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Consider your typical Dyson Sphere scenario. You are standing on the inner side of the sphere. You look up and see the shining sun far away.

My question (or thought experiment rather) is: Completely negating the gravitational forces, mass, etc. of the sun, if you could be placed anywhere within the radius of the Sphere, is there a point within the Sphere where you would float freely?
The center seems the most obvious answer, but is it the only one? Keep in mind: standing at any point on the ground you have a "small" amount of mass beneath your feet, and FAR much more mass "above" you. Does this mass pull you more than the ground beneath your feet?

Qmechanic
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Garrettfromhp
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    So basically you want to know what is the gravitational force acting on you if you are inside an hollow shell (no star inside)? Because in that case Gauss' Theorem tells you that the force you would experience is $0$ everywhere inside the shell. – valerio Jul 14 '16 at 20:29
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    See Newton's shell theorem. Possible duplicates: http://physics.stackexchange.com/q/150238/2451 and links therein. – Qmechanic Jul 14 '16 at 20:34
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    As Qmechanic says, it's Newtons Shell Theorem, with the caveat that, as you probably know, it is physically impossible to build a complete shell, so you would need to space the discrete elements of the sphere very carefully to allow you to float at any location within the sphere. –  Jul 14 '16 at 20:46
  • @valerio92 Doesn't using the term "hollow shell" imply zero thickness? Which, even if the DS is far larger than its theoretical thickness, it's not completely negligible right?

    That being said, I think zero force everywhere is still correct :)

     – Garrettfromhp Jul 14 '16 at 21:15
  • @Garrettfromhp No, I just meant a shell with nothing inside. The thickness can be arbitrary. – valerio Jul 14 '16 at 21:18
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    @Garrettfromhp The shell's thickness and density must be uniform, though. – valerio Jul 14 '16 at 21:22

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Exactly. You have a little mass below you, and far more above you. But the mass above you is also in far more distance.

The result is the - for the first spot, surprising - fact, that these two exactly balance eachother.

Thus, if somehow you don't feel the gravity of the Sun, then there is a weightlessness in the whole sphere.

Actually, you can imagine that so, that consider looking into any direction and a small $d\omega$ solid angle. In that direction, the wall of the sphere is in $r_1$ distance. In the exactly opposite direction (with the same solid angle), you see the wall in $r_2$ distance.

It doesn't matter, what is $r_1$ and $r_2$, because the wall size you see on the depends quadratically on them, thus the wall mass depends also qudratically. But their gravity to you is proportional to the quadratical recipe of the distances.

The result is that these opposite areas off the wall have exactly the same force on you, but into the opposite direction.

On the language of the formulas:

