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Density of dark matter in the Sun's orbit (Sun rotates with speed υ ≈ 220 km/s at a distance r = $2.57\times 10^{20}$ m from center of Milky Way ): ρ☼=$8.737\times 10^{−22}$ $kg/m^3$ = $78.5\ \mu J /m^3$ or 522000 protons/m$^3$. To learn more, you can find in mini-novel "Flippon" at the Flippin Theory website: ...


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There has been a number of search for possibly engineering projects in space. The one I remember the most is a paper running a search for Dyson spheres from astronomical data, some of the results are listed here : http://home.fnal.gov/~carrigan/infrared_astronomy/Fermilab_search.htm


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Don't forget that the inert core is still very hot, and doesn't necessarily have to start cooling down, because it's underneath the hydrogen-burning shell, which produces heat. It does start contracting later, but this is probably because the fluid can no longer configure itself to support the envelope without producing a bit more pressure, either by ...


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My guess, and I hope you get a better answer, is that gravity is the only force involved, and that does not have complicated laws, unless it's a many body problem, which I don't think you are referring to here, as you want to treat cosmological objects as distinct, and far away from each other. Moreover, does it even make sense to talk about a speed ...


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The knee is believed to be due to one of the following reasons: 1) the reduced efficiency of the galactic magnetic field to confine the cosmic ray particles with energies above the knee within galaxy, 2) the knee corresponds to the maximum energy that protons can have under diffusive shock acceleration in supernova remnants, 3) the contribution from a ...


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The math in that article is based in Cartesian space. Note specifically figure 4, where a portion of a Cartesian plane is pinched in at one side to show the supposed warping due to gravity. Using the shown transformation, she concludes that space is compressed near a black hole rather than stretched. The diagrams after that along with the process ...


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The origin of this equation is reasonably well explained in Abramowicz (1991). If you take a relativistically expanding enevelope and only consider Thomson scattering, then as the electron scattering cross-section in the co-moving frame $\sigma_T$ is independent of frequency, then the mean free path of a photon in the co-moving frame is independent of the ...


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The core of the Sun is radiative. That means that energy is transported outwards primarily by photons and is stable to convection. This means, to first-order, that the centre of the Sun is not mixed up by convection. As the hydrogen in the core is burned, it forms helium, which has a larger atomic mass. The helium sinks towards the middle and the core fills ...


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The image below represents the Sun's density gradient, which shows how the density changes with the radius. The ground we stand on should have a density between 2 to 3 $g/cm^{3}$. That should put you just above the water point on the vertical axis. The corresponding radius is then about 0.45 of the solar radius. Note that the vertical axis is in a ...


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As @KyleKanos point out, "the answer is a google search away", but it's not quite as simple as he suggests. The mean density doesn't answer your question (the mean density turns out to be about 1.4 times the density of water by the way). An ill-defined idea in your question is the "surface" of the sun. Where is the "surface" of the sun given that none ...


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In about four billion years the sun will have used up most of the hydrogen in its core and fuse helium into heavier elements. The fusion of helium releases more energy than the fusion of hydrogen. This will cause the sun to expand to, at least, the orbit of Venus and likely the Earth. Both planets will, of course, be vaporized.There was a similar question ...


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In a dilute gas, the photon density should be the same as inside an evacuated black box of the same temperature (independent of the gas density). Edit: In other words, the massive particles in the Sun have a temperature based on their kinetic energies. The photons must be distributed as black-body radiation at the same temperature. The density of photons in ...


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As I understand it, a tide is basically a wave with a wave-length = 1/2 the circumference of the earth. The peak to trough is 1/4 the circumference, see pretty picture: That's why lakes don't have measurable tides. The east to west distance isn't sufficient on lakes. Only oceans have significant tides. The Pacific is already large enough to have a ...


