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Luminosity of a star

  • Thread starter Morpheus
  • Start date
  • #1
Morpheus
Here are a few more questions that I really need help on, thanks!




3. A Cepheid variable star is observed in the nearby galaxy M31. Its period of variation
is measured to be 40 days. Using the period - luminosity relation given in the table
below, what is the luminosity of the Cepheid in solar luminosities?
Period Luminosity
(days) (solar luminosities)
2.5 645
4.0 1120
6.3 2000
10.0 3550
15.8 6310
25.1 11500
40.0 20900
63.1 38000

4. For the Cepheid in M31, what is its luminosity in ergs/sec? (Recall that the solar
luminosity is 3.85 x 1033 ergs/sec.)


5. The Cepheid in M31 has its apparent brightness measured at 1.06 x 10-12 ergs/sec/cm2.
The inverse square law of light may be written as
b = L ÷ 4 pi r^2
Where b is the apparent brightness, L is the luminosity and r is the distance.
How far away is the Cepheid in centimeters? Convert your answer to parsecs by
knowing that one parsec contains 3.09 x 1018 centimeters.
 

Answers and Replies

  • #2
Morpheus
My biggest problem with these is that I'm totally lost. My professor assigns these problems, but our book and his notes don't show how to do them.
 
  • #3
HallsofIvy
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What? You mean your professor actually expects you to think for yourself? How evil of him.

Okay, that bit of sarcasm off my chest, in problem 3 you are given a table:
(days) (solar luminosities)
2.5 645
4.0 1120
6.3 2000
10.0 3550
15.8 6310
25.1 11500
40.0 20900
63.1 38000

and asked to determine the "luminosity" if "days"= 40. This is purely a test of how you can read a table? (emphasis mine.)

Once you know the number of "solar luminosities" you are told that each solar luminosity is 3.85 x 1033 ergs/sec. Okay, convert your solar lumnosities to ergs/sec. (Hint: if you knew each hamburger cost $1.50, how would you find the cost of 6 hamburgers?)

Finally you are given a formula: b = L ÷ 4 pi r^2
You know b (apparent brightness) is 1.06 x 10-12 ergs/sec/cm2.
You know L (luminosity) from problem 3 and you know (I hope!) that pi is approximately 3.1416. Plug them into the equation and solve for r.

By the way, do you recognize 4 pi r^2 as the formula for surface area of a sphere? In other words, that formula says that after the light has gone a distance r, the light, L, has spread out over the surface of that sphere giving the apparent brightness, b, for someone at one point on that sphere.
 
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