Week 4: WS7 #3

3. One of the most common ways to measure the distance to a star is using the "trigonometric parallax." This works by measuring the angular distance a star appears to move, with respect to a background field of much more distant stars, as the Earth moves one quarter of an orbit (i.e. as the Earth translates "sideways" by a distance b = 1AU). 
There are 60 arcminutes in a degree and 60 arcseconds in an arcminute. What is the distance, measured in cm and light years, of a star that moves by 1 arcsecond when the Earth moves by 1AU? Give a general formula relating distance d to the angle \(\theta\), moved by the star. 

First we can convert the angle to radians.
\(1"\frac{1'}{60"}\frac{1^\circ}{60'}\frac{\pi}{180}=4.8\times10^{-6}\)
This is a really small angle, so we can use the small angle approximation to say that the distance from the star to the Sun is the same as the distance from the star to the Earth.

This also allows us to say that \(tan\theta=\theta\). 

\(tan\theta=\theta=\frac{1AU}{d}\)

Plugging in \(\theta\) and solving for d, we get \(d=3.13\times10^{18}cm=3.29ly\). This makes sense, because these are approximately the values of a parsec--a unit that's defined as 1AU/1".