# Tag Info

9

You can look at databases of binary stars to tell you what the range of orbital periods/separations of stars currently are. Unfortunately that isn't going to answer your question because many short-period binary systems have evolved to be that way - e.g. the short-period cataclysmic variable stars or the "contact" binaries known as W UMa systems, where the ...

8

There is a database of visual binaries - that's a good place to start. It includes the following plot of period and eccentricity: The bottom left corner represents a system with a period of $10^{-1.6}\approx 0.025$ years or just over 9 days. Now from Kepler's Laws, we have that the square of the period scales as the cube of the average distance. If the ...

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 ...

-2

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 ...

5

The image is that of the entire sky above the telescope taken with some kind of fish-eye lens. It is quite common at observatories to use such arrangements to monitor for cloud cover. However you seem to have found a particularly poor example - possibly using a CCD imager with very few pixels. It is also possible that there is some light cirrus over the ...

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 ...

5

Yes, there are theoretical models of stellar evolution that tell us what to expect. Broadly, we expect that stars with an initial mass less than 8 solar masses ($8M_{\odot}$) will end their lives as white dwarfs. So I think there is a misconception in your question - the progenitors of white dwarfs and neutron stars are usually a lot more massive than what ...

0

You can't see the light coming towards you. But you can see it as soon as it reaches your position. That means, that if you're in space and there is a star which is 10 times far than the speed of light in seconds. And if that star emits the light "now", then you won't be able to know that whether the light is emitted "now" or not. Instead, after 10 seconds, ...

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

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 ...

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