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Plumb line down mineshaft

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Drop a particle from the Earth's surface down a mineshaft to a depth h. Show that in this case there is no southerly deflection due to the variation of gravity and that the total southerly deflection is only
    [tex]\frac{3}{2}\frac{h^2\omega^2}{g}sin(\lambda)cos(\lambda)[/tex]


    2. Relevant equations

    g is defined at ground level.

    The southerly deflection from above ground is given by:
    [tex]C_i\frac{h^2}{g}\omega^2sin(\lambda)cos(\lambda)[/tex]
    for individual forces where:
    The Coriolis force to second order: [tex]C_1=\frac{2}{3}[/tex]
    The variation of centrifugal force with height: [tex]C_2=\frac{5}{6}[/tex]
    The variation of gravitational force with height: [tex]C_3=\frac{5}{2}[/tex]

    3. The attempt at a solution

    I figured out that without the gravitational force attached, the equation given clearly turns into the equation that is desired.

    However, I have been unable to figure out how the gravitational force would effect the deflection, and why falling through the earth (reduction of gravity) would cause a cease of southernly deflection as opposed to falling to the earth (increase of gravity).
     
  2. jcsd
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