The supergiant star Betelgeuse has a measured angular diameter of 0.044 arcsecond. Its distance has been measured to be 427 light-years. What is the actual diameter of Betelgeuse?

Dec 13, 2015

$861 , 000 , 000 \text{ km}$

Explanation:

This is a pretty straight forward trigonometry problem. We can set up a diagram showing that the distance to Betelgeuse and the radius of Betelgeuse make a right angle.

Therefore, we can use the $\sin$ function to find the radius of Betelgeuse. Since $\theta$ is very small, we can use the small angle approximation, $\sin \left(\theta\right) \approx \theta$ if we convert $\theta$ to radians.

$.044 \text{ arc seconds" = 2.13 xx 10^-7 " radians}$

Since $\theta$ is the total diameter of Betelgeuse, we want to use $\sin \left(\theta \text{/} 2\right)$ to calculate the radius.

$r = d \sin \left(\frac{\theta}{2}\right) \approx d \frac{\theta}{2}$

But the radius is $1 \text{/} 2$ of the diameter, $D$, so we have;

$\frac{D}{2} = d \frac{\theta}{2}$

Canceling the $2$s leaves;

$D = d \theta$

Now we have an expression for the diameter, we can plug in what we know.

$D = \left(427 \text{ light years}\right) \left(2.13 \times {10}^{-} 7\right)$

$D = 9.10 \times {10}^{-} 5 \text{ light years}$

Light years are not the most practical units for measuring the diameter of a star, however, so lets convert to $\text{km}$ instead.

$D = \left(9.10 \times {10}^{-} 5 \text{ light years")(9.46 xx 10^12 " km / light year}\right)$

$D = 8.61 \times {10}^{8} \text{ km}$

This is about 600 times the diameter of the sun!