Q53P - Fundamentals Of Physics

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.

                     

Step-by-Step Solution

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 1: Given data

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

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

Step 2: Understanding the unit conversion

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

Step 3: (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 $4.9\times {10}^{-6} \mathrm{parsecs}$ .

Step 4: (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 $1.69\times {10}^{-5} \mathrm{light} \mathrm{year}$.