gravitation inside a shell: which way is down?

[beskeptical]

OK. I have not had time to read the links from Phobos so no need to answer yet. But for the moment, common sense says it is not zero Gs at all points inside a hollow sphere. The density inside differs from the shell. Now if you were talking atmospheric pressure or something, I could understand. How thick is the shell. Does it matter? How big can this sphere be? How dense or not dense does the inside atmosphere need to be? How uniform? Are there no air currents? Temperature variations? and so on.

I will hold to my common sense until I find something more logical.

[jumbo]

For a uniform hollow sphere there is zero g inside the sphere. Common sense may not imediately say this but the maths does. If you are on the inner surface of the sphere you have a comparatively small mass below your feet which is very near but the gravitational influence of this is cancelled out by the far larger mass above you which is further away. The thickness and density of the shell is irrelevant if the sphere is uniform. (Dense and thick material below is matched by the dense and thick material above us).The size of the sphere makes no difference either. This is because a large sphere will have a larger mass but all of the distances involved increase too, thus negating the increase in mass. If you have a material inside the sphere you only need to consider the gravitational influence of that material because the sphere is no longer a factor.

[John Kierein]

If there's no gravity at the center of the earth, why is it so dense there? I think it has something to do with buoyancy.

[jumbo]

There is zero g only at the very centre of a spherically symetric uniformly solid sphere (gravity from a spherically symetric mass distribution pulling in all directions).
You feel the gravity due to the mass interior to your distance from the centre of the planet. In a hollow sphere this mass is zero so no gravity. In a solid body like the earth this mass is positive hence there is gravity inside the earth except for the special case at the centre of the earth. At this point there is zero mass interior to you so no gravity.

[jumbo]

Envelope maths alert!!
The value of g inside the solid body is given by g=-(4/3)G*PI*Density*r
For a uniform sphere, where r is the distance from the centre.
This could be used to calculate the gravity in a gas filled hollow earth scenario i guess. As long as the gas is unformly dense you should just be able to use the density of the gas of your choice.

[traztx]

On 2002-06-20 08:59, John Kierein wrote:
If there's no gravity at the center of the earth, why is it so dense there? I think it has something to do with buoyancy.

Probably caused by several effects.

I'm imagining the dirty nebula that gave birth to our system. Matter is attracted to the thicker clumps, forming asteroids, which collide to form larger ones.

Violent collisions knock lighter pieces out faster than heavier pieces. The heaviest pieces return to the asteroid 1st (only because they weren't moving away as fast). Therefore each time violent collisions shake up an asteroid, they ultimately reorganize its matter so that heavier parts go to the center.

As an asteroid becomes uncommonly large, there is less chance of violent reorganization of its pieces.

But when it is large enough such that the surface cannot radiate the heat from the collisions fast enough, then the material with the lowest melting point melts first inside, causing still-solid pieces to settle down. Even though there is 0 force of gravity at the center, there is non-0 force everywhere else, so the solids near the center would be pushed down by solids above them, causing a pressure on the center that squeezes out anything liquid.

I don't know... does this all sound reasonable to you all?
--Tommy

[roidspop]

[quote]
On 2002-06-20 14:00, traztx wrote:

But when it is large enough such that the surface cannot radiate the heat from the collisions fast enough, then the material with the lowest melting point melts first inside, causing still-solid pieces to settle down. Even though there is 0 force of gravity at the center, there is non-0 force everywhere else, so the solids near the center would be pushed down by solids above them, causing a pressure on the center that squeezes out anything liquid.

I don't know... does this all sound reasonable to you all?
--Tommy

It sounds good to me. I would offer the observation that there doesn't seem to be any particular reason for liquids to be squeezed out since during melting, there would have been stratification by density...those would be some HEAVY liquids, like iron/nickel and the crust (silicates) would be lighter by far. Also, in a system forming from the debris of recent supernovas, you'd have material enriched in short-halflife radioisotopes, and they'd contribute markedly to internal heating. I wonder if there'd be less differentiation in bodies forming in old dust clouds?

