Chapter 33, Problem 17P

To determine

Relative sizes of two stars.

Answer to Problem 17P

Star with lower wavelength is 0.15 times in diameter as compared to the other star.

Explanation of Solution

Given:

Distance between star and Earth is same in each case that is $d_1=d_2$ .

Wavelength of first star is $500\text{ nm}$ and other one is $700\text{ nm}$ .

The ratio of their apparent brightness is 0.091.

Formula used:

The Stefan-Boltzmann equation is

  $P=\beta A{T}^4$

Calculation:

To find the size, relationship between temperatures is found using Wien’s law

  $\begin{array}{l}{\lambda }_{P}T=\alpha \\ \text{So,}\\ {\lambda }_{P1}{T}_1={\lambda }_{P2}{T}_2\\ \frac{T_2}{T_1}=\frac{{\lambda }_{P2}}{{\lambda }_{P1}}=\frac{\text{500 nm}}{\text{700 nm}}=\frac{5}{7}\end{array}$

Now, using relation of absolute luminosity to deduce a relation between their sizes

  $\begin{array}{l}{l}_1=0.091\times l_2\\ \frac{L_1}{4\pi d_1^2}=0.091\frac{L_2}{4\pi d_2^2}\\ \frac{d_2^2}{d_1^2}=\frac{0.091L_2}{L_1}=\frac{0.091P_2}{P_1}\end{array}$

Now, introducing concept of Stefan-Boltzmann equation in the above equation,

  $\begin{array}{l}\frac{d_2^2}{d_1^2}=\frac{0.091A_2{T}_2^4}{A_1{T}_1^4}=\frac{(0.091)4\pi r_2^2{T}_2^4}{4\pi r_1^2{T}_1^4}=0.091\frac{T_2^4}{T_1^4}\frac{r_2^2}{r_1^2}=1\\ \text{So,}\\ \frac{r_1}{r_2}=\sqrt{0.091}{\left(\frac{5}{7}\right)}^2=0.15\end{array}$

Therefore,

  $\begin{array}{c}\frac{D_1}{D_2}=\frac{2r_1}{2r_2}\\ =\frac{r_1}{r_2}\\ =0.15\end{array}$

Conclusion:

Thus, star with the wavelength $500\text{ nm}$ is 0.15 times in diameter as compared to the other star.