133

The Sun is actually a THIRD generation star. What I mean by this is that there are chemical elements in the Sun that were made inside another star, but that star itself can only have made those elements because it had material in it that must also have been made inside previous, second generation stars. Eventually we get back to the first generation stars, ...


42

Models for the future behaviour of the Sun do vary, mainly as a result of uncertainty of mass loss during the red giant (H shell burning) and asymptotic red giant (H+He shell burning) phases. A highly cited paper by Schroeder & Smith 2008 claims that the Sun will reach its maximum size of about $256 R_{\odot}$ (1.18 au) at the very tip of the red giant ...


28

Most of the galaxy's gas is not incorporated into stars and remains as gas and dust. This is not really my area of expertise, but papers such as Evans et al. 2008 and Matthews et al. 2018 seem to suggest that in the Giant Molecular Clouds where most stars in the Milky Way Galaxy form, the star formation efficiency is about 3-6%. So the vast majority of the ...


23

The heaviest elements known in nature are forged deep within stars. No, the heaviest elements are made on Earth in scientific laboratories, or in the extreme gravity of a neutron star's crust. These elements are made possible by the high densities/temperature/pressures within the stars. Many of the larger elements can be made in supernovae and neutron ...


20

I think you've answered your own question. if 1st and 2nd stars generation burned hydrogen to helium and more heavier elements, then should it be like 90% of all universe hydrogen already converted to helium and something else? If yes, then there should not be enough hydrogen to make the Sun. Clearly the Sun does have enough Hydrogen to form and the ...


19

Does a star fuse helium to beryllium on the main sequence? Stars don't fuse helium to beryllium except as a very, very short intermediate step toward carbon. Helium-helium fusion to form beryllium is endothermic: It consumes energy. To make matters worse, the beryllium-8 that results has an extremely short half-life, less than $10^{-16}$ seconds. Helium ...


19

I don't think there is an accepted definition of a "black dwarf" - it is not a term used in the scientific literature. A popular definition that appears to circulate on the internet is that it is a white dwarf that has cooled down to the extent that it no longer emits any radiation in the visible part of the spectrum. But this is an unworkable theoretical ...


19

TL; DR Somewhere between now and a few hundred billion years time. (For a co-moving volume) Now read on. If stellar remnants are included, then the answer is very far in the future indeed, if and when the constituents of baryons begin to decay. So let's assume that "stars" means those things that are undergoing nuclear fusion reactions to power their ...


17

Arcturus is a RGB star, probably fairly similar how the sun will look when it becomes a red giant. Arcturus is slightly more massive than the sun ($m_{\rm Arc}=1.08 m_{\odot}$), but the main difference is the lower metallicity of $[Fe/H]\approx-0.5$. This low metallicity reduces the opacity in the stellar radiative zone (which fills a significant portion of ...


17

A relevant paper here is Laughlin, Bodenheimer & Adams (1997) "The End of the Main Sequence". From the abstract: We find that for masses $M_\ast < 0.25\ M_\odot$ stars remain fully convective for a significant fraction of the duration of their evolution. The maintenance of full convection precludes the development of large composition ...


16

The total entropy actually increases, as the molecular cloud shrinks under gravity. It may seem that as the molecules are getting closer, they are more ordered, which means less entropy. That is however only one part of the process. The second (important) part is: when the molecules are closer, they also have higher kinetic energy (since they descended into ...


16

Although various astronomers have speculated that the Sun was a star (some were imprisoned or even burnt alive for such heresy), this was not known definitively until 1838 when Friedrich Bessel used parallax to calculated the distance to 61 Cygni. In the late 19th century, Lord Kelvin provided rather small (less than 100 million years) for the ages of the ...


14

Starting from a protostar, one would hope to be able to predict everything about its future development if we knew its initial mass, chemical composition and angular momentum. Mass is fundamental because it determines how much fuel the star will have and the pressure at its core. Composition is key because among other things it determines the opacity of the ...


13

Naked eye nova are fairly common, several per year. Here's one. Naked eye supernova are far rarer. SN1987a in the large Magellanic cloud was naked eye visible (vid). From this list, it appears the supernova in 1987 was the most recent naked eye supernova. There was a naked eye gamma ray burst in 2008, but I don't think anyone actually got outside in time to ...


13

I think what you need is here on the Wikipedia. In section "Radiation and cooling," it says "The rate of cooling has been estimated ... After initially taking approximately 1.5 billion years to cool to a surface temperature of 7140 K, cooling approximately 500 more K ... takes around 0.3 billion years, but the next two steps of around 500 K ... take first 0....


