Vicar Philip

Think about this: let's say we were able to install a piece of 4" conduit, perfectly straight, from one side of the earth to the other, going through the exact center of the earth.

Now, disregarding temperature variations and the impossibility of actually installing such a pipe, take a nice shooter marble and drop it in the exact center of the conduit.

The marble will accelerate to its maximum velocity. Here is my question: What happens to the marble as it approaches the very center of the earth? Will it slow down and eventually stop, hovering in mid-air? Or, will it pass the center, reverse direction, and fall the other direction? And repeat the process again and again until it stops?

Once it stops, would it be immoveable?

Undercurrent

Yes, that's pretty much it. Neglecting air resistance, the marble will forever go back and forth between the two ends of the hole. With air resistance, the oscillations with decrease in amplitude over time until the marble is stopped at the centre of the earth, hovering in mid-air (although really, it will probably be pulled to one side of the tube since the earth doesn't have perfectly uniform density.)

It wouldn't be "immovable" though. With a suitably long stick you could poke it out the other side no problem.

Friar Bellows

The marble would oscillate about the centre, with the amplitude of the oscillation decreasing with time because of friction.

It's fairly simple to analyse the idealised case of zero friction and a uniform density of the earth. The gravitational force exerted on a marble of mass, m, at a distance, d, from the centre of the Earth is given by:

F = - G * M(d) * m / d^2

where M(d) is the mass of the earth interior to d. The mass of the earth exterior to d has no effect on the marble (i.e. its gravitational effects cancel out). The mass of the earth interior to d is given by:

M(d) = p * 4 * pi * d^3 / 3

where p is the uniform density of the earth. Substituting this expression into the force equation gives:

F = - (4 * pi * G * p * m / 3) * d

Or:

F = - k * d

Whenever you have a force that is proportional to the displacement, d, but oppositely directed (hence the minus sign), you have a simple harmonic oscillator. The period, T, of this oscillation is:

T = 2 * pi * sqrt(m / k) = sqrt(3 * pi / G * p)

When you plug the numbers in you get a period of roughly 84 minutes. Note that the period of oscillation is independent of the mass of the marble, and the earth for that matter (pardon the pun).

Friction would produce damped harmonic oscillation. I think a non-uniform density (the earth actually gets denser nearer the centre, and not in a very smooth way) would still probably produce oscillation but of a non-sinusoidal variety, but I'm not sure.

fando

Also consider that when you're at the center of Earth, you're being pulled by all the mass around you in a uniform way. That is, you'll be floating there as if in free fall. Someone can push you out of this state with a suitably long stick, just as you can push a fellow astronaut in space with the shuttle arm.

thebeave

I remember solving that problem in Junior level Physics in college (Dynamics) about 20 years ago. In our class, the problem was represented as a new, proposed high speed train thru the earth. I think in our problem, the train didn't go thru the exact center, but some distance above it. Same outcome; oscillating back and forth.

Of course, that liquid outer core would present some engineering difficulties, to say the least.

StrongMan

Man, that is totally F=-k*d up!

Feather

Hahah! I love it when people make jokes with physics. That one's a keeper.

Rousseau_CHN

Gravity does not exist; the earth really sucks