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Thinking outside the box - Gravity problem

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posted on Aug, 7 2005 @ 03:48 AM
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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.

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 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 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.

[edit on 7-8-2005 by Simon666]




posted on Aug, 7 2005 @ 07:09 AM
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There was a thread about this a long time ago on ATS.


www.abovetopsecret.com...

I even chimed in with some info and a purty graphic. But everyone was all caught up in the whole "Maglev" portion of the topic that nobody noticed my nifty "use gravity as your main engine" two cents.


I still think it's genious and will be used some day...

[edit on 7-8-2005 by CatHerder]



posted on Aug, 8 2005 @ 07:40 AM
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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."

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.

[edit on 7-8-2005 by Simon666]

This got me thinking. The force of gravity should reduce the closer that you come to the Earth's core. Wouldn't you reach a point where gravity = drag and you would stop like achieving neutral buoyency in water?



posted on Aug, 8 2005 @ 08:44 AM
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The drag force is always equal or as good as equal to the gravity force when going just about terminal velocity. But you're right, I didn't think of buoyancy yet. If the air pressure gets high enough near the center, the air density could become greater than the density of water so you'd already start floating a lot sooner. Not sure however whether that would actually be the case. Someone with too much time on his hands could calculate that one.

[edit on 8-8-2005 by Simon666]



posted on Aug, 8 2005 @ 11:11 AM
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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.


Simon666,
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 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.


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. This is only a description of gravity approaching zero as you near the center. I said nothing about bouyancy. Are you under the impression that the gravitational acceleration, constant at the surface, will remain so as you approach the center of gravity? If it does not, then the increase in air pressure will not occur in the manner you describe.


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.


Describe the behavior of rho in this equation in light of the fact that g is decreasing as you fall farther and farther into the hole. I maintain that rho will decrease as the air pressure decreases. For what is pressure if not the weight of air, and is not the weight of air decreasing in the same way that the weight of the person falling is decreasing?

This is why I said before that it's not as simple as some of the posts here have made it out to be.

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.

Harte



posted on Aug, 8 2005 @ 11:27 AM
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This is turning into an interesting discussion.

For the sake of argument. Let’s take a step back a bit and imagine a huge long tube out in the middle of space somewhere, filled with air, but with open ends.

With no gravitational effects, but that from the tube walls and the gravitational attraction of the gas for itself, what would happen?

I thing that the answer would depend on the length of the tube. If it was only 100 ft long, the air would disperse out the open ends.

Now what if the tube were a 100,000 miles long? How about a million? 100 million miles? At some point, the air would start to form it’s own “center” much like a star forms out of an interstellar gas cloud.

The question would be, at what point would the gravitational attraction of the gas molecules start to supercede the forces of dispersion (I’m thinking that would be a function of the entropy, is that right?)

Adding the mass of the Earth around the tube would complicate things a bit, but I’m guessing that the basic principles would be the same.



posted on Aug, 8 2005 @ 11:31 AM
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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 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.

The net force however will still be directed towards the center as there is still more mass below than above. I don't see why this is a problem, after all I've consistently implied the gravity force gets less towards the center of the earth. And like I said, this simply means the increase in pressure with depth will lessen as you go deeper, not the pressure itself.



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.

You seem to think there will be a decrease in pressure. For an incompressible fluid or even solid, say lava, the equation for pressure would be (in the previous expression I wrote down, there was a small error):

p(h) = p0 + integral (rho * g(h) * dh)) from h = h0 = 0 m to h = hr = earth radius

If we state the pressure at the surface p0, and knowing the average density of the earth, and the dependancy of g versus h (g is a linear function that goes from g0 = 9,81 m/s² at h0 = 0m to gr = 0 at hr = radius of the earth -> g(h) = g0 * (hr-h)/hr ). This is a simple integral of a linear function in h, allowing us to approximate the pressure in the earth. It is an approximation because of compressibility and an increase in density towards the center of the earth, the earth's core contains iron in high concentrations. The result would be a net tremendous pressure at the center, which everyone knows, yet you seem to think the pressure would be zero because the gravity force is zero there?

