gravitation inside a shell: which way is down?

[roidspop]

I'm embroiled in a long-running discussion with a hollow-earther and I find myself needing some mathematical assistance.

The assertion that this fellow's guru, Lamprecht, seems to be making is that the effective center of mass of a thick spherical shell is within the shell, not in the actual CG. I can see how this might come about...local gravitational effects dominating the global ones so that, right up against its inner surface, the shell might SEEM like an independent, discrete body and local 'down' would be against the "ceiling".

I also see that no amount of logic will put this bed...it needs to be done mathematically. Unfortunately, I also see that it's a calculus problem and that discipline is nearly forty years in my past. I can't ask someone to do the math, but if you could suggest a reference in which a kindly author has already produced an nice, algebraic expression for net gravitational force at any point within a hollow sphere, boy, would I be grateful!

I have to hand it to this guy...a hollow planet would be cool! But man! What a wacky mess to have to believe in! He does it without any problems, though, polar holes, fission mini-sun and everything!

[Phobos]

On 2002-06-17 10:53, roidspop wrote:
I'm embroiled in a long-running discussion with a hollow-earther and I find myself needing some mathematical assistance.

The assertion that this fellow's guru, Lamprecht, seems to be making is that the effective center of mass of a thick spherical shell is within the shell, not in the actual CG. I can see how this might come about...local gravitational effects dominating the global ones so that, right up against its inner surface, the shell might SEEM like an independent, discrete body and local 'down' would be against the "ceiling".

I also see that no amount of logic will put this bed...it needs to be done mathematically. Unfortunately, I also see that it's a calculus problem and that discipline is nearly forty years in my past. I can't ask someone to do the math, but if you could suggest a reference in which a kindly author has already produced an nice, algebraic expression for net gravitational force at any point within a hollow sphere, boy, would I be grateful!

I have to hand it to this guy...a hollow planet would be cool! But man! What a wacky mess to have to believe in! He does it without any problems, though, polar holes, fission mini-sun and everything!


I had this discussion with someone elsewhere on the board. Unaware that the maths had already been done I had to revert to intuition (not always accurate). Intuitively I said that if you were inside a hollow planet but not in the centre, then you would drift towards the inner surface.

Later I was notified that when the maths are applied the surprising result is that the gravitational forces of a hollow sphere (as applied to an object within the sphere) effectively cancel each other out. It was then stated that inside a hollow planet you would experience zero g.

So my intuition was wrong about the force moving you towards the inner surface, but I was not yet satisfied so I started using this new information in my thoughts.

I then realised that the idea that when inside a hollow planet you would experience zero g is WRONG for the for the following 2 reasons:

reason 1 - Planets (even hollow ones) would orobably have the mass unevenly distributed. Whilst the effect of distribution would be very small it cannot be fully discounted.

reason 2 - Even hollow planets would not be filled with vacuums. The calculations only involve the masses of the outer shell of the planet. If we assume that the effective gravitational force of the planets hollow shell is zero, then we next have to ask ourselves what about the gases contained inside a hollow planet ?

As a result of reason 2 the correct answer to the question "what direction is the force of gravity for an object inside a hollow planet ?" is "towards the centre of the planet".

Ok, if you follow the logic so far, hold on to your seat because the ride gets a bit bumpy now...

We have only considered the force of gravity from inside the hollowed shell of a hollow planet. What about forces inside the shell and on the surface ?

On the surface the force of gravity would be towards the centre. If we now travel underground towards the inner shell we will notice that the force of gravity decreases until we reach the inner shell. This dropping off of gravitational force continues all the way to the centre of the planet, but only when we are at the exact centre of gravity will we have zero g - well almost ...

Even at the centre of the hollow planet we will not be totally weightless. if our belly button was exactly in the middle of the planet, our body parts would all experiece different forces (where both our head and our feet would experience g, but in opposite directions towards our belly button).

Phobos

<font size=-1>[ This Message was edited by: Phobos on 2002-06-17 11:27 ]</font>

[traztx]

Here's my take on it:

Assuming a uniform shell, objects inside the shell are accelerated towards each other as if there were no shell. Objects outside the inner edge of the shell are accelerated towards the center of the planet by gravity.

If a gas is inside the planet, then at any given altitude, the acceleration by gravity is towards the center as if the only mass was the gas at lower altitudes (I am discounting turbulence).

As you move towards the center, the air pressure increases but the acceleration from gravity decreases. As the pressure increases, you become more buoyant. If there is enough pressure, then buoyancy could prevent you from reaching the center.

--Tommy
http://www.tommyraz.com

[p9107]

Is the strength of gravity determined by mass, density or phyical size?

[Wiley]

On 2002-06-17 12:55, p9107 wrote:
Is the strength of gravity determined by mass, density or phyical size?


