The answer really is 42

[pokerfan]

Anyone remember this from Differential Equations?

Suppose you could drill a hole through the Earth and then drop into it. How long would it take you to pop up on the other side of the Earth?



http://hyperphysics.phy-astr.gsu.edu/Hbase/mechanics/earthole.html">Journey through the center of the Earth

[Turbineguy]

1. I knew it!

[lapfog_1]

2. What is 6 x 9????

(helps to know how many fingers are on your favorite alien's hands)

[pokerfan]

3. Exactly!

[ET Awful]

4. How many roads must a man walk down?

:P

[Clear Blue Sky]

5. The calculus is the easy part.

Getting a hole through the center of the earth will be a little bit trickier...

[Lerkfish]

6. my physics professor pointed out that due to the relative mass

of you and the earth, that once you fell to the center, you would never have enough velocity and mass to travel back "up" to the other side. In other words, you'd go the center and stay there. Of course, you couldn't do this anyways since the center is molten.

[pokerfan]

9. due to friction?

Without drag the object would oscillate indefinitely.

[Lerkfish]

10. to be honest, I don't remember ALL of his point, but he was railing that

hypothetical "problems" like these always conveniently ignore all the variables, and just concentrate on those that seem to be important.
He mentioned friction, and the fact that the mass of the SPHERE is cubed, due to the properties of volume of three dimensional objects, not merely linear (as most calculations assume) when it surrounds you. You would essentially be submitted to the entire mass of the planet, minus only the tube down the center.

There were other variables, but I cannot recall all of them, this was back in 1980.

[The_Casual_Observer]

13. He was wrong, you would go back & forth like a pendulum.

[Lerkfish]

14. well, that could be: I wasn't a physics major.

I was an art major. This was beginning physics.
If you're a more reliable physics authority than that professor and my memory, then I"ll have to go with what you say.

[The_Casual_Observer]

21. He was probably right. There is a problem with the speed

you could practically get to due to "terminal velocity" of a falling body.

[Lerkfish]

23. LOL! now I"m confused. have you changed your mind or am I misunderstanding you?

[The_Casual_Observer]

24. I changed my mind after thinking about your comment some more.

[Lerkfish]

25. hey, that's cool.

but like I said, I"m not a physics major. :)

[TechBear_Seattle]

16. You are forgetting gravitational drag

Gravity is center-of-mass to center-of-mass. As you approach the center of mass, gravitational attraction decreases. Very soon after falling into the hole, you hit a terminal velocity were gravity decreases faster than you can accelerate, and you remain at a speed less than "normal" terminal velocity.

After you pass the earth's center of gravity, gravity starts to increase again; only now, it is pulling you back to the center. You cannot reach the other end of the hole because your velocity is not sufficient to break the pull of gravity. You get pulled back again, but now you start significantly closer to the center of mass, which means your terminal velocity is lower. You will fall back on the other side even closer to the center, and so forth.

Consider this problem: You have a ball y centimeters in diameter. You drop it from a height of x meters, it bounces back up to 98% of its height. It bounces again to 98% of that 98%, and so on. How long until the ball comes to rest?

Mathematically, the ball will bounce forever, each time reaching a height of 98% of the previous bounce. But any engineer will tell you that the ball effectively stops bouncing once the height is within the bounds of the ball; after that, it is quivering, not bouncing. Same thing with the "falling body through the center of the earth" puzzle.

[pokerfan]

17. I guess I "forgot" it because I've never heard of it

you hit a terminal velocity were gravity decreases faster than you can accelerate

I realize that the problem ignored a lot of details just to turn it into a simple calc problem, but I've never heard of gravitational drag. Is it like a tidal force, i.e. a gradient?

[caraher]

29. Assumptions are important

The original problem is really about a non-rotating sphere and a tunnel in which there is no friction or air resistance. Given those assumptions energy will be conserved and an object dropped into the tunnel would indeed oscillate with the stated period - and do so with undiminished "amplitude" (that is, you would make it all the way through to the opposite side and back; repeat forever).

Realistically... well, technologically physically constructing such a tunnel is a non-starter. But lets suppose that - somehow - there exists such a tunnel. Is there air in the tunnel? That causes drag; the sum of kinetic and gravitational energy decreases steadily, and you don't make it through.

The case considering rotation is rather interesting... if the tunnel were truly frictionless, rotation would not matter! As someone mentioned, you would hit a wall because of this, but if you assume a frictionless tunnel that doesn't matter. As you fell, without friction the sum of kinetic and gravitational energies would be constant. The rotational speed of Earth would have to increase as you approached the center, but not by a noticeable amount. (It's like considering how jumping moves the location of the center of mass of Earth...)

But this business about "gravitational drag" in the previous post is incorrect. It is correct that the net gravitational attraction decreases as you approach the center of Earth, where it is zero, and after that it increases as you move out. But absent any "losses" the point at which you reach the extreme of your travel is when you are the same distance from the center as the distance from which you started (assuming you fall from rest). It's simple conservation of energy.

The argument outlined about "terminal velocity" is only valid in the context of air drag. Terminal velocity is by definition the speed at which the force due to air resistance is equal in magnitude and opposite the direction of weight. Weight does decrease as you move inward, but if you are considering an air-filled tunnel not making it to the other side can be explained simply by noting that that drag is a "loss" like friction.

I'm also not sure if the terminal velocity would change in quite the way you think, because the density and pressure of air probably wouldn't be constant as a function of height, even if you somehow kept the tunnel at the same temperature. (Though off the top of my head I'm not quite sure exactly how it would change; I'd expect pressure to increase all the way down, but at a diminishing rate thanks to the diminishing value of "g" as you approached the center.)