[beskeptical]

On 2002-06-20 06:36, jumbo wrote:
For a uniform hollow sphere there is zero g inside the sphere. Common sense may not imediately say this but the maths does. If you are on the inner surface of the sphere you have a comparatively small mass below your feet which is very near but the gravitational influence of this is cancelled out by the far larger mass above you which is further away. The thickness and density of the shell is irrelevant if the sphere is uniform. (Dense and thick material below is matched by the dense and thick material above us).The size of the sphere makes no difference either. This is because a large sphere will have a larger mass but all of the distances involved increase too, thus negating the increase in mass. If you have a material inside the sphere you only need to consider the gravitational influence of that material because the sphere is no longer a factor.

I assume I am wrong but I am reminded of the experiment where the group (in on the deal) says a line is equal when it isn't and the testee eventually gives in and says the same despite the obviously incorrect situation.

If you measured this gravity at all points inside this sphere and it was 0 everywhere, what is stopping a person inside from being affected by the mass of the shell? If one sticks to the top of the surface, why wouldn't the same be true inside? I sort of get the gist of the math you are talking about, but I fail to see how the mass of the shell that is a diameter away has an equal effect on a body that is next to the shell, regardless of how many air molecules are also being acted upon. There is space between the molecules so they would be compressible. I see no reason gravity inside would be so different from outside.

What happens if there is a tiny hole in the sphere? What if the hole is big? What if there are lots of holes? Don't those silly Hollow Earthers think you can get inside? Do they need to shut the door in a hurry? [img]/phpBB/images/smiles/icon_lol.gif[/img]

[jumbo]

If you measured this gravity at all points inside this sphere and it was 0 everywhere, what is stopping a person inside from being affected by the mass of the shell?

The mass of the shell surrounds the person inside the shell giving g in all directions.The person inside is affected by the mass of the shell but is affected in all directions equally by it,so the value of g in each direction is equal. Imagine your feet are strapped to the inner surface of the airless hollow sphere. Below your feet you are very close to the sphere but there is only a little of its mass below your feet. If you look up you would see the rest of the sphere. It is further from you than the mass below you but there is a lot more of it. The geometry of the situation means that the distant large mass has the same gravitational effect as the lower close mass so they cancel each other out. Giving a net force on you of zero.
The maths proving this are at the link i gave earlier.(Its better than my explaination because it gives instructions for a helpful diagram)

If one sticks to the top of the surface, why wouldn't the same be true inside?

Now imagine you are stood on the outside of the hollow sphere. Below your feet you now have the whole mass of the sphere, above you there is nothing. In this case there is no matter above you to cancel out the effect of the matter below your feet so you stick to the outer surface, but not the inner surface of the sphere(for the reasons above).

If there are holes then the sphere no longer has uniform density and the problem becomes more difficult to tackle mathematically. In this case there will be some gravity inside the shell due to an uneven mass distribution. The direction of these forces will depend upon where the holes are. There may still be points within that sphere that all of the forces equal each other and cancel themselves out leaving zero g.

[traztx]

Another interesting property of a uniform spherical shell is apparent when you imagine it orbiting a star.

If there is a single object inside the shell and otherwise vacuum, then (as stated previously) the object has no net gravitational force in relationship with the shell.

However, it does gravitate towards the star.

If the object is at the center of the shell, then its orbit will coincide with the shell's orbit around the star. Otherwise the orbit will not coincide and some force will be required to keep it from drifting in relation with the shell.
--Tommy
http://www.tommyraz.com

[roidspop]

On 2002-06-21 11:24, traztx wrote:

If the object is at the center of the shell, then its orbit will coincide with the shell's orbit around the star. Otherwise the orbit will not coincide and some force will be required to keep it from drifting in relation with the shell.
--Tommy
http://www.tommyraz.com

So, such an internal object's position is unstable, subject to external perturbation? One element of this "hollow earth" creed is that there is a small, fission-powered sun in the cavity. The information you've provided shows that this would be impossible, given the tidal effects produced by the sun and moon. In a moonless system, would inhomogeneities in the shell have an effect on the central body?

