This is a very interesting question, with an even more fascinating answer, as I think you'll agree. Fortunately, I don't have to answer with a theory, but rather with a scientific fact, assuming certain idealizations.
If the hole were friction free, and the Earth non-rotating, you'd continuously accelerate downwards, passing through its centre with your highest velocity, then decelerate until you reached your own personal antipodal point (the point exactly on the other side of the centre). Unless you were then able to step out of the hole, you'd start back again, doing exactly what you did before. In other words, you'd oscillate back and forth like that for eternity.
The time each trip to the other side of the Earth would take depends upon the details of how the Earth's density (assumed spherically symmetric) is radially distributed. A precise answer can be given if one assumes the Earth to be a UNIFORM SPHERE, with its density everywhere equal to its mean density of ~ 5.5 grams per cubic centimetre. Each one-way trip would then consist of one "one-way swing" of perfect simple harmonic motion (SHM), and take ~ 42 1/4 minutes. The complete SHM, round-way trip would take ~ 84 1/2 minutes.
It is NO COINCIDENCE that this latter time is the circular orbital period of the LOWEST POSSIBLE Earth-orbiting satellite. (For years my better students have been able to show why this is so --- even under examination conditions.)
What is even more intriguing but much less well known is this:
THE HOLE DOESN'T HAVE TO BE DRILLED THROUGH THE CENTRE !!
That's right --- idealized friction-free travel could be performed to ANYWHERE on Earth's surface, just under gravity, in a time of order 3/4 of an hour, with round-trip times of about an hour and a half. (My very best students have been able to prove this, also.) There's only be about enough time to eat your lunch during this ultimate "free lunch" --- or should we say "free launch?" !!
Live long and prosper.
P.S. By the way, more scientific facts:
(1.) As Isaac Newton himself showed, there is NO gravitational force whatsoever at the very centre of ANY spherically symmetrical distribution of matter. So it's completely wrong to assert that "gravity" (per se) would "crush you."
People often get confused about the separate consequences of gravity and pressure. The PRESSURE in the surrounding material there (arising from the total weight of the overlying material) would certainly CRUSH YOU, but ONLY if you were encased in that material of course. But since you're falling through a HOLE --- without that direct embedding --- that's of no concern to you. This leaves aside the great engineering challenge of building and maintaining the walls of the hole under that same pressure!
(2.) If the Earth were of UNIFORM DENSITY, the gravitational acceleration would indeed reduce linearly with your distance from the centre. It's that LINEAR dependence that is responsible for the motion being SIMPLE HARMONIC MOTION, in that case, and in that kind of case only.
2007-01-20 08:57:08
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answer #1
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answered by Dr Spock 6
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When you jump down the hole, you will go almost all the way to the other side of the earth, but then you will fall back toward the center and keep going like that until eventually you are suspended at the very center of the earth. This is because the center of the earth is where it's gravity is the strongest, so once you become suspended in the middle it will be nearly impossible to get back out.
2007-01-20 08:41:36
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answer #2
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answered by sportsfreak 2
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The question may be responded if we make here assumptions: (a million) The earth isn't spinning. (2) The earth is of uniform density throughout. (3) the hollow varieties a diameter of the earth (i.e., it passes contained in the path of the earth's midsection). the hollow must be contained in the path of the midsection; otherwise, the gravity from the earth might pull you into the element of the hollow ultimately; your action does no longer exist alongside a promptly line for the comparable reason that a bullet shot promptly forward falls in the direction of the exterior. (4) there is not any air resistance. The gravitational rigidity vector will constantly component in the direction of the midsection of the earth alongside the path of the hollow no rely what path you're shifting in and its value would be proportional to your distance from the earth's midsection. In symbols, the rigidity would be of the type ok(c - v), the place v is your place vector and c is the region vector of the earth's midsection. this might reason you to execute harmonic action--you will oscillate lower back and forth. you will possibly attain the different end of the hollow at which component you will quickly end and then commence falling contained in the different path. you will possibly proceed to pass lower back and forth between the ends of the hollow retaining a relentless finished mechanical capability.
2016-10-07 11:23:18
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answer #3
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answered by ? 4
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Well, assuming the walls of the hole through molten magma can be kept back, you would be totally cooked. If you talk about a ball of titanium that would survive the heat, it would oscillate back and forth, losing "height". Depending on the path of the hole (NS, angled, equatorial) lots of friction from rotation of earth.
2007-01-20 08:42:42
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answer #4
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answered by Mike1942f 7
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You see the Earth has gravity, the core of the gravity is, well, in the core. So you would be squished together from all sides. And die. If you find a way around it you would probably have to force yourself to the top. But you would probably get a really big hesdache
2007-01-20 09:26:38
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answer #5
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answered by Anonymous
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my guess is that you would probably accidently sever some
channel causing the sea to be drained
depends how big the hole
if it is not that wide you might get away with it
as a whale or
a shark would probably block it.
it would be a close one though
hope that helps
2007-01-20 08:58:37
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answer #6
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answered by nick m 2
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You would probably wind up stuck in the middle..
2007-01-20 08:42:43
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answer #7
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answered by Anonymous
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