Cecil says “I was going to calculate how long this would take, but twenty minutes of computation has produced no useful result.”
This came up in a physics exam I wrote ages ago.
As it happens the answer is the same as the period of a satellite in orbit at the same altitude – in this case zero, sea-level. So something less than an hour and forty minutes if I remember correctly.
The column being referred to is What if you fell into a tube through the earth?
Well, the calculation that you’re referring to is fairly straightforward. The answer is T = 2 pi sqrt(R[sup]3[/sup] / GM) = 84m28s for the Earth, assuming a uniform density, which is actually a pretty bad approximation, as is ignoring resistance.
Since this takes a bit less than twenty minutes to compute, I believe Cecil was referring to the much more difficult problem of determining how long it would take to come into equilibrium.
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Referece to…
http://www.straightdope.com/classics/a1_165.html
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I’m reminded of one of my favorite amusement-park rides, which insurance liability lawyers have long ago caused to vanish forever. Only this amusement worked in reverse, using centrifugal force pulling out rather than gravity pulling in. It consisted of a cylindrical room about thirty feet in diameter, with smooth padded walls. People would enter the room and stand against the walls, and at the beginning of the ride the cylinder would begin to spin around a fixed pole in the center. The centrifugal force would press you against the wall and then the floor would be retracted about eight to ten feet, leaving everyone stuck to the wall with only the force to hold them there.
That was a lot of fun, and I’ve seen others like it in America. But this one was in Italy and the best feature, which makes it relevent to this thread (and also unique to that particular ride), was at the end, when the floor was brought back up and the the cylinder allowed to slow slightly. One by one, people would attempt to leave the wall and reach the center pole. This is unbelievably difficult to do, because the centrifugal force diminishes with each step toward the pole. The first step out is against heavy force, the next is much lighter, etc. “Normal” would be achieved as you reach the pole, but remember that the next step past the pole everything changes from “pulling against” to “restraining from”. Unless you can time yourself just right you will run right past the post and be caught up in a hilarious (and futile) attempt to keep from slamming into the padding on the opposite wall! :eek: (unlike the Earth-tube, the ride adds the need to compensate laterally as well; the pole isn’t a straight shot in the spinning room – you need to calculate a spiral path to get anywhere near it). Everyone laughs, and the next fool tries. After remaining at this reduced speed for awhile, the operator lets the ride gradually slow down so that eventually everyone (or at least most) can manage to catch the post.
It was great fun, but apparently too dangerous to be allowed at any U.S. park these days. Maybe then, too (late 1960’s), since this was in Italy.
As you fall down the tube, you will find yourself pushed against the side of the tube that is in the direction of the earth’s rotation. This is due to the “coriolis” force experienced by an object that moves radially while also moving circumferentially.
So there are actually two types of “frictionlessness” required here:
And, thus, you’re not so much “falling” down the tube, you’re “sliding” down it.
Likewise, on the “other side” of the center, as you rise back up the tube, you’ll be pushed against the side of the tube that is opposite the direction of the earth’s rotation… which is the same side you were pushed against on the way down.
Note that this is also related to the “spiral path” required to reach the center of the amusment park ride described in a previous posting.
But wouldn’t the mass or the gravity pushing down on all sides of you just crush you?
What I said in the other thread: we’re assuming a smooth, frictionless tube through the center of the earth that is uniform heat… and you’re worried that it’s not realistic because of the coriolis effect?
Welcome to the Straight Dope Message Boards, tmalaher, it’s a great first post (my satiric remark aside) and we’re glad to have you here!
And welcome to you, too, Hypo, glad to have you with us. I’ll let someone else comment on your comment, but I’d guess we’re assuming some sort of gravity-proof, heat-proof suit that allows our falling fellow to breathe despite the high velocity, wind resistance, etc.
The real question: this fall will take some time, did anyone give the poor soul a lunchbag?
I don’t think so. You would have only the mass below you pulling you down right? The mass above you would be subjected to the same gravity as you, so it wouldn’t affect you. I can’t think of exaactly what this is called. maybe Achernar and his genius math could help. but, as you go further into the tube the radius of the mass affecting you decreases. So, there is no mass “pushing against you” from all sides.