[SeanF]

On 2002-06-21 11:24, traztx wrote:
Another interesting property of a uniform spherical shell is apparent when you imagine it orbiting a star.

If there is a single object inside the shell and otherwise vacuum, then (as stated previously) the object has no net gravitational force in relationship with the shell.

However, it does gravitate towards the star.

If the object is at the center of the shell, then its orbit will coincide with the shell's orbit around the star. Otherwise the orbit will not coincide and some force will be required to keep it from drifting in relation with the shell.
--Tommy
http://www.tommyraz.com

Interesting . . . of course, the sphere itself would be subject to tidal forces in that situation and some force would be required just to keep it a perfect sphere, no?

BTW, when an object (say, a pen) is released and allowed to "float" inside ISS, which way does it drift?

[traztx]

On 2002-06-21 11:51, roidspop wrote:
So, such an internal object's position is unstable, subject to external perturbation? One element of this "hollow earth" creed is that there is a small, fission-powered sun in the cavity. The information you've provided shows that this would be impossible, given the tidal effects produced by the sun and moon.

I agree that there is no way the earth is hollow. Seismic data alone contradicts this.

Perhaps a "type II" civilization could construct an earth-sized spherical object that was hollow and place a fission device in the center. They would have to be careful to make sure that anything emitting from the device was balanced to keep it from jetting from position. As long as they did that, then such an object would follow the same orbital pattern as the shell, regardless of moons orbiting the shell.

In my opinion (and I admit I'm not a professional expert), such an object would not look like the earth, which is not a sphere. Oceans and mountains would present a huge challenge in balancing the forces inside (see below).

On 2002-06-21 11:51, roidspop also wrote:
In a moonless system, would inhomogeneities in the shell have an effect on the central body?

Interesting question. The way I see it (still assuming no atmosphere or other medium filling the shell): Even if the center of mass of the inner body coincided with the center of mass of the shell, the inhomogeneities would cause a non-0 gravitational effect on parts of the inner body that are not located at its center. So areas of the inner body would experience a non-0 net gravitational acceleration in relation with the shell, pulling it out of the center. This drift would be enhanced as it went into a new orbit around the star.

Note: A "type II" civilization was presented in "Cosmos" as one with the technology to harness the total power of a star. Type III is capable of harnessing the power of a galaxy. We have no evidence that either exists in the universe.
--Tommy

[traztx]

On 2002-06-21 12:39, SeanF wrote:

On 2002-06-21 11:24, traztx wrote:
Another interesting property of a uniform spherical shell is apparent when you imagine it orbiting a star....

Interesting . . . of course, the sphere itself would be subject to tidal forces in that situation and some force would be required just to keep it a perfect sphere, no?

True, the material would already have to be rigid enough to prevent it from collapsing. It would need to be even more rigid to deal with the effect you mention and any centrifugal effect from the shell's rotation. But as long as the shape and homogeneity is maintained, the gravitational effects will be maintained.

BTW, when an object (say, a pen) is released and allowed to "float" inside ISS, which way does it drift?

I dunno... towards the nearest air return vent? [img]/phpBB/images/smiles/icon_smile.gif[/img]
--Tommy

[DaveC]

The principle of zero net gravitational force inside a sphere doesn't have to be postulated upon. The exact same mechanism is what makes it impossible to electroplate the inside of a sphere - or the inside of a tube. There is zero electrochemical force between the anode and the cathode. It seems odd that ions wouldn't move to the electrode with an opposite charge, but they don't. They are each drawn equally in all directions and therefore don't move. As a result, there is no dissolution of the metal anode and no deposition of metal ions on the cathode - that is no electroplating on the inside.

[roidspop]

[quote]
On 2002-06-21 12:54, traztx wrote:

I agree that there is no way the earth is hollow. Seismic data alone contradicts this.

