If it was possible to dig a hole that went from one side of the earth to the other

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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).