/sci/ - Science & Math

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Anonymous Sat Feb 5 14:53:42 2011 No.2480499 [Reply] [Original]

I pose a challenge:

Let's say it were possible to drill a hole several meters wide straight through the center of the Earth at the equator and reinforce it against heat and pressure through some means. This hole will be open at both ends, but with each opening contained in some sort of way so that the entire length of the tube can be removed of any atmosphere. I will then, wearing appropriate protective gear, fall into the opening. Assuming I make no contact with the tube walls and that there is no friction inside the tube, how long will it take me to reach the other opening?

>>	Anonymous Sat Feb 5 14:55:42 2011 No.2480513 Twice as long as it took to reach the middle.

>>	Anonymous Sat Feb 5 14:56:42 2011 No.2480522 just as long as it would take in a straight line from any point on the surface to any other point

>>	Anonymous Sat Feb 5 14:57:42 2011 No.2480532 /thread

Anonymous Sat Feb 5 14:58:42 2011 No.2480536

This question has a definite numerical answer, but I can't remember it, and I don't feel like calculating it right now.

But here's an interesting fact: Even if you drill that hole at an angle, it would take an object the same amount of time to reach the surface again, regardless of the angle at which the hole was drilled. This is assuming no friction, of course. I can't remember how to prove this, but one of my professors did it once... or maybe I simply misunderstood what he was saying; I'll leave it to some dedicated /sci/entist to prove or disprove this claim.

>>	Anonymous Sat Feb 5 14:58:42 2011 No.2480538 lol im pretty sure i saved the physics hw that asked to express the frequency of oscillation.....its also in my diff Eq book somewheres, its not that difficult

>>	Anonymous Sat Feb 5 14:59:42 2011 No.2480545 42 minutes.

>>	Anonymous Sat Feb 5 14:59:42 2011 No.2480550 The cool thing is that you are probably going to appear on the other side at the same velocity that you fell. Means you will be almost stopping and falling again, but you step forward instead. Also, don't fall again or you're screwed.

>>	Anonymous Sat Feb 5 14:59:42 2011 No.2480552 42 minutes

>>	Anonymous Sat Feb 5 15:00:42 2011 No.2480556 >>2480499 just over two hours assuming acceleration due to gravity is a constnat 9.8 m/s/s which it would not be

>>	Anonymous Sat Feb 5 15:00:42 2011 No.2480557 I think OP forgot to mention that, in order to solve this problem with ease, you have to assume that the Earth has a uniform density (which, in reality, it surely does not). Are we allowed to make this assumption, OP?

Anonymous Sat Feb 5 15:02:55 2011 No.2480569 [DELETED]

>>2480556
> assuming acceleration due to gravity is a constnat 9.8 m/s/s

Yeah, that would give you a much different answer.

At each point of descent, the acceleration must be calculate based only on the gravitational pull from pieces of the Earth that are closer to the center than you are. The concentric "shells" of Earth that lie "above" you will no longer affect your acceleration.

Basically, this means you have to set up an integral to solve this properly.

Anonymous Sat Feb 5 15:03:42 2011 No.2480577

>>2480556
> assuming acceleration due to gravity is a constnat 9.8 m/s/s

Yeah, that would give you a much different answer.

At each point of descent, the acceleration must be calculated based only on the gravitational pull from pieces of the Earth that are closer to the center than you are. The concentric "shells" of Earth that lie "above" you will no longer affect your acceleration.

Basically, this means you have to set up an integral to solve this properly.

>>	Anonymous Sat Feb 5 15:04:42 2011 No.2480580 >>2480552 Yeah, I'm pretty sure it's 42 minutes between any two seal-level points on the Earth's surface so long as gravity is the only acting force.

>>	Anonymous Sat Feb 5 15:04:42 2011 No.2480581 >>2480556 also all I did was a simple conservation of energy then v = at then doubled the time

>>	Anonymous Sat Feb 5 15:05:42 2011 No.2480588 >>2480557 Yeah, I forgot to mention that. We can assume that for simplicity.