13

Stellar remnants are completely different from planets. The Earth was never a star and fusion has never occurred in the Earth's core at any time in its history. When a small to medium sized star dies, and the outer layers are lost, what remains is a white dwarf. It is still much more massive than the Earth, and is very hot. It is crushed by its own gravity, ...


12

The scenario you describe may occur. On the other hand it may actually be that neutronisation in a white dwarf is the trigger for a thermonuclear type Ia supernova. You may be misunderstanding the Pauli Exclusion Principle (PEP).The PEP states that no two fermions can occupy the same quantum state, not that they cannot occupy the same space or be compressed ...


11

H$\delta$ absorption is formed when hydrogen in the level $n=2$ is excited to $n=6$. To get strong H$\delta$ absorption lines you need large amounts of hydrogen in the first excited state $n=2$ and a radiation field that contains large numbers of photons with an energy equal to the difference between the $n=6$ and $n=2$ states. These requirements are ...


11

The solar neutrino luminosity is about 2.3% of its electromagnetic luminosity (i.e. light). So the extra mass lost in the form of neutrino energy is 2.3% of your original calculation. The average mass loss in the form of a wind and coronal mass ejections is about $4\times 10^{16}$ kg/year, but varies with the solar cycle (and from cycle to cycle) (Mishra et ...


10

(This is somewhat simplified but I hope it gets the idea across.) The reactions stop in the core because it runs out of fuel. During the main sequence, the star is supported by the fusion of hydrogen into helium. Eventually, the hydrogen runs out at the centre, so hydrogen fusion is no longer possible there. Why doesn't it start fusing helium into carbon ...


10

Supernova create huge spikes in neutrino emissions. Since neutrinos pass through a stellar mass mostly unimpeded, they're visible up to 3 hours before the shockwave even starts to affect the star's surface. Since neutrinos travel at the speed of light, they will always keep their 3 hour head start. Thus, unless you have a neutrino detector buried a few ...


10

The thing you are not taking into account is that our predictions for the future evolution of the Sun are not based on our understanding and observations of the Sun alone. They're based on our understanding and observations of all the stars we've ever looked at. The thing is, stars of the same mass and composition must necessarily evolve in the same way. So ...


10

It's all to do with the relationships between mass, spectral-type and luminosity and the initial mass function of stars. I think your explanation of points 1 and 2 are completely correct. O and B stars are rarely born and short-lived; so even though they have enormous luminosities relatively few make it into a list of stars ordered by apparent brightness. A ...


10

What defines the main sequence? Main sequence stars are characterized by hydrogen fusion in their cores, either through the proton-proton chain (for lower-mass stars) or the CNO cycle (for stars more than about 1.5 times the Sun's mass). Outside the core, no significant fusion takes place; the outer layers are involved in radiative or convective energy ...


10

$$\frac{dP}{dr} = - \rho g,$$ is the equation of hydrostatic equilibrium, where $\rho$ and $g$ are the local density and gravity, $P$ is pressure and $r$ is the radial coordinate. This can be rewritten as $$\frac{d\rho}{dr} \frac{dP}{d\rho} = -\rho g.$$ Since $\rho$ and $g$ are positive numbers, the pressure gradient is negative. For all types of matter $P = ...


9

Indeed conservation of angular momentum dictates that in a single star like the the Sun, rotation should be much slower when it becomes a red giant. This is because at the present time the Sun does not rotate at vastly different rates with depth, thus when it expands, the moment of inertia increases drastically and convection in the outer envelope will ...


9

A succinct summary of supernova types is given in the following image based on Heger et al. (2003): Image courtesy of Wikipedia user Fulvio 314 under the Creative Commons Attribution-Share Alike 3.0 Unported license. The graph is based on the graph in Fig. 1 of the linked paper. The pair instability realm is upwards of ~100 solar masses, though it is ...


9

There is no general consensus on this. Different evolutionary models give different results. The factors (in addition to the initial mass of the star) that effect the final black hole mass would be the rotation rate of the progenitor, its composition (or metallicity) and whether it was in a binary system or not and whether that binary system was able to ...


9

I think the most important part of any answer is that, as Rob Jeffries said, "black dwarfs" aren't really a thing in the astronomical literature, and I suspect that's the reason that you get different answers about how long it takes to become one. Different people come up with different thresholds for becoming one. I would argue that 3000 K is too hot to ...


8

Nuclear fusion rates in the core of a star have very non-linear and strong dependences on temperature, pressure and density (for temperature it's like T^40 for some processes - http://www.astro.soton.ac.uk/~pac/PH112/notes/notes/node117.html). So, as the star gets bigger, denser and hotter at the centre, the rates of fusion rise much faster and the star ...


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