(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Drop a particle from the Earth's surface down a mineshaft to a depthh. 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 fromabove groundis 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).

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# Homework Help: Plumb line down mineshaft

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