What would happen if...

[deleteduser]

Say you drilled a hallow hole through the center of the earth and out the other side. Now say you droped a marble down the hole, what would happen to the marble when it reached the center? What would happen to the particles if it were crushed? Just a thought that crossed my mind...

[Andrew Karpinski]

It would be melted by lava probably.

[deleteduser]

It is a whole which is reinforced, there is nothing in this whole and it goes from one side of the earth to the other no lava...Think of it as a tunnel through the earth that will not buckle under gravity or be melted by lava....

[surftheiop]

All i do know on the subject is at the very center of the earth there will be very little effect of gravity because the mass of the earth will be in equilibrium with itself in every direction.

[Matt R.]

Given your conditions, the marble would oscillate back and forth from one side of the hole to the other. It would reach it's maximum speed at the center of the earth and then slow back down until it reached the other side of the earth -- precisely opposite to where it was first dropped. Think of it like the motion of a stretched spring when released, or the horizontal component of a pendulum's motion.

[Lestat]

Here's a related question I've always had:

In which direction does water go down a drain directly on the equator?

[yukoncpa]

Here's the answer from the Straightdope. (http://www.straightdope.com/classics/a1_165.html)

[hmkpoker]

I assume that the marble would be more or less free floating. Think about it, each point of the earth is pulling the marble towards it, but because an equal amount of earth exists in the opposite direction as the initial force, all gravitational forces cancel each other out, leaving the marble weightless.

However, the further the body drifted from the center point, the more gravity would exist in the direction its moving against, so let's say you fell in and were trying to climb out, you would begin to do so with incredible ease (weightless), but as you got further from the centerpoint, you'd gradually begin to weigh more.

[Sharkey]

If you fell into such a hole, you would be weightless for the whole trip.

[theweatherman]

How much air pressure would you experience at such a depth? Would it be enough to crush you?

[Brom]

[ QUOTE ]
I assume that the marble would be more or less free floating. Think about it, each point of the earth is pulling the marble towards it, but because an equal amount of earth exists in the opposite direction as the initial force, all gravitational forces cancel each other out, leaving the marble weightless.

[/ QUOTE ]

The earth isn't exactly spherical though, more like a symmetric egg. So would the marble actually end up floating in the middle of nothing, or would it be drawn to one side of the tunnel and end up being stuck to a wall?

[Sharkey]

[ QUOTE ]
How much air pressure would you experience at such a depth?

[/ QUOTE ]

Barometric pressure would be zero.

[Borodog]

[ QUOTE ]
Say you drilled a hallow hole through the center of the earth and out the other side. Now say you droped a marble down the hole, what would happen to the marble when it reached the center?

[/ QUOTE ]

The behavior of the marble on the way down would depend on a number of things. Eventually, the marble would end up floating weightless at the center of the Earth, after losing all of its gravitational potential energy.

The detailed behavior would depend on how the tunnel was bored. If it was bored along the rotational axis the behavior is very different than if it were drilled from one side of the equator across to the other side, for example.

It would also depend on the atmosphere in the tunnel. If you sealed it and pumped all the air out to create a vacuum, the behavior is different than if the tunnel is filled with air.

You can actually show some fairly nifty things about these different cases if you do the math. The axial vacuum case has a particularly elegant and familiar result. The equatorial vacuum case, if you assume that the collisions the marble will have with the walls are perfectly elastic, has a surprising result. Any case with air is relatively boring.

[hmkpoker]

[ QUOTE ]
The behavior of the marble on the way down would depend on a number of things. Eventually, the marble would end up floating weightless at the center of the Earth, after losing all of its gravitational potential energy.

The detailed behavior would depend on how the tunnel was bored. If it was bored along the rotational axis the behavior is blah blah blah...



http://img119.imageshack.us/img119/6878/boobs4ag1.jpg



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Sorry, that's what I read :P

[theweatherman]

[ QUOTE ]
[ QUOTE ]
How much air pressure would you experience at such a depth?