  • You see $d\omega$ solid angle in both direction
  • The distance of the wall is $r_1$ and $r_2$
  • Let the "surface density" (i.e. the mass of a single $m^2$ wall) $\rho$
  • Then you see $d\omega\cdot\rho\cdot r_1^2$ and $d\omega\rho\cdot r_2^2$ masses in the both directions,
  • Their gravititational acceleration ($G\frac{m}{r^2}$) are thus $G\frac{d\omega\cdot\rho\cdot r_1^2}{r_1^2} = Gd\omega\rho$. It doesn't depend on $r$!
peterh
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    Apologies if you know this already, or have taken it into account in your answer but: Dyson replied, "A solid shell or ring surrounding a star is mechanically impossible. The form of 'biosphere' which I envisaged consists of a loose collection or swarm of objects traveling on independent orbits around the star http://en.wikipedia.org/wiki/Dyson_sphere –  Jul 14 '16 at 20:50
  • @count_to_10 Thanks. I didn't know this (until now), but I knew that the Dyson spheres are instabile. On this reason, Dyson's concept was attacked from many directions, which he toke probably seriously. On my opinion, although the concept wouldn't work on a naive way, the idea is still revolutionary. Compare this to the Tsiolkovsky rocket equation. Tsiolkovsky lived in the Russia of the 1910's. Most of the population wasn't even literate there and couldn't even imagine flying. – peterh Jul 14 '16 at 20:57
  • @peterh This answer was my thinking too! It's an interesting question I thought would be fun to discuss. So, now, if we DO consider the sun's gav effects, one would be pulled into the sun at all points within the sphere? – Garrettfromhp Jul 14 '16 at 21:12
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    I only remembered it from a previous question, complete suprise to me too. It's ruined at least 100 SF stories I've read. –  Jul 14 '16 at 21:14
  • @Garrettfromhp Yes. You would feel only the gravity of the Sun. And, there are two further problems: 1) A planet is stabilized on its orbit around the Sun by the centrifugal force. But in the case of a sphere, it wouldn't work on the poles. 2) There is nothing which would stabilize the sphere. I.e. any minor change in its location/form would lead to yet bigger changes due to the gravity of the Sun.| On these reasons, the sphere would require some like an adaptive gravitational stabilization, which is an unthinkable technology on our current level. – peterh Jul 14 '16 at 21:17
  • @count_to_10 I think it is not impossible if there is a technology to stabilize the gravitation in the sphere. Well, quantum gravity is currently a little bit crazy, but radioactivity was also a crazy thing a century earlier. And anyways, thinking on these is surely not a bad thing, maybe the next Einstein is a child now and reads scifi. It would be important if (s)he would choose an academical career, instead of going into the business world to optimize stock shares (for much higher wage). – peterh Jul 14 '16 at 21:22
  • @peterh Ah, you mean you cannot rotate the DS? What is the problem with the poles, if you should try to rotate it?

    Or are you referring to the orbiting planet? If that's the case, since Fg within the sphere is zero, shouldn't the orbit be normal and stable?

    And any nonuniform change to its surface would destabilize it, of course, because youre changing it's presumably-perfect form.

     – Garrettfromhp Jul 14 '16 at 21:23
  • @Garrettfromhp I mentioned the poles of the Dyson sphere. You can't compensate the gravity of the Sun by simply rotating the sphere (as the planets do), because it wouldn't work on the poles. If you somehow find a way to compensate it, you will feel inside the sphere only the gravity of the Sun (i.e. everything will fall into the Sun). – peterh Jul 14 '16 at 21:27
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    Ha! You discovered my plan, the first part worked ( I read lots of science fiction as a kid) but the Einstein gene ain't in me. Dyson is a genius (apart from his climate change stuff) but not an engineer, so anythings possible. –  Jul 14 '16 at 21:29
  • @peterh Ah! But a rotating DS leads to another interesting question!

    If the sphere is rotating at, say, a speed needed to maintain 1G (at the equator), at what latitude, on the surface, would you need to be at to float? Too near the equator, and you're pulled to the surface, too near the poles and you head towards a fiery death.

    No doubt a difficult, I dont want to even think about. XD

     – Garrettfromhp Jul 14 '16 at 21:33
  • @Garrettfromhp It won't work. The Dyson sphere doesn't work on its naive sense, but it could work if we have some currently unknown technology for that. For example, if the dark energy is a field, maybe its excitations are antigravitating elementary particles. If we can somehow trap them... and, the sphere is an extreme big thing, but if we have enough robots, they can build it for us. And robots can be produced by other robots. – peterh Jul 14 '16 at 21:53
  • @Garrettfromhp Btw, you don't need 1g to balance the gravity of the Sun, the Earth's acceleration on the Sun's gravitational field is only $0.004 \frac{m}{s^2}$ (it requires a half year to revert its $30 \frac{km}{s}$ into the opposite direction). But, on the poles will be the centrifugal force exactly zero. If a material would exist which can hold it, it would be much bigger wonder as the DS itself. Probably even crystallic neutronium weren't enough for that. – peterh Jul 15 '16 at 06:28