This HEer named Jan Lamprecht has fiddled around with the density gradient of the shell to the point that, he claims, it would produce the same seismological effects that we see in the solid earth "model". He doesn't go so far as to claim these effects were derived in a rigorous fashion (his graphics seem to be just artwork), but it made me think of the old submariner's trick of hiding under a thermocline from a pursuer's sonar. I think a seismic wave would reflect strongly from a rock-vacuum boundary, but I can't say for sure that some sort of density gradient might not mask this effect. Fiendishly clever, these hollow earthers!

[beskeptical]

I get it jumbo. Don't sell yourself short. Your explanation is comprehensible, no diagrams needed. I'm going to have to ponder for a while though, because the 'perfect sphere' that would be needed seems to be like a perfect line or point. We only have them in math theory, but not too many real geometric objects are that perfect.

I'm going to look into what DaveC last posted. On paper, maybe, but it just seems to clean for reality. But from the looks of it, "the truth is out there". [img]/phpBB/images/smiles/icon_wink.gif[/img] I'll be back after a few more physics chapters.

[GrapesOfWrath]

On 2002-06-20 08:59, John Kierein wrote:
If there's no gravity at the center of the earth, why is it so dense there? I think it has something to do with buoyancy.

There is a high pressure. The pressure gradient points to the center, for the most part, so pressure increases as you go down, because of the mass above--that is a result of gravity.

As to the idea of zero gravity inside a hollow sphere--of course that is only as good an approximation as the sphere is a perfect sphere. But we can do reasonable calculations and come up with reasonable answers, and they will be reasonably close to the actual situation! So, it may come as a surprise that a point next to the inside wall of a sphere experiences just as much gravity from the nearby wall as it does from the whole rest of the sphere, but it is an easy calculation, and it's true.

Something related that is interesting, I think, is that in the case of the Earth, as you go down, gravity pretty much stays the same, instead of decreasing, all the way down to the core-mantle boundary, because of the density jump at the CMB. If there were a constant density inside the Earth, gravity would decrease linearly with distance from the center, as jumbo's formula shows.

[John Kierein]

The hollow sphere problem is related to the gravitational equivalent of Olbers paradox. The number of massive objects in the universe increase as the cube of the distance but their gravitational force only falls off as the inverse square. One would think we would be pulled apart by this infinite source of force. But the solution is identical to the hollow sphere case. The rest of the universe is a big series of spherical shells where their masses cancel vectorially. This works really well in a Euclidean universe, but may not be so easy to show in a non-Euclidean one.

[beskeptical]

On 2002-06-20 08:59, John Kierein wrote:

If there's no gravity at the center of the earth, why is it so dense there?

On 2002-06-22 12:47, GrapesOfWrath wrote back:

There is a high pressure. The pressure gradient points to the center, for the most part, so pressure increases as you go down, because of the mass above--that is a result of gravity.

As to the idea of zero gravity inside a hollow sphere... a point next to the inside wall of a sphere experiences just as much gravity from the nearby wall as it does from the whole rest of the sphere... in the case of the Earth, as you go down, gravity... stays the same, instead of decreasing, all the way down to the core-mantle boundary, because of the density jump at the CMB. If there were a constant density inside the Earth, gravity would decrease linearly with distance from the center, as jumbo's formula shows.

Allow me to dig myself deeper into stupidness here.

Why would the gravity of a hollow sphere differ from the gravity of a solid one, if we are using that hypothetical perfect sphere? Density shouldn't matter, atmosphere or iron, (excluding the CMB for the moment).

According to the math, gravity is zero G inside. It seems logical that each molecule affected by the gravity would exert pressure on the molecules below thus creating a pressure gradient. But you are saying there is a gravity gradient, which is not consistent with zero G throughout.

If gravity is zero, then the mass just below the surface of the sphere shouldn't experience any further pull toward the center. If the mass continues to be pulled toward the center, it makes better sense to me that gravity within the sphere would be equal to the gravity of the whole sphere, and attraction would have to be oriented toward the center. You cannot have a pressure gradient without gravity.