Unless I’m completely wrong. Then, just disgrace me and move on. :dubious:
I vaugely remember a discussion (on the web? in a physics textbook?) of a Russian guy who proposed building a perfectly straight railway tunnel between to distant cities. The idea was that the train would basically be gravity powered, taking 84 minutes or so to reach the other city. Somehow the author tied this discussion to Lewis Carroll and the idea of the rabbit hole that Alice falls down.
Because the time taken to “fall” from one end of such a tunnel to the other is independent of the length of the tunnel. A tube through the center of the earth is just a special case.
The reason this is true is that, while a tube from New York to London, for example, is shorter than one right through the center, you don’t end up being pulled as fast by gravity (because it’s at an angle). Amazingly, it all evens out, so you end up taking the same amount of time (84 minutes, I belive is correct… but from memory so don’t quote me.)
HypocrisyofCake asked:
“But wouldn’t the mass or the gravity pushing down on all sides of you just crush you?”
Gravity doesn’t push. It pulls.
So imagine you’re sitting at the center of the earth. A rock on the surface is pulling “up” on you (towards it). But another rock on the opposite side of the planet is also pulling up on you… in the opposite direction. So they cancel out. Likewise, for every chunk of mass on the planet that’s pulling on you, you can find an equal and opposite mass that cancels it out. So you’d be weightless at the center, as Cecil said.
You may, however be thinking of the pressure from all those Gazillions of tons of rocks pushing down on each other. But Cecil also said we’re going to ignore how you’d actually build such a tube, so we can assume that the pressure and temperature problem has been taken care of. A “Small Matter of Engineering”.
Thanks, tmalaher, that’s what I was trying to say. I just didn’t have a complete enough understanding of the “why.”
About 42 minutes, actually. It’s the same as half an orbit.
I am curious …
Is this statement … “This process would continue forever.” meant to be serious ? Wouldn’t this mean a “perpetual motion” mechanism ? I would think the gravity on each cycle would slowly but surely sap some of the energy away from the moving body so that eventually it would come to rest in the center. Maybe this would take a long time but wouldn’t that occur ?
the newbie coherentlight
Remember, we said “frictionless”. So, yes. It would continue forever.
Note that a completely friction-free setup is impossible.
Even if it were possible, if you tried to extract any energy from the system, you’d simply slow it down and it would eventually stop. So you still would not have a perpetual motion machine.
You have to check the original assumptions in cases like this.
I think that even if you did manage to remove all friction, that the gravitational waves emitted by the moving mass (you falling up and down the hole) would eventually bring you to a stop at the center. This might take a while. 
I started thinking about the case where we DO include air friction. The column mentions terminal velocity. Well, as you fell, with the net effect of gravity diminishing, would not the terminal velocity also decrease? So once you reached terminal velocity, you would begin to gradually slow down. Now I don’t know the calculations to run, but might this not cause you to come to a complete stop right at the center of the earth the FIRST time? In other words, no repeated oscillations past the center.
Anybody?
Air resistance also decreases with velocity. I do not think you’ll reach the zero state.
Coriolis force is irrelevant here, since the hole is stated to be from pole to pole. One thing which is a confounding factor, though, is that the simple analysis assumes that the Earth is constant density, which it’s not. As I recall, the density varies in such a way that acceleration due to gravity remains approximately constant until you get all the way to the core.
For a discussion of gravity inside of an object, by the way, see my Staff Report on Planet Doughnut, particularly the last paragraph. To sum up, it’s only the mass “below” you (that is to say, closer to the center) which matters.
Right, like I mentioned in my post. 
The higher density toward the center would decrease the travel time. Since I don’t have the density distribution of the Earth handy, and the differential equation is a pain anyway, I don’t know exactly how much it would decrease the trip time, but no more than 10%. 10% is the amount you’d decrease it if the acceleration were constant all the way down.
Wouldn’t the air pressure at the center be extreme? There is a huge column of air pressing down on it, many, many times the depth of the atmosphere at the surface, although the decreasing gravity would make the air pressure not quite as great as it would otherwise be. I would not be surprised if the air wasn’t a gas at all in the center. Slamming into liquid or solid nitrogen would be a real downer during your gravi-bungee jump.
Since the original problem has the tunnel going pole-to-pole, I don’t think there would be a Coloris effect.