[edit on 8-8-2005 by Simon666]



posted on Aug, 8 2005 @ 12:40 PM
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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.



posted on Aug, 9 2005 @ 01:58 AM
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Well, the centrifugal force lessens gravity a bit, on top, due to this rotation the earths radius r is also a bit larger at the equator than at the poles, so that at the poles you are closer to the mass center of the earth. Since gravity is proportional to 1/r², both these effects ensure that at the poles you weigh around 0,5% more.

curious.astro.cornell.edu...

[edit on 9-8-2005 by Simon666]



posted on Aug, 9 2005 @ 12:54 PM
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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.


Simon 666,

This part clears up my confusion quite well, thank you. You were correct in that I was mixed up concerning the behavior of a gas vs. that of a solid under these circumstances. It makes perfect sense that the pressurized gas at the surface (and below it) would maintain a pretty high pressure, and thus density, near and at the core.

Sorry for my fuzzy thinking.

Harte



posted on Aug, 10 2005 @ 01:48 AM
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No problem, mistakes are easily made by all.



posted on Aug, 11 2005 @ 07:49 AM
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a couple of posts back some one made a remark about ether and gravity and spinning force and such.
ether is a made up thing to explain what electromagnetic energy was in the 1800's.
gravity is around weather the object spins or not. gravity is what causes an object to start spinning in the first plkace



posted on Aug, 11 2005 @ 08:41 AM
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It's of little use to reply to the most obvious crackpots. Although creationists tempt me.

[edit on 11-8-2005 by Simon666]



posted on Aug, 11 2005 @ 09:55 AM
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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.



I remember reading somewhere that the Earth's gravity is not constant every where. This was first proven when they were trying to calculate the precise orbits of satellites. I don't remember where i read it at and will try to find the reference. I know that I looked it up and remembered it because of the gravity displacement device that was used for navigation in Clancy's Hunt for Red October. I wanted to see if it was possible.



posted on Aug, 11 2005 @ 10:50 PM
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interesting topic, but there is a truelly gravity zero??? i mean in the center the force acts with components to all directions, that is comparable with the free-fall effect of the space ship???



[edit on 12-8-2005 by grunt2]



posted on Jan, 30 2006 @ 08:23 PM
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you will fall to the south pole then rise back up to the north pole, you will continue this trend until your path becomes shorter placing you right in the center of the earth.....gravity will essentially allow you to float



posted on Jan, 30 2006 @ 10:27 PM
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Nice question! But i would not use the Earth...
Take an old rock planet with not atmosphere and not active magma core
but with the same g of the earth (same mesurements ) and not tourning around imself...just a quiet rocky planet. Drill it from side by side... How would be the force needed for the drill machine to escape and win the zero point gravity in the center?

second question: how would interact the matter in this zero gravity spot?

third question: gravity exist because matter exist.so we can imagine a spot
in the space where matter (gaz or other sort of "system") would create a sort
of field, in the middle of this field would we have a zero point gravity? and
would be a singularity or a more big zero area gravity where things are
stuck there for eternity?

Now i can sleep better....
bonne nuit
Darkiss

lol i dont have look at the post date...

[edit on 30-1-2006 by GothicDj]



posted on Jan, 31 2006 @ 10:14 AM
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Classical Dynamics of Particles and Systems by Thorton, this is basically a problem in that book.

The only way this question works is if you're considered a particle, Earth has a uniform density, and neglect rotational effects. If this is all true, the motion will be simple harmonic. I'll solve the problem here too:

F = Gm/r^2 * (4*pi*r^3*p/3) [p is the density of Earth]

F = m(d^2(r)/dt^2) = - Gm/r^2 * (4*pi*r^3*p/3)

m(d^2(r)/dt^2) + w^2r = 0 where w^2 = 4*pi*G*p/3

There's the simple harmonic oscillator equation, now I can show the period knowing:

T = 2*pi/w = squareroot(3*pi/(G*p)) ... This should give around 84 minutes.

Hope this answers your question!




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