Um, yeah, you, um . . . decelerate. And jeez, you 've got a lot of time to decelerate. I hadn’t realized that even under constant gravity the trip would be slowed to 44 hours. Air drag, on a humanoid form, is more powerful than I thought.
So it sounds like if you fall into a deep enough well, you live. The prolonged air friction from rapid travel near the start of the trip might be more of a problem than the bump at the end. Wind burns, and all that.
Let me get this straight.
We are going to ignore the facts of physics that make it impossible for the hole to exist at all. We are going to ignore the facts of physics that make it impossible for the hole to stay open. We are going to ignore the facts of physics that make it impossible for our path to be a straight line.
And then we are going to selectively rely on an arbitrary subset of those physical facts to ascertain the time it takes us to fall through the imaginary hole.
Hey, it’s eight thousand miles. Let’s say our average speed is a thousand miles an hour. (Why not? It’s an nice round number, and there are no real facts of physics involved in the problem, so what the heck, lets go for round numbers.) So, eight hours for the trip through, we grab the rim on the other side. If you want a different answer, pick a different set of made up numbers. It’s all fantasy, so it doesn’t matter what numbers you pick.
Tris
Sure, but the OP said
So flurb is still a smear 
If it were drilled at the equator, what tunnel path shape would be necessary to prevent a collision with the wall? This tunnel would not be as fun because you couldn’t see all the way through. Maybe if we put mirrors in it or something…
Triskadecamus, sometimes made-up and unrealistic problems can help us learn. If you do not understand how this works, it might be an edifying thread for you to start somewhere else.
I wasn’t trying to say that the thought experiment itself was pointless, only that the implication that a mathematical formula determines precise times based on the laws of physics was a bit silly.
The hole at the equator would have to be curved, with the rate of curvature determined by that same elusive factor, the speed of fall. As you fall, you move with the rotational speed of the surface which remains constant for you, and accelerate toward the center of the Earth. The wall of the hole moves at a rotational speed that decreases with the distance from center as you fall. Once you pass center, the walls are moving opposite directions from your original rotation, at a rate that increases as you move to the opposite end.
And unless something changes your original vector, you will then crash into the opposite wall as you move back down, since that hole has its shape determined by a theoretical object falling the other way. So, what you really need is a pair of intersecting curved holes, each precisely aligned for the speed of an object falling in one direction. Only one problem with that. Either something acts on you to change your basic inertial velocity with respect to the center, or on the next pair of trips through the hole will be misaligned, and you end up a red smear on the imaginary walls.
Tris
Howzabout we coat the wall of the hole with a frictionless material and have the body actually be a ball made of same material (the ball being the same weight as a person, say, 160 lbs.)?
Help any?
Is air still less dense than a human being at the pressures experienced 4000 miles down? Otherwise you’ll stop and float at some point. (And be squashed to death, of course.)
.148 metres per second means it takes you a whole second to drop the final six inches. You’d hurt yourself more walking downstairs.
What facts are those? Sure the Earth is filled with magma but that’s not a law of physics, that’s an arbitrary geological situation. Let’s posit the same question on an Earth size body whose core has cooled (and hasn’t been destroyed by an expanding sun). Or did you just that it’d be expensive?
Which facts? The magma again?
Can somebody explain this one to me? People are saying you would hit the wall (except at the poles where I guess you’d be expected to just spin around) because of angular momentum. But doesn’t someone on the Earth already have the same angular momentum? Isn’t is conserved?
And even assuming your angular momentum becomes disparate, wouldn’t the Earth’s gravitational pull from both sides as you got deeper act to keep you centered in the hole?
Well, I know, while reading about the possibility of drilling tunnels, that any tunnel drilled through the earth straight through, even at an angle, will take something like 27 minutes to pass through. It doesn’t matter if you drill a tunnel at a one degree angle to a point a couple of miles away or if you drill it at a 90 degree angle to the exact opposite side. So if you half that, then you get your time. The speed is likely to be really quick.
I’m questioning the assumption that the air will get denser the farther down you go.