>>	Anonymous Sat Feb 5 15:06:42 2011 No.2480592 >>2480577 >the sheels that are above you don't effect your acceleration yes they do they pull you the other way.

>>	Anonymous Sat Feb 5 15:06:42 2011 No.2480593 >>2480581 Well that's not good enough. "a" changes as a function of your depth, so you need an integral.

>>	Anonymous Sat Feb 5 15:06:42 2011 No.2480594 It's impossible to not make contact with the tube walls. At the surface, the Earth gives you a certain speed at which you circulate the planet. When you go down, the lower parts cycle on much lower speeds and therefore making you hit the wall for sure.

>>	Anonymous Sat Feb 5 15:07:42 2011 No.2480604 >>2480594 Okay. Then you can sit out while we do this solvable conceptual problem.

Anonymous Sat Feb 5 15:09:42 2011 No.2480620

>>2480592
> yes they do they pull you the other way.

Nope.

A hollow, spherical shell of uniform density will have no gravitational effect on objects which lie inside of it.

If you're in the center of that hollow sphere, the reason for this is obvious. If you're on the edge, however, you'll still feel no gravitational pull because, while you're closer to one side of the sphere, most of the mass is now on the other side with respect to your position.

It can be proven mathematically, and I'm too lazy to do it. Feel free to argue about it while I go get lunch.

Anyway, if you think of the Earth as being composed of an infinite number of concentric spherical shells, you can see conceptually that, if you're below the surface of the Earth, only those shells "below" you will exert any gravitational force on you.

>>	Anonymous Sat Feb 5 15:09:42 2011 No.2480613 this has been done over and over http://www.scientificamerican.com/article.cfm?id=would-you-fall-all-the-wa

>>	Anonymous Sat Feb 5 15:09:42 2011 No.2480612 >>2480593 I addressed that fact in my original post

>>	Anonymous Sat Feb 5 15:09:42 2011 No.2480619 >>2480592 learn to Gauss law of gravitation

>>	Anonymous Sat Feb 5 15:11:42 2011 No.2480628 >>2480612 Okay, well you basically said "it's 2 hours which it's not." So why bother posting anything at all?

Anonymous Sat Feb 5 15:11:42 2011 No.2480631

http://www.scientificamerican.com/article.cfm?id=would-you-fall-all-the-wa

http://www.scientificamerican.com/article.cfm?id=would-you-fall-all-the-wa

http://www.scientificamerican.com/article.cfm?id=would-you-fall-all-the-wa

http://www.scientificamerican.com/article.cfm?id=would-you-fall-all-the-wa

http://www.scientificamerican.com/article.cfm?id=would-you-fall-all-the-wa

http://www.scientificamerican.com/article.cfm?id=would-you-fall-all-the-wa

>>	Anonymous Sat Feb 5 15:13:42 2011 No.2480648 >>2480631 But does anyone in this thread actually know how to set up that integral?

>>	Anonymous Sat Feb 5 15:14:42 2011 No.2480659 >>2480628 because I felt like it?

>>	site witent Sat Feb 5 15:14:42 2011 No.2480662 File: 15 KB, 555x559, pdg_inside_earth.png [View same] [iqdb] [saucenao] [google] T=2π*sqrt(R/g)=84.44 min so roughly 42 minutes but it's not that easy, see pic

>>	Anonymous Sat Feb 5 15:15:42 2011 No.2480674 >>2480662 fascinating graph is fascinating

>>	Anonymous Sat Feb 5 15:16:42 2011 No.2480681 >>2480659 Well then it was a curious move to defend your original, meaningless, post.

>>	Anonymous Sat Feb 5 15:20:42 2011 No.2480704 http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

>>	Anonymous Sat Feb 5 15:23:42 2011 No.2480725 >>2480620 >if you're below the surface of the Earth, only those shells "below" you will exert any gravitational force on you. lolno. The mass above you exerts a force as well.

>>	Anonymous !wM5G0xiB/Q Sat Feb 5 15:25:42 2011 No.2480741 >>2480725 see >>2480620 Actualy .. is that right? I cant see it happening. Maybe in the center, but not close to the surface.

