# Hollow earth physics **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [September 23, 2011, 8:53pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385 "2011-09-23T20:53:08Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![sailorman](https://avatars.discourse-cdn.com/v4/letter/s/e99b99/32.png) [@sailorman](https://boards.straightdope.com/u/sailorman)\ **Post date:** [September 23, 2011, 8:53pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/1 "2011-09-23T20:53:08Z")
In today’s [Hollow Earth column](http://www.straightdope.com/columns/read/3015/how-much-would-i-have-to-hollow-out-the-earth-to-make-the-days-longer), 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? ---
**Author:** ![Canadjun](https://avatars.discourse-cdn.com/v4/letter/c/76d3ee/32.png) [@Canadjun](https://boards.straightdope.com/u/Canadjun)\ **Post date:** [September 23, 2011, 9:03pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/2 "2011-09-23T21:03:40Z")
> [@sailorman](#): > > In today’s [Hollow Earth column](http://www.straightdope.com/columns/read/3015/how-much-would-i-have-to-hollow-out-the-earth-to-make-the-days-longer), 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. ---
**Author:** ![Telemark](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/telemark/32/372_2.png) [@Telemark](https://boards.straightdope.com/u/Telemark)\ **Post date:** [September 23, 2011, 9:07pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/3 "2011-09-23T21:07:37Z")
> [@sailorman](#): > > 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. [Shell theorem - Wikipedia](http://en.wikipedia.org/wiki/Shell_theorem) ---
**Author:** ![Una\_Persson](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/una_persson/32/346_2.png) [@Una\_Persson](https://boards.straightdope.com/u/Una_Persson)\ **Post date:** [September 23, 2011, 9:59pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/4 "2011-09-23T21:59:16Z")
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. ---
**Author:** ![sailorman](https://avatars.discourse-cdn.com/v4/letter/s/e99b99/32.png) [@sailorman](https://boards.straightdope.com/u/sailorman)\ **Post date:** [September 23, 2011, 10:34pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/5 "2011-09-23T22:34:31Z")
> [@Telemark](#): > > The math is correct. [Shell theorem - Wikipedia](http://en.wikipedia.org/wiki/Shell_theorem) 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. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [September 23, 2011, 11:32pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/6 "2011-09-23T23:32:01Z")
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. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [September 24, 2011, 1:50am UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/7 "2011-09-24T01:50:27Z")
> [@sailorman](#): > > 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. > [@](#): > > 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. ---
**Author:** ![Powers](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@Powers](https://boards.straightdope.com/u/Powers)\ **Post date:** [September 24, 2011, 1:58am UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/8 "2011-09-24T01:58:51Z")
So here’s the part I don’t get: > [@](#): > > 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. Powers &8^] ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [September 24, 2011, 2:48am UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/9 "2011-09-24T02:48:16Z")
Look at Slug’s drawing. The full quote is > [@](#): > > 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. ---
**Author:** ![Rickymouse](https://avatars.discourse-cdn.com/v4/letter/r/34f0e0/32.png) [@Rickymouse](https://boards.straightdope.com/u/Rickymouse)\ **Post date:** [September 24, 2011, 2:54am UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/10 "2011-09-24T02:54:29Z")
But the earth’s not hollow ---
**Author:** ![spenczar](https://avatars.discourse-cdn.com/v4/letter/s/858c86/32.png) [@spenczar](https://boards.straightdope.com/u/spenczar)\ **Post date:** [September 24, 2011, 3:18am UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/11 "2011-09-24T03:18:00Z")
> [@Rickymouse](#): > > But the earth’s not hollow Yes it is. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [September 24, 2011, 8:59am UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/12 "2011-09-24T08:59:33Z")
> [@sailorman](#): > > 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/r[sup]2[/sup] 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. ---
**Author:** ![Jake](https://avatars.discourse-cdn.com/v4/letter/j/f475e1/32.png) [@Jake](https://boards.straightdope.com/u/Jake)\ **Post date:** [September 24, 2011, 3:30pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/13 "2011-09-24T15:30:02Z")
> [@spenczar](#): > > Yes it is. Of course it is. That’s where the Hobbits live…😉 ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [September 24, 2011, 5:54pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/14 "2011-09-24T17:54:02Z")
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. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [September 24, 2011, 9:55pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/15 "2011-09-24T21:55:57Z")
> [@Rickymouse](#): > > 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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [September 24, 2011, 11:56pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/16 "2011-09-24T23:56:15Z")
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. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [September 24, 2011, 11:59pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/17 "2011-09-24T23:59:59Z")
> [@Chronos](#): > > 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 [File:EarthGravityPREM.jpg - Wikipedia](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). ---
**Author:** ![sailorman](https://avatars.discourse-cdn.com/v4/letter/s/e99b99/32.png) [@sailorman](https://boards.straightdope.com/u/sailorman)\ **Post date:** [September 26, 2011, 1:10pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/18 "2011-09-26T13:10:50Z")
> [@naita](#): > > 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/r[sup]2[/sup] > 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/(2R[sup]2[/sup]) – 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. ---
**Author:** ![Quercus](https://avatars.discourse-cdn.com/v4/letter/q/7ab992/32.png) [@Quercus](https://boards.straightdope.com/u/Quercus)\ **Post date:** [September 26, 2011, 1:12pm UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/19 "2011-09-26T13:12:38Z")
> [@sailorman](#): > > 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. ---
**Author:** ![AargleZymurgy](https://avatars.discourse-cdn.com/v4/letter/a/ec9cab/32.png) [@AargleZymurgy](https://boards.straightdope.com/u/AargleZymurgy)\ **Post date:** [September 30, 2011, 3:59am UTC](https://boards.straightdope.com/t/hollow-earth-physics/597385/20 "2011-09-30T03:59:08Z")
So which “hollow earth” article is correct? > **[How much would I have to hollow out the earth to make the days longer?](https://www.straightdope.com/21344110/how-much-would-i-have-to-hollow-out-the-earth-to-make-the-days-longer)** [http://www.straightdope.com/columns/read/154/what-if-you-fell-into-a-tube-through-the-earth](http://www.straightdope.com/columns/read/154/what-if-you-fell-into-a-tube-through-the-earth) hmm? [Next page](https://boards.straightdope.com/t/hollow-earth-physics/597385.md?page=2)