Question

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.

Short answer

1. The Earth-Sun distance is $4.9\times {10}^{-6}\mathrm{persecs}$.

2. The Earth-Sun distance is $1.6\times {10}^{-5}\mathrm{light}\mathrm{years}$.

Step-by-step solution

Given data

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

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

Understanding the unit conversion

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

(a) Determination of Earth-Sun distance in parsecs

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 $4.9\times {10}^{-6}\mathrm{parsecs}$.

(b) Determination of Earth-Sun distance in light years

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$1.69\times {10}^{-5}\mathrm{light}\mathrm{year}$.