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Moment of inertia of earth

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data

    The density of Earth, at any distance r from it's center is approximately

    p(r) = [14.2 - 11.6 (r/R)] * 10^3 kg*m^-3

    where R is the radius of the Earth. Show that the density leads to a moment of inertia I = 0.330 MR^2 about an axis through the center, where M is the mass of the Earth.

    2. Relevant equations

    I = [tex]\int[/tex] r^2 dm
    I of a hollow sphere = 2/3MR^2
    I = [tex]\int[/tex] dI

    3. The attempt at a solution

    So what I believe is right is to sum up the contributions of the moments of inertia of infinitesimally thin hollow spheres, with limits of integration 0 to R.

    So, I have a hollow sphere with thickness dr a distance r from the center of the Earth, whose moment is given by 2/3MR^2 so I have:

    I = 2/3 [tex]\int[/tex] [R,0] r^2 dm

    dm would be p(r) dV

    dV should be 4/3[tex]\pi[/tex](r+dr)^3 - 4/3[tex]\pi[/tex]r^3 because the all the mass is concentrated at the outside of the shell of thickness dr. This equals 3r^2dr.

    So. 2 [tex]\int[/tex] p(r) * 3r^2 * r^2 dr = I

    integrating with limits [R,0] and substituting in the actual values i got about 2 * 10^37 kg * m^2, but the answer is 8 * 10^37 kgm^2.

    What did I do incorrectly?
     
  2. jcsd
  3. Dec 6, 2009 #2

    diazona

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    Homework Helper

    That's not correct. Check your math on that differential. (I mean: you're right that [itex]\mathrm{d}V = \frac{4}{3}\pi(r + \mathrm{d}r)^3 - \frac{4}{3}\pi r^3[/itex], you just computed the difference incorrectly.)
     
  4. Dec 7, 2009 #3
    Oh, I forgot to move the 4pi/3 down. Wow. It came out to the right answer after that. Thanks.
     
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