A Hole in the Ground

I'd say about 6 ft....here

I guess we're going to find out....

"Seeking the elusive 'Moho'

Scientist said this week they had drilled into the lower section of Earth's crust for the first time and were poised to break through to the mantle in coming years.

The Integrated Ocean Drilling Program (IODP) seeks the elusive "Moho," a boundary formally known as the Mohorovicic discontinuity. It marks the division between Earth's brittle outer crust and the hotter, softer mantle.

The depth of the Moho varies. This latest effort, which drilled 4,644 feet (1,416 meters) below the ocean seafloor, appears to have been 1,000 feet off to the side of where it needed to be to pierce the Moho, according to one reading of seismic data used to map the crust's varying thickness."

http://www.livescience.com/6959-hole-drilled-bottom-earth-crust-breakthrough-mantle-looms.html

I read about this in Popular Science in 1963.

Haven't made much progress since then.

Thats sounds about right! Where I am it is only sand and I would guess 1 feet is maximum but this is 2 feet wide hole. Always have some fun while planting to keep the hole open long enough until the compost has filled it.

By the way the sticky compost is less likly to cave in like the sand does.

I am not trying your hole size. Did you look for the bedrock? Or digging for diamonds?

IS

Neither, buryin' treasure...shhh

The bullet would vaporize, somewhere on the trip to the center of the earth.

You'd need a Cargonite™ drill and casing from LynDoor just to get close.

The bullet would bounce back and forth through the earth forever, assuming no outside forces but gravity were in play.

The force due to gravity will vary with depth.....

For deep holes check out the Kola hole in Russia. You are right the rock acted like plastic at about 180 C and simply closed once the drilling stopped and rods were retrieved to change bits. It was some 40,000 feet or 12200 meters.

Here is another old problem along the same lines:

Hole Through the Earth

Assume that a hole is drilled from one point on the globe
through the center of the Earth to the Antipodes, as illustrated,
and a steel ball is dropped into the hole at point A.
Ignoring any external influences such as air resistance,
friction and conditions of the Earth's core, answer the following
questions:
1. As the ball is traveling from A to the Earth's center, does
its velocity increase, decrease, or stay the same?

2. Will the ball weigh less or more when it reaches the center

of the Earth?

3. Will the ball's mass change during its journey?

4. At what point will the ball be subject to zero gravity?
5. If the ball fell through such a cylindrical hole through the

center of the moon, would the one-way journey take more

or less time than it would on Earth?

The tunnel or hole A to B need not pass through the center of the earth and can be any hole tangent to the center. Assume no friction and no air resistance. How long will it take the ball to fall? No matter what the length of hole the time will be the same. This is an old problem best solved with simple calculus.

Cool....

http://en.wikipedia.org/wiki/Kola_Superdeep_Borehole

1/4. The ball accelerates all the way to the centre, but the acceleration rate falls as it drops further. When it is exactly at the centre, it stops accelerating. As it ascends from the centre, it decelerates slowly and then decelerates faster as it gets nearer the surface.

2/3. Its mass remains the same at all times. However, its weight decreases as it approaches the centre. At the centre its weight is zero. Its weight then increases as it ascends.

5. Mass goes up as the cube of the radius of the moon/earth. The attractive force between the moon/earth and an object on its surface goes down as the square of the radius. So g increases linearly with the radius (assuming density is same for earth and moon and constant).

Under constant g, v*v = 2*g*s. So over a given distance, the velocity increases as the square root of g. The ball will go thru the moon faster. At least that is my gut feeling, I can't work out the remaining logic, requires an integral, and it's time for tea.

My guess...

1. Increase initially, then decrease.

2. That's relative.

3. Yes. It will become less dense.

4. In reality, never.

5. Less. Even considering the smaller diameter.

The answer can get quite complicated but we have taken out all the factors of resistance including the difference in temperature, spin, and gravity as you pass through the earth. Gravity may be 9.8 m/sec squared at the surface but is 10.3 m/sec squared at the core (a relationship with density). This we ignore to simplify the answer and not bruise myself.

