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Free-fall Acceleration with Centrifugal Force

  1. Feb 23, 2008 #1
    1. The problem statement, all variables and given/known data

    From Taylor's "Classical Mechanics", problem 9.15:

    On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude [tex]g = g_0[/tex] at the North Pole and [tex]g = \lambda[/tex] [tex]g_0[/tex] at the equator (with 0 [tex]\leq[/tex] [tex]\lambda[/tex] [tex]\leq[/tex] 1). Find g([tex]\theta[/tex]), the free-fall acceleration at colatitude [tex]\theta[/tex] as a function of [tex]\theta[/tex].

    2. Relevant equations

    [tex]g = g_0 + (\Omega \times R) \times \Omega[/tex]

    3. The attempt at a solution

    The north pole corresponds to a colatitude of 0 degrees, and the equator corresponds to 90 degrees. Using these values, and a trial-and-error approach, I arrived at the equation

    [tex]g(\theta) = g_{0}(Sin(\theta) + \lambda Cos(\theta))[/tex]

    the book gives the following answer:

    [tex]g(\theta) = g_{0}\sqrt{Sin^{2}(\theta) + \lambda^{2} Cos^{2}(\theta)}[/tex]

    And I'm not really sure where to start with this. The equation I've given is the only one given for free-fall acceleration, but I can't see any way to manipulate it into anything like this answer. Any suggestions would be appreciated.
  2. jcsd
  3. Feb 24, 2008 #2
    You can compute [tex]g[/tex] as a function of theta with [tex]g = g_0 + (\Omega \times R) \times \Omega[/tex]

    this is a vector equation and [tex]g_0[/tex] doesn't point in the same direction as [tex](\Omega \times R) \times \Omega[/tex]
    Once you've done that you can find [tex]\Lambda[/tex] by looking at what happens at the equator.
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