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Can quark stars form under an event horizon? This is an excellent question that I have often thought about myself. Personally, I agree with Graeme Heald that quark stars can indeed form under an EH. Why do I agree? Well, quarks are fermions, and fermions must obey the Pauli Exclusion Principle. This results in quark degeneracy pressure at some high curvature ...

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There is some reason to believe that the Moon's presence has been crucial to the development of life on Earth. Many decades ago, people thought that perhaps the periodic wetting and drying of tidal ponds may have been important to the chemistry that led to the emergence of life. But there is another reason the Moon may have been crucial. The moon's early ...

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I believe that the temperature of earth would drop some amount due to the loss of friction earth normally gets from gravitational forces effected by the moon. other than some sort of cascading butterfly effect type of deal, I'm finding it difficult to see how the moon adds any value to earth. It is, however, very useful for leaving earth. I think the whole ...

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This is a complicated question as it involves plasma physics. But the concept you're looking for is magnetocentrifugal acceleration and concepts like Alfven surface etc. are involved. What we will be looking at is the centrifugal-gravitational potential $$\Phi(r,z) = - \frac{GM}{\sqrt{r^2+z^2}} - \frac{GM}{2r_0^3}r^2$$ where $r,z$ ...

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I did not read this book but I believe it will definitely suit you Extragalactic Astronomy and Cosmology: An Introduction by Peter Schneider Here is the first edition https://www.amazon.com/Extragalactic-Astronomy-Cosmology-Peter-Schneider/dp/3642069711 Here is the link for the second edition https://www.amazon.com/Extragalactic-Astronomy-Cosmology-Peter-...

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It's partially convention, but I think it's much more that Gaussian units are nicer to work with for a lot of the common problems in astrophysics. It's not that we prefer cm, g, and erg over m, kg, and J; that's just to signal that we're using Gaussian units. To me, it's that Maxwell's equations have constants with intuitive physical meaning that we use all ...

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The cosmological constant or dark energy is present everywhere and at all scales according to theory. (Experimentally we don't know that it is, though it certainly doesn't clump very much if at all.) But it doesn't, even in theory, cause a time variation of orbital parameters. The easiest way to understand its effect is to think of it as matter with a ...

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They all need tunneling; you just need to work out the required temperatures to penetrate the Coulomb barrier. The "most likely" reaction would be getting two protons together. If you argue that they need to get within $r\sim 10^{-15}$ m of each other to feel the strong nuclear potential then the Coulomb barrier is of height:  E_C = \frac{e^2}{4\...

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Both $^{14}$N and $^{15}$N are produced as part of the CNO cycle during the hydrogen-burning main sequence phase of stars more massive than the Sun. However $^{15}$N reacts rapidly with protons to (re)form $^{12}$C and an alpha particle, whereas the much slower $^{14}$N$(p,\gamma){}^{15}$O reaction allows $^{14}$N abundances to dominate when the CNO cycle ...

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I offer several suggestions. 1.For z=1100, Ωr << 1, so it can be ignored. 2.For z>=9, ΩΛ << Ωm/(1+z)^3, so it can be ignored in this range. 3.I have used a spreadsheet to numerically calculate this integral with respect to a (not z) for a=0.1 to a=1. I used da = 0.01, da = 0.005, and da = 0.0025, and the differences were very small. When you ...

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As the hydrogen fuel is exhausted, the core heat that resists gravity diminishes, so gravity takes over and compresses the star. If the star was big enough to start with, then the extra pressure from the compression is enough to enable helium fusion and the heat generated by that is enough to blow parts of the outer envelope off the star and expand it into ...

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First of all, the Schwarzschild metric is the most general spherically symmetric vacuum solution to the Einstein field equations. A Schwarzschild black hole is a black hole that, having neither electric charge nor angular momentum, is described by the Schwarzschild metric. The Kerr metric has a few "problems" and cannot be used to describe ...

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Presumably you want to know the mass flow that is equivalent to the energy of the light emitted by the sun. I won't do the calculation for you (it is easy), but here is the approach. Look up the amount of light energy emitted by the sun, in joules per second. Divide that value by the square of the speed of light in meters per second. The result is the ...

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As far as I understand, rocky planets can only grow up to a certain size. This has to do with planetary formation period. A planet cannot grow indefinitely. It can grow only as long as there are particles around the star that can contribute to its increase of mass. During the formation period, dust particles collide and coalesce to form chunks, which further ...

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No signal from any spacetime point inside the black hole can reach any spacetime point outside it. You can't evade that fact by relaying signals between devices. That's just another form of signaling. There are situations in which speeds of the form $c-v$ can be meaningful (there are some examples in Einstein's original 1905 paper), but this isn't one of ...

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One could argue that the rotation curves of galaxies are intra-galactic evidence, since it requires the somewhat continuous presence of dark matter across the galaxy. If you want smaller scales, the smallest systems where dark matter is firmly established are dwarf galaxies; these are dark matter dominated. Finally, there are tidal streams. Some of those ...

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I found the answer! So the continuous spectrum is formed is usually formed from a heated body (i.e a heated filament or a star) and when the continuous spectrum passes through a cooler gas, the electrons in the gas absorb the energy of a photon with a specific wavelength (which is why is shows black lines against a continuous spectrum). The electrons then ...

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