Chapter 33, Problem 35GP

To determine

The distance to the nearest star.

Answer to Problem 35GP

The distance to the nearest star is $5.0\text{ly}$ .

Explanation of Solution

Given:

The intrinsic luminosity of sun is $L=3.90\times {10}^{16}\text{W}$ .

The value of distance between earth and sun is $1.496\times {10}^{11}\text{m}$ .

The brightness of star is ${10}^{11}$ fainter than Sun.

Formula used:

The expression for brightness of Sun is

  $l=\frac{L}{4\pi d^2}$

Here, $d$ is the distance between Earth and Sun.

Calculation:

The brightness of the Sun is,

  $\begin{array}{c}l=\frac{3.90\times {10}^{16}}{4\times \pi \times {\left(1.496\times {10}^{11}\right)}^2}\\ =1.387\times {10}^{-7}{\text{W}/\text{m}}^2\end{array}$

The luminosity of star is,

  $\begin{array}{c}\text{Luminosity}\text{of}\text{star}=\frac{1.387\times {10}^{-7}}{{10}^{11}}\\ =1.387\times {10}^{-18}\text{W}/{\text{m}}^{\text{2}}\end{array}$

The distance of star is,

  $\begin{array}{c}{d}^2=\frac{L}{4\pi \left(\text{Luminosity}\text{of}\text{star}\right)}\\ d^2=\frac{3.90\times {10}^{16}}{4\times \pi \times 1.387\times {10}^{-18}}\\ d^2=0.223\times {10}^{34}\text{m}\\ d=\sqrt{0.223\times {10}^{34}}\text{m}\\ d=4.731\times {10}^{16}\text{m}\times \left(\frac{1.057\times {10}^{-16}\text{light}\text{year}}{1\text{m}}\right)\\ d=\text{5}\text{.0}\text{ly}\end{array}$

Conclusion:

The distance to the nearest star is $5.0\text{ly}$ .