Falling through the Earth
What would happen if you drilled a shaft through the center of the Earth all the way to the other side and dropped something into it (or jumped into it just for the hell of it)? Would the object (or you) continue to fall past the other side of the Earth, floating on forever into empty space? Thankfully, the answer is "no" (if the answer was "yes", those of you who would have been foolish enough to jump into the shaft would be screwed now, wouldn't you). We can see why this is so both qualitatively and mathematically.
Qualitatively, we need only to examine the situation from an energy viewpoint. Since we are dealing with gravity (and I'm assuming gravity only, there's no air resistance or anything else), the total mechanical energy must be conserved; the object cannot gain or lose energy. Thus, if we initially have the object at the surface of the Earth, it is some distance away from the center of the Earth (the center of gravity) and hence, has some definite gravitational potential energy (GPE). This is going to be equal to the total mechanical energy (kinetic plus potential) the object will always have once it is dropped into the shaft. So, if it is initially at the surface of the Earth, all of its mechanical energy is in the form of GPE. As it falls into the shaft, that GPE is converted into kinetic energy (KE); the object gets closer to the center of the Earth so its GPE decreases, but it is speeding up as it falls deeper, which is increasing its KE. At the instant that the object is at the center of the Earth, it has no GPE because its distance from the center of the Earth is zero. The object's KE though is at its maximum (which is equal to its initial GPE), thus conserving energy. As the object continues past the center of the Earth, it begins to slow because the pull of gravity is in the direction opposite to its movement. Thus, the object's KE is decreasing while its GPE is increasing (because it is getting farther from the center of the Earth). Eventually, it will reach the point where it has no KE (it has zero speed) and a maximum GPE. This case corresponds to when the object is at the surface of the Earth, except this time on the other side of the shaft. At this point, the whole situation looks just like what we started with, which means that the whole process of falling into and out of the Earth will repeat itself forever, barring any outside influences, of course. This means for those of you who would have jumped into the shaft, you would be able to get back onto solid ground again (assuming you would survive a trip through the center of the Earth. You know how those mole people can be ;).
Looking at the whole situation mathematically is just as easy (assuming you know how to solve a second-order ordinary linear differential equation). Using Newton's laws, the gravitational force acting on the object will produce an acceleration on it proportional to its mass:
F(r) = -GMm/r2 = ma
where F is the gravitational force, G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the object, r is the object's distance from the center of the Earth, and a is its acceleration. Now, we must actually modify what we mean by M. It really is the mass responsible for the pull on the object. But since the object will be inside the Earth as it falls, the mass responsible for the pull on the object in this case is actually the mass inside the sphere whose radius is the distance from the center of the Earth to the object. (This is a consequence of Gauss' Law.) So M itself is a function of r:
M(r) = r4pr3/3, r is the (uniform) density of the Earth
= [M / (4pR3/3)] (4pr3/3), R is the radius of the Earth
= Mr3/R3
Putting this into the force equation above, replacing a with the second time derivative of r, and canceling the common m from both sides gives:
-GMr/R3 = d2r/dt2
The solution to this differential equation is a cosine:
r(t) = R cos[(GM / R3)1/2 t]
Thus, the motion of the object is just the simple harmonic motion of a mass on a spring. The period of oscillation is:
T = 2p/(GM / R3)1/2 = 2p/(g/R)1/2
With g = 9.8 m/s and R = 6.4 million meters, T = 84.6 minutes.