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 $\left(pc\right)$ 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, travelling through a vacuum with a speed of $186000mi/s$ , would cover in $1.0$ year. Express the Earth-Sun distance is light-years

Answer 1

Hint :Length of an arc $\ell =r\theta$
Where $r$ is the radius and $\theta$ is the angle subtended by an arc on it’s centre.
 $\theta$ should be in radian. In case $\theta$ is given in some other units then it needs to be consented first into radian.

Complete Step By Step Answer:
 Distance between Earth and Sun $=92.9\times {10}^6$ mi
A parsec is the distance at which a length of $1AU$ subtend an angle of $1$ second of arc so the distance at which $1AU$ subtend only $1$ second of arc $r={\displaystyle \frac{\ell }{\theta }}$
But first we need to convert $1$ second of arc into radian.
 $1$ sec of arc $={\displaystyle \frac{1}{60}}$ min of arc
 $1$ min of arc $={\displaystyle \frac{1}{60}}$ degree
 $1$ degree $={\displaystyle \frac{\pi }{180}}$ radian
Therefore $1arc\sec =({\displaystyle \frac{1}{60}}\times {\displaystyle \frac{1}{60}}\times {\displaystyle \frac{\pi }{180}})$ radian
 $=4.85\times {10}^{-6}$ rad
thus $1$ parsec $={\displaystyle \frac{1AU}{4.85\times {10}^{-6}}}=2.06\times {10}^{5}AU...............\left(i\right)$
speed of light $=186000{}^{mi}/{}_s$
thus in one year distance travelled by light
 $1\ell y=186000\times$ (seconds in $1year$ )
 $=186000\times (365\times 24\times 60\times 60)$
 $1\ell y=5.9\times {10}^{12}mi$
Now because $1AU=92.9\times {10}^{6}mi$
Given and in $1Mi={\displaystyle \frac{1}{5.9\times {10}^{12}}}\ell y$
So $1AU=92.9\times {10}^6\times {\displaystyle \frac{1}{5.9\times {10}^{12}}}\ell y$
 $=1.57\times {10}^{-5}\ell y$
And from equation (i)
 $1AU={\displaystyle \frac{1}{2.06\times {10}^5}}par\sec$
 $=4.85\times {10}^{-6}par\sec$
Thus distance between earth and sun $=1AU=92.9\times {10}^{6}mi=1.57\times {10}^{-5}\ell y$
 $=4.85\times {10}^{-6}par\sec$ .

Note :
In the formula $\ell =r\theta$ , students must be careful that $\theta$ should be kept in radian. Students should practice the unitary method for conversion questions.
 $1miles=1.609$ kilometre
Students if want to solve the question in M.K.S unit standard, they can but it will just increase the number of steps. So students should be careful while working in other unit standards.