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The Science of a Bottomless Pit
StartsWithABang writes It's the ultimate dream of many children with time on their hands and their first leisurely attempt at digging: to go clear through the Earth to the other side, creating a bottomless pit. Most of us don't get very far in practice, but in theory, it should be possible to construct one, and consider what would happen to a very clever test subject who took all the proper precautions, and jumped right in. Here's what you would have to do to travel clear through the Earth, come out the other side, and make the return trip to right back where you started.
24 hours *if* you have air resistance. And then you're moving so slow that you barely get past the center.
Note that no vacuum is perfect so you will lose velocity. Their scenario should have started the person off at the south pole, not the north, for the extra altitude.
Note that the heat isn't really the materials problem that they make it out to be - it's an energy problem. You don't need a material that can withstand 4000, you just need cooling. And not linearly high cooling, but an exponential decline. The longer you cool the rock down to your target temperature, the deeper your effect on the rock temperature behind your tunnel walls, and thus the shallower the temperature gradient, and thus the lower the rate of heat loss. It's like trying to cool a hot house - the air conditioner really struggles in the beginning but it gradually becomes easier with time as the walls and everything inside the house cool down.
Now, the pressures, those are insane, and the normal approach to pressure maintenance on deep drilling - filling with a heavy mud - obviously wouldn't work here.
We gotta go to a crappy town where I'm a hero.
No its not. Because the problem is not central force motion between two point objects. The attracting mass effectively decreases as the object descends. A simple application of Gauss's Law. The motion of an object falling through a bottomless pit is harmonic, not Keplerian.
Mathematically it's an example of a degenerate orbit with one zero semi-axis, and the orbital period can be simply calculated from Kepler's laws. What's more interesting, it even holds true if you do not move through the center of the Earth! For example, a train from any place on Earth to any other place on Earth will move all by itself and always arrive at destination in about 45 minutes (neglecting the oblateness of the Earth and need to compensate for Coriolis forces and friction) if you put it inside a completely straight tunnel.
Nope, this is not "an example of a degenerate orbit with one zero semi-axis" and Kepler's first and third laws do not apply. Kepler's laws do not apply when you are falling through a sphere (or ellipsoid) that has its mass spread throughout its volume.
The orbit is not elliptical. Because the acceleration is not at all proportional to inverse of the squared distance. But if you plot the orbit, it does look a lot like an ellipse with a small semi-axis.
Kepler's second law applies due to conservation of angular momentum.
The calculation of the orbit is made more difficult because the density of the Earth varies from about 3 g/cm^3 to 13 g/cm^3. (We should be able to compute it pretty easily with Runge-Kutta.) To computer the orbit, we could reference the acceleration graph on the "Structure of the Earth" Wikipedia page.
http://en.wikipedia.org/wiki/S...
Cyberax's second comment about the train (on a frictionless track) is really cool. I wonder how much the Coriolis force would affect the travel time.
Mathematically it's an example of a degenerate orbit with one zero semi-axis, and the orbital period can be simply calculated from Kepler's laws.
No, it can't; it's not a Keplerian problem. You could calculate the period using Kepler's laws if the Earth were a point mass. But it's not. You could calculate the period using the Brachistochrone calculation if the Earth were a uniform sphere. But it's not. The Earth is layered, with the density changing as you go closer to the center. Only way to solve the problem correctly is numerical integration.
(I'd actually be interested in seeing the calculation done in the article.)
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