4 ms·

The derivation in the original article (Paul W. Cooper, Through the Earth in Forty Minutes, Am. J. Phys. vol. 34 (1966) p. 68) relies on the assumption of const

The derivation in the original article (Paul W. Cooper, Through the Earth in Forty Minutes, Am. J. Phys. vol. 34 (1966) p. 68) relies on the assumption of constant density. (It's hard to say this for sure because some of the details are left out, but Cooper at least states he's making this assumption.) In the case where all the earth's mass is concentrated at the center, a point mass starting on the surface of the Earth would just go to the center and stay there. In the opposite extreme where all the mass is concentrated on the surface, (that is, the Earth is a hollow shell) it actually turns out that the gravitational acceleration at any point inside the shell is zero, so it wouldn't work in that case either. I suspect that for some reasonable class of spherically symmetric mass distributions (that is, the density only depends on the distance to the center of the Earth), tunnel systems like this are mathematically possible. But I'd be surprised to learn that there are mass distributions other than the uniform one for which the travel time doesn't depend on distance. But I'm not going to work this out because I have Real Work to do. (Now I wish I were teaching calculus so I would have an excuse to work out this problem...)

eru 18y ago
> In the case where all the earth's mass is concentrated at the center, a point mass starting on the surface of the Earth would just go to the center and stay there. Why? That would violate conservation of energy, wouldn't it? In a naive model of a point mass you'd get a singularity at the center. But using standard techniques (e.g. numeric pertubation, or Lebesgue integration) one gets an objects that swings back and forth like in the other scenarios.
- jibiki 18y ago
  Hmm, unless I'm going crazy here, we get the differential equation x'' = -(x^-2) x = (kt)^(2/3)(with k = (2/9)^(-3/2), not that it matters) seems to be a solution? I guess that doesn't help with the singularity, but neither does looking at energy (since you have infinite kinetic energy at the center and infinite potential energy everywhere else.)
  - eru 18y ago
    That differential equation only holds for positive x. Try something like the following: x'' = - signum(x) * (x^-2)
- madcaptenor 18y ago
  That's a good point. I don't know what I was thinking.

eru 18y ago

> In the case where all the earth's mass is concentrated at the center, a point mass starting on the surface of the Earth would just go to the center and stay there. Why? That would violate conservation of energy, wouldn't it? In a naive model of a point mass you'd get a singularity at the center. But using standard techniques (e.g. numeric pertubation, or Lebesgue integration) one gets an objects that swings back and forth like in the other scenarios.

jibiki 18y ago
Hmm, unless I'm going crazy here, we get the differential equation x'' = -(x^-2) x = (kt)^(2/3)(with k = (2/9)^(-3/2), not that it matters) seems to be a solution? I guess that doesn't help with the singularity, but neither does looking at energy (since you have infinite kinetic energy at the center and infinite potential energy everywhere else.)
- eru 18y ago
  That differential equation only holds for positive x. Try something like the following: x'' = - signum(x) * (x^-2)
madcaptenor 18y ago
That's a good point. I don't know what I was thinking.