The strength of the field is determined by the mass and it's distrubition.

It's been a while, so I may get some details wrong. The gravitional potential (F_g = grad P) obeys Gauss's law: if we take an imaginary sphere and if there is no mass enclosed by this sphere, the potential is zero (and thus no force) everywhere with in the sphere. Newtonian gravity satisfies Laplace's equation, right?

[roidspop]

I follow the arguments (LaPlace? Gauss? Well...), but the result I'm seeking may involve fidgety stuff hiding in the undergrowth of the math. I agree with Phobos about the overall elaboration of this problem, but I think there is a chance that at a certain shell thickness and planetary diameter, there might possibly be a local, shellward gravitational effect, in which the local mass acting at the shell's surface MIGHT be stronger than the net attraction acting from the center of mass of the entire body. I can't resolve this without math.

I like the buoyancy deal...what would the density and pressure be at the center, I wonder?

I also would like to be able to analyze the strength of the shell and see at what point it would collapse of its own weight...This guy proposes a shell about 800 miles thick. A real crude calculation showed the pressure at that depth would far exceed the strength of rock, ignoring the effect of heat and so forth. This is another area where I could use some help. Let's not even get started on the fission-powered mini-sun that supposed to be floating in the void!

[Kaptain K]

Back of the envelop calculation. Anyone who feels the need for a more rigorous treatment, feel free to correct me.

Air at the Earth's surface has a density of approx. 0.6mg/cc.

Earth's density, approx. 5.5g/cc.

Therefore:
Mass of an Earth sized ball of air is about 1/100,000th that of the Earth.

Since the gravitational attraction is proportional to the masses.

The maximum gravitational force inside a hollow Earth would be 0.00001g. May not be "zero" g, but definitely "micro-gravity".

_________________
When all is said and done - sit down and shut up!

<font size=-1>[ This Message was edited by: Kaptain K on 2002-06-17 14:17 ]</font>

[traztx]

Other points to bring up with your friend:

- The Earth is not a sphere. If it were hollow, then the poles would attract each other. What force keeps them apart?

- Also, what keeps the tidal forces from pulling the middle out and turning the hollow earth into a disk?

- If the Earth were hollow, our seismic data would be very different! We measure the subterranian material by the effect density has on the speed and attenuation of sound.

- If the Earth were hollow, then what could cause the Himalayas to form? There wouldn't be the convection system to move the plates SIDEWAYS. Likewise, what could spread out the floor in the mid-atlantic ridge?

[justncredible]

think on this,,, as you go down the mass from your starting point would get smaller, untill you hit the center, at the center, you would have half the gravity forces pulling you back to your starting point, while at the same time the rest of this half force will be pulling you in all directions, you would not feel like your in zero-g but more like being pulled apart, if you weighed 100lbs on the surface, at the center you would feel 50lbs of force in every direction, lets say your 3/4 of the way to center,,, you would weigh 75 lbs on a scale, yet your mass will not change,,, lets say you lived there,, you would have weak bones, and weak muscles,, maybe you think not much,,, but put on another 25lbs overnite, and cut your muscles down by 25% and you won't be doing much on the surface, kinda like a one of those boneless chickens flopping around on the ground. the mass from the other side of this planet gives us our gravity, and this is a big rock... but maybe i'm wrong

[Phobos]

On 2002-06-17 19:10, justncredible wrote:
think on this,,, as you go down the mass from your starting point would get smaller, untill you hit the center, at the center, you would have half the gravity forces pulling you back to your starting point, while at the same time the rest of this half force will be pulling you in all directions, you would not feel like your in zero-g but more like being pulled apart, if you weighed 100lbs on the surface, at the center you would feel 50lbs of force in every direction, lets say your 3/4 of the way to center,,, you would weigh 75 lbs on a scale, yet your mass will not change,,, lets say you lived there,, you would have weak bones, and weak muscles,, maybe you think not much,,, but put on another 25lbs overnite, and cut your muscles down by 25% and you won't be doing much on the surface, kinda like a one of those boneless chickens flopping around on the ground. the mass from the other side of this planet gives us our gravity, and this is a big rock... but maybe i'm wrong


Not quite true - Specifically this - "you would not feel like your in zero-g but more like being pulled apart". At tht centre of the Earth the opposing forces would cancel out, and you would not be able to tell the difference from the zero g in the ISS.

As stated earlier there would be tiny gravity differences about your centre, but they would be too small for you to detect.

Phobos

<font size=-1>[ This Message was edited by: Phobos on 2002-06-17 19:26 ]</font>

[Karl]

On 2002-06-17 14:14, roidspop wrote:


I also would like to be able to analyze the strength of the shell and see at what point it would collapse of its own weight...This guy proposes a shell about 800 miles thick. A real crude calculation showed the pressure at that depth would far exceed the strength of rock, ignoring the effect of heat and so forth.


