An astronomical unit (AU) is equal to the average distance from Earth to the Sun, about $92.9\times {10}^6\mathrm{mi}$. A parsec (pc) is the distance at which a length of 1 AU would subtend an angle of exactly 1 second of arc (Fig. 1-8). A light-year (ly) is the distance that light, traveling through a vacuum with a speed of $186000\mathrm{mi}/\mathrm{s}$, would cover in 1.0 year . Express the Earth – Sun distance in (a) parsecs and (b) light-years.

$1\mathrm{AU}=1\mathrm{parsec}\times 1\mathrm{arcsec}$

The average distance between the sun and earth is $92.9\times {10}^6\mathrm{mi}$.

Chain-link conversions are used for unit conversion in which given values are multiplied successively by conversion factors.

Using the given conversion factor, convert the distance from astronomical units to the parsecs or to light-years.

The length of the arc is,

$S=R\times \theta$

Therefore,

$1\mathrm{parsec}=\frac{1\mathrm{AU}}{1\mathrm{arcsec}}$

But, $1\mathrm{AU}=92.9\times {10}^6\mathrm{mi}\mathrm{and}1\mathrm{arcsec}=4.85\times {10}^{-6}\mathrm{rad}$.

Then,

$$\begin{gathered} 1\mathrm{par}\sec =\frac{92.9\times {10}^6}{4.85\times {10}^{-6}} \\ =19.15\times {10}^{12}\mathrm{mi} \end{gathered}$$

Substitute the values,

$$\begin{gathered} 1\mathrm{mi}=\frac{1}{19.15\times {10}^{12}} \\ =0.052\times {10}^{-12}\mathrm{parsecs} \\ 1\mathrm{AU}=92.9\times {10}^6\mathrm{mi}\times 0.052\times {10}^{-12} \\ =4.85\times {10}^{-6}\mathrm{parsecs} \\ \approx 4.9\times {10}^{-6}\mathrm{parsecs} \end{gathered}$$

Thus, the Earth-Sun distance is $4.9\times {10}^{-6}\mathrm{parsecs}$.

It is given that,

$1\mathrm{A}.\mathrm{U}.=92.9\times {10}^6\mathrm{mi}$It is given that,

However,

$$\begin{gathered} 1\mathrm{light}\mathrm{year}=186000\frac{\mathrm{mi}}{\mathrm{s}}\times \left(3.16\times {10}^7\right)\mathrm{s} \\ =5.9\times {10}^{12}\mathrm{mi} \end{gathered}$$

So,

$$\begin{gathered} 1\mathrm{mi}=\left(\frac{1}{5.9\times {10}^{12}}\right) \\ =1.69\times {10}^{-13}\mathrm{light}\mathrm{year} \end{gathered}$$

Now,

$$\begin{gathered} 1\mathrm{A}.\mathrm{U}.=\left(92.9\times {10}^6\right)\left(1.69\times {10}^{-13}\right)\mathrm{light}\mathrm{year} \\ =1.57\times {10}^{-5}\mathrm{light}\mathrm{year} \\ \approx 1.69\times {10}^{-5}\mathrm{light}\mathrm{year} \end{gathered}$$

Thus,Earth-sun distance is$1.69\times {10}^{-5}\mathrm{light}\mathrm{year}$.