# Thread: gravitation inside a shell: which way is down?

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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!

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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>

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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

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

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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<sub>g</sub> = 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?

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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!

7. 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".

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

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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?

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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

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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>

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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.

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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?

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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>

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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

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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.

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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

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

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

19. 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/Num...ng/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.

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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................

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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.

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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.

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

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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?

or better still

Earth's Center and Gravity

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

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>

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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

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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....

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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

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Hey Phobos - Wrong thread [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

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On 2002-06-18 15:32, SpacedOut wrote:
Hey Phobos - Wrong thread [img]/phpBB/images/smiles/icon_biggrin.gif[/img]
No, I have already been witty over there and thought this thread could do with some of the same !

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

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

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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?

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