Ok, for many years I have thought about this question: If there was a hole straight through the center of the earth (nevermind the lava and heat) and you threw a rock straight down, How would it stop, or would it? Would it pass the center of the earth and fall into china then come back down through the hole? Or would it Fall to the center of the earth and stop at the center and be ripped apart by the gravitational pull? And if it fell to china, how far into the sky would it get until it came back down? Ive always wondered these things…

It would fall to the center and stay there.

The Earth is constantly spinning and its very center technically has no gravitational pull. However, keep in mind that since it would be deep inside the Earth and the rock is very fragile, it might crumble under the extreme pressure.

This may help–Cecil’s column on the subject: What if you fell into a tube through the earth?. There have been other threads that have discussed it as well.

Unless you’re living the middle of the South Pacific, it wouldn’t come up in China anyways. If you are living in the US, it would come up in the Indian Ocean.

Actually it would oscillate back and forth by the middle, slowing down eventually to a stop. (Actually, unless drilled your hole at the North (or South) Pole, the rotation of the earth would drive the rock into the side of the hole. Then again, actually, the question is so “theoretical”, in the sense of “not ever going to happen in a bazillion years”, that I can’t believe I answered it.)

There is zero gravitational pul at the center of the earth so, no, it will nto be ripped apart.

Um, there is half a G from every direction at the center of the Earth. I don’t know where you got that idea that there is 0 G, but think about it: You’re at the center of the Earth. There is half the matter in every direction than when you’re standing ON the Earth. There is a gravitational pull, but it’s from every side at once, at equal force. Hence the fact that you could theoretically float in the middle of the Earth, because half the gravitational pull would tug on you from all sides at once. One thing for sure, it wouldn’t feel like 0 G, it would feel pretty weird, maybe like being inflated…

No. The forces cancel out. There is no gravity at the very center of the Earth (well, there’s gravity from the sun and moon, and stuff, but nothing significant). The gravity does not act only on the side of you that’s closest, but on every point of your body. And at each point, the forces cancel each other out.

oops!

yeah, right… (no sarcasm intended)

Well, there IS a gravitational effect involved there, but nothing perceptible, I guess…

I remember this problem from differential equations. If you drilled a hole through the Earth and evacuated it and the sides were made of frictionless material (i.e. remove all possible friction), it would take about 80 minutes to reach the other side. The cool thing was that if you drilled the hole at an angle (say from Asheville, North Carolina to Hong Kong) it would take the exact same amount of time reach the other side. Or say you had a table with all point on the table top perfectly coplanar. Of course this table is also made of a frictionless material and in a vacuum. Because the table top is a plane, if positioned correctly the, the center of the table will be closer to the center of the Earth than the table’s edges. If you place an object on the edge of this table, 80 minutes later the object will have crossed the table.

If you think of gravity as a force that comes from the presence of matter a la Isaac Newton, it makes sense that at the center of the earth the forces pull in all directions and therfore cancel out. You would not experience the forces.

But in Einstein’s theory, gravity is not a force that comes from matter; it is the warping of space time in the presence of matter. And it seems to me that (I’m not sure of this) that the warp increases as you go deeper into the earth. Or does it stay the same? In either case, the gravity at the center would not be zero. It would either be greater than or equal to the gravity at the surface.

Can anyone help me out here. I’m confused.

The point I made in the above post is backed up by what happens when we fall into a black hole. I have always heard that there is a point called the Schwartzschild radius that lies a certain distance from the center of the black hole. Once we cross this line, we are unable to turn back and escape from the black hole. This is because the gravity gets ever stronger, i.e. space-time becomes more and more warped, the closer we come to the center of the black hole. At the ceneter of the black hole, space-time is infinitely warped (something which may be a problem for relativity theory).

Likewise, it seems to me that same should be true for the center of the earth. Gravity hould increase as we approach the center of the earth.

The calculations of the time it would take to fall through such a hole usually assume that the density of the earth is a constant–so the force of gravity would continually decrease as you neared the center of the earth. But because of the earth’s dense core, the force of gravity stays pretty much about the same all the way down to the core-mantle boundary, even increasing a little. Then, it decreases to zero as you near the center of the earth.

Gravity is everywhere (and may be equivolent to spacetime itself). When the forces cancel out, there is no net acceleration. You may **feel** weightless (i.e., “no gravity”) but, technically, it’s still there.

More curvature of space where there is more mass.

There is no such thing as zero gravity in a technical sense, but for human perception, we can feel weightless and call it zero gravity.

In a black hole, the gravity keeps on increasing as you approach the center because all of the mass of the hole is concentrated at a single point (or at least, it acts like it is). In that sense, you never pass the “surface” of the object, or at least, not until after you pass the event horizon, after which all bets are off. All of the mass is always “below” you.

With the Earth, by contrast, the gravitational field increases until you reach the surface (in exactly the same way it would for a Earth-mass black hole, BTW), but once you reach the surface and start falling into that hole, the mass “below” you is no longer constant, since some of it is now above you. Assuming that everything’s spherically symmetric, you can completely ignore the mass of all layers “above” you (at a larger radius), and get the right answer for the net force, in Newtonian or Einsteinian gravity.

Said rock would melt, duh.

**Daniel Shabasson:** If you’re inside something, you feel no net force of gravitation (or charge). It’s the shell thereom. In the center of the Earth, there are forces acting on you, but there is no *net* force since all forces cancel out. You are in equilibrium (stable, I think).

So if the radius of the earth was R, and you were at (1/4)R, the shell of the Earth greater than (1/4)R would have no net gravitational effects on you. All of the forces cancel out. Gravitational forces increase the closer you get to the source (by the inverse of the square of the distance), but keep in mind that once you’ve started going in your tunnel through the Earth, you’ve *passed* part of the mass of the Earth that was attracting you.

All bets are off for a black hole, really. I doubt your molecules would even be intact by the time “you” got that close to a black hole.

**Chronos:** Everything can be considered a point-mass, given that you’re not inside said mass. And the event-horizon, or Schwartzschild radius is a distance outisde of the black hole, not on it or in it. As someone said below, it’s pretty much the point of no return.

Actually, the event horizon is usually considered to be the “surface” of a black hole. We don’t know if it is or not; all we know is that the mass is contained entirely witin it, and that it’s distributed spherically symmetrically. Anything *spherically symmetric* can be considered a point massas long as you’re outside it, and you’re definitely outside a black hole until you cross the event horizon.

Yes, for a stellar-mass black hole, the tidal forces would kill you long before you reached the event horizon, but for, say, the black hole at the center of the Galaxy (much larger, and thus much lower tidal forces), you probably wouldn’t even notice the tidal forces at the horizon.

You’re dead-on about the shell theorem; that’s what I was trying to get at in my last post.

I feel a strange urge to add to this discussion a few things like black holes. All singularities are black holes, all black holes are not singularities. A star collapsing into a black hole becomes a black hole once the star is so dense that the escape velocity at its surface exceeds c. There is a non-zero amount of time after this happens when said star/black hole has a non-zero radius. At this point of becoming a black hole the event horizon coincides with the surface of the star. As the star continues to become more and more dense its radius decreases while the radius of the event horizon pretty much stays where it is. Thus there is a difference between the event horizon and the actual massive part of the black hole. Arguing which is the surface of the black hole is largely meaningless.