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Old 09-23-2011, 04:53 PM
sailorman sailorman is offline
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Hollow earth physics

In today's Hollow Earth column, Cecil claims that someone suspended inside a hollow spherical shell of an earth would experience zero net gravity regardless of position. This seems very unintuitive to me, given that presumably I would experience something like normal gravity if standing on the outside of the shell, and moving just a short distance so that I'm just inside the shell shouldn't make much difference in the net forces on my body.

In fact the math presented does not make much sense to me since it only addresses the mass of the sphere close to a line defined by my body and the center of the sphere, and ignores the effects of off-axis mass.

But I'm not a physicist. What do the real scientists think?
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Old 09-23-2011, 05:03 PM
Canadjun Canadjun is offline
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Originally Posted by sailorman View Post
In today's Hollow Earth column, Cecil claims that someone suspended inside a hollow spherical shell of an earth would experience zero net gravity regardless of position. This seems very unintuitive to me, given that presumably I would experience something like normal gravity if standing on the outside of the shell, and moving just a short distance so that I'm just inside the shell shouldn't make much difference in the net forces on my body.

In fact the math presented does not make much sense to me since it only addresses the mass of the sphere close to a line defined by my body and the center of the sphere, and ignores the effects of off-axis mass.

But I'm not a physicist. What do the real scientists think?
I'll let someone else provide the mathematics to show it works out evenly, but keep in mind that if you are on the outside of a hollow sphere then the whole sphere is pulling you more or less down (either directly down or down at an angle, but no part pulls up). On the inside you are pulled in both directions, and it turns out the pulls balance.
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Old 09-23-2011, 05:07 PM
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In fact the math presented does not make much sense to me since it only addresses the mass of the sphere close to a line defined by my body and the center of the sphere, and ignores the effects of off-axis mass.
The math is correct. http://en.wikipedia.org/wiki/Shell_theorem
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Old 09-23-2011, 05:59 PM
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What's not quite emphasized in the column is that differences in density, and the fact that the Earth is not a sphere, will cause minor internal gravity effects. Possibly not enough to keep you from bouncing away with a small hop.
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Old 09-23-2011, 06:34 PM
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Ok, I will take Wikipedia as authoritative. Suppose however that I am a Lilliputian point mass resting on the exterior of this enormous, thin sphere. The Wikipedia math seems to indicate that without loss of generality we can treat the shell as having zero thickness. The result seemingly implies that I simultaneously have my normal weight (since I'm outside the sphere) and zero weight (since I'm zero distance from the inside). Is there an intuitive way to resolve this apparent paradox?

Math question: Is the Wikipedia math valid for the case r=R corresponding to a point mass on the surface of the shell? Some terms appear to be undefined (e.g. zero divided by zero) in this case.
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Old 09-23-2011, 07:32 PM
Andy L Andy L is online now
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A helpful way to think about it is to imagine oneself anywhere inside the hollow sphere. If you hold your thumb at arms length in any direction, your thumb is covering the mass that would pull you in that direction. The amount of mass covered up by your thumb is proportional to the square of the distance to the shell, and inversely proportional to the square of the distance, so no matter what direction your thumb is pointing you feel the same force. Since that applies to all directions, there can't be a preferred direction to be pulled towards, as long as you're inside the shell.
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Old 09-23-2011, 09:50 PM
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Originally Posted by sailorman View Post
Ok, I will take Wikipedia as authoritative. Suppose however that I am a Lilliputian point mass resting on the exterior of this enormous, thin sphere. The Wikipedia math seems to indicate that without loss of generality we can treat the shell as having zero thickness. The result seemingly implies that I simultaneously have my normal weight (since I'm outside the sphere) and zero weight (since I'm zero distance from the inside). Is there an intuitive way to resolve this apparent paradox?
If the shell has the same mass as the Earth, and has zero (or very small) thickness, it would have to have infinite (or very large) density. So there's still a lot of mass at your feet. And that mass is very close to you, so it makes a difference which side of you that mass is on.

Quote:
Math question: Is the Wikipedia math valid for the case r=R corresponding to a point mass on the surface of the shell? Some terms appear to be undefined (e.g. zero divided by zero) in this case.
You'd need to take the limit as r approaches R in that case, and you'd get different results if r -> R from above or from below.
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Old 09-23-2011, 09:58 PM
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So here's the part I don't get:

Quote:
for any two masses on opposite sides of you, the smaller but closer mass A and the larger but more distant mass B pull on you with precisely equal force.
Surely there's a qualification missing here? A skydiver closer to an airplane than to the Earth doesn't float.


