R}}_{\odot }\end{array}$

Thus, radius of large star is $\text{13}{\text{.1R}}_{\odot }$.

The ratio of radius of star 1 and 2 is calculated as,

    $\begin{array}{c}\frac{R_1}{R_2}=\frac{\text{0}{\text{.082R}}_{\odot }}{\text{13}{\text{.1R}}_{\odot }}\\ =0.0063\end{array}$

Thus, the ratio of radius of star 1 and 2 is $0.0063$.