Rock? To match the density of the earth with an 800 mile thick shell requires a material with the density of lead or higher, considering that the upper few kilometers has been cored and verified as being rock.

[roidspop]

While we're theorizing about this problem, I recall a nifty idea for a transportation system, outlined back in the seventies I believe. The idea was that you'd bore parabolic tunnels between distant points on the earth and pump all the air out (we don't worry about little engineering details like keeping a tunnel from collapsing at a depth of hundreds of kilometers...). A levitated train in this tunnel would fall at increasing speed, pass the bottom of the arc and start upward, slowing all the way. At the other end it would come to a stop after a 45 minute trip. And the real trick was that it didn't matter how far apart the ends were...the trip would take 45 minutes. Period. NY to London? 45 minutes. Brisbane to Paris? 45 minutes. The extreme case was a vertical shaft through the core to the opposite side of the planet...drop the train through that and the trip takes...45 minutes. We REALLY need to learn how to do this!

...and I still don't know how to solve this problem, but I think Wiley is correct about the gravitational potential in the shell.

Back to the gravity problem...I think a special case of this might be a sort of croissant-shaped asteroid... the center of mass is out in free space between the 'horns'. What's the gravity like on the surface just in the center the limbs? If you released a rock at that point, would it fall up? Is it too much to expect that for some objects like this (if any exist) that there might be 'gravel piles' of debris trapped in an external center of mass?

[Silas]

On 2002-06-17 21:36, roidspop wrote:
Back to the gravity problem...I think a special case of this might be a sort of croissant-shaped asteroid... the center of mass is out in free space between the 'horns'. What's the gravity like on the surface just in the center the limbs? If you released a rock at that point, would it fall up? Is it too much to expect that for some objects like this (if any exist) that there might be 'gravel piles' of debris trapped in an external center of mass?


The problem is, since gravity falls off as the square of the distance, you'd have to perform an integration of the distance between all points on the crescent-shaped body and the point that you're investigating. Judging by a typical croissant, the thick part of the body, being close, would overwhelm the pull from the horns, which are distant, and thus a person in the middle would feel gravity toward the body of the...um...body. However, if the crescent is much thinner, well...other solutions might apply.

There's a comic book (I'm kicking myself all around the room for failing to remember the name!) where there is a "Bowl Shaped World." Take a "Hollow Earth," and then cut it in half, and throw away one half. The center of gravity is inside the bowl. Quite amusing...

Silas

edited to add: Nexus, by Mike Baron!




<font size=-1>[ This Message was edited by: Silas on 2002-06-17 22:22 ]</font>

[thkaufm]

While we're theorizing about this problem, I recall a nifty idea for a transportation system, outlined back in the seventies I believe. The idea was that you'd bore parabolic tunnels between distant points on the earth

Yes I remember this too. But you wouldn't need a curved tunnel. a straight line between any two points on earth's surface would produce a tunnel that should work.
From our perspective on the surface it would seem curved though.

Tom

[beskeptical]

Forgive me if my comments are naive.

How thick is the shell? If it were large enough to generate gravitational attraction that would hold me on the surface, how much of the surface is actually interacting with my mass? Since gravity decreases the farther away from the mass I go, I would be too far away from the far side of the Earth for that part of the mass to be attracting me with much force, (compared to the mass on the side I was close to).

So if I were inside, I should be attracted to the shell, assuming the volume of air were not incredibly dense in the center compared to the upper layers of atmosphere. The average gravitational force might be close to zero G, but the average force would not be equal for every point inside except the exact center.

[Phobos]

On 2002-06-18 01:14, beskeptical wrote:
Forgive me if my comments are naive.

How thick is the shell? If it were large enough to generate gravitational attraction that would hold me on the surface, how much of the surface is actually interacting with my mass? Since gravity decreases the farther away from the mass I go, I would be too far away from the far side of the Earth for that part of the mass to be attracting me with much force, (compared to the mass on the side I was close to).

So if I were inside, I should be attracted to the shell, assuming the volume of air were not incredibly dense in the center compared to the upper layers of atmosphere. The average gravitational force might be close to zero G, but the average force would not be equal for every point inside except the exact center.


considering the gravitational force of the shell allone for a observer placed inside it, and near the inner surface -

The force of gravity from the atoms above you is a lot stronger than the force below you (due to them being nearer). However, because you are away from the centre there are more atoms below you than above. When all atoms are considered and the total forces are calculated you will find that for a uniform hollow sphere they will exactly cancel out no matter how thick the sphere is or where within that sphere you are - provided you are within the inner shell you should feel no graviational force from the planet itself (only the other stuff inside).