Sure, that makes sense to a point. But that discounts to gravitational gradient as you approach the center of the earth. For a tunnel only a few kilomaters deep we can ignore the gradient. Gravity pulls air down into the tunnel, and the weight of the air above compresses the air below. But at the center there’s no gravity. So at the center there’s no effective weight of air above, and therefore the air at the center can’t be compressed. So I imagine the air would get denser and denser, then at some point become thinner and thinner. Although the dense middle air would expand into the thinner center, not because of gravity, but just the gas laws. Hmmm.
I’m not quite that brain-fried… YET. I just didn’t want to work out the amplitude and the frequency and solve for the time. And, hey, good thing I didn’t bother, 'cause Freddy’s right, it’s mathematically equivalent to the projection of the low-earth orbiter.
Incorrect. The air at the center exerts no weight, but the air above it still has weight and is pressing downward.
The momentum is conserved, but as you travel deeper, the parts of the earth you pass through have a lower angular momentum so you get smeared, not by the earth catching up to you, but by you catching up to earth on the other side of the hole. As far as I understand it.
So if we shot a projectile straight up (at less than escape velocity) on the equator of an atmosphere-less rotating planet(oid), we can expect it to never land on the same spot it was launched from? Or would it form some sort of loop, falling behind horizontally on the way up but catching up again on the way down?
I won’t bore you with the math but basically you’d be eaten by dinosaurs.
A projectile fired straight up at the Equator will land a bit to the west of the launch point. On the other hand, a projectile dropped straight down from above the Equator will land a bit to the east of the drop point. A projectile fired or dropped elsewhere on the Earth (except at the poles) will also land somewhat east or west, but the motion will be more complicated. In the rotating coordinate system, this is explained in terms of the Coriolis effect (the same thing which determines the rotation direction of cyclones, and which is alleged to determine the direction a toilet flushes in).
Yes, a projectile shot upwards won’t fall back to the same spot.
Before you shoot the projectile it has a certain sideways velocity due to the rotation of the body. That velocity is independent of it’s upward velocity. So, you shoot the projectile up. It’s traveling at whatever speed up, and whatever speed sideways, which is equal to the rotational speed at body’s surface. However, as the projectile goes higher, to remain directly overhead it would need to sweep the same arc that the surface does. But this is impossible, since it would require the projectile to go faster than the surface. So the projectile falls behind.
An additional few questions, assuming that there is air and therefore a terminal velocity: as you approach the center of the Earth, the amount of mass “below” you decreases, correct? And therefore, since the force of gravity exerted upon you is decreased, terminal velocity also decreases, right? And all the while, the air is acting as a brake, slowing your descent, so wouldn’t there be a point just before the center where your terminal velocity (and actual velocity, due to the slowing by the air) would be slower than tortoise speed? So, wouldn’t you just sort of gently coast to a stop at the center, rather than going past it at all?
And just how dense is the air in the center of the tunnel going to be? I still can’t shake my suspicion that there isn’t going to be a smoothly increasing air density. I know gravity has to pull the air in the tunnel down…but eventually it’s not being pulled down, so how can it be pulled down?
I suppose it’s GOT to be most dense in the center, I wouldn’t expect a ball of gas like a star to have a lower density region in the center. But still…
It’s not getting pulled, it’s getting pushed. The air near the center doesn’t know or care about any gravitational field. What it does care about is the fact that there’s a bunch more air adjacent to it (above it, in fact, though it doesn’t know that) that’s pushing on it.
Max, sciguy already answered your question. At the center, your speed would be a mere .148 m/s, equivalent to a fall from a height of a millimeter or so.
The acceleration of gravity is 32.17 ft/sec[sup]2[/sup] at the surface. Assume a homgeneous earth and an air free hole and forget all about Coriolus and that sort of stuff.
As you go down the acceleration will decrease by the ratio of the area of a circle having the earth’s radius minus the area of that part of the circle above you to the area of the whole circle reaching zero at the center. That makes the acceleration a function of the distance you have fallen.
Use that acceleration in the formulta t = √(radius of earth/acceleration)
Working that all out when the distance you have fallen equals the radius of the earth I got 2 hours 38 minutes and 53 seconds.