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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"If the momentum of an object increases by #20%#, what will be the percent increase in its kinetic energy?"
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!