Phobos

[John Kierein]

The "pushing gravity" cause of gravity makes this hollow sphere all seem easier to visualize; in my mind anyhow.
http://shop.alpmicro.com/apeiron/

[Phobos]

On 2002-06-18 08:38, John Kierein wrote:
The "pushing gravity" cause of gravity makes this hollow sphere all seem easier to visualize; in my mind anyhow.
http://shop.alpmicro.com/apeiron/


John, are you pushing gravity or book sales ?

Phobos /phpBB/images/smiles/icon_wink.gif

[jumbo]

I did a back of the envelope kind of calculation and ended up with the same result as the one at (drum roll please)
http://www.grc.nasa.gov/WWW/K-12/Numbers/Math/Mathematical_Thinking/grvtysp.htm
Of course the earth isnt uniform so there will be a slight deviation towards clumps of mass in the shell. Add in an internal atmoshpere and you will get a force towards the centre of mass of the atmosphere which will be above your feet thus pulling you off of the inner surface of the sphere.

[justncredible]

Phobos, how does it "cancel" out??? gravity from two opposing bodies does not cancel out each other... if it did then the moon and eath would not affect each other. the zero G on the ISS is not a "free fall",,, they still have gravity affectting them,, but are falling away from earth, i still hold that at the center of earth which by the way is not hollow, gravity still pulls on you, you might be able to find a place of equlliberuim, where the forces are balneced out, but you would still have the forces of gravity acting upon your body, remember gravity has not been proven to be a wave, if it was then yeah it would "cancel" itself out, but since it is not, we can only assume it permantes thru everything in all directions and acts as a force in all directions, but i'm open to other ideas as well................

[Paul Unwin]

On 2002-06-17 21:36, roidspop wrote:
Back to the gravity problem...I think a special case of this might be a sort of croissant-shaped asteroid... the center of mass is out in free space between the 'horns'. What's the gravity like on the surface just in the center the limbs? If you released a rock at that point, would it fall up? Is it too much to expect that for some objects like this (if any exist) that there might be 'gravel piles' of debris trapped in an external center of mass?

There's some confusion at work here. "Center of mass" doesn't correspond with the "location toward which gravity is directed" except in special cases like spheres. In fact, a sphere might be the only shape for which gravity points toward the center at every location.

[justncredible]

paul,,,, does gravity "point",,, i have not heard this before,,, gravity acts in all directions, and is dependant on mass, the more mass under you then the more gravity you have to deal with, the moon has less mass so it has less gravity, at the center of the earth you still have a radius of the planet in all directions exerting gravity upon you in the center, these hollow spheres ya'll are talking about must be small and have a thin shell, and are in a unknown dimesion that has no gravity,,,, anything that exist has gravity, with that thought, think of this,, you would never find a spot in the center of a planet or star that has 0 gravity,,, why,,, the rest of the universe exerts its gravity upon that spot, so it would move about in a dance so fast that would be impossible to find, the spinning planets the galaxys, the whole universe spins so the "spot" would move fast, never be able to find it, and it would be very small, my point is: gravity has no direction, is not a wave, is mass dependant.

[justncredible]

paul, i misunderstood your point and your right, you would be pulled towards the center of a sphere,,, my bad

[Phobos]

On 2002-06-18 11:58, justncredible wrote:
Phobos, how does it "cancel" out??? gravity from two opposing bodies does not cancel out each other... if it did then the moon and eath would not affect each other. the zero G on the ISS is not a "free fall",,, they still have gravity affectting them,, but are falling away from earth, i still hold that at the center of earth which by the way is not hollow, gravity still pulls on you, you might be able to find a place of equlliberuim, where the forces are balneced out, but you would still have the forces of gravity acting upon your body, remember gravity has not been proven to be a wave, if it was then yeah it would "cancel" itself out, but since it is not, we can only assume it permantes thru everything in all directions and acts as a force in all directions, but i'm open to other ideas as well................


Seems like I need to convince you on this one ...

Is gravity more or less intense as you move towards the centre of the Earth? (http://www.sciencenet.org.uk/database/Physics/9812/p01207d.html)

or better still

Earth's Center and Gravity (http://newton.dep.anl.gov/askasci/env99/env002.htm)


At the center of the earth, you would not feel any gravity. This is
because the gravitational pull from every region of the earth is exactly
counteracted by the gravitational pull from the corresponding region on the
opposite side of you. This all adds up to a great bug zero.

After a bit more digging I finally found a link which completely illustrates what I mean.