1. The ball's velocity would steadily increase From zero at point A
to a maximum at the center of the Earth. It would steadily
decrease thereafter to zero at point B, taking 42 minutes for the
complete trip. This fascinating speculation goes back to Plutarch.
Even Francis Bacon and Voltaire argued about it. Galileo
gave the authoritative answer which is generally accepted:
The ball would fall faster and faster, though with decreasing
rate of acceleration, until it reached maximum velocity, about 5
miles per second, at the Earth's center. It would then decelerate
until its speed reached zero at the far end of the hole. If air
resistance is ignored, it would oscillate back and forth, like a pendulum,
ad infinitum.
2. At the center of the Earth, there is no overall gravitational force
on the ball. This is because the Earth's pull on the ball will be
equal in all directions. Therefore, the weight will be zero.
3. The weight will change but the mass will not.
4. The ball would be in free-fall throughout the entire trip, so it
would always be in a state of zero gravity.
5. More time. The trip would take about 53 minutes. Although the
distance is much shorter than on Earth, the moon's gravity is

only about one-sixth of that of the Earth.

Question and answer from Super Science Problems.

The time to travel through any tunnel tangent to the center would remain 42 minutes on the earth. So we could travel from London to New York in 42 minutes or any other two points on the earth meeting the criteria. A little longer on the moon.

A complete answer can be found here with the formulae.

I meant to say more time on 5 considering only about 1/3 the diameter but with only 1/6 the gravity. And as far as my answer on 2, I answered it's relative because depending on where you placed the scale you would probably record some weight. In other words you would never really be able to get the ball to an exact center. And in 3, I still believe the mass would be affected because of the ball being pulled apart in all directions. I concede 4, but I think what I was getting at was that the ball would probably end up being attracted to one side or the other before it reached the center.

Good effort. And yeah reality can suck. In real life, the ball would touch the wall and spiral down the first leg and up the second. Other factors of resistance would enter into the problem including variances in gravity, wind, and friction. I was trying to keep it simple because I admit that I could not do the calculations without going to a lot of effort and even then I would likely get it wrong. It is just a fun problem with a remarkable truth if we could ever build such a tunnel. Maybe we could in short lengths of several hundred kilometers to avoid sever problems with plasticity of the bedrock at depth. So we could travel from New York to Philadelphia in 42 minutes or something a little longer time wise, and then to Pittsburgh in 42 minutes and so on. Kind of like a cell call passed from tower to tower. The cost to travel in terms of energy would be cheap but the capital cost would be enormous and may take centuries to build. Crossing an Ocean with a tunnel would present serious barriers. These ideas will be Scifi for as far in the future as I can dream. Many fun ideas out there to dream about. Keep smiling.

Self-gravitational strain would tend to close the hole. As the material squeezes into the center, there is horizontal force due to friction as well as the radial force of gravity. Deep in the lithosphere, there is a zero displacement depth where the ratio k of horizontal stress to radial stress remains constant. Above this depth, k increases nonlinearly. Because of the free outer surface of the self-gravitating sphere, the material is able to expand in portions near the surface as compared with the material somewhat deeper. This might paradoxically raise material, opposing the force of gravity and leading to uplift. A deep enough hole will create a local free surface where the expansion due to self-gravitational strain would start to fill it up.

I'm not sure what you mean by "horizontal force due to friction". Are you simply talking about pressure from material above?

Horizontal stress acts perpendicular to the shaft. Radial stress can be toward the core, due to gravity in the usual way, but it can also be in the opposite direction, due to self-gravitational strain uplifting material. Squeezing material toward the surface would block the hole from the bottom.

Imagine the Earth as a multitude of wedges converging at the core. Friction between the wedges as they slide along each other, squeezed together by gravity, causes horizontal stress simultaneously as gravity causes radial stress. The pressure from material above relates, but indirectly.

