parsec

logistic_guy

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Using the definitions of the parsec and the light-year, show that \(\displaystyle 1 \ \text{pc} = 3.26 \ \text{ly}\).
 
Using the definitions of the parsec and the light-year, show that \(\displaystyle 1 \ \text{pc} = 3.26 \ \text{ly}\).
Let \(\displaystyle d = 149.6 \times 10^6 \ \text{km}\) be the average distance between the earth and the sun.

Then, one parsec is defined as:

\(\displaystyle 1 \ \text{pc} = \frac{d}{1''}\)

where \(\displaystyle 1''\) is one arcsecond.

Let us write \(\displaystyle 1 \ \text{pc}\) in terms of meters.

\(\displaystyle 1 \ \text{pc} = \frac{d}{1''} = \frac{149.6 \times 10^{9} \ \text{m}}{1''} = \frac{149.6 \times 10^{9} \ \text{m}}{1'' \times \frac{1^{\circ}}{3600''} \times \frac{\pi}{180^{\circ}}} = \textcolor{indigo}{3.086 \times 10^{16} \ \text{m}}\)

Let \(\displaystyle 2.998 \times 10^{8} \ \text{m/s}\) be the speed of light, and let \(\displaystyle 365.25\) be the numbers of days in one year.

Then,

\(\displaystyle 1 \ \text{light-year} = 2.998 \times 10^{8} \ \text{m/s} \times 365.25 \ \text{days} = 2.998 \times 10^{8} \ \text{m/s} \times 365.25 \times 24 \times 3600 \ \text{s} = \textcolor{indigo}{9.46 \times 10^{15} \ \text{m}}\)

Then,

\(\displaystyle 1 \ \text{pc} = \frac{3.086 \times 10^{16} \ \text{m}}{9.46 \times 10^{15} \ \text{m}} \ \text{light-year} = \textcolor{blue}{3.26 \ \text{ly}}\)
 
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