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Originally posted by Harte
This part ain't necessarily true. It is erroneous to assume that the air pressure would increase in a manner similar to the way water pressure increases in the ocean depths. After all, the ocean is on the Earth's surface, the hole is not.
Originally posted by Harte
As anything (air or a person) flows farther into the depths of the hole, more and more of the Earth's mass is exerting an upward force. Hence, at the center, the air pressure would be zero.
Originally posted by Harte
The variation of air resistance against the acceleration of a falling body versus distance into the hole makes for an interesting differential equation. Your terminal velocity would certainly not decrease due to this. F=Ma remember. Any "F" will result in an "a."
Originally posted by Simon666
That has no relevance to the nature of the problem. Air is a fluid just like water, and hence if air would be incompressible you could calculate the increase in pressure with the simple formula p = p0 + ρ * g * h, with p0 the pressure at depth 0 (planet surface), ρ the density of the fluid, g the earth's gravity constant and h the depth to the surface. Since air is compressible, the increase in pressure would actually go a bit faster than linear as the density increases with pressure, and also g is not constant when going significantly up or down from the surface of the earth. This involves solving p = p0 + integral (ρ(h) * g(h) * h) for h = 0 -> earth's radius.
The variation of air resistance against the acceleration of a falling body versus distance into the hole makes for an interesting differential equation. Your terminal velocity would certainly not decrease due to this. F=Ma remember. Any "F" will result in an "a."
Originally posted by Simon666
Originally posted by Harte
This part ain't necessarily true. It is erroneous to assume that the air pressure would increase in a manner similar to the way water pressure increases in the ocean depths. After all, the ocean is on the Earth's surface, the hole is not.
That has no relevance to the nature of the problem. Air is a fluid just like water, and hence if air would be incompressible you could calculate the increase in pressure with the simple formula p = p0 + ρ * g * h, with p0 the pressure at depth 0 (planet surface), ρ the density of the fluid, g the earth's gravity constant and h the depth to the surface. Since air is compressible, the increase in pressure would actually go a bit faster than linear as the density increases with pressure, and also g is not constant when going significantly up or down from the surface of the earth. This involves solving p = p0 + integral (ρ(h) * g(h) * h) for h = 0 -> earth's radius.
Originally posted by Simon666
Originally posted by Harte
As anything (air or a person) flows farther into the depths of the hole, more and more of the Earth's mass is exerting an upward force. Hence, at the center, the air pressure would be zero.
This is mambo jambo. "The earth's mass is creating an upward force"? Hello, we're talking about a tunnel, not how someone floating in lava would go up by buoyancy or something. I don't even know how to interpret this paragraph, that's how little sense it makes.
Originally posted by Simon666
Originally posted by Harte
The variation of air resistance against the acceleration of a falling body versus distance into the hole makes for an interesting differential equation. Your terminal velocity would certainly not decrease due to this. F=Ma remember. Any "F" will result in an "a."
Actually, it is a very simple one, the drag equation is F(drag) = Cd * ρ * v² * A, this needs to be equal to F(gravity) = m * g plus F(inertia) = m * a. When solving this, in the limit of time t -> infinite your acceleration a will go to zero, so you can find the terminal velocity by equalling F(drag) = F(gravity). This results in the equation v(terminal) = square root (2 * m * g / Cd * ρ * A ). As you can see in this formula, your terminal velocity will go down as the density of air increases with depth when nearing the surface and it would even go down more as g decreases with depth. G actually becomes zero at the exact mass center of the earth.
Originally posted by Harte
Are you saying that at the core, where gravity is zero, there would still be air pressure? Isn't air pressure created by gravity?
Originally posted by Harte
Sorry if I was unclear. What I mean is the portion of the Earth's mass that the "falling" person has passed causes an "upward-directed" gravitational force.
Originally posted by Simon666
Imagine instead of a person falling into the hole, we start with just atmosphere falling into the hole. Describe what happens to the air pressure as the atmosphere "falls" toward the center of gravity, where there is no gravitational acceleration.
Originally posted by Simon666
Originally posted by Harte
Are you saying that at the core, where gravity is zero, there would still be air pressure? Isn't air pressure created by gravity?
Yup. Ask yourself, how is it that the lava in the center of the earth is under such tremendous pressure, while the gravity at the exact center is zero? Answer is: because of the pressure of whatever lies above. When reaching the center of the earth, the increase in pressure with depth will indeed diminish due to decreasing gravity, yet the pressure itself will not decrease.
Originally posted by TxSecret
Interesting thread indeed, still getting caught up on it though. One thing however: someone mentioned earlier that the Earth's rotation had something to do with it's gravity.. From what I understood it does not. Maybe centrifugal force lessens it's effect. Would I weigh less at the equator than at the poles? Someone please elaborate.