Nagging Question

[Ryback]

This seemingly simple question has been on my mind for while and I think I know the answer. But I would like to know if I am right.

Quite simply, suppose you drill a shaft through the earth. You start from one side and drill straight down to and emerge from the other side. What will be the air pressure at the center of the earth? Will it be a vacuum since the net force of gravity is zero?

Thank you!
Ryback

[George]

Quite simply, suppose you drill a shaft through the earth. You start from one side and drill straight down to and emerge from the other side. What will be the air pressure at the center of the earth? Will it be a vacuum since the net force of gravity is zero?

The gravity acts on all the air in the column. This pulls the air toward the middle creating an increasing pressure at the middle. The total pressure will be the summation of the changing amount of the gravitational force at each distance from center.

Consider the opposite. If you have a heavy weight in each hand and start spinning, the net force might seem to be zero, but it is not, as your body clearly will indicate to you. :wink:

[Jens]

It's a good question, and I'm looking forward to hearing an answer too. I know the answer beforehand -- no, it won't be zero, in fact it will be enormous -- but unfortunately, I can't figure out why that is. Actually, if you imagine gravity as a deformation of space, like a well, it appears obvious that it would be enormous. But I'm not sure why that should be true.

[George]

It's a good question, and I'm looking forward to hearing an answer too. I know the answer beforehand -- no, it won't be zero, in fact it will be enormous -- but unfortunately, I can't figure out why that is. Actually, if you imagine gravity as a deformation of space, like a well, it appears obvious that it would be enormous. But I'm not sure why that should be true.

You can start with 14.7 psi at the top.

[Saluki]

You would have great difficulty actually doing this on many levels, but assuming you had drill bits/shafts and support structures made of unobtaium, the pressure should continue to increase the deeper you go. You are confused abiut the statics analysis of a point at the center of a gravity well (the resultant force is zero). This does not mean that no forces act on the point. Forces act on the point from all directions. The fact that their resultant is zero means that they will not cause the point to accelerate in any direction.

[George]

FWIW, Ryback, you will only have to drill halfway through the Earth.

I trust you know this drillilng idea is impossible with anything remotely related with current technology, or any reasonably imaginary.

However, if you must make the attemp, be sure you drill from the top of the World downward rather than from the bottom upward. [Yes, I'm joking ]

The net pressure is complicated further due to temperature issues. The hotter the air, the less the density, the less the force, the less the incremental pressure at that point, the lower the final pressure in the middle. Then there is the issue of changes in molecular concentrations which effects the density, which effects the force, etc.

[Ryback]

Quite simply, suppose you drill a shaft through the earth. You start from one side and drill straight down to and emerge from the other side. What will be the air pressure at the center of the earth? Will it be a vacuum since the net force of gravity is zero?

The gravity acts on all the air in the column. This pulls the air toward the middle creating an increasing pressure at the middle. The total pressure will be the summation of the changing amount of the gravitational force at each distance from center.

Consider the opposite. If you have a heavy weight in each hand and start spinning, the net force might seem to be zero, but it is not, as your body clearly will indicate to you. :wink:

Hey you guys are great, it is much clearer to me now. Thanks for you help!

Anyone have an rough estimation of the pressure at the center?

[George]

Anyone have an rough estimation of the pressure at the center?

The accuracy would depend on how many variables you want to consider.

The incremental change in pressue, dP, would be equal to the density, D, times the gravitational acceleration, g, times the incremental change in elevation, dz. This site walks through it.

so, dP = Dg dz

then there is the gas law of PV = RT, which tells you the change in pressure for a change in temperature if your volume is constant.

Density if simply Mass/Volume. (D = M/V). M is molecular weight of air.

It gets ugly from here and over my head. Changes in temperature, molecular weight, density, gravitational acceleration value and elevation all contribute to your final value.

If you fix the molecular weight and the value of g (compromising real accuracy), then...

dP = (M/V) g dz and V = RT/P, so....dP/P = (Mg/RT)dz

integrating.... lnP=(Mg/RT) z

so, maybe if my math is not messed-up....

