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17

It's because the value of the gravitational field at the center of a star is not the relevant quantity to describe gravitational collapse. The following argument is Newtonian. Let's assume for simplicity that the star is a sphere with uniform density $\rho$. Consider a small portion of the mass $ m$ of the star that's not at its center but rather at a ...


13

The analogy is facile. Helium fuses at a temperature ($10^8\ \text{K}$) roughly ten times higher than hydrogen ($10^7\ \text{K}$), so a better analogy would be alcohol and thermite. That higher temperature is achieved only by massive gravitational contraction after hydrogen fusion [EDIT: in the core] is exhausted. EDIT: To expand, different mass stars ...


12

Well, you're right that a particle sitting at the centre of a star (or generally the centre of any spherical distribution of matter) feels no net gravitational force. So, in the absence of other forces, it will simply continue to sit at the centre. But every other particle in the spherical distribution will feel a gravitational force pulling it toward the ...


11

I'll chip in here because I'm a research student and I work with a stellar evolution code (the Cambridge STARS code) more-or-less daily. Regarding some of the comments to the question, stellar evolