I was noodling around with my son on the classic question of what happens if you jump in a hole that goes all the way though the earth. It’s pretty clear if you ignore air resistance. But if you have air in the hole, but ignore the heat of the core, what is the air pressure at different depths. My initial guess is that it is a function of the weight of the air column above you, taking into account the fact that the force of gravity decreases as you descend. But at the same time the air will compress, making it more dense and thus heavier.
My best guess is that at the midpoint of the hole the air would be weightless, but at very high pressure and that the pressure would increase from the endpoints to the midpoint. Is that right?
Just the opposite - the force of gravity will increase. The gravitational attraction between two bodies (i.e. the earth and you) is inversely proportional to the square of the distance between them. As you approach the earth’s center of mass, that distance decreases, resulting in a greater attractive force the deeper you go.
Xema’s point that there’s no net gravitational force at the center of the earth is true because once you reach it, you won’t be dislodged by anything less than an enormous force.
No, as you get inside the earth, more of the mass is above you pulling the opposite way. Gravity will decrease as you get towards the center. At the center, the mass of the earth is pulling you equally in all directions.
No. Gravity does not really act from the “center of mass,” it acts from where the matter is. “Center of mass” is just a useful mathematical fiction for when you are calculating the force that is exerted on a body outside the Earth (and some distance away). At the center of the Earth, gravity would be pulling you up, but as it would be pulling equally in all directions you would experience zero net force.
For a uniformly dense Earth, the force is directly proportional to the distance from the center until you reach the surface, at which point it goes down as the inverse square of the distance from the center.
The air pressure would go up exponentially as you go down, since the density is proportional to pressure and the rate of change of the pressure is proportional to the density. The doubling distance is about 5 miles (roughly the height of Mt. Everest), I’ve forgotten the exact number. Thus at 25 miles it would be about 32 times denser. Very soon you would reach the pressure at which air liquifies ahd the exponential behavior would end.
And for the actual Earth, because of density variation with depth, the force of gravity is roughly constant until about halfway down, then roughly linear the rest of the way.
The best way to think about it is when you are beneath the earth you have a “shell” surrounding you inside of which there is no net gravitational force and you are standing on a “sphere” the radius of which is the distance to the center of the earth. The mass of that sphere, and the distance to the center, determine the force of gravity.
Except in the classic thought experiment as posed in the OP, one would need to have a rigid tunnel in which to fall through. Physical pressure due to the density of the earth would instead be resisted by the unobtanium structure of the tunnel, not equalized by the atmospheric pressure.
The only force driving atmospheric pressure would then be gravity, so as you descend into the hole, air pressure would reduce along with the gravitational force, and you’d be in an effective vacuum upon reaching the center.
Actually, scratch what I just said…I’m afraid I was thinking about it backwards and neglecting the cumulative column of air. I can definitively say that my mental math isn’t up to the task and I need to hit it with Matlab
The gravity does approach zero as one approaches the center, but the matter at the center must bear the weight of all the matter above it. So even though gravity isn’t pulling the air at the center in any particular direction, the air just above it is being pulled slightly toward center; the center air has to support that, plus the weight of the air just above that other bit of air, which is being pulled toward center with slightly more gravitational force…on up to the surface of the earth. The pressure at the top of the hole (at sea level) is 14.7 psi; something has to hold that air up, along with all the other air in the hole that isn’t at the center.
Has anyone done the calculations to see what effect the hole would have on the air pressure at the surface? While the volume of the hole is not large, the air in it would be compressed so it should take a lot to fill.
Yeah, the center of mass is a convienient way to consider a body as a point so that the math is easier, but it doesn’t affect the fact that in real life a body is not a point.
The center of mass is really useful when you are talking about forces acting between external bodies that are fairly uniform in shape, such as planets orbiting stars, such that the actual gravitational vectors even out, and even then, it isn’t perfect. I think we had a discussion on here a while back about a cube shaped planet - read that.
When you are inside the earth, you are not an external body and gravity vectors are pulling you in all directions. In a mine, yer not gonna notice it since the net pull toward the center is going to be exactly what it is at the surface plus or minus some extremely small amount that you probably can’t measure, let alone feel with the body. At the center, you probably will.
That’s simple enough to compute. The radius of a hole that would contain the entire atmosphere is about sqrt(24000miles5miles*d), where d is the ratio of the density of air at atmospheric pressure to liquid air, which is about a factor of 1,000. Thus, the hole would need to be more than 10 miles in diameter to consume the atmosphere.
