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?

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    $\begingroup$ 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. $\endgroup$
    – valerio
    Jul 14 '16 at 20:29
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    $\begingroup$ See Newton's shell theorem. Possible duplicates: physics.stackexchange.com/q/150238/2451 and links therein. $\endgroup$
    – Qmechanic
    Jul 14 '16 at 20:34
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    $\begingroup$ 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. $\endgroup$
    – user108787
    Jul 14 '16 at 20:46
  • $\begingroup$ @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 :) $\endgroup$ Jul 14 '16 at 21:15
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    $\begingroup$ @Garrettfromhp The shell's thickness and density must be uniform, though. $\endgroup$
    – valerio
    Jul 14 '16 at 21:22

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 $r$!
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    $\begingroup$ 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 en.wikipedia.org/wiki/Dyson_sphere $\endgroup$
    – user108787
    Jul 14 '16 at 20:50
  • $\begingroup$ @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. $\endgroup$
    – peterh
    Jul 14 '16 at 20:57
  • $\begingroup$ @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? $\endgroup$ Jul 14 '16 at 21:12
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    $\begingroup$ I only remembered it from a previous question, complete suprise to me too. It's ruined at least 100 SF stories I've read. $\endgroup$
    – user108787
    Jul 14 '16 at 21:14
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    $\begingroup$ 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. $\endgroup$
    – user108787
    Jul 14 '16 at 21:29

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