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Old 09-23-2011, 10:48 PM
ZenBeam ZenBeam is offline
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Look at Slug's drawing.

The full quote is
Quote:
In fact, if we examine the illustration that the gifted Slug Signorino has been kind enough to provide, and assume hollow earth is a spherical shell of uniform thickness and density, we see (via equations suppressed here but viewable by the curious at the Straight Dope website) that for any two masses on opposite sides of you, the smaller but closer mass A and the larger but more distant mass B pull on you with precisely equal force.
The part you're missing is that mass A and mass B both subtend the same solid angle, so their masses aren't arbitrary. If Little Ed is twice as far from mass B as from mass A, mass B is four times mass A.
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Old 09-23-2011, 10:54 PM
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But the earth's not hollow
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Old 09-23-2011, 11:18 PM
spenczar spenczar is offline
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But the earth's not hollow
Yes it is.
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Old 09-24-2011, 04:59 AM
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Originally Posted by sailorman View Post
The result seemingly implies that I simultaneously have my normal weight (since I'm outside the sphere) and zero weight (since I'm zero distance from the inside). Is there an intuitive way to resolve this apparent paradox?
Where is that implied? Outside the sphere gravitation is related to the square of your distance from the center. Inside the sphere it's 0. Intersecting the sphere I suspect equals being outside the sphere since your point mass only tangentially intersects the sphere.

So for various distances r from the center of a sphere with radius R we have:

r >= R, gravity proportional to 1/r2
r < R, gravity 0

This covers all r's with no paradox.

It's possible r = R is a special case different from r > R, but there's no paradox as gravity 0 only applies to r < R and not r = R.
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Old 09-24-2011, 11:30 AM
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Yes it is.
Of course it is. That's where the Hobbits live...
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Old 09-24-2011, 01:54 PM
John W. Kennedy John W. Kennedy is offline
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One might add that electrostatic attraction follows the same inverse-square law that Newtonian gravity does, and it is well established that there is no electrostatic attraction inside a hollow sphere. This is basic high-school stuff, guys.
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Old 09-24-2011, 05:55 PM
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Originally Posted by Rickymouse View Post
But the earth's not hollow
Actually the fact that a hollow earth would have no interior gravity provides evidence that the Earth is not hollow. If the Earth was a hollow shell, say 1 mile thick, gravity would be noticeably lower in tunnels/caves/wells. A non-hollow earth, with increasing density towards the center, would have (slightly) higher gravity at the bottom of wells/tunnels - and I'm pretty sure this effect is measurable and has been measured.
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Old 09-24-2011, 07:56 PM
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If the Earth were of uniform density, then the gravitational field strength would decrease linearly as you approached the center: Halfway down, it'd be 4.9 m/s^2 instead of 9.8 m/s^s, and so on.

As it happens, the Earth is not of uniform density: It gets denser as you go deeper. In fact, it gets denser in just such a way that the gravitational field is close to constant with depth, all the way down to the boundary of the core. The core is, more or less, uniform density, so at that point it does start decreasing linearly.
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Old 09-24-2011, 07:59 PM
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If the Earth were of uniform density, then the gravitational field strength would decrease linearly as you approached the center: Halfway down, it'd be 4.9 m/s^2 instead of 9.8 m/s^s, and so on.

As it happens, the Earth is not of uniform density: It gets denser as you go deeper. In fact, it gets denser in just such a way that the gravitational field is close to constant with depth, all the way down to the boundary of the core. The core is, more or less, uniform density, so at that point it does start decreasing linearly.
This chart http://en.wikipedia.org/wiki/File:EarthGravityPREM.jpg shows a small increase with depth in the upper mantle, but yeah it's pretty close to constant, and definitely different than what you'd get with a hollow shell (a much more rapid decrease towards zero) or with a uniform Earth (the linear decrease you describe).
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Old 09-26-2011, 09:10 AM
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Where is that implied? Outside the sphere gravitation is related to the square of your distance from the center. Inside the sphere it's 0. Intersecting the sphere I suspect equals being outside the sphere since your point mass only tangentially intersects the sphere.

So for various distances r from the center of a sphere with radius R we have:

r >= R, gravity proportional to 1/r2
r < R, gravity 0

This covers all r's with no paradox.