Planet Earth: On the pull (http://www.newscientist.com/lastword/answers/388earth.jsp?tp=earth2)


Newton found that the attractions of the materials in the hollow spherical shell of radius greater than R will all cancel one another out--a beautiful mathematical consequence of the fact that gravity decreases as the square of the distance. You feel only the pull of the mass in the sphere below you.

So that at least explains it as far as Isaac Newton was concerned.

Phobos

<font size=-1>[ This Message was edited by: Phobos on 2002-06-18 13:21 ]</font>

[DoctorDon]

On 2002-06-18 11:58, justncredible wrote:
Phobos, how does it "cancel" out??? .... gravity has not been proven to be a wave, if it was then yeah it would "cancel" itself out, but since it is not, we can only assume it permantes thru everything in all directions and acts as a force in all directions...


The force of gravity is a vector field. That is, at any point in space, it has a magnitude and a direction associated with it. The net gravitational force on an object at a given point (let's assume a point-like object for the moment, just to avoid confusion) is the sum of all the contributions from all available sources of gravity. Since vectors have both magnitude and direction, opposite and equal vectors do cancel each other out, leaving a net force of zero at that point. It may well be that just a hair's-breadth away from that point, it's non-zero, but at that point, the gravitational force is zero.

And it just so happens that for every point inside a spherically symmetric shell of matter, the net gravitational force is indeed zero. It's easy to see this is true at the exact center of the shell, but it's also true for any point inside the shell. In a sense, you can say that for an off-center point, the force from the closer part of the shell is stronger, but there is more of the shell that is further away, and these two effects cancel exactly. If you know how to do volume integrals over vector quantities, you can do the integral yourself and see that it does actually give you zero magnitude.

Yes, a given source of gravity acts in all directions, but at a given point, the total gravitational force can only point in one direction.

Hope that helps,

Don

[justncredible]

okay Phobos, the 1st link has no explaination, the second link i agree with and the 3rd, almost qoutes you, i think it is the symantics being used that is confusing me, i know that gravity exist in the center, i THINK you would still feel a pulling effect in the center, to use math to prove it does not exist, does not PROVE that there is no effect of gravity at the center, the one guy on the 2nd link says that since all the forces would be pulling in all directions you would not feel it and you would appear to be weightless,,, kinda makes sense, but it might be if you take the R^2 for gravity and since your in the middle then it would be double your own weight at the center,, you would have both of the mass of the radi to contend with,,,, well maybe,,,, and just who is this newton guy??? if they mean that dead guy from the 1700's,,, that explains alot,,, I mean he did get hit on the head with an apple,,, it might have caused some damage....

[Phobos]

and just who is this newton guy??? if they mean that dead guy from the 1700's,,, that explains alot,,, I mean he did get hit on the head with an apple,,, it might have caused some damage....


It could have been worse just think what could have happened if he was hit by a monitor as well (Apple didn't have those plasma screen jobs in those days).

Phobos

[SpacedOut]

Hey Phobos - Wrong thread (http://www.badastronomy.com/phpBB/viewtopic.php?topic=1519&forum=1&21) /phpBB/images/smiles/icon_biggrin.gif

[Phobos]

On 2002-06-18 15:32, SpacedOut wrote:
Hey Phobos - Wrong thread (http://www.badastronomy.com/phpBB/viewtopic.php?topic=1519&forum=1&21) /phpBB/images/smiles/icon_biggrin.gif


No, I have already been witty over there and thought this thread could do with some of the same !

Phobos /phpBB/images/smiles/icon_biggrin.gif

<font size=-1>[ This Message was edited by: Phobos on 2002-06-18 18:06 ]</font>

[roidspop]

Man, ya gotta love this place!

Jumbo, that link was super; Thanks! Doctor Dan, you put it all in exactly the nutshell that I was hoping to find. Thkaufm...right! Straight tunnels! I appreciate it all, folks.

This principle of feeling gravitational attraction only from the shells beneath one ought to apply to the cosmos. But as there is no definable center, it seems that the same summing of gravitational forces would cancel out everywhere within the universe. I think (we'll use that term for the moment, lacking anything vaguer) of the "shell" of the original fireball, visible in all directions; it appears we're inside a universal shell, hence no gravitational effects from the Cosmic Egg should be apparent. But I suppose you could imagine shells in 4-space, so there might be some sort of gravitational differential through spacetime... How do you know where "down" is in a 4-dimensional shell?

[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]

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

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? /phpBB/images/smiles/icon_lol.gif

[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? /phpBB/images/smiles/icon_smile.gif
--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]

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

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". /phpBB/images/smiles/icon_wink.gif 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.

/phpBB/images/smiles/icon_biggrin.gif


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. /phpBB/images/smiles/icon_biggrin.gif

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. /phpBB/images/smiles/icon_smile.gif

[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. /phpBB/images/smiles/icon_smile.gif


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 /phpBB/images/smiles/icon_smile.gif
--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 /phpBB/images/smiles/icon_smile.gif
--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?

