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Initial Mass Function problem

  1. Dec 30, 2008 #1
    Hi, I've got a solution to this problem but I don't know if it's correct. My lecturer hasn't given us any examples but I had a go and the answer seems fine. Here it is:

    The IMF of a cluster of stars is: dN[tex]\propto[/tex]m-2.5dm
    There are 5 stars in the cluster with mass greater than 10 solar masses.
    What is the number of stars with mass greater than 2 solar masses?

    I put in 5 for dN, and solved the integral from 10 to infinity, to get the proportionality constant: 237.
    Now using the constant, I solved the integral from 2 to 10 to find N in this range of masses. Then I added the 5 stars which are more massive than 10 solar masses.
    Final answer: 56 stars.

    Seems plausible to me, but I have no way right now of checking the answer.
    Thanks a lot for any help!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 30, 2008 #2


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    Welcome to PF :smile:

    I agree with your math, though I am not familiar with mass distributions of stars and will take your word for it that part is valid.

    FYI, you could also do the integral from 2 to ∞, and that will include the 5 stars with m>10.
  4. Dec 30, 2008 #3
    thank you for your opinion. I've been working away at a few different problems and I seem to have them sorted out now :approve:
    I found some examples on my physics department website which helped.

    I did agree with you about solving the integral from 2 to [tex]\infty[/tex] but I tried it and it doesn't give the right answer. I think this is since the question only says there are 5 stars above 10 solar masses, and so these could be very massive or only a little more than 10 solar masses. So I think the method is to integrate from 2 to 10 then simply add on the remaining 5 to give the correct answer :smile:

    thanks again :smile:
  5. Dec 31, 2008 #4


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    That's weird, I definitely get 56 using 2 to ∞ for the integral.

  6. Dec 31, 2008 #5
    Oh, I probably made a silly mistake when I did the integral. At least it confirms my answer of 56 :smile:
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