Question #169319
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.
1
Expert's answer
2021-03-08T08:24:31-0500

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,    ASB=2θ.AB = 2a, \;\;\angle ASB = 2\theta. So a=1.5108km,θ=1.a = 1.5\cdot10^8\,\mathrm{km}, \theta = 1''.

The height d of an isosceles triangle is d=atanθ=1.5108kmtan1.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.

11206265rad    d1.5108km2062653.11013km.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 c3105km/s,c\approx 3\cdot 10^5\,\mathrm{km/s}, and there are 86400365.243.1610786400\cdot365.24 \approx 3.16\cdot10^7 seconds in a year, so the light-year is 3.16107s3105km/s=9.51012km.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 3.110139.510123.3\dfrac{3.1\cdot10^{13}}{9.5\cdot10^{12}} \approx 3.3 light years.


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