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Speed of an object thrown down a hole to the center of the earth

  1. Feb 23, 2004 #1

    qpt

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    I am supposed to calculate how fast an object is moving at the center of the earth if thrown down some hole that was drilled through the earth. I am supposed to set up a dr/dv differential equation and solve it. I am supposed to use Gauss's law. I am to assume the Earth is uniformly dense.

    Here is what I have so far, and you will see why I am puzzled:

    int(g . da) = Menc / G

    (g is the force, G is the universal gravitational constant).

    So g = Fg/m = GMe/r^2, no problem there. I can take it out of the integral by symmetry:

    GMe/r^2 int(da) = Menc / G

    Now the da is what is getting me. A = 4 pi r^2, so da = 8pi*r dr, but then I have int (8pi*r dr) and then my dr's vanish. I am supposed to get this in terms of a DE with dr/dv so I can't have my dr's vanishing.

    One thing I thought about doing, and I don't know if this is right or not:

    int(da) = 4pi* (dr)^2

    And Menc = int(k*dv) (v is volume, k is some constant),
    Menc = 4/3 (Pi*(dr)^3).

    When I collect the Menc and the da, all but one dr cancels, leaving me with.

    g = (4/3)Pi*dr / (G)

    Is this the right approach?

    My other problem, assuming that what I have is true, is that I can rewrite g in terms of Force, which has an acceleration, which has a dv/dt term in it. But now I have 3 differentials; dv/dt and dr. dt is the odd guy out:

    [some coefficient here] dv/dt = (4/3)Pi*dr / G.

    I can think of no way to "chain rule" dt out. Is there something obvious I'm missing.

    Thanks in advance.
     
  2. jcsd
  3. Feb 23, 2004 #2
    This is one of my favorite problems, actually. However, I never used the approach you are. I used a conservation of energy approach.

    However, for Gauss's law,

    [tex]\int_\textrm{surface}F \cdot dA = \frac{M_\textrm{earth}}{G}[/tex]

    you're probably going to want to convert the integral into spherical coordinates. When you do that, you must replace [itex]dA[/itex] with [itex]r^2\textrm{sin}\theta dr\,d\theta\,d\phi[/itex] where r is the radius, [itex]\theta[/itex] is the angle off the z-axis, and [itex]\phi[/itex] is the angle in the xy-plane.

    Once you evaluate the integral, there should be no differentials left.

    Just out of curiosity, why do you want to know the gravitational field? Do you intend on using it to calculate potential?

    cookiemonster
     
    Last edited: Feb 23, 2004
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