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By "consume" we mean "convert into helium." That $6\times10^{11}\ \mathrm{kg}$ of hydrogen is part of the Sun (specifically it is found in the core of the Sun), and it is converted into $6\times10^{11}\ \mathrm{kg}$ of helium. The Sun doesn't need to suck up material from space. Note that this amount of material is miniscule compared to the $2\times10^{30}\ ...


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It's a bit of a puzzling question. I'll try to work it out, but one of the tricky parts is that atoms absorb and emit photons all the time, higher temperatures emit higher wavelengths. Photons created in the sun (per second) can be estimated, but those are fusion gamma rays. The sun burns about $564$ million tons of hydrogen per second. (Source), and 1 ...


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The number of atoms in the sun is on the order of $10^{57}$, see here. The number of photons emitted per second is on the order of $10^{44}$, see here. The difference is on the order of $10^{13}$. So if photons emitted for $10^{13}$ seconds, 315,00 years, the number of photons would begin to overtake the number of atoms in the sun. The sun is much older ...


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Molecular clouds are the most common places where stars are born, so it's best to discuss them. Richard Larson (who has studied molecular clouds in depth) discusses two processes in an overview of these structures in The Evolution of Molecular Clouds: Random collisions between particles in the interstellar medium (ISM) grow over time until eventually ...


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As a distant observer we can watch the shadow of the black holes forming in front of the background stars. According to a nice little paper by Daisuke Nitta, Takeshi Chiba, and Naoshi Sugiyama ("Shadows of Colliding Black Holes, 2011") the answer is yes. To a distant observer in a finite span of time two black holes form a shadow that is indistinguishable ...


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As I see it, and, correct me if I'm wrong, but there is a way to do it. The problem with using pure Uranium or any other readily fissionable element is that, as one element decays, that releases 2 neutrons which can speed up the decay of nearby elements. If you had a ball of U235 or U238, say, the size of a planet or even a small moon, you'd have a ...


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In principle yes, though it would be a highly contrived situation and not one likely to arise naturally. A star works because hydrogen to helium fusion is energetically favourable. But the process has a huge activation energy so you need an environment as hot and dense as a star's core to provide that activation energy. Likewise, for any sufficiently heavy ...


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Answer from astrophysicist. Point 1. The problem with eternal shining is that star loses energy with photons (and some material) and it'll need income of energy from somewhere. There are brown dwarfs, black dwarfs, black holes etc., which are just remaining of stars and don't 'shine' (or, in case of white dwarfs, fade out to the point we cannot detect ...


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(By stars I'm assuming you're implying stars like the Sun, which are a majority of the stars we see. @Dirk Bruere's answer about Black Dwarves is correct. ) No, I don't think they can. The primary process that 'fuels' stars is nuclear fusion. In the process of nuclear fusion, lighter elements fuse together, releasing a tremendous amount of energy (because ...


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The word thing you are looking for is "Black Dwarf" stars. Which are White Dwarf stars which have cooled to match the temperature of the cosmic background. Since this is likely to take more than the current age of the universe, there aren't any. These will exist forever, unless hypothetical proton decay finishes them off or a hypothetical Big Rip due to Dark ...


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If you crush hydrogen to very high densities, the Pauli exclusion principle will prevent electrons in the material from occupying the same quantum states. The electrons will obey Fermi-Dirac statistics and they will fully occupy states up to the Fermi energy - which increases as the density gets higher. Presumably in metallic hydrogen what happens is that ...


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Rickman (2014) reviews the relationship between the Oort cloud and long period comets, but does touch on the origins of the Oort cloud. The review of Jewitt (2010) deals with the Oort cloud, Kuiper belt and comets (basically, icy bodies in the solar system), highlighting what we don't know. The Annual Reviews articles by Mumma & Charnley (2011) and by ...


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The lithium test has an ambiguity because it also depends on the age of the objects. Li is depleted once the cores of fully convective objects reach temperatures of about $3\times 10^{6}$ K. The time at which this occurs depends on the contraction timescale of newly formed objects. This timescale is longer for lower mass objects. Thus we can say that ...



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