>>	Anonymous Sat Feb 5 15:26:42 2011 No.2480743 assuming a vacuum inside the tube, the center of mass is still at the center of earth, so you will just be going simple harmonic motion, so the exact same time. you'll oscillate forever

>>	Anonymous Sat Feb 5 15:32:42 2011 No.2480784 The deeper you fall the less gravity there is...

>>	Anonymous Sat Feb 5 15:33:42 2011 No.2480792 >>2480784 gtfo you idiot

Anonymous Sat Feb 5 15:36:42 2011 No.2480811

You would go into a perfect oscillation like he said. The time it would take you would be irrelevant since you will never perfectly make it to the other side. the amount of gravity acting upon you will change due to your relation to the center of the earth

>>	Anonymous Sat Feb 5 15:37:42 2011 No.2480822 yeah son you will fall one way tell gravity starts pulling ya the otherway. or else if gravity is stronger than we think. You will be torn in half at the center of earth and your legs will end up chili. and your torso in Kazakhstan

>>	Anonymous !wM5G0xiB/Q Sat Feb 5 15:38:42 2011 No.2480825 >>2480811 Asuming perfect conditions, no friction and perfect sphere. He would arive exacly at same distance how he started. So if he started 1m over surface, he would "stop" 1m over surface on the other side.

Anonymous Sat Feb 5 15:40:35 2011 No.2480842 [DELETED]

\oint_{S}\vec{g}\cdot \vec{dA}=-4\pi GM[/spoiler]

M=\rho V[/spoiler]

V=\frac{4}{3}\pi R^3[/spoiler]

\vec{g}4\pi%20R^2=-4\pi G(\rho \frac{4}{3}\pi R^3)[/spoiler]

\vec{g}=-\frac{4}{3}\pi G\rho R[/spoiler]

\frac{\partial^2 R}{\partial t^2}=-\frac{4}{3}\pi G\rho R[/spoiler]

solving the differential equation:

R=R_0cos(t\sqrt{\frac{4}{3}\pi G\rho}+\phi )[/spoiler]

The period is:

T= \sqrt{ \frac{3 \pi}{G \rho}}[/spoiler]

The time to get to the other side is half the period, so under ideal conditions it would take:

t=\frac{1}{2}\sqrt{ \frac{3 \pi}{G \rho}}[/spoiler]

Simply plug in the approximate density of the Earth and evaluate:

http://www.wolframalpha.com/input/?i=1%2F2+sqrt%28%283pi%29%2F%28%285540+kg%2Fm^3%29*G%29%29

So it would take about 42 minutes.

>>	Anonymous Sat Feb 5 15:41:42 2011 No.2480846 File: 27 KB, 500x404, 1258678517015.jpg [View same] [iqdb] [saucenao] [google] another science challenge for dr tyson. http://www.teachersdomain.org/asset/oer08_vid_gravitynsn/

Anonymous Sat Feb 5 15:43:42 2011 No.2480867

\oint_{S}\vec{g}\cdot \vec{dA}=-4\pi GM[/spoiler]

M=\rho V[/spoiler]

V=\frac{4}{3}\pi R^3[/spoiler]

\vec{g}4\pi R^2=-4\pi G(\rho \frac{4}{3}\pi R^3)[/spoiler]

\vec{g}=-\frac{4}{3}\pi G\rho R[/spoiler]

\frac{\partial^2 R}{\partial t^2}=-\frac{4}{3}\pi G\rho R[/spoiler]

solving the differential equation:

R=R_0cos(t\sqrt{\frac{4}{3}\pi G\rho}+\phi )[/spoiler]

The period is:

T= \sqrt{ \frac{3 \pi}{G \rho}}[/spoiler]

The time to get to the other side is half the period, so under ideal conditions it would take:

t=\frac{1}{2}\sqrt{ \frac{3 \pi}{G \rho}}[/spoiler]

Simply plug in the approximate density of the Earth and evaluate:

http://www.wolframalpha.com/input/?i=1%2F2+sqrt%28%283pi%29%2F%28%285540+kg%2Fm^3%29*G%29%29

So it would take about 42 minutes.