It's possible r = R is a special case different from r > R, but there's no paradox as gravity 0 only applies to r < R and not r = R.
Actually, redoing the math for r=R seems to give an answer of GmM/(2R2) -- the average of the "inside" and "outside" limits at that point.

Perhaps paradox is not the right word, but I'm surprised to learn that my weight would change discontinuously as I moved an arbitrarily small distance. Of course the discontinuity seems to derive from the assuming a shell of zero thickness but positive mass, and from assuming that I am a point mass. Neither is possible, of course. Nevertheless, the result is surprising, at least to me.
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Old 09-26-2011, 09:12 AM
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Originally Posted by sailorman View Post
Ok, I will take Wikipedia as authoritative. Suppose however that I am a Lilliputian point mass resting on the exterior of this enormous, thin sphere. The Wikipedia math seems to indicate that without loss of generality we can treat the shell as having zero thickness. The result seemingly implies that I simultaneously have my normal weight (since I'm outside the sphere) and zero weight (since I'm zero distance from the inside). Is there an intuitive way to resolve this apparent paradox?

Math question: Is the Wikipedia math valid for the case r=R corresponding to a point mass on the surface of the shell? Some terms appear to be undefined (e.g. zero divided by zero) in this case.
To answer: Yes, there is (depending on your personal intuition of course). And No it's not valid.

Bottom line: you can't both treat the shell as having no thickness and at the same time say that you're within the shell. If you do that, you're naturally going to end up with a paradox. Is that intuitive enough for you?

And of course in the real world shells can't be infinitely thin, so there's always a smooth transition from being outside of the shell to being inside the shell.
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Old 09-29-2011, 11:59 PM
AargleZymurgy AargleZymurgy is offline
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STILL confused

So which "hollow earth" article is correct?

http://www.straightdope.com/columns/...he-days-longer

http://www.straightdope.com/columns/...ough-the-earth

hmm?
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Old 09-30-2011, 09:33 AM
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I'll let someone else provide the mathematics to show it works out evenly, but keep in mind that if you are on the outside of a hollow sphere then the whole sphere is pulling you more or less down (either directly down or down at an angle, but no part pulls up). On the inside you are pulled in both directions, and it turns out the pulls balance.
This is the best answer to the OP - a correct and intuitive explanation that he was looking for.
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Old 09-30-2011, 12:39 PM
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Both.

I'm guessing the confusion is that in the Tube Through the Earth case, you're only weightless in the center, while in the Hollow Earth case, you're weightless everywhere inside. The difference is that in the first case, almost all of the Earth is still present, pulling you towards its center. In the second, all of the Earth's interior has been removed and placed on the surface.
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Old 09-30-2011, 01:41 PM
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Agreed.
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Old 10-01-2011, 09:07 AM
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The conditions of the math in question are that the mass distribution is radially symmetrical. That is, the density at a given point can be determined entirely from your distance from the center of mass. This is obviously not true in the case of a hole bored through the planet, but will be true (at least to a close approximation) when the planet has been fully hollowed out.
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Old 10-01-2011, 12:41 PM
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The conditions of the math in question are that the mass distribution is radially symmetrical. That is, the density at a given point can be determined entirely from your distance from the center of mass. This is obviously not true in the case of a hole bored through the planet, but will be true (at least to a close approximation) when the planet has been fully hollowed out.
It will also be true *to a close approximation* with the hole through the planet. The hole is narrow and the missing mass doesn't make much difference to the gravity field.

Last edited by Kurt N; 10-01-2011 at 12:42 PM.
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Old 10-01-2011, 04:24 PM
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Both.

I'm guessing the confusion is that in the Tube Through the Earth case, you're only weightless in the center, while in the Hollow Earth case, you're weightless everywhere inside. The difference is that in the first case, almost all of the Earth is still present, pulling you towards its center. In the second, all of the Earth's interior has been removed and placed on the surface.
Yes.
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Old 10-01-2011, 06:34 PM
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It will also be true *to a close approximation* with the hole through the planet. The hole is narrow and the missing mass doesn't make much difference to the gravity field.
Well, yes--the point is, you won't be weightless everywhere inside the hole. Gravity will slowly decrease as you get closer to the center of the planet, as the mass outside the narrow hole will continue to exert gravitational force on you.
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Old 10-01-2011, 06:49 PM
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^ Right. I'm just saying that radial symmetry isn't the point of difference. The key difference is the amount of mass that's closer to the Earth's center than you are. For the hollow Earth scenario that amount is zero. For the tunnel case it's nonzero.
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