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Homework Help: Star diameter - momentum

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A star of mass 1.81×1031 kg and diameter 8.10E+9 m rotates with a period of 25.0 days. Suddenly the star changes size, and rotates with a new period of 18.0 days. Assuming a uniform density both before and after the size change, what is the new diameter of the star?

    2. Relevant equations

    Volume = 4/3*3.14*R^3
    Density = M / VOL
    I =2/5*M*R^2

    3. The attempt at a solution

    R(initial) = 8.10*10^9/2=4.05*0^9
    Volume(initial) = 4/3*3.14*(4.05*10^9)^3=2.782*10^29
    Density= (1.81*10^31/(2.782*10^29)=65.061

    the computer said that this answer is wrong and i don't know why
  2. jcsd
  3. Jan 15, 2010 #2


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    Homework Helper

    Hi nnokwoodeye! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    It's very difficult to read what you've done, but I think you're misreading the question …

    the mass is constant, not the density …

    the question isn't saying that density is the same before as after, only that the mass is always evenly distributed.
  4. Jan 15, 2010 #3
    I did not do the calculations but I noticed the units you used for the period were days (as given in the problem); converting to seconds might also make a difference.
  5. Jan 15, 2010 #4
    o.k, so how do i calculate the new density? i need it to find the radius and solve the question.

    to Gear300: you are right i forgot to convert the days to seconds, but it dosn't matter because i am using days in both side of the equation so the unit conversion would have been canceled out.
  6. Jan 15, 2010 #5
    o.k. I succeeded in solving the question
    Thanks for the help
    Last edited: Jan 15, 2010
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