Question

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

Answer 1

The objective of this problem is to convert the Earth Sun distance to parsecs and light years.

To relate parsec (pc) to $AU$ , we note that when $\theta$ is measured in radian,

it is equal to the arc length $s$ divided by the radius $R$.

For a very large radius circle and small value of $\theta$, the arc may be approximated as the straight line segment of length $1AU$. Thus,

$\theta =1arcsec=\left(1arcsec\right)=\left(\frac{1arcmin}{60arcsec}\right)\left(\frac{1^o}{60arcmin}\right)\left(\frac{2\pi rad}{{360}^o}\right)=4.85\times {10}^{-6}rad$
Therefore, one parsce is

$1pc=R_0=\frac{s}{\theta }=\frac{1AU}{4.85\times {10}^{-6}}=2.06\times {10}^{5}AU$

Next, we relate $AU$ to light year (ly). Since a year is about $3.16\times {10}^{7}s$, we have

$1ly=(186,000mi/s)(3.16\times {10}^{7}s)=5.9\times {10}^{12}mi$

Since $1pc=2.06\times {10}^{5}AU$, inverting the relationship gives

$R=1AU=\left(1AU\right)\left(\frac{1pc}{2.06\times {10}^{5}AU}\right)=4.9\times {10}^{-6}pc$