Build a bottomless pit

Recently I enjoyed reading Ethan Siegel article on falling into bottomless pit and after reading it I got struck by the question: which shape should a bottomless pit at the equator have? Sure if it is a straight line you will hit the walls of the pit, but if you make the pit shape curved you should go fine … what is the shape of that pit? Well it turns our that the answer is more fascinating than expected and in the way to it a lot can be relearned…

Assumptions…

The problem is quite complicated… temperature, air drag, pressure, density changes, friction… to tackle the problem we will make some (false yet can be used as first approach) assumptions:

• No friction, no air, only force acting is gravity
• Earth is perfect circle at the equator
• Earth density is constant respect to the radius

Gravity as we fall

You might be familiar with Newton gravity (good aproximation so no need to use GR) to calculate orbits, what you might not be so familiar is with the shell theorem basically… outer layers of Earth do not affect you gravitationally once you are below them, only the sphere formed by the layers below.

Sooo Whaaat?

Well instead of the classical:

You get…

Where k is a constant that depends on that assumption regarding constant density of earth.

Looks strange but is actually the same formula of the force of a spring the more you strech it the more force it does.

Central force field

So we have a force pointing to the earth center depending linearlly on the distance to that center. That is a special case of central force field that is described by:

All the dots represent first (velocity) and second (acceleration) of the radial (r) and angular (theta) position.

Gladly Nicholas Wheeler did that so we can see a nice solution of it in chapter 4 regarding central force problems

Figure 9: Typical centered elliptical orbit of an isotropic harmonic oscillator, showing circles of radii rmax = a and rmin = b. The isotropic oscillator is exceptional (though not quite unique) in that for this as for all orbits the angular advance per radial oscillation is ∆θ = π: all orbits close after a single circuit.

So the solution is that of an ellipse but instead of having one of the focus at earths center it(the ellipse)is centered around earth center.

Good thing is that gravity force acts like a spring, it can be decompossed into x-y forces quite easy and see that the period depends only on K, indepently of starting point and initial speed:

Digging the pit

We are almost there but we can not dig the pit as that ellipse as earth is rotating. We need to account for it otherwise we would hit the oposite side of the pit.

So we need to put some numbers… how long will it take to go to the opposite side of the earth? Or how long will it take yo go back?

We can obtain K and T and substituting some values we can get a rough estimate of 84 mins.

So in 42 mins aprox we are at the other side of earth.

Keeping in mind that earth rotates once every 24h we have a displacement of around 15° so we end up quite far from where we began, more than 1660km to the west!!

Conclusion

The pit can be digged but return path is different from initial path. Also having a variable density makes hard to calculate, but could be done too.

I hope you have enjoyed this. Remember that something being difficult doesn’t make it impossible!