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?

Question

Hunter Arthur · Accepted Answer

#861,000,000 " km"#

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(theta) ~~ theta# if we convert #theta# to radians.

#.044 " arc seconds" = 2.13 xx 10^-7 " radians"#

Since #theta# is the total diameter of Betelgeuse, we want to use #sin(theta"/"2)# to calculate the radius.

#r = d sin(theta/2) ~~ d theta/2#

But the radius is #1"/"2# of the diameter, #D#, so we have;

#D/2 = d 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 = (427 " light years")(2.13 xx 10^-7)#

#D = 9.10 xx 10^-5 " light years"#

Light years are not the most practical units for measuring the diameter of a star, however, so lets convert to #"km"# instead.

#D = (9.10 xx 10^-5 " light years")(9.46 xx 10^12 " km / light year")#

#D = 8.61 xx 10^8 " km"#

This is about 600 times the diameter of the sun!

Hunter Arthur · Answer

undefined $$ \text{Actual Diameter} = \text{Angular Diameter} \times \text{Distance} \times \frac{\pi}{180 \times 3600} $$
$$ \text{Actual Diameter} = 0.044 \, \text{arcseconds} \times 427 \, \text{light-years} \times \frac{\pi}{180 \times 3600} $$
$$ \text{Actual Diameter} \approx 1.09 \, \text{million kilometers} $$