Question

Take the diameter AB of the Earth's orbit as 3 x 108 km and consider a star S at a distance d, such that SA = SB and the angle ASB = 2 arcseconds. Calculate d. This is the distance unit of one parsec. Relate it to one light year.

Answer 1

Due to the symmetry of the triangle SAB we may calculate the distance from the Sun to the star as the height of the triangle.

Let $AB = 2a, \;\;\angle ASB = 2\theta.$ So $a = 1.5\cdot10^8\,\mathrm{km}, \theta = 1''.$

The height d of an isosceles triangle is $d = \dfrac{a}{\tan\theta} = \dfrac{1.5\cdot10^8\,\mathrm{km}}{\tan 1 ''}.$

For small angles their tangents are approximately equal to the angle expressed in radians.

$1'' \approx \dfrac{1}{206265}\,\mathrm{rad} \; \Rightarrow \; d \approx 1.5\cdot10^8\,\mathrm{km}\cdot 206265 \approx 3.1\cdot10^{13}\,\mathrm{km}.$

The speed of light is $c\approx 3\cdot 10^5\,\mathrm{km/s},$ and there are $86400\cdot365.24 \approx 3.16\cdot10^7$ seconds in a year, so the light-year is $3.16\cdot10^7\,\mathrm{s}\cdot3\cdot10^5\,\mathrm{km/s} = 9.5\cdot 10^{12}\,\mathrm{km}.$

Therefore, 1 pc is equal to $\dfrac{3.1\cdot10^{13}}{9.5\cdot10^{12}} \approx 3.3$ light years.