91Ó°ÊÓ

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

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

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,

$S=R×\mathrm{θ}$

Therefore,

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

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

Then,

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

Substitute the values,

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

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

It is given that,

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

However,

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

So,

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

Now,

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

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