18
$\begingroup$

The Hollow Earth theory is/was a pseudoscientific idea that our world is actually on the inside of a large sphere. The "sky" points inward towards the center, where the "Sun" (a light source) is, while the "ground" points outward.

Let's assume that we have a planet the size of Earth, except that it is a hollow Earth. In reality, it's simply an Earth-sized cavity inside a larger body, notably, some sort of artificial megastructure. I had assumed that said megastructure was spherical, but clearly I should have stated it explicitly, so I'll do so now. The structure will have spherical symmetry and be as uniform as possible.

Is it possible to put a moon inside the sphere - somewhere between the central light source and the "ground" - and have it move in an "orbit" around the center? Would the moon crash into the ground, or would it be stable?

I'm almost positive that the moon can't be as big as Earth's moon, but I don't know a reasonable size. I'm fine with anything bigger than, say, Janus or Epimetheus.

Bonus question (not necessary to answer): Is the setup possible if the hollow Earth is non-spherical, i.e. ellipsoidal?

|improve this question
$\endgroup$
  • $\begingroup$ Related question, by the way. $\endgroup$ – HDE 226868♦ Oct 12 '15 at 13:02
  • $\begingroup$ Can't understand exactly where the moon in question is being proposed to be put. $\endgroup$ – Youstay Igo Oct 12 '15 at 13:10
  • $\begingroup$ @YoustayIgo It's orbiting around the center, inside the sphere. $\endgroup$ – HDE 226868♦ Oct 12 '15 at 13:11
  • $\begingroup$ An in, a hollow earth with a moon in the center? Something like Edgar Rice Burroughs presented in his At The Earth's Core series? $\endgroup$ – Youstay Igo Oct 12 '15 at 13:12
  • 2
    $\begingroup$ worth pointing out (though there are already a couple of answers to this effect). The inside of a spherical shell has 0 gravity iff the shell is of uniform density and thickness. We can't assume this is the case for this world (though it might be) $\endgroup$ – Joe Bloggs Oct 13 '15 at 11:45
18
$\begingroup$

No.

Bodies composed of known materials the size of the Earth and Moon are in "hydrostatic equilibrium",

This occurs when external forces such as gravity are balanced by a pressure gradient force.[1] For instance, the pressure-gradient force prevents gravity from collapsing Earth's atmosphere into a thin, dense shell, whereas gravity prevents the pressure gradient force from diffusing the atmosphere into space.

Hydrostatic equilibrium is the current distinguishing criterion between dwarf planets and small Solar System bodies, and has other roles in astrophysics and planetary geology. This qualification typically means that the object is symmetrically rounded into a spheroid or ellipsoid shape, where any irregular surface features are due to a relatively thin solid crust. There are 31 observationally confirmed such objects (apart from the Sun), sometimes called planemos,[2] in the Solar System, seven more[3] that are virtually certain, and a hundred or so more that are likely.

What this means in your case is that the strength of the materials is insufficient to support the mass above them. So they flow like molten plastic/liquid and fill all voids.

So in your Scenario the "Earth" outer shell would collapse under its mutual gravitation.

The strength of materials required to prevent this from happening would be quite high and I don't have the time to perform the necessary calculations for you.

|improve this answer
$\endgroup$
  • 12
    $\begingroup$ "Let's assume that we have a planet the size of Earth, except that it is a hollow Earth" - under this assumption (which was a part of the original question), of course it is possible. Let's not be distracted by trifles like non-existing materials :-) (after all, if we can discuss ringworlds...) $\endgroup$ – Radovan Garabík Oct 12 '15 at 18:48
  • 4
    $\begingroup$ I think we have to assume whomever built this hollow Earth has the material science to hold it together. This is to allow the more interesting question of "well how do you explain the Moon"? $\endgroup$ – Schwern Oct 12 '15 at 19:58
  • $\begingroup$ The theory generally held that there wasn't a shell, but a rigid substance which continued either infinitely or certainly for some considerable distance. Of course that has its own problems with gravity, but the theory rejected Newtonian models of gravity too (and Einsteinian even more so, often conflated with the people holding the theory being very often extremely anti-Semitic). $\endgroup$ – Jon Hanna Oct 13 '15 at 11:26
  • $\begingroup$ How is this any different from the idea of a dyson sphere? Those are considered plausible in many sci-fi settings, even though we certainly don't have the technology to build one today. $\endgroup$ – Ajedi32 Oct 13 '15 at 14:52
  • 1
    $\begingroup$ @RadovanGarabik, the problem is that even if you handwave away the strength of materials problems, you still have a stability problem with both designs (a solid shell Dyson Sphere or Moon inside the Earth). A body inside a spherical shell is statically stable but dynamically unstable (ball at the top of the hill). A tiny perturbation causes the inner body to crash into the shell unless active controls are used to maintain the configuration. $\endgroup$ – Jim2B Jun 6 '16 at 15:51
9
$\begingroup$

