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11

Lithium and other light elements (e.g. beryllium) can be formed indirectly from supernovae via cosmic ray spallation, a process where protons and neutrons are ejected when a cosmic ray collides with another atom. The nucleons can then form new elements. Nakamura & Shigeyama (2004) were able to calculate yields for 6Li, 7Li, and isotopes of Beryllium and ...


8

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 ...


7

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}\ ...


4

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 ...


4

(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 ...


3

Short answer: They don't. Contact binaries are a possible evolutionary phase of close binary systems, where both stars fill and/or overflow their Roche limit, i.e. 'kiss' each other at the L$_1$ point. As a consequence, they lose matter to a common envelope. This exerts a drag onto the stars and leads to further shrinking of the binary, which eventually ...


3

I agree with Walter, they don't. However, in addition to the common envelope drag and mass exchange a very important feature of their evolution is the loss of angular momentum through a magnetised wind. The loss of angular momentum from the orbit also leads to orbital shrinkage and closer contact, until presumabaly at some point they truly merge. This ...


3

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 ...


2

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 ...


2

If you have $L$ and you have $T$, then nothing more complicated than Stefan's law is required. If $T$ is the effective temperature of the star then this gives an exact answer. $$ R = \left(\frac{L}{4\pi \sigma_B T^4}\right)^{1/2}$$, where $\sigma_B = 5.67\times 10^{-8}$ in SI units. If on the other hand you are trying to solve the structure from first ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


1

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 ...


1

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 ...


1

Even if your ship traveled at (almost) the speed of light, you would have to follow the trajectory of a ray of light traveling from your point of origin to the star; not from the star to your point of origin. As described, you keep following the trajectories of many different rays of light (emitted at different times) backwards, putting yourself on a curved ...


1

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 ...


1

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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