Why Is There No Southerly Deflection When Dropping a Particle Down a Mineshaft?

[Math Jeans]

Homework Statement

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

Homework 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: $$C_3=\frac{5}{2}$$

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).

 

[Jhero]

Thank you for your interesting question. I would like to provide you with an explanation for why there is no southerly deflection when a particle is dropped down a mineshaft on Earth.

Firstly, it is important to understand the concept of the Coriolis force. This is a fictitious force that appears to act on objects moving in a rotating reference frame. On Earth, this force is responsible for the rotation of large-scale weather patterns, such as hurricanes.

Now, let's consider the scenario of dropping a particle down a mineshaft on Earth. At the surface, the particle experiences a gravitational force (due to the mass of the Earth) and a centrifugal force (due to the rotation of the Earth). These two forces, along with the Coriolis force, contribute to the southerly deflection of the particle.

However, as the particle falls deeper into the mineshaft, the gravitational force decreases due to the reduction in the distance to the Earth's center. This decrease in gravitational force is greater than the decrease in centrifugal force, resulting in a net decrease in the southerly deflection.

At a certain depth, the gravitational force becomes equal to the centrifugal force and the net deflection becomes zero. This is why there is no southerly deflection when a particle is dropped down a mineshaft on Earth.

To further understand this concept, let's consider an extreme case where the mineshaft is dug all the way to the center of the Earth. In this scenario, the gravitational force becomes zero and the centrifugal force is the only force acting on the particle. As the particle falls towards the center, the centrifugal force pulls it away from the center, resulting in a net southerly deflection.

In summary, the variation of gravitational force with height plays a crucial role in determining the southerly deflection of a particle dropped down a mineshaft on Earth. The equation provided in the forum post takes into account the effects of both the Coriolis force and the variation of gravitational force with height.

I hope this explanation helps to clarify your doubts. If you have any further questions, please do not hesitate to ask.