Will Bart make it to the other side?

Question

A perfectly straight tunnel is drilled trough the Earth, from the North pole to the Equator, as seen in the link (curtesy of "Jack"):

http://s226.photobucket.com/albums/dd11/jackphotos123/?action=view¤t=Cat3b.jpg

Let's say the tunnel is exactly 9020 km long.

The tunnel's floor is flat and coverd with solid Skatelite material.

Bart decides to try his new skateboard in the tunnel.

He stands at the very edge of the tunnel, which is at a 45 degree angle from his skateboard, he slides right in.

Lisa is waiting on the other side with a long rope, tied to a solid tree, 1 km from the hole, just in case.

If Lisa needs to throw the rope in to help her brother out of the tunnel how long a rope would she need?

Is the straight tunnel the equivalent of a 9020 km halfpipe?

Would its "equivalent shape" be the same as the arc of the Earth over the lenght of the hole?

The tunnel is perfectly straight, by "equivalent shape" I meant, would the gravitational force along such a tunnel be the equivalent of the gravitational force acting on a skateboarder riding on a halfpipe?

Mr Aviron, OK, let's imagine the Earth is not rotating, (you got me on that one).

Also, lets imagine a 45 degree edged halfpipe that runs over 9020 km.

Can Bart make it to the other side without Lisa's help?

Oh great Locust, thank you for the wiki link about the train.

I almost picked your answer.

Thank you, to all three answerers, that I admire greatly, for answering my question.

Answer 1

Ok, we need to make some assumptions here:

1. The Earth is not rotating, so we can ignore centrifugal force
2. There is no atmosphere and no frictional forces
3. The Earth is homogeneous and completely solid
4. The Earth is perfectly spherical

If we can make these assumptions, then no Lisa would not need to rescue Bart using a rope. Bart would reach Lisa right at the hole's entrance, because this is equivalent to riding a roller coaster from the top of a hill and then stopping exactly at the top of another identical hill (another name for this tunnel is - linear oscillator).

So in this sense, the straight tunnel is equivalent to a halfpipe, because Bart's initial velocity is zero, but has maximum potential energy, and as he goes through the tunnel, his velocity increases and eventually reach a maximum at the center of the tunnel (here PE = 0, and KE = max), and then as he comes out of the other end, his PE returns to maximum and KE = 0. This is precisely what happens with a skater riding a halfpipe.

Answer 2

Just imagine Bart shooting out of that hole shouting "To infinity and beyond."

No, No. Wrong cartoon. That line came from Toy Story. But the little guy would get some major air when he came out the far side.

You see, the earth is spinning pretty fast at the equator and not at all at the pole. Bart would be accelerated by the hole. Centrifigual force. I'm going to assume that this acceleration would more than make up for the fact that the earth bulges at the middle. Pretty sure I'm right, but If I'm wrong, Lisa would need about 15 km of rope to make up for the fatter earth.

Is it the equivalent of a 9020 halfpipe? No, it has 45 degree, not 90 degree edges. The lack of a vertical segment would really screw up your tricks.

Yea, the equivalent shape of the tunnel would be pretty close to the shape of an inverted 90 degree arc, which is the shape of the earth's arc over the holes.

Answer 3

There's a wiki article on this, "Gravity Train". See link. Also see the pdf paper. The more interesting question is what's the equivalent shape "above a flat ground"? Before answering that, the force on Bart in the tunnel is proportional to the distance to the halfway point, which explains why we get the "spring differential equaion", x''(t) = kx, with sinusoidal solutions. The simplest physical analogy "above ground" is the pendulum, where the arclength is proportional to angle θ, and the force on the skater on the halfpipe (eighths pipe?) is proportional to Sin(θ), or θ for small angles. This is the reason why clock pendulums also have nice sinusoidal behavior, but only for small angles. However, for real half pipes, this breaks down. So, is there an half-pipe like curve where the force on the skater is proportional to the arclength from the bottom of the trough? Hold on and see I can get back to you on this.

Addendum: Read Alexander's excellent answer to the matter of earth's oblateness and centrifugal force, in which he says they cancel out and so is irrelevant to this problem.

http://answers.yahoo.com/question/index?qid=20071214090337AA8G1uG&r=w