Take a small hollow sphere in space, the mass would be weightless. Given equal air pressure inside and out, shouldn't the ball collapse as the interior molecules migrated to the center and created a pressure gradient?

I completely understand the gravity of the bigger side that is farther away is equal to the smaller side that is closer, no need to rehash that. But how do you get a pressure gradient if the gravity isn't centered.

And if it is centered, then it might be equal throughout, but not zero.

And if the sphere is hollow but has an atmosphere, what would stop everything from being pulled toward the center, since air is more compressable than iron? You wouldn't expect to float as if weightless. I don't think you could maintain a breathable atmosphere.

I'll stop here, my head is spinning.

[Phobos]

Perhaps this helps you visualise the difference between gravity and pressure.

Imagine you are playing rugby with very bad BO! for half the game everyone avoids you and it is as if you are playing on your own. No forces are pushing you in any direction.

Suddenly the ball falls into your hands and one of the opposing players is forced to tackle you. Your body is now pushed into a different position because you received a force on one side only (and because you now have the ball).

The game now seems more interesting to you so you decide to hold on to this ball. Suddenly all the other players decide to tackle you. You are being crushed by the other players from all directions. Despite feeling multiple forces in multiple directions the net force is zero and you do not move in any direction. But you can feel the pressure from the other players.

Then a rumour spreads amongst the supporters that the ball is actually worth a fortune and they all decide they want it. They join in and you feel the pressure of more people, but still the net force is zero.

Hope that helps.

Phobos

[Donnie B.]

On 2002-06-30 01:13, beskeptical wrote:

Why would the gravity of a hollow sphere differ from the gravity of a solid one, if we are using that hypothetical perfect sphere? Density shouldn't matter, atmosphere or iron, (excluding the CMB for the moment).

According to the math, gravity is zero G inside. It seems logical that each molecule affected by the gravity would exert pressure on the molecules below thus creating a pressure gradient. But you are saying there is a gravity gradient, which is not consistent with zero G throughout.

If gravity is zero, then the mass just below the surface of the sphere shouldn't experience any further pull toward the center. If the mass continues to be pulled toward the center, it makes better sense to me that gravity within the sphere would be equal to the gravity of the whole sphere, and attraction would have to be oriented toward the center. You cannot have a pressure gradient without gravity.

Take a small hollow sphere in space, the mass would be weightless. Given equal air pressure inside and out, shouldn't the ball collapse as the interior molecules migrated to the center and created a pressure gradient?

I completely understand the gravity of the bigger side that is farther away is equal to the smaller side that is closer, no need to rehash that. But how do you get a pressure gradient if the gravity isn't centered.

And if it is centered, then it might be equal throughout, but not zero.

And if the sphere is hollow but has an atmosphere, what would stop everything from being pulled toward the center, since air is more compressable than iron? You wouldn't expect to float as if weightless. I don't think you could maintain a breathable atmosphere.

I'll stop here, my head is spinning.

One problem with thinking about hypothetical situations like this is that they're highly idealized, and therefore we tend to have bad intuition because we're used to situations that are not idealized and have multiple interacting forces.

For example, we're used to life on the surface of the Earth, where there's a nice comfortable 1-atmosphere of pressure. That pressure is, of course, due to gravity (the total lumped gravity of the whole Earth), and is easily conceptualized as the weight of the atmosphere above us.

Of course, pressure is not always, or necessarily, due to gravity. The space shuttle cabin is pressurized but in microgravity.

So what of our hypothetical hollow Earth? The thought experiment includes an unstated assumption: that the material of our sphere is strong enough to keep from collapsing under its own weight. In other words, a cavity at the center would (in the real Earth) experience the full weight of the entire planet above it, and would certainly collapse instantly and with horrific consequences for any brainy apes up on the surface.

