# Plumb line down mineshaft

1. Nov 28, 2008

### Math Jeans

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
$$\frac{3}{2}\frac{h^2\omega^2}{g}sin(\lambda)cos(\lambda)$$

2. Relevant equations

g is defined at ground level.

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

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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