>>	Anonymous Sat Feb 5 15:43:42 2011 No.2480870 File: 83 KB, 670x580, jhjjj.jpg [View same] [iqdb] [saucenao] [google] Masses pull you.

Anonymous Sat Feb 5 15:44:42 2011 No.2480875

>>2480741
>Actualy .. is that right?
lolno. Everything exerts gravity on everything else. It's just that some things can be neglected or considered point sources in some assumptions. This is not one of those times. The entire mass of the planet and its location relative to you is important.

>>	Anonymous Sat Feb 5 15:45:42 2011 No.2480883 >>2480681 I wasn't really defending my post, just explaining how I got that answer.

>>	thinkfag Sat Feb 5 15:47:42 2011 No.2480904 You won't make it to the other end of the tube. Unless you bring along some climbing gear, lol. You'd still have to consider it taking you less than a week.

Anonymous Sat Feb 5 15:52:42 2011 No.2480932

>>2480725

No, you're an idiot.

http://en.wikipedia.org/wiki/Shell_theorem

> Isaac Newton proved the shell theorem saying that:
> 1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
> 2. If the body is a spherically symmetric shell (i.e. a hollow ball), no gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
> 3. Inside a solid sphere of constant density the gravitational force varies linearly with distance from the centre, becoming zero at the centre of mass.

Read number two.

>>	NASA Senior Sat Feb 5 15:53:42 2011 No.2480942 >>2480725 Sure it does. But there's an equally potent counter-force produced by the rest of the shell.

Anonymous Sat Feb 5 15:54:42 2011 No.2480948

>>2480875
> Everything exerts gravity on everything else.

But inside a hollow sphere, the contributions to net gravitational force exerted by each part of that sphere will cancel out. The net gravitational force due to a hollow sphere will be zero for an object located at any point inside that hollow sphere.

This is a fact.

Arli Individual Sat Feb 5 16:10:42 2011 No.2481025
File: 13 KB, 663x309, g_inside_outside_earth.jpg [View same] [iqdb] [saucenao] [google]

So there is no gravitational acceleration at the center, but that doesn't mean there is no gravity. Bcause at the center, the aether densi.. *cough* (sorry) I mean gravitational potential is 3/2 times that on the surface. Which means your clock will run slower. How much so?

>>	Anonymous Sat Feb 5 16:11:42 2011 No.2481035 >>2481025 a bit slower

>>	Anonymous Sat Feb 5 16:12:42 2011 No.2481049 >>2480932 >>2480942 >>2480948 >implying the earth is hollow. I really have no face at your collective faggotry.

>>	Anonymous Sat Feb 5 16:13:42 2011 No.2481051 >>2481025 > relativistic effects Does this mean that the answer to OP's question will differ significantly depending on whether time is measured by the participant or by an outside observer?

>>	Anonymous Sat Feb 5 16:15:42 2011 No.2481077 >>2480932 >>2480932 > implying earth is a hollow sphere

>>	Anonymous Sat Feb 5 16:18:42 2011 No.2481088 >>2481049 >>2480948 >>2480942 >>2480932 Why are you faggots still arguing? >>2480867 already solved it.

Anonymous Sat Feb 5 16:18:42 2011 No.2481095

>>2481077
>>2481049
Stop derailing the thread. No one said the Earth is hollow.

He was just trying to prove that Earth's gravitational pull varies linearly with distance to the center. When you're inside Earth, every layer of Earth above can be thought of as a hollow shell and its contribution to gravitational pull is 0.

>>	Anonymous Sat Feb 5 16:19:42 2011 No.2481103 >>2480932 Durp what? That's a shell. Not a sphere.

>>	Anonymous Sat Feb 5 16:19:42 2011 No.2481102 >>2481051 lolno. relativistic effects can be ignored.

Anonymous Sat Feb 5 16:20:42 2011 No.2481106

>>2481049

You're such a moron.