[traztx]

On 2002-07-01 23:48, roidspop wrote:
Now, does somebody have a way of calculating the density gradient of this airball? How would you account for temperature?


There are ways to simplify this:
1) Assume the "strongium" shell is a uniform temperature. Let's say the rocky layer is a good insulator and the shell conducts heat well enough to make it uniform. This prevents us from having to consider convection. Now we can assume the air has uniform temperature.
2) Assume the planet is not rotating. Now we can assume no centripetal force.
4) Assume the air stays mixed. Thus we don't have to consider water precipitating out or oxygen settling deeper or anything.

Then pick a temperature and mass.

Then we can at least talk about the mass of air within points occupied by an arbitrary shell between 2 radii. This allows a chance at integrating it, but I'll need more time to figure out how (hey I'm just the dummer in this band!). The gravitational vectors at each point seem to depend on the density gradient, which seems to depend on the gravitational vectors! Perhaps we should employ an iterative numerical analysis instead of good old calculus?

Food for thought.
--Tommy

[beskeptical]

The more this goes on the less dumb I feel. I do understand physics, I do understand physics, I do..... /phpBB/images/smiles/icon_smile.gif

What I mean to say is at first I thought I wasn't grasping the logic of basic physics. Now I see that I hadn't thought of a point or two, but I wasn't totally out there.

[jaydeehess]

roidspop wrote " While we're theorizing about this problem, I recall a nifty idea for a transportation system, outlined back in the seventies I believe. The idea was that you'd bore parabolic tunnels between distant points on the earth and pump all the air out (we don't worry about little engineering details like keeping a tunnel from collapsing at a depth of hundreds of kilometers...). A levitated train in this tunnel would fall at increasing speed, pass the bottom of the arc and start upward, slowing all the way. At the other end it would come to a stop after a 45 minute trip. And the real trick was that it didn't matter how far apart the ends were...the trip would take 45 minutes. Period. NY to London? 45 minutes. Brisbane to Paris? 45 minutes. The extreme case was a vertical shaft through the core to the opposite side of the planet...drop the train through that and the trip takes...45 minutes. We REALLY need to learn how to do this! "

A hollow vacuous shell of an Earth would make this an easier engineering feat. Still it seems to me the deepest hole ever drilled into the crust is 10 miles. That doesn't come close to what would be required.

[beskeptical]

On 2002-07-07 00:16, jaydeehess wrote:
roidspop wrote " While we're theorizing about this problem, I recall a nifty idea for a transportation system, outlined back in the seventies I believe. The idea was that you'd bore parabolic tunnels between distant points on the earth and pump all the air out (we don't worry about little engineering details like keeping a tunnel from collapsing at a depth of hundreds of kilometers...). A levitated train in this tunnel would fall at increasing speed, pass the bottom of the arc and start upward, slowing all the way. At the other end it would come to a stop after a 45 minute trip. And the real trick was that it didn't matter how far apart the ends were...the trip would take 45 minutes. Period. NY to London? 45 minutes. Brisbane to Paris? 45 minutes. The extreme case was a vertical shaft through the core to the opposite side of the planet...drop the train through that and the trip takes...45 minutes. We REALLY need to learn how to do this! "

A hollow vacuous shell of an Earth would make this an easier engineering feat. Still it seems to me the deepest hole ever drilled into the crust is 10 miles. That doesn't come close to what would be required.



The heat, the pressure, the G-force of going the Earth's diameter in 45".... I can't take it, ahhhhhh!!!!!

[Donnie B.]

On 2002-07-07 04:40, beskeptical wrote:


On 2002-07-07 00:16, jaydeehess wrote:
roidspop wrote " While we're theorizing about this problem, I recall a nifty idea for a transportation system, outlined back in the seventies I believe. The idea was that you'd bore parabolic tunnels between distant points on the earth and pump all the air out (we don't worry about little engineering details like keeping a tunnel from collapsing at a depth of hundreds of kilometers...). A levitated train in this tunnel would fall at increasing speed, pass the bottom of the arc and start upward, slowing all the way. At the other end it would come to a stop after a 45 minute trip. And the real trick was that it didn't matter how far apart the ends were...the trip would take 45 minutes. Period. NY to London? 45 minutes. Brisbane to Paris? 45 minutes. The extreme case was a vertical shaft through the core to the opposite side of the planet...drop the train through that and the trip takes...45 minutes. We REALLY need to learn how to do this! "

A hollow vacuous shell of an Earth would make this an easier engineering feat. Still it seems to me the deepest hole ever drilled into the crust is 10 miles. That doesn't come close to what would be required.