But this is a thought experiment. We'll just line the cavity with UltraStrongium. Now the load of the overlying planet is carried in our liner, much like an arch carries the load of a brige (except in three dimensions -- so it's more like a dome, I suppose, or a perfect eggshell). Now we can do anything we like inside the cavity -- pump out all the molten iron, leaving a vacuum, or fill it with air at any pressure we choose. The pressure due to the planet's gravity is borne by our UltraStrongium liner.

Now, we have our ideal free-fall chamber inside the Earth. But wait! If that chamber is large enough, and full of air, then the air itself has mass and will experience mutual attraction. So we do have gravity inside, and it does point to the middle, and we will have a pressure gradient (less on the outside, near the shell, and more at the center). But the magnitude of this gravity is small, and the pressure gradient is weak... in fact, it's exactly the same as you'd find in the same volume of air out in deep space.

If our internal atmosphere is being held at standard temperature and pressure, then the thermal-kinetic pressure of the air would probably overcome its self-gravity, and you wouldn't notice the gradient... just as you'd expect the same volume of gas at stardard temperature and pressure out in space to dissipate rather than remaining in a sphere.*

That's really the key here: if you have a hollow sphere that can support itself against its own gravity, then what goes on inside (gravitationally) is just what would happen to the contents of the cavity if the shell weren't there at all.

*[This would hold true for small cavities, lesser in volume than, say, a gas giant planet. An interesting exercise would be to calculate how large a volume of "room-temperature" air you'd need for it to hold together as a planet. You'd have to equate the high end of the thermal kinetic range with the escape velocity, and make some assumptions about how long something has to last to be considered a planet.]

[beskeptical]

On 2002-06-30 12:00, Donnie B. wrote:
Of course, pressure is not always, or necessarily, due to gravity. The space shuttle cabin is pressurized but in microgravity.

Yes, but no gradient

So what of our hypothetical hollow Earth? The thought experiment includes an unstated assumption: that the material of our sphere is strong enough to keep from collapsing under its own weight. In other words, a cavity at the center would (in the real Earth) experience the full weight of the entire planet above it, and would certainly collapse instantly and with horrific consequences for any brainy apes up on the surface.

Just as I thought.

But this is a thought experiment. We'll just line the cavity with UltraStrongium.

[img]/phpBB/images/smiles/icon_biggrin.gif[/img]

Now we can do anything we like inside the cavity -- pump out all the molten iron, leaving a vacuum, or fill it with air at any pressure we choose. The pressure due to the planet's gravity is borne by our UltraStrongium liner.

If that chamber is large enough, and full of air, then the air itself has mass and will experience mutual attraction. So we do have gravity inside, and it does point to the middle, and we will have a pressure gradient (less on the outside, near the shell, and more at the center). But the magnitude of this gravity is small, and the pressure gradient is weak...

If our internal atmosphere is being held at standard temperature and pressure, then the thermal-kinetic pressure of the air would probably overcome its self-gravity, and you wouldn't notice the gradient...

An interesting exercise would be to calculate how large a volume of "room-temperature" air you'd need for it to hold together as a planet. You'd have to equate the high end of the thermal kinetic range with the escape velocity, and make some assumptions about how long something has to last to be considered a planet.

OK. That I totally understand. But you'd only have zero G when you had a vacuum or the gas thermal-kinetic pressure counteracted the gravity. [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

And, technically, without the thermal-kinetic pressure, the answer to the question that opened this thread is the attraction would be toward the center.

Ta-dahhh!

[roidspop]

But you'd only have zero G when you had a vacuum or the gas thermal-kinetic pressure counteracted the gravity

To pressurize the interior of such a shell would not counteract the gravity of the mass of the atmosphere inside the shell. You would have a sphere of gas with some particular mass which would exert a net gravitational force on any particle inside that volume depending on its position wrt the center. If the gas cloud in the shell is massive and can cool off somehow, it ought to develop a density gradient. If it is very hot and thin, maybe it wouldn't. Regardless, though, if we are to believe what we've learned here, the gas would produce a gravitational field inside the chamber. Throw in some water and you'd get a spherical ocean in the center, I suppose. One big window through the UtraStrongium wall, and you'd have a sun (the incandescent material surrounding the chamber). This could get interesting.