[/ QUOTE ]

Barometric pressure would be zero.

[/ QUOTE ]

Why? Unless this is a vaccum tunnel wouldnt there bean extra 3k miles (or whatever the radius of the earth is) of air on top of you?

[Hopey]

[ QUOTE ]
Here's a related question I've always had:

In which direction does water go down a drain directly on the equator?

[/ QUOTE ]

It goes down the in the same direction it does in either hemisphere. It's a myth that water drains differently in each hemisphere.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
How much air pressure would you experience at such a depth?

[/ QUOTE ]

Barometric pressure would be zero.

[/ QUOTE ]

Why? Unless this is a vaccum tunnel wouldnt there bean extra 3k miles (or whatever the radius of the earth is) of air on top of you?

[/ QUOTE ]

Air pressure would decrease as you descended.

The local gravitational escape speed at the center of the Earth would be zero. Any gas molecules there would fly “up” the shaft in either direction.

[Borodog]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
How much air pressure would you experience at such a depth?

[/ QUOTE ]

Barometric pressure would be zero.

[/ QUOTE ]

Why? Unless this is a vaccum tunnel wouldnt there bean extra 3k miles (or whatever the radius of the earth is) of air on top of you?

[/ QUOTE ]

I'm glad you responded to this, or else I wouldn't have seen it. The pressure at the center is most emphatically NOT zero. In fact it will be tremendously high. The actual calculation is complex, and would require a computational solution (if there is one; see below).

The condition for hydrostatic equilibrium in the tunnel is that the net force on any gas element is zero. There are 3 forces acting on any gas element in the tunnel: gravity pulling toward the center, the pressure of the gas element above our element pushing down, and the pressure of the gas element below pushing up. So the condition for hydrostatic equilibrium for a gas element at radius r is:

GM(r)rho(r)/r^2 + dP/dr = 0,

where G is the gravitational constant, M(r) is the mass of the earth contained within a sphere of radius r, rho(r) is the density of a gas element at radius r, and dP/dr is the radial gradient of the pressure.

M(r) is itself a complex function, since the density of the Earth is not uniform for all radii.

Furthermore, you need an equation of state for the gas that tells us how the pressure, density, and internal energy are related.

You'd also need some boundary conditions and have to say something about the temperature of the gas. Is the tunnel perfectly insulated? Or is the gas in local thermal equilibrium at all points with the surrounding Earth?

In fact, I doubt there is any equilibrium solution that can be calculated except under nonphysical assumptions (not that the whole problem doesn't count as nonphysical!). For example if you assume the gas in the tunnel is isothermal with a sound speed equal to the speed of sound at the Earth's surface and an outer boundary condition that the gas pressure is 1 atmosphere (which we require so that air will not flow into or out of the tunnel), you get that the pressure at the center is an astounding 1.2x10^125 atmospheres! Clearly unphysical.

[Sharkey]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
How much air pressure would you experience at such a depth?

[/ QUOTE ]

Barometric pressure would be zero.

[/ QUOTE ]

Why? Unless this is a vaccum tunnel wouldnt there bean extra 3k miles (or whatever the radius of the earth is) of air on top of you?

[/ QUOTE ]

Actually, you have asked a very interesting question. So interesting, in fact, it could be suitable for another “airplane on a conveyor belt” series. (Not.)

On thinking about the air pressure problem further, I realized I had neglected the total weight of the columns of air. Then I realized that air pressure in an enclosed volume is not only a function of the kinetic energy of the molecules but also of the walls of the container elastically reflecting those molecules. It further occurred to me that the internal pressure sustained by a conventional container cannot exceed the maximum pressure the walls can exert inward. In the case of the shaft, the equivalent maximum pressure would the 1 atmosphere present at the surface. So, perhaps the air pressure at the center would be 1 atmosphere. It could be something to ponder, in any case.

The heat of the magma, rotation of the earth, etc are ignored.