The heat, the pressure, the G-force of going the Earth's diameter in 45".... I can't take it, ahhhhhh!!!!!

Heat and pressure, we assume, you'd be protected from by the vehicle (more Strongium?). As to the G-forces, for the drop through the center, they would be exactly... zero. For other trajectories, they would be somewhere between zero and one g. You might need motion sickness pills, but nothing more.

[beskeptical]

On 2002-07-07 09:11, Donnie B. wrote:
As to the G-forces, for the drop through the center, they would be exactly... zero. For other trajectories, they would be somewhere between zero and one g. You might need motion sickness pills, but nothing more.


If I go one Earth diameter in 45 minutes, granted, once I get up to speed, there will be no G-force, but what about during acceleration and deceleration? I've been on that roller coaster, you can't fool me. /phpBB/images/smiles/icon_smile.gif

[roidspop]

If you drop from pole to pole, you're in free-fall all the way (assuming maintenance did its job and pumped all the air out of the passage). If you drop from any other point, though, your capsule would be be forced to change from tootling along to the east at about 1000 mph due to the Earth's rotation to tootling along in the opposite direction. That's a 2000 mph change in 45 minutes, only about 0.03 gee.

Dropping along a chord, you'd experience some fraction of a gee all the way through, depending on the steepness of the chord. There would also be a component from rotation, but it would be minor.

On amusement park rides, you have all this extra input to your senses that amplifies the experience and gives you an intense sensation of gaining speed, changing direction and so forth. You know you're really hauling. In one of these hypothetical transportation tubes, presumably you'd have very little sensation except that of diminished weight. The man who invents UltraStrongium controls the world!

[Donnie B.]

What roidspop said... /phpBB/images/smiles/icon_smile.gif

The through-the-center route would be free-fall, except for that small sideward deflection force (which I ignored as being negligible). You may define this free-fall as a G-force, of sorts, but I'd quibble with that. (Of course, at the very end, you'd go from zero-G to one-G when the vehicle was "caught" at the far end of the tube.)

It occurs to me that there would eventually be a lot of tubes trying to pass right through the center, and maybe even intersecting there. Imaging the traffic control headaches! All those capsules -- maybe thousands a day -- whizzing through the center past each other at orbital speeds... wow!

[roidspop]

I think you've hit on something there! I can also imagine that you have to have the equivalent of a space elevator to support the construction and maintenance end of the project. "We got a magma leak at 2.5K, Charlie. Get your crew down there and patch it asap!"

Also, it's not as if your tubes are anchored in anything that is perfectly rigid...we keep hearing about these flows in the mantle; given time, they'd really do a job on your gravity tunnel. UltraStrongium had better be Hyperstrong!

<font size=-1>[ This Message was edited by: roidspop on 2002-07-09 00:05 ]</font>

[traztx]

On 2002-07-09 00:01, roidspop wrote:
Also, it's not as if your tubes are anchored in anything that is perfectly rigid...we keep hearing about these flows in the mantle; given time, they'd really do a job on your gravity tunnel. UltraStrongium had better be Hyperstrong!


Not to mention the problem of your competitor inventing a worm-hole generator and rendering your whole endeavor obsolete. /phpBB/images/smiles/icon_wink.gif

[Donnie B.]

On 2002-07-09 12:04, traztx wrote:
Not to mention the problem of your competitor inventing a worm-hole generator and rendering your whole endeavor obsolete. /phpBB/images/smiles/icon_wink.gif

Don't you mean our hole endeavor?

[jaydeehess]

It occurs to me that there would eventually be a lot of tubes trying to pass right through the center, and maybe even intersecting there. Imaging the traffic control headaches! All those capsules -- maybe thousands a day -- whizzing through the center past each other at orbital speeds... wow!

No problemo, we could hire Swiss air traffic contollers. (let me know if I crossed the line on taste with that one)

[Dunash]

Geocentrist Bouw's list of papers showing that a shell of stars going round a stationary Earth every 24 hours will produce EXACTLY the same phenomena as if the Earth was rotating on its axis every 24 hours.

http://www.geocentricity.com/papers.htm

[Kaptain K]

Dunash,

And your point is?

[beskeptical]

On 2002-07-07 23:26, roidspop wrote:
If you drop from pole to pole, you're in free-fall all the way (assuming maintenance did its job and pumped all the air out of the passage). If you drop from any other point, though, your capsule would be be forced to change from tootling along to the east at about 1000 mph due to the Earth's rotation to tootling along in the opposite direction. That's a 2000 mph change in 45 minutes, only about 0.03 gee.

Dropping along a chord, you'd experience some fraction of a gee all the way through, depending on the steepness of the chord. There would also be a component from rotation, but it would be minor.