The ball analogy is not applicable. A real ball deforms when it hits the ground. This deformation stores energy, some of which is released as the ball rebounds but some of which converts to thermal energy. The random thermal energy is not available for conversion into either kinetic or gravitational energy, so it effect it is "lost." That's the difference between dropping a rubber ball and a lump of clay - the clay deforms irreversibly, while the ball's deformation is largely reversible. The surface from which the ball bounces also experiences this deformation and (usually imperceptible) heating.

In addition, the ball experiences air drag, whether or not it is perfectly elastic. These two "loss" mechanisms are why real balls eventually stop bouncing.

The one thing I may be missing would be some kind "drag" that might come up through general relativity. I don't know whether there's something there for me to miss; this is simply a small disclaimer!

[TechBear_Seattle]

30. I'm an engineer, not a theoretician

My major was in applied mathematics, not abstract mathematics. :hi:

I had a math professor who moonlighted in the Philosophy department teaching logic and critical thinking, and he tended to combine the two fields in all the classes he taught. He encouraged us to think outside the situation, and it was a running joke in his math classes to take word problems out of the book and discuss why it was "purely theoretical" and completely unrelated to any kind of real-world situation. This wasn't to criticize the problems, which served well to teach us how to derive a mathematical solution from a given situation; this was to remind us that real-world situations were much messier than text-book illustrations and that if we were going to be engineers or scientists, we needed to remember that.

So when I see problems like this, my first thought is, "Assume the technology is there to create a stable hole through the center of the earth. Given the vast difference in mass, gravity must certainly play a part in there somewhere, as we are using the earth and a human being, not a hollow sphere and a null point."

But even there, I strayed into the box by imagining the hole going through the axis of rotation, where Coriolis forces are null. Mea culpa.

[caraher]

31. Speaking of wacky physics questions...

Have you seen this?.

I've seen a few articles, I think in the American Journal of Physics, devoted to the practical impossibility of situations used in elementary texts. For instance, there was one that involved the force between a pair of 1.0 Coulomb charges placed one meter apart... there's a pretty long list of reasons that would be hard to pull off!

[krispos42]

19. Not to mention Coriolis force

The Earth rotates. You will not fall straight down, you will drift eastward into the wall of the tunnel.

Talk about road rash...

You'll lose a lot of energy to friction. No oscillation, really. You'll go past the center, be pushed up against the west wall of the tunnel as you rise, then you'll stop, fall down again (sliding down the east wall), etc., until you come to a complete stop.

In fairly short order, too.

[FiveGoodMen]

20. Actually, you'd fall into a bunch of very hot goo before you got close to the center

Just saying that figuring out whether you'd hit a wall that you could never build to begin with...

diminishing returns and all that

[benld74]

7. 3 - as in how long does it take to get to the center of a tootsie pop?

[Lithos]

8. As I recall

42 was the number of varieties of tea available at the tea shop frequented by Mr. Adams.

L-

[MindMatter]

11. You might come out a little crispier than you might like

It is kinda warm down there.

[The_Casual_Observer]

12. 42.4 minutes

Pi*sqrt(r/g)

r =4000 mi

[Orsino]

27. And you'll still be told by TSA to be at your terminal two hours in advance. n/t

[TechBear_Seattle]

15. I remember a similar problem

What is the minimum volume of a cylinder drilled along a diameter of a solid sphere x units in radius?


Zero, as it is mathematically acceptable to have a cylinder with a width of zero radius, which makes its volume zero as well.

[pokerfan]

18. Yeah, there's a second solution

with the cylinder being as wide as the diameter of the sphere which means that it will have a height of zero.

[muriel_volestrangler]

22. Meh - the cool bit of the calculus is that the oscillation period is the same for any tunnel

whether or not it goes through the centre of the earth. If you have a frictionless tunnel, so that the component of the force of gravity parallel to the tunnel is accelerating/decelerating you, while the rest is just a normal force at right angles, the time taken from one end to another of any tunnel is exactly the same - the maximum velocity is greatest when going through the centre of the earth, but so is the distance you need to cover.

No, I can't prove this any more (I remember it as a question back as school), but I think there will be a proof somewhere on the 'net.

On edit:

There's a related problem. In Chapter 7 of Lewis Carroll's 1893 book Sylvie and Bruno. The fictional German professor, Mein Herr, proposes a way to run trains by gravity alone. Dig a straight tunnel between any two points on Earth (it need not go through the Earth's center), and run a rail track through it. With frictionless tracks the energy gained by the train in the first half of the journey is equal to that required in the second half. And also, in the absence of air resistance and friction, the time of the journey is about 42 minutes (84 for a round trip) for any such tunnel, no matter what the tunnel's length.

Martin Gardner discussed this in one of his Scientific American "Mathematical Games" columns.

The period in the frictionless hole:

Restating the question: The period for the circular "skimming" orbit (of radius equal to the Earth's radius, R) is

T = 2p (R/g)1/2

Show that this is the same as the period of motion in frictionless straight hole through the center of the stationary Earth.

Partial Answer:

....

http://www.lhup.edu/~dsimanek/scenario/insight.htm

[Jim__]

26. Thanks for that link.

The problems and solutions on that page look interesting and now I have another website to read when I have some time to kill.

[pokerfan]

28. Yeah

It's also the period of the lowest possible orbit. Well, if the atmosphere didn't exist.

[eridani]

32. But do you know where your towel is? n/t