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8

You can create a massive sphere of cold iron up to about 1.1 solar masses that could be supported by electron degeneracy pressure. The exact point of instability is likely to be caused by inverse beta decay reducing the electron number density (it could also be modified upwards slightly by extremely rapid rotation). [NB: This is lower than the ...


0

Short answer: It would get crushed. Long answer: See today's "What If" on http://what-if.xkcd.com/138/


2

There have been plenty of studies on the connection of long-duration GRB rates and star formation (e.g., Robertson & Ellis 2011, Trenti, Perna & Tacchella 2013 and Wang 2014, all arXiv links). The relation comes from observations of star formation history, $\dot{\rho}_*(z)$, and the number of gamma-ray bursts, $dN/dz$, with $z$ being redshift in both ...


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Do gamma ray bursts play a role in cosmic evolution? I don't really know the answer to that. But I imagine black hole jets in general play some kind of role. Have you ever read any articles like Black Hole Creates a Galaxy? "The astronomers think the black hole is powering star formation in the nearby galaxy by spraying its jets of high-energy ...


1

There are probably duplicates/variants of this question on this site. Expressing the problem in mass terms seems a bit odd to me though, as it's been put in energy terms every other time I have seen it asked, unless I have misunderstood your question. Here is an answer (to an identically worded question) based on the link: Predicted Mass of Quantum Vacuum ...


1

In the early days of the solar system, it is generally believed , as the comment above implies, that it was a place of constant collisions. The moon may have been part of the Earth, there is strong evidence for that, and an impact is the most probable way that it would be have been able to became a separate orbiting body. For the giant planets, Jupiter ...


7

The Wikipedia article refers to a fireball, but as Wikipedia itself explains the word fireball has many meanings and doesn't necessarily literally mean a fire as in the combustion of a material in oxygen. In this case it means a ball of very high temperature gas. The gas is heated by the impact and gets hot enough to emit light just like the gas heated in a ...


3

My intuition is that creating H gas in the lab is very hard (as opposed to H$_2$ gas). Not at all; any sufficiently hot hydrogen plasma will have a greater abundance of H than of H$_2$. To see why this is so, it is sufficient to consider the energies of the molecular bond relative to the ionization energy. The energy of the bond in a hydrogen molecule ...


2

We don't need to "observe" a star's internal structure to know if they will end as white dwarves or neutron stars. its only a question of finding the mass of their progenitor stars. I think you might be confused about the Chandrasekhar limit, which only gives you the upper mass limit of the white dwarf or the lower mass limit of the neutron star. Your ...


0

In astrophysics, the mass–luminosity relation is an equation giving the relationship between a star's mass and its luminosity. The relationship is represented by the equation: $$ \frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^a $$ where L⊙ and M⊙ are the luminosity and mass of the Sun and 1 < a < 6.[1] The value a = 3.5 is commonly used ...


2

Bad journalism, bad economics, and bad engineering. Bad journalism: Someone played telephone with kilograms and metric tons. This kind of bad scientific reporting happens all the time. It's just sad. Bad economics, part 1: The current price of platinum is rather depressed. It's much more fun to use the \$1497/troy ounce value of platinum from a year ago ...


2

The simple answer is that the 90 to 100 million tons is wrong. It's kilograms. Let's assume 300 x 300 x 600 meters, that's a volume of 54 x $10^6$ $m^3$. Note that an early estimate was 500 x 500 x 1000 meters, for a volume of 250 x $10^6$, and that is what should be used. Assume it is an LL chondrite with a density of 3200 kg/$m^3$, and the mass is 800 x ...


2

The bounce, as opposed to the supernova which is still somewhat mysterious, is caused by a drastic hardening of the equation of state (EOS - the relationship between pressure and density). Prior and during core collapse, the EOS is dominated by relativistically degenerate electrons - a relatively soft EOS. The beginning of core collapse (initiated either by ...


1

Gravity assist might be used in some cases, but generally it is incompatible with the mission required. The space shuttle was only used in low earth orbit. There is nothing in that region that can be used for gravitational assist. Missions to the moon could launch toward the moon on free-return trajectory. But they had to leave that trajectory to line ...


0

Last I heard there was no direct experimental evidence to pinpoint the onset of the accelerated expansion. Evidence for this will probably come from very precise measurements of the Sn1a luminosities, but at the moment the data isn't precise enough. However the indirect evidence is compelling. We have established that general relativity gives a good ...


0

Actually, there are papers that argue that deuterium fusion can appear in planets as small as Jupiter. http://arxiv.org/abs/1506.03793 The argument is that although Deuterium fusion is thought to happen in the order of magnitude of 10 jupiter masses, the fusion of Deuterium can be facilitated by electrons "screening" the protons of Deuterium. If a Deterium ...


1

As explained above, the dominant theory of stellar formation, and thus the formation of the Sun, is through the collapse of gas. The theory goes that a large cloud called the giant molecular cloud will start "clumping" through a combination of gravity and shock waves, and these clumps will eventually collapse into stars. A giant molecular cloud can form many ...


3

Star formation isn't completely answered, but it is well believed that a solid core is not necessary. However if the sun did form around a planetary-sized solid core we would not know the difference. Due to the very high temperature of the sun, the result is not meaningfully different from colliding with planetary bodies early on (which is plausible given ...


1

The sun definitely does not have a solid core. The temperature and pressure is way too high to maintain such a close-proximity atomic structure, especially since it is a hydrogen fusion reactor. Solids have atoms with nuclei that remain in a constant position (not like, the object doesn't move, but it's relative position is fixed), but that would mean that ...


2

The Sun did not form around a solid core. Rather, it seems to have formed from a cloud of collapsing gas that may have been further enriched by matter from a nearby supernova. Gravitational force caused the collapsing cloud to start spinning, and the spinning compressed it into a disc with a bulge in the center that became the Sun. Here is a better ...


10

No, the Sun is not thought to have formed around a solid core, and solids would not exist at the temperatures and pressures at the centre of the protosun. The Sun formed simply from the gravitational collapse of a large cloud of gas. The situation for Jupiter is different because far out in the circumstellar disc of the forming solar system, it was cool ...


11

As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium. Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by ...


3

First, note that Population III stars are expected to be massive, not tiny, with masses upwards of $10^6\,M_\odot$. The reason for this is due to the Jeans criteria, where the mass follows $$ M_J\propto T^{3/2} $$ In the early universe (~ 1 Myr), the temperature was around 10,000 K; so in a pure-hydrogen environment, no cloud with a mass less than about a ...


5

Enzo is fundamentally a grid-based finite-volume hydrodynamics code. That is, the domain is divided into cells, each is assigned various fluid properties (density, velocity, etc.), and at each timestep fluxes of those quantities across the interfaces between cells are used to update the quantities in the cells. It has a choice of particular methods for ...


5

Checking for electron degeneracy is a matter of comparing the Fermi kinetic energy with $kT$. If $E_F/kT \gg 1$, then you may assume the electrons are degenerate. The central density of the Sun is around $\rho=1.6\times 10^5$ kg/m$^3$ and the number of atomic mass units per electron is around $\mu_e =1.5$. The number density of electrons is therefore ...


3

The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal. The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average ...



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