It's not possible without active intervention.

The sphere's gravity is zero on the inside. So there is no force acting on the moon from the sphere. This would allow the moon to orbit a mass at the center of the planet.

However any perturbation of the sphere would not be transmitted to the moon. Hence nothing prevents the moon, and whatever it is that is in the center for the moon to orbit, to pick up velocity relative to the sphere.

|improve this answer
$\endgroup$
  • $\begingroup$ What if the light source was fixed at the center, though, in relation to the sphere (as was sometimes used in hollow Earth theories)? Would that minimize the effects of motion of the outer sphere, because it would therefore impact the orbit of the moon? $\endgroup$ – HDE 226868♦ Oct 12 '15 at 13:46
  • $\begingroup$ +1 And things would get very complex if/when there are mountains on the planet and/or the moon. $\endgroup$ – Youstay Igo Oct 12 '15 at 13:46
  • 1
    $\begingroup$ @HDE226868 Yes, as long as the light source were sufficiently massive. $\endgroup$ – Alex S Oct 12 '15 at 13:48
  • $\begingroup$ @HDE226868 I have absolutely no idea. I would suppose it depends on the size of the hill sphere of the central body, although I am unsure about which body you should caculate that hill sphere in relation to. $\endgroup$ – Taemyr Oct 12 '15 at 13:49
  • 2
    $\begingroup$ @SJuan76 It's a shell, so there's zero gravity everywhere inside the shell. Again, see the shell theorem. $\endgroup$ – HDE 226868♦ Oct 12 '15 at 20:28
8
$\begingroup$

Certainly - the sphere's gravity at the inside will be zero, so the composition and thickness of the crust is irrelevant. The moon has to orbit the central light source, and the mass of the light source should be much bigger than the moon's, and you'd better keep vacuum in the cavity (since there is no gravity, you cannot expect the air to stick close to the crust). Other than that, it's possible.

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ +1 I was going to say it was impossible, but I'd forgotten about the mass of the light source. However, I wonder how big the both of them could get before their orbits of each other would end up slamming them into the walls (or each other). $\endgroup$ – DaaaahWhoosh Oct 12 '15 at 13:31
  • 1
    $\begingroup$ You can't have the moon orbiting the light source if there is air in the cavity; air resistance will drain energy from the system until they collide. $\endgroup$ – pjc50 Oct 12 '15 at 15:17
  • 3
    $\begingroup$ The moon would soon become unstable, any amount of instability that shifts the moon ever so slightly will increase the pull between the moon and the outside on one side and decrease it on the other. This means that soon the moon will crash into the outer shell (or the other way around). $\endgroup$ – Thijser Oct 12 '15 at 15:57
  • 2
    $\begingroup$ @Thijser Why would that be the case? The force of gravity at any point inside the sphere - not just the center - is zero. $\endgroup$ – HDE 226868♦ Oct 12 '15 at 15:59
  • 2
    $\begingroup$ @Thijser The gravitational force due to the moon is immaterial here. The force of gravity on the moon due to the sphere is the same as the force on the sphere due to the moon. $\endgroup$ – HDE 226868♦ Oct 12 '15 at 16:03
2
$\begingroup$

Not without artificial gravity or some other outside source.
Your mega structure large enough to have an earth sized pocket is going to have a lot of mass. Something that big is going to need futuristic materials and technologies to keep from collapsing.