[beskeptical]

On 2002-06-30 22:42, roidspop wrote:
To pressurize the interior of such a shell would not counteract the gravity of the mass of the atmosphere inside the shell.

Just when I was starting to get it. I'll ponder this and get back later. [img]/phpBB/images/smiles/icon_smile.gif[/img]

[traztx]

On 2002-07-01 15:08, beskeptical wrote:

On 2002-06-30 22:42, roidspop wrote:
To pressurize the interior of such a shell would not counteract the gravity of the mass of the atmosphere inside the shell.

Just when I was starting to get it. I'll ponder this and get back later. [img]/phpBB/images/smiles/icon_smile.gif[/img]

Yes, adding air does change things, eh?

If you take a sphere as large as the Earth (radius 6378km), subtract 400km for the shell (inner radius 5978km).

The volume of a sphere:
V= (4/3) pi r^3

Were looking at over 890 trillion cu meters of air (correct me if I'm wrong here).

Now consider the density of air at 15C at sea level: 1.224 kg/cu m

How much mass does 890 trillion cu m of air at that density have?

Were talking about an object with over 1089 trillion kg mass!

And it's all just a bunch of air [img]/phpBB/images/smiles/icon_smile.gif[/img]
--Tommy

[Wiley]

On 2002-07-01 15:57, traztx wrote:
If you take a sphere as large as the Earth (radius 6378km), subtract 400km for the shell (inner radius 5978km).

The volume of a sphere:
V= (4/3) pi r^3

Were looking at over 890 trillion cu meters of air (correct me if I'm wrong here).

I get 8.95 x 10^20 cubic meters.

V = (1.333)*(3.141)*(5.978 x 10^6)^3

This assumes that I did not have another brain lapse - they are occuring more frequently now.

Now consider the density of air at 15C at sea level: 1.224 kg/cu m

How much mass does 890 trillion cu m of air at that density have?

Were talking about an object with over 1089 trillion kg mass!

My calculation gives about 1.1 x 10^21 kg. This is the same mass as the asteriod Ceres and about 600,000 times the mass of Mars's moon Deimos. And about 1/6000 of the solid Earth.

Interesting calculation, Tommy.

[traztx]

On 2002-07-01 17:24, Wiley wrote:

I get 8.95 x 10^20 cubic meters.

V = (1.333)*(3.141)*(5.978 x 10^6)^3

This assumes that I did not have another brain lapse - they are occuring more frequently now.

Oops... I cubed the kilometers. I should have turned them into meters 1st. My bad [img]/phpBB/images/smiles/icon_smile.gif[/img]
--Tommy

[Donnie B.]

When I said that the pressure of the air would counteract its self-gravity, I was thinking of the case of a very small (house-sized) chamber at the center of the Earth. I agree that in the case of a larger cavity (most of the Earth hollow) then you'd get a density gradient -- absolutely true.

I think your calculation of the mass of that internal atmosphere is a bit oversimplified. You assumed a constant air pressure similar to that at sea level on Earth -- but in fact, that density gradient would mean that the center would be much denser than that. But maybe you meant that that was the average density. In that case, we can indeed compute the total mass of the airball (meow!)... I wonder how hard it would be to determine what the density would be at the center and at the shell. Would it be self-supporting (essentially zero atmospheres of pressure at the shell) or would it pressurize the cavity?

[roidspop]

I based my calculations on a shell thickness of 1288 km (because the HE folks keep citing this for some reason) and an average density of 0.0013 g/cm3, which I believe is about right for air density at sea level. I get a "surface gravity" just under the shell of 0.00072 gees and an orbital velocity (get that air outta the way!) of about 0.42 m/s.

Now, does somebody have a way of calculating the density gradient of this airball? How would you account for temperature?