How thick would a Dyson Sphere have to be for a person standing on the inside to experience gravity equal to standing on the Moon? Equal to standing on the Earth?
Lets assume it’s constructed of the densest material we have the technology to manipulate (no black holes :D).
It doesn’t matter how thick it is.
There’s a net zero gravitational force inside a uniform sphere.
I’ve never understood how any of the Dyson spheres in science fiction were supposed to stay up – they’d have to be built of super-strong Unobtainium in order to stay up against the inward pull of gravity themselves, let alone holding anything else. All Dyson spheres in SF seem to be the work of other, possibly vanished super-civilization. Bob Shaw’s Orbitsville, in the book of the same name and its sequel, had gravity “woven into” its structure. Everybody else pretty much ignores it.
I don’t think Dyson himself ever saw a Dyson sphere as any sort of solid structure like this – IIRC, he was saying that any high-tech interplanetary civilization would completely enclose its sun, in order to grab every last scrap of powrer it produced. Exactly what held it up (light pressure plus solar wind?) I never learned. But I don’t think it was supposed tio be something you could walk on.
In any event, in the ansence of super-science, i can’t imagine anything realistic that would hold you down – the sun would pull you toward the center, while the sphere itself would exert no force on you. If you spun it, anfd it kept its shape, you could hae centrifugal “gravity” along the equator. That was the idea behind Larry Niven’s Ringworld. But he still had to imagine super-strong scrith to make up the body without coming apart.
And even then it was dynamically unstable, as MITSFS nerds pointed out at a WorldCon. Which is why he put attitude jets on it in the sequel.
I suspect a solid Dyson Sphere might have the same kinds of problems. In other words, it’s hatrd enough to build a workable, solid Dyson sphere without imagining what would happen to keep you stuck to it on the inside.
To expand on beowulff’s answer a bit, and trying to span the range from Schoolhouse Rock to calculus:
Every object in the universe pulls on every other object. The bigger the object, the stronger the pull; but the greater the distance between the objects, the weaker the pull becomes. For something really big, like the Earth, the gravity you feel is the accumulation of all the various bits of the Earth pulling on you.
Which is where the really cool part kicks in. Assume you got a thin, hollow sphere. For an object on the outside, if you add up the pull to all the little bits of the sphere, it’s exactly the same as if there were a single, equivalent mass at the exact center. But, for an object inside the sphere, the pull from all the bits cancels out. And that’s true if you’re at the center, or anywhere inside the sphere.
Actually, I think we both got it wrong - you’ll be gravity neutral at the center, but you’ll experience increasing gravity the further you get from center. Where I got it wrong was the direction that gravity will pull you. Standing on the inside of the sphere, you’d be pulled towards the center, wouldn’t you? Even gravity generators wouldn’t help, not if they were evenly distributed about the outside of the sphere.
Hmph.
To expand on this - I’m imagining a straight hole through the center of the Earth. Standing next to the hole, you’ve got 1 g. Drop into the hole, and you’ll be pulled towards the center, but the gravitational effect of the Earth surrounding you will diminish, until you’ve got 0 g at the middle, right?
Now do the same thing, but make the Earth hollow. Standing on the outside surface, you’ve got X g., right? You’re going to have nearly X g just inside the surface, right?
Or am I getting something horribly wrong?
No.
This was one of the first examples in one of my College Physics classes. Regardless of where you are inside the sphere, the net gravitational force is zero. If you are close to the skin, you are attracted strongly by all the atoms near you, but that is balanced by the weaker attraction of the far greater number of atoms in the far side of the sphere. You can solve this with triple integration.
ETA: Gauss's law for gravity - Wikipedia
Search for the word “Hollow”.
Right.
Nope. At any point inside the sphere, there is no net gravitational force. You would be weightless anywhere inside the sphere.
That doesn’t include the gravity from the star, does it? I presume when we’re talking about Dyson Spheres, we’re talking about a hollow sphere plus a solid center.
Nope – see my post above.
The shell theorem article on Wikipedia has a mathematical proof that a uniform spherical shell does not exert gravity on anything inside it. This holds true in general relativity as well, according to Birkhoff’s theorem.
Same story here; worked through it once in a Physics class. The result is completely non-intuitive, and the math to get there is pretty gnarly, but the result is so beautifully simple that I’ve never forgotten it. Anywhere inside a hollow sphere, the gravity of the sphere pulls equally in all directions and cancels out.
The sphere will cancel itself out, so the only net gravitational force is that from the star.
I think Dyson’s original idea amounted to a bunch of free-orbiting power satellites, arranged densely enough that all the light from the star would hit some satellite or another. The sphere wouldn’t be a single object, but a cloud of them.
If you do somehow make a complete rigid sphere (using materials far beyond anything we have, of course), you don’t have to worry about “holding it up”. By the same calculation that the shell has no gravitational field inside, and Newton’s Third Law, we can see that nothing inside the sphere can exert any net gravitational force on it, either. You’ve got a situation of neutral stability, like a marble sitting on a tabletop: If it starts rolling, it’ll keep on rolling, but it won’t speed up. You would of course need some station-keeping thrusters to correct for the odd perturbation (meteor impacts, other masses passing by, etc.), but you wouldn’t need too much station keeping.
That said, at least one instance of Dyson spheres in science fiction, the buuthandi from *Schlock Mercenary, is made of a very thin, lightweight fabric, supported by light pressure. If the fabric is light enough, you could (nearly) balance the light pressure with the weight, and be able to get away with material of plausible strength (much less than scrith or the like).
What would happen if you put the Dyson sphere on a treadmill?
I maintain that you still need something to maintain the Dyson sphere in its shape – otherwise it’ll bend and billow with any influence, like a thin plastic garment bag in the wind. That’s why I suggested light pressure or solar wind (which you’ll have, in any case) in my OP. But a random sphere out in space somewhere is not going to maintain its shape in the presence of random forces (molecular impact, or micrometeorites, or even light pressure from a nearby star) unless you have something keeping it spherical.
The math would be beyond me but the end result of basically no pull in any direction towards the inner surface of the sphere makes perfect sense to me.
This is why I’m not a physicist, I guess. That’s so nonintuitive!
But what’s the gravitational attraction of a Dyson Ball? :dubious:
I don’t know, but my D17 really sucks!
Really.