Anything on the inside of the sphere is going to be drawn toward the center of the structure. If the sphere is off center then everything will be pulled toward one side. If it's in the center of the structure then things would just fall off the surface toward the "sky", unless...
To get gravity on the inside surface of a sphere without artificial gravity it needs to be spinning, meaning the structure needs to be spinning around the sphere, or a mechanism needs to be spinning the earth size sphere independently of the structure.

Spinning the whole structure is not a good idea, since if you have 1g at the surface of the sphere, you'll have even more the further out you go.

Angular Velocity: 0.0118 rotations/minute

Earth Radius: 3959 miles
Gravities: 1g

Radius: 5000 miles
Gravities: 1.26g

Radius: 8000 miles
Gravities: 2g

Stress on the structure would be increased exponentially the further out from the center of gravity you get.

If you just spin the sphere independently you still need super materials like ringworld scrith and you better hope it's really well balanced.

So you'll have to pick your magic: Artificial gravity or impossible building materials.

Edit:
Without artificial gravity you'd need a gravity point source in the center to orbit a moon around, since centripetal force wouldn't work to orbit inside a sphere.

Any variations in mass in the outside structure would tend to destablize the system, similar to the three body problem, requiring constant corrections to keep the moon from crashing.

With artificial gravity, getting a moon to do anything you want is easy.

|improve this answer
$\endgroup$
  • $\begingroup$ This is thorough, but doesn't address the actual question. $\endgroup$ – HDE 226868♦ Oct 12 '15 at 17:46
  • $\begingroup$ @HDE226868 You're right! I got caught up in the rest of it and forgot my original point. Fixed. $\endgroup$ – AndyD273 Oct 12 '15 at 17:59
2
$\begingroup$

You're describing a Dyson Sphere. It is theoretically possible to still have orbital bodies inside the sphere in their orbit around the sun. Assuming we live on the inside of the Dyson shell and can process the entirety of the sun's output without frying, we wouldn't have a moon since the only orbital bodies are the planets further in and the Sun. There's no center of gravity for a moon to orbit around, since the Dyson sphere's center of gravity is the sun itself. That doesn't stop the Venus or Mercury from having moons of their own.

|improve this answer
$\endgroup$
  • $\begingroup$ My concept is similar to a Dyson Sphere but is obviously not identical, being Earth-sized. As such, planets aren't a problem. $\endgroup$ – HDE 226868♦ Oct 12 '15 at 22:33
1
$\begingroup$

I will answer your bonus question: If the hollow earth is oblate or prolate spheroidal, and the two body system of moon and light source remains in the plane on which the cross section of the hollow earth is still circular, then stable orbits will exist as they would otherwise by symmetry. For a more complicated case, I expect nothing but a series of rather intense numerical simulations could determine the answer to whether there exist any periodic orbits at all.

|improve this answer
$\endgroup$
  • $\begingroup$ I think you might not have finished the second sentence: If the hollow earth is oblate or prolate spheroidal, and the two body system of moon and light source remains in the plane on which the cross section of the hollow earth is still circular . . . then? $\endgroup$ – HDE 226868♦ Oct 12 '15 at 13:54
  • $\begingroup$ @HDE226868 Thanks, getting ahead of myself. $\endgroup$ – Alex S Oct 12 '15 at 13:58
  • $\begingroup$ Great, thanks. Would there still be gravity inside the sphere, though - in other words, does the result of the shell theorem still hold? $\endgroup$ – HDE 226868♦ Oct 12 '15 at 14:00
  • $\begingroup$ the shell theorem would not hold. There would be a gravitational field attracting toward the narrowest parts of the shell. But in the plane I have described, there would still be zero gravity, which is why this works. $\endgroup$ – Alex S Oct 12 '15 at 14:01
1
$\begingroup$

Yes, but you need:

  • a very dense central light source, and this density is in contrast with what we know about sufficiently powerful energy sources, i.e. nuclear fusion. Of course you could go with an artificial light source, such as a degenerate matter sphere which reflects light beamed from the inside of the shell.

