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I'm going to assume you have some familiarity with linear algebra, as the math becomes much less tedious than trying to do three-dimensional trigonometry with the $x$, $y$, $z$-coordinates directly. You are looking for a function describing the line of the horizon. Since it is a circle and thus a one-dimensional object, I'm going to call it $\vec{h}(t)$, ...

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This is to add up a little more detail to the discussion. Water as it exists in the form of $H_3O^+$ and $OH^-$ ions are bound together by "Van der waals forces" which is "the sum of the attractive or repulsive forces between molecules other than those due to covalent bonds, or the electrostatic interaction of ions with one another, with neutral molecules, ...

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A Community Wiki answer to make some other people's comments permanent and tie some loose ends up. To add to Mark Mitchison's Answer, the reason that the prevailing shape is the one that minimises surface energy as he states is that, in the case of water, the liquid's total energy is an (almost) constant offset (the potential and kinetic energy of the ...

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Another way to look at it is the following. The main force on the molecules will come from other water molecules and be due to cohesion. The system will try to minimize it's energy and bond the molecules together as much as possible. This means minimizing the surface which results in a sphere.

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The droplet wants to minimise its surface energy. This energy is proportional to its surface area. So the equilibrium shape is that which minimises the surface area for fixed volume (the bulk density is fixed by the temperature and pressure).

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Unfortunately the answer to "is the structure gravitationally stable?" is "most definitely not." Anything planet-sized pretty much has to be close to a sphere, unless it's spinning very rapidly, because the gravitational forces increase with the body's size, whereas the electromagnetic forces holding atoms together don't, so the material's strength will ...

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As mentioned by Qmechanic, the answer to your questions is no. However, assuming space-time is oriented, we have the following: For any pseudo-Riemannian metric $g$, there exists a normalized time-like one-form $h^0$ and a Riemannian metric $g^R$ so that $$g = 2h^0\otimes h^0 - g^R$$ This yields a locally Euclidean topology compatible with the manifold ...

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Clarifying Wolphram jonny's drawing / equations a little bit...: You can write the following equations: $$\tan\theta_1 = \frac{d}{2D}\\ \tan\theta_2 = \frac{d}{2(L+D)}$$ For small angles ($d<<L$), the observed angle $\theta_1 = 2\theta_2$ when $L=D$. So if I correctly interpret your statement "the internal end of the tube that is farthest away is ...

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If you know the distance from your eye to the closest end of the tube you can, otherwise you can't. The reason if that you need that distance to calculate the angle of aperture and thus the distance to the far end. Because the ratio of perceived size depends on this distance. Hope the drawing helps (My drawings are becoming worse with time, I apologize) ...

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It could exist in the sense I believe you are thinking. It will not exist based on what we know about physics, or made of the particles we already know. But it is possible to imagine a parallel 2D universe that for some reason collides with ours (some people define universe as the whole, so in that case it would be part of our universe, but just initially ...

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