P = e^[(Mg/RT)z].......Maybe. #-o

Your best bet is to google for someone who has done this already. :-?

[Ilya]

The net pressure is complicated further due to temperature issues. The hotter the air, the less the density, the less the force, the less the incremental pressure at that point, the lower the final pressure in the middle. Then there is the issue of changes in molecular concentrations which effects the density, which effects the force, etc.

Even assuming a constant temperature, the deeper you go, the smaller increase in pressure occures with every unit of distance. Near the surface, descending 8 km increases air pressure by 1 atm. Halfway to Earth's center, effective force of gravity is determined by the new radius (1/2), and the mass of the Earth still below you (1/8 volume, but rather more than 1/8 mass, because Earth is much denser near the center). So effective force of gravity at halfway level is somewhere between 1/2 G and 1 G, therefore it takes more than 8 km descent to add 1 atm of air pressure...

Overall, it is a VERY hairy problem

[George]

The net pressure is complicated further due to temperature issues. The hotter the air, the less the density, the less the force, the less the incremental pressure at that point, the lower the final pressure in the middle. Then there is the issue of changes in molecular concentrations which effects the density, which effects the force, etc.

Even assuming a constant temperature, the deeper you go, the smaller increase in pressure occures with every unit of distance. Near the surface, descending 8 km increases air pressure by 1 atm. Halfway to Earth's center, effective force of gravity is determined by the new radius (1/2), and the mass of the Earth still below you (1/8 volume, but rather more than 1/8 mass, because Earth is much denser near the center). So effective force of gravity at halfway level is somewhere between 1/2 G and 1 G, therefore it takes more than 8 km descent to add 1 atm of air pressure...

Overall, it is a VERY hairy problem

Yes. That's the way I see it but, admittedly, my calculus is horribly rusty.

FWIW, I was giving examples of the problems in the quote.

I attempted to qualify the variables in the next post and mentioned g is a variable (not a constant).

Where I might be overally concerned is in the molecular weight. I would guess the current value might not be a bad approximation for this scenario since it is existing air which we will assume would fill the pipe. This assumes we not allow time to let very heavy molecules, in excess of the norm, get down there.

[hhEb09'1]

So effective force of gravity at halfway level is somewhere between 1/2 G and 1 G, therefore it takes more than 8 km descent to add 1 atm of air pressure...

Overall, it is a VERY hairy problem

Interestingly enough, it's just a little bit more than 1 G, at the core-mantle boundary, which is about halfway. It's very nearly constant, down to there.

[George]

Interestingly enough, it's just a little bit more than 1 G, at the core-mantle boundary, which is about halfway. It's very nearly constant, down to there.

That is interesting and it does make sense. If we slice the earth in half (why not, we just drilled to the center :wink: ) and consider the c.g. of each, as we travel toward the center from the surface, we get closer to both c.g.'s which gives us a strong g value.

With this analogy, wouldn't g actually increase in value for a little more than half the way down before decreasing in value? [edit: ok, that's what you were saying ops:]

I'm afraid to ask, but....is that calculus of merit? 8-[

[hhEb09'1]

[edit: ok, that's what you were saying ops:]

well, not quite what I said--it pretty much does not increase (nor does it decrease), it stays about the same.

That's against the usual rule--if earth had a constant, or more uniform, density, the force of gravity would decrease regularly all the way down. In general, as you tunnel into a solid sphere, the effect of the material above you (that is, everything above your radius--which includes material on the other side of the earth from you) disappears. That's a well-known result from calculus--there is no force of gravity inside a hollow shell. So, the force of gravity depends only on the material beneath your feet, which gets closer (the squared term) but is smaller (a cubed term), so the force would decrease linearly.

However, for purposes of illustration, we can take the core to be about half the diameter of the earth, and its density to be about twice earth as a whole. That's not a bad approximation. So, at the surface, the acceleration due to gravity is GM/r², but at the core the radius is half, so the volume is an eighth while the density is twice, which means the acceleration is G x 2 x (M/8)/(r/2)², which is of course again GM/r²

[George]

[edit: ok, that's what you were saying ops:]

well, not quite what I said--it pretty much does not increase (nor does it decrease), it stays about the same.