Imagine that the Earth is composed of an infinite number of concentric spherical shells. The gravitational contributions from each of these shells can be calculated separately. Each shell below you will affect you gravitationally as though all of its mass were concentrated at a single point located at the center of the Earth. Each shell "above" you -- that is, each shell which now surrounds you -- will have no net gravitational affect on you. Individual pieces of the Earth will exert a force on you regardless of their location, but the net gravitational force exerted by any spherical which surrounds you will be zero, regardless of your position within that spherical shell.

Do you get it now?

These posts are factually correct:
>>2480577
>>2480620
>>2480932
>>2480948
Deal with it.

>>	Anonymous Sat Feb 5 16:22:42 2011 No.2481120 >>2481103 > That's a shell. Not a sphere. Stop trolling. It clearly says "spherically symmetric shell." It's both.

>>	Anonymous Sat Feb 5 16:25:42 2011 No.2481140 >>2481106 >Calculating a sum when an integral can be performed. The faggotry is strong with you.

>>	Anonymous Sat Feb 5 16:40:42 2011 No.2481241 >>2481140 No one said anything about doing a summation. We have an infinite number of infinitely thin shells; this implies that an integral will be used instead. Stop trolling unless you're going to do it well.

>>	Anonymous Sat Feb 5 16:43:42 2011 No.2481264 >>2481140 >>2481241 Gauss' Law for Gravity. >>2480867 uses it in his solution.

>>	Anonymous Sat Feb 5 16:52:42 2011 No.2481322 Wouldn't one be under the influence of .5gs from all directions at the center?

Manisons general Sat Feb 5 16:56:42 2011 No.2481349

>>2481102
>relativistic effects can be ignored.

Ok, I'll drop that question and try to solve it on my own.

But the graph I posted earlier (>>2480662) would allow us to arrive at a more realistic approximation by assuming constant g for the first 3000km followed by a constant slope to zero g at the center. Shouldn't be that difficult.

>>	Anonymous Sat Feb 5 17:01:42 2011 No.2481370 >>2481322 When you're OUTSIDE the earth: <span class="math">g=- \frac{GM}{r^2}[/spoiler] When you're INSIDE the earth: <span class="math">g=-\frac{4}{3}\pi G\rho R[/spoiler] where <span class="math">\rho[/spoiler] is the density.

>>	Anonymous Sat Feb 5 17:12:42 2011 No.2481419 >>2481322 Inside a uniform hollow sphere, there is no net gravity. You only have to worry about the radius of earth that is between you and the center.

>>	Anonymous Sat Feb 5 17:19:42 2011 No.2481456 >>2481370 I love this board.

>>	Anonymous Sat Feb 5 17:20:42 2011 No.2481467 Guys, we have to account for quantum teleportation!

>>	uallay number Sat Feb 5 17:43:42 2011 No.2481603 >>2481456 Don't we all love sterilized pseudo science? Ignore objective reality, parrot your math!

>>	Anonymous Sat Feb 5 17:48:42 2011 No.2481629 84 minutes was the period for one oscillation, I think. So you'd reach the other opening in 42 minutes.

>>	Anonymous Sat Feb 5 17:57:42 2011 No.2481685 >>2481629 >Fails to realize this was already posted: >>2480867

>>	Anonymous Sat Feb 5 18:28:42 2011 No.2481811 >>2481370 http://en.wikipedia.org/wiki/Shell_theorem

>>	Anonymous Sat Feb 5 18:49:42 2011 No.2481921 >>2480592 >>2480725 >>2481077 >>2481103 >>2481140 Trolls, trolls everywhere.

Anonymous Sat Feb 5 18:54:42 2011 No.2481946

What's cool is that when you hit the other side, you'll start slowing down and eventually go the other way. Then you'll just spring back and forth, going less and less before eventually coming to a complete stop in the core.

I always find that idea amusing.

>>	Anonymous Sat Feb 5 22:42:42 2011 No.2483245 >>2481946 Actually, under perfect conditions you would continue to oscillate at the same amplitude.

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