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?

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Zack M. Share
Dec 13, 2015

Answer:

#861,000,000 " 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.

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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!

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Trippy Share
Dec 13, 2015

Answer:

It has a diameter of 861744026,4 km, so #~~#862 million km.

Explanation:

In astronomy, the angular diameter is given by the following formula:

#delta=(206265d)/D#.

#delta# - Angular Diameter
#d# - diameter of the object
#D# - distance between the observer and the object

206265 is simply the number of arcsecs in a radian.

So before we chuck in the values we have to rearrange the equation in a way that we isolate #d#, which is what we want, so let's do just so:

#delta=(206265d)/D##hArr##delta*D=206265d# (I moved #D# to the other side here).

#delta*D=206265d##hArr##(delta*D)/206265=d#, by moving 206265 to the left hand side, we get #d# isolated. So now we chuck in the values:

#(0.044*428)/206265=d##hArr##d=9,1*10^-5# light years, because we were using light years in our #D#, we get the result in light years.

Now to turn our #d# into km, we need to know how many km's there is in a light year. A light year is equal to #9.4607*10^12# km.

#9.4607*10^12*9.1*10^-15=861744026,37367578# which is #~~##861744026,4# km

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