But what I'm thinking is that as you get closer to the core you'll start having conditions where there is more mass above you to the point that their would be less attraction to the core itself causing less horizontal forces.

"If you drill a hole through a planet, and shoot a bullet through the hole, it will take less time than one fired on a parallel course of the same length,through empty space."

Ignoring drag, Coriolis, etc, the bullet through the hole will beat an Earth-skimming satellite in circular orbit by quite some margin. I have not worked out by how much, but considering that a ball dropped from rest will fall through in the same time as half an Earth-skimming orbit, the bullet with its starting velocity will easily win.

"It would appear this way to an outside observer, but with measuring rods changing length and clocks changing speed, how would the bullet see it."

The speeds are not very relativistic, but in principle the bullet will see a slightly 'flattened Earth' (Lorentz contraction of the diameter of Earth in direction of hole's length). But that's not the reason for beating the satellite.

-J

I wondered how far down one could drill without having gravity close it up, I don't mean dropping debris down the shaft, but a squeeze.

It seems to me that gravity would tend to expand the hole rather than squeeze it shut, especially near the center of the earth. Anybody else agree?

Me thinks it's the pressure in the rock as you go deeper that will close it. Granted, gravity causes the pressure, but that's cumulative from the top. Highest pressure at the center.

I was thinking of the gravity perpendicular to the shaft. You are talking of the gravity parallel to the shaft. Yours will likely win out if there are any angular cracks in the rocks that will cause a vector force to the shaft. The liquid core will surely close.

Are you thinking of the tidal force that tries to "squeeze" objects in a direction perpendicular to the gravitational force vector? If so, it's negligible for earth on the scale of a hole diameter, I think. Further, the tiny tidal force on the hole diameter will be maximal at the center.

Pressure works in all directions, though...

I think what Standards Guy as well as myself are thinking, is that by the time you approach the center you'll have enough mass on all sides to negate the initial pull of gravity. You might say reach a point of zero gravity if it weren't for the forces now pulling in all directions.

"I think what Standards Guy as well as myself are thinking, is that by the time you approach the center you'll have enough mass on all sides to negate the initial pull of gravity."

OK, but what I'm thinking is that the hole walls are not subjected to any substantial perpendicular gravitational forces anywhere; not near the surface and also not near the center.

The gravitational force is very nearly vertical, parallel to the hole wall faces. It creates increasing pressure with depth, with resulting omnidirectional pressure forces, which are maximal at the center. Isn't this what will make the hole cave in at some depth?

-J

I'd say yes near the surface.

To clarify, if you were in dead center, the radius of the earth perpendicular to the shaft would pull the closest wall out, and the same on the other side, but it would also pull the far wall inward with almost as much force, so they almost cancel but not quite.

I think your last paragraph is correct for a fluid, so the hole will close in the center.

I will agree that there is a point of zero gravity at the center or at the center of gravity. All gravity will be negated for only that one point. The force of gravity will be positive if a ball were dropped, zero at the center, and negative on the trip back up to the surface.

That's how I see it.

The Kola drill hole in Russia (see Solar Eagle's highlighted site) had encountered temperatures of 180 degrees C. At that temperature the rock was actually flowing or plastic. Their target depth would have seen temperatures of 300 degrees C and at that temperature even their drill bits would start to melt. The hole will close and the large pressures would increase the temperature directly with the depth drilled. The plasticity of the rock will increase and gravity will just help the hole to become more unstable. Drilling into the mantle is very difficult and may not be possible without some form of continuous cooling and even then drilling much deeper is doubtful.

My father worked underground and the mine (Kerr Addison in N. Ontario) was over one mile deep. He always told us it was very hot and well over 100 degrees F. I can only imagine a mine at 20,000 feet. That would be where robots would be needed but only if they wouldn't melt.

That's what I'd say. At least past half way to the center.