# ASTRONOMY HOMEWORK

Here is a set of real deep-space spectra that include some stars (and galaxies!)

http://skyserver.sdss.org/dr16/en/tools/getimg/plate.aspx?choosesurvey=sdss&P=1125899951184832512&S=1868993899549190144&B=4037477152865658880&A=apogee.apo25m.s.r12.10001.57372

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All you have to do is find spectra of two stars (called “STAR” on the page) where the temperature of one of them is clearly greater than the other.

Then upload a screenshot of each star (this lab allows 2 file uploads) and indicate which one is the hotter and which is cooler. The point of this exercise is to make sure you can correctly compare the temperatures of two stars based on their spectra. Then continue below to complete the assignment.

We’ll need to use the equation

L=
R
2
T
4
L=R2T4

in solar units, and solving for the radius we’ll get

R=
L
T
4

R=LT4

So what we need to find is the temperature (T) of Arcturus and its luminosity (L). The temperature is easy! Suppose we see a spectral peak for Arcturus at 700 nm.

(A) Use Wien’s Law to find the temperature, and divide by the Sun’s temperature (5800 K) to find it in “solar units”

(B) Now to find L: We know from parallax measurements that Arcturus is 11.26 pc away. Remember that each parsec is about 200,000 AU, so calculate how many AU away Arcturus is (in other words, how many times farther away is Arcturus than the Sun?)

(C) Almost there! We know that brightness decreases as the square of the distance, so square your answer to (B) to find how many times dimmer the Sun would appear to be at that distance.

(D) We can measure that Arcturus is actually 30 billion (30×10^9) times dimmer than the Sun appears to us, so divide (C) by this number to see how many times brighter Arcturus “really” is as compared to the Sun at the same distance (this is its luminosity, in solar units)

(E) Ok, now plug the T (from part (A)) and this L into the above equation to find how big Arcturus is (your answer will be in solar units)

(just enter your answers for (A) through (E) in the text box)

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