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...And you jumped in.

What would happen when you got to the middle of the Earth? Would you gradually slow down, until you got to the middle and once you were in middle would every direction feel like it was up?

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This Stack question is closely related: physics.stackexchange.com/q/2481 - does it answer this one? –  Roy Simpson Mar 21 '11 at 11:16
    
@Roy: I don't think so, since the answers to that other question don't address the details of the oscillation that would occur. –  David Z Mar 21 '11 at 15:59
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@David, I agree it doesnt handle oscillation, but the question is another "gravity physics at centre of the Earth" topic, so it is worth pointing that out. –  Roy Simpson Mar 21 '11 at 16:03

3 Answers 3

And you jumped in.

No, I'd refuse, You should do that Yourself. :=(

Do You want to "solve" this with or without friction by air?

Without friction, You would fall and reach maximum speed in the center of earth, going on until You reach the antipods, where You would stand still for a fraction of a second, then You would go down again. Very boring indeed and You should take some sandwiches and some drink with You.

With friction You wold accelerate up to a speed around 300 km/h, then fall down with this constant speed, slowing down when approaching the center, and go on some distance, fall down to center again and do that oscillation until all Your potential/kinetic energy is consumed by friction.

Because about a dozen of such holes were drilled here in physics.SE alone, You might find some company at the center of earth.

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smile :). It was one of the first problems I had to solve in mechanics back when. –  anna v Mar 21 '11 at 12:27
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Unless the hole goes pole-to-pole, you'll have to worry about effects of Earth's rotation as well as friction. Coriolis forces will press you against one side of the tube most of the time. As long as you're ignoring that friction as well as air resistance, the results are qualitatively the same. –  Ted Bunn Mar 21 '11 at 14:15
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One other fun exercise for those who haven't done it: Suppose that instead of going through the center of the Earth, the hole goes from one point to a non-antipodal point along a straight line (i.e., a chord of a circle rather than a diameter). Show that the time to fall from one end to the other is independent of the points chosen. (Ignore Earth's rotation and assume frictionless sliding.) –  Ted Bunn Mar 21 '11 at 14:17
    
The last sentence just made me laugh.. –  Sachin Shekhar Mar 15 at 15:23

I seem to remember Dr Karl on the Triple J radio station (Karl Kruszelnicki) explaining this one. In a zero friction environment, you would oscillate back and forth. It would take about 40 minutes to travel from one side of the earth to the other. An interesting side note was that if you were to dig the hole from any point in the Earth to any other point, it would still take the same amount of time to travel from one side to the other.

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Read the comments from Ted Bunn "below". –  Georg Mar 21 '11 at 18:58

Let the radius of the earth be $R$. By Newton's shell theorem, after travelling a distance $r$, You will feel the gravitational pull only of the part of the earth which has radius of $R-r$. Assuming the earth has a constant (same throughout time), uniform (same throughtout space) density,

$$\vec g(r)=-G\frac{4\pi\rho\left(R-r\right)^3}{\left(R-r\right)^2}\hat{e}=-4\pi G\rho\left(R-r\right)\hat{e}$$

$$\frac{\mbox{d}^2x}{\mbox{d}t^2}=-4\pi \rho G\left(R-x\right)$$

Yay! A differential equation which resembles the simple harmonic motion differential equation! Just to make it clear, make the change of variables: $$s=R-x$$ let the constants $$k=4\pi\rho G$$

So that, then, $$\frac{\mbox{d}^2s}{\mbox{d}t^2}=ks$$ $$s^{(2)}-ks^{(0)}=0$$

The characteristic equation is: $$y^2-k=0$$ $$y=\pm\sqrt k$$

So, the solution is, in general,

$$s=Ae^{\sqrt kt}+Be^{-\sqrt k t}$$

Now, apply the initial conditions, which I will leave to you to do. I suggest that you do that during the long and boring journey through the Earth.

Edit:

The below (which I had wrote before) is actually false, as pointed out by Bernhard in the comments:

P.S. Air resistance will be very low at the Earth's centre so you don't need a very long rope (around 50 km would be fine) to get out of the Earth once you are very near to the other side. Oh, but you need a little oxygen cylinder with you. Unless you are one of those sea mammals which can hold their breath for a long time (Let's hope that when you are substituting the initial conditions while falling, you don't start panicking because the ride will take even longer than you can hold your breath for, or because you forgot to bring a pencil and a paper to do the calculations). 
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I don't agree with your statement about air resistance. If a human gets dragged in, why would air not? Is this $50km$ based on facts? –  Bernhard Jun 17 '13 at 12:05
    
@Bernhard: Good point that I forgot (I only thought abut the fact that the gravity at the centre is 0.) Edited it away... –  Dimensio1n0 Jun 17 '13 at 12:16

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