  • some next-to-uncompressible material with which to build the shell, or some mechanical means of counteracting the gravity of the inner sphere and preventing it from making the outer shell collapse (the late Paul Birch suggested a network of rails where heavy carriages would travel at orbital velocities, thus exterting an outward pressure capable of counteracting gravity. Of course, this leaves us with the problem of energizing the network itself.

  • some way of keeping the outer shell centered on the light source. The net force exerted by the shell on the central sphere would be zero, which means there's nothing to stop small perturbations to make the central sphere drift against the shell (it's the same reason Niven's Ringworld has stabilizing engines)

  • the internal volume should be kept in vacuum, or the Moon's orbit would quickly degrade.

If the perturbations are small enough, it could be feasible to stabilize the shell (and maybe the moon too) using the pressure of the light being beamed from the shell to the "Sun".

However, the central volume would be uninhabitable, as gravity would push towards the "Sun" and the whole cavity would be in vacuum. People on the inside of the shell would need to receive light through thick, airtight glass floors.

|improve this answer
$\endgroup$
-1
$\begingroup$

So I learned a lot playing with this idea, and the question lends itself rather well to a thought experiment.

If we built a perfectly uniform shell around the earth and it's moon, the net change in gravitational effects inside the shell would be zero. We could build any size shell around any size orbital system and not disrupt the existing orbit. The gravity of each mass object inside the shell continues to behave exactly as it did before.

So the real question becomes, could you develop a working orbital mechanic inside a planet sized space? And of course the answer is yes, but the objects inside the space are going to have to be significantly smaller than the space. And while each orbiting body could theoretically have an atmosphere, I suspect that orbiting bodies inside of a thing the size of the earth won't be able to sustain a breathable atmosphere, unless they were surprisingly massive.

Of course the outside of the shell gets to be an interesting place. Instead of just having the single center of mass of the shell, it has a vector composite center of gravity depending on the location of each internal orbiting body. The shell itself could support an external atmosphere, but if internal bodies had enough mass, the shell's external atmosphere could compress and develop wild weather patterns in response to internal mass shifts, especially if the orbiting bodies get very close to the shell.

And then there's probably a star to contend with in the center of the system. The shell has no correcting affect on anything inside of it, so the shell and all of its contents must independently have developed a stable orbit around that star, and the composite mass of all object needs to also have a stable orbit.

Would this develop naturally? No way. Could you build it with a lot of energy and unobtainium? Absolutely.

|improve this answer
$\endgroup$
-1
$\begingroup$

A mass at the centre and a uniform superstructure of sufficient mass such that the gravitational attraction on any body between the central mass and the superstructure by the central mass equal the attraction due to the superstructure in that direction. There's your answer. ;)

Think of gravity as a set of elastic ropes. The one rope pulling a "moon" towards the central must be balanced by ropes extending from the superstructure in every direction. Calculating these forces will be an infinite series, as the rest of the superstructure will also exert a gravitational attraction towards a moon in between. It is doable, but obviously nothing could live there, unless you build a set of biospheres along the superstructure.

Tidal forces will be a major concern, as will Coriolis forces is the superstructure is spinning. The poles might be liveable; the equator, between the moon-based tides and the structure's own spin, would be like the inside of a washing machine.

|improve this answer
$\endgroup$
  • $\begingroup$ Um... you mean it would be silvery and riddled with holes? $\endgroup$ – Burki Jul 25 '16 at 14:42
  • 1
    $\begingroup$ @Burki: No, I mean Turbo-Clean (tm) with cyclone wash (r) and 6-motion direct drive(c) :P $\endgroup$ – nzaman Jul 25 '16 at 15:33
  • $\begingroup$ Just to clarify in case my last comment wasn't obvious: the superstructure's rotation would create angular momentum outwards, resulting in liquid being forced towards the poles. The slip caused by the "moon" would create a current going the other way, but perpendicular to the other. The result would be the kind of turbulence you'd expect inside a good washing machine. $\endgroup$ – nzaman Jul 26 '16 at 14:07
  • $\begingroup$ It ws clear (to me), but i simply couldn't resist :-) $\endgroup$ – Burki Jul 27 '16 at 7:11
  • $\begingroup$ It's no fun to have to explain a joke ;) $\endgroup$ – nzaman Jul 27 '16 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.