That's against the usual rule--if earth had a constant, or more uniform, density, the force of gravity would decrease regularly all the way down. In general, as you tunnel into a solid sphere, the effect of the material above you (that is, everything above your radius--which includes material on the other side of the earth from you) disappears. That's a well-known result from calculus--there is no force of gravity inside a hollow shell. So, the force of gravity depends only on the material beneath your feet, which gets closer (the squared term) but is smaller (a cubed term), so the force would decrease linearly.

ops: . I see the error of my way now. Splitting the Earth to form two c.g.'s was ok, but getting closer to the first would, obviously, not increase the gravitational force. I don't what I was thinking. :-?

However, for purposes of illustration, we can take the core to be about half the diameter of the earth, and its density to be about twice earth as a whole. That's not a bad approximation. So, at the surface, the acceleration due to gravity is GM/r², but at the core the radius is half, so the volume is an eighth while the density is twice, which means the acceleration is G x 2 x (M/8)/(r/2)², which is of course again GM/r²

Wow! Not intuitive but clearly and colorufully correct. You are saying if the outer 7/8's of Earth (by vol.) vanished and we were left standing on top of the core, we would still be feeling almost the same force of gravity (assuming the core didn't swell and we were somehow still around to feel it. :wink: ).

[hhEb09'1]

Wow! Not intuitive but clearly and colorufully correct. You are saying if the outer 7/8's of Earth (by vol.) vanished and we were left standing on top of the core, we would still be feeling almost the same force of gravity (assuming the core didn't swell and we were somehow still around to feel it. :wink: ).

I'm also saying that that would be the force of gravity if we were to somehow tunnel down to the core, like they would do in the OP. Gravity acts so that it would be as if the rest vanished, even though it hadn't.

[George]

Wow! Not intuitive but clearly and colorufully correct. You are saying if the outer 7/8's of Earth (by vol.) vanished and we were left standing on top of the core, we would still be feeling almost the same force of gravity (assuming the core didn't swell and we were somehow still around to feel it. :wink: ).

I'm also saying that that would be the force of gravity if we were to somehow tunnel down to the core, like they would do in the OP. Gravity acts so that it would be as if the rest vanished, even though it hadn't.

Yes. I realized it was what you meant but I got a little distracted with the effects of the loss of 7/8 of our planet. :wink:

I suppose g would diminish as the inverse square law from this point to the center (assuming uniform density).

[hhEb09'1]

I suppose g would diminish as the inverse square law from this point to the center (assuming uniform density).

No, quit thinking about massive disintegration of the planet. I never said that.

It would be linear in r. It's GM/r², but M is a function of r, M = d x 4pi/3 x r^3, where d is the (constant) density. So, it's GM/r² = G(d 4pi/3 r^3)/r^2 = (4Gdpi/3) x r

[snowcelt]

Check out a thread from a couple of years ago, there was a similar conversation.

"Bowling Ball, Empty Hole, Pole to Pole, and Poll."

[hhEb09'1]

Check out a thread from a couple of years ago, there was a similar conversation.

"Bowling Ball, Empty Hole, Pole to Pole, and Poll."

Naw, the closest we got to it was my post, which refers back to another thread, about the movie The Core. Not The Core though, not Re: More really obvious Bad Stuff in "The Core" either. Maybe it was Gravity in "The Core"

[George]

I suppose g would diminish as the inverse square law from this point to the center (assuming uniform density).

No, quit thinking about massive disintegration of the planet. I never said that.

I think I might make a good "Harry Carry" for the BA just as in that SNL skit I heard on slacker astronomy. :wink:

It would be linear in r. It's GM/r², but M is a function of r, M = d x 4pi/3 x r^3, where d is the (constant) density. So, it's GM/r² = G(d 4pi/3 r^3)/r^2 = (4Gdpi/3) x r

I would have assumed M is constant but you are saying the mass outside our distance to center is removed from the equation. Another not so intuitive, but interesting, aspect of Newton's formulae. 8)