On amusement park rides, you have all this extra input to your senses that amplifies the experience and gives you an intense sensation of gaining speed, changing direction and so forth. You know you're really hauling. In one of these hypothetical transportation tubes, presumably you'd have very little sensation except that of diminished weight. The man who invents UltraStrongium controls the world!


I understand all the hypothetical low to zero G stuff. I understand UltraStrongium. But somewhere in this hypothetical situation you have to go from your zero MPH to that free fall. If you leave the shuttle orbiting Earth for a space walk you might not experience the speed, but you went from a vehicle going the same speed as you before going on your walk.

Since I'm not one density and I have different parts, my little brain might not get up to speed as fast as my skull, or my stomach contents may decide to travel outside my body. There would still have to be acceleration and deceleration effects.

(You are welcome to give up on me at any time if I'm wearing you out here.)

[Donnie B.]

On 2002-07-11 05:37, beskeptical wrote:
I understand all the hypothetical low to zero G stuff. I understand UltraStrongium. But somewhere in this hypothetical situation you have to go from your zero MPH to that free fall. If you leave the shuttle orbiting Earth for a space walk you might not experience the speed, but you went from a vehicle going the same speed as you before going on your walk.

Since I'm not one density and I have different parts, my little brain might not get up to speed as fast as my skull, or my stomach contents may decide to travel outside my body. There would still have to be acceleration and deceleration effects.

(You are welcome to give up on me at any time if I'm wearing you out here.)

Of course there's acceleration, one G at first and diminishing all the way to the center and, inversely, all the way from there to the other end of the tunnel. But you, the vehicle, your skull, your brain, and (hopefully) your stomach contents would all be accelerating together at exactly the same rate (re: Galileo and the Tower of Pisa, apocryphal or not).

So you wouldn't feel the acceleration, as you would in, say, an elevator or rocket sled. You would certainly feel it in a different way, as you would float around the cabin during the whole trip if you're on a through-the-center route, or would feel less weight than normal on a chord route.

[traztx]

On 2002-07-11 05:37, beskeptical wrote:
Since I'm not one density and I have different parts, my little brain might not get up to speed as fast as my skull, or my stomach contents may decide to travel outside my body. There would still have to be acceleration and deceleration effects.

Jump on a trampoline and close your eyes. You feel weightless the moment your feet leave the surface. You are decelerating, then reach 0 velocity, then accelerate back towards the trampoline. You feel no difference in weight until your feet again touch the trampoline.

But other cues tell you something is going on. In a capsule with no window in a vacuum tube, those cues are gone.

[beskeptical]

I will have to meditate on this and get back later. If I close my eyes on a swing, my stomach knows it's past the peak and is on its way back down. And, acceleration going from x to y mph in z time should still be different than going from x to b mph in z time, but no need to respond until I think it through a bit more.

[roidspop]

I used to give my students a taste of freefall...I had them hop down from their chairs to the floor. Whoopee (index finger held upward and rotated). Well, still. It was freefall, even if only for a fraction of a second. And to be all scientific about it, they'd hold a spring scale with a weight suspended from it. As they began falling, the scale would clearly register zero. If you imagine every particle in your body connected to its neighbors with springs, the sensation that you call "weight" is the stretching of those springs as every part of the whole edifice attempts to fall downward but is held back by its connection to its neighbors and ultimately, to some external source of resistence. Fall, and that resistence goes away, the springs all relax and that subliminal sensation of the drag of gravity vanishes. One of the things that happens is that your suspensory system, the mesentery, rearranges your internal organs a bit...the sensation of your stomach rising may be real. The drop through the earth would be the equivalent of falling around the earth...a straight orbit instead of a curved one.

[beskeptical]

Yes, I am beginning to see gravity and acceleration are tied together differently than inertia and the force that would change inertia. OK the fuzz is clearing a bit, I'll keep meditating on this one.

[traztx]

On 2002-07-12 06:37, beskeptical wrote:
Yes, I am beginning to see gravity and acceleration are tied together differently than inertia and the force that would change inertia. OK the fuzz is clearing a bit, I'll keep meditating on this one.


Yes, the difference is that gravity is a force acting on all the inner parts equally. Any force you try to apply only acts on the part of the object you are in contact with. Inertia is still there, though. Otherwise we wouldn't have craters /phpBB/images/smiles/icon_smile.gif

[beskeptical]

On 2002-07-12 10:24, traztx wrote:

Yes, the difference is that gravity is a force acting on all the inner parts equally. Any force you try to apply only acts on the part of the object you are in contact with. Inertia is still there, though. Otherwise we wouldn't have craters /phpBB/images/smiles/icon_smile.gif


That is a very good explanation, I like it. /phpBB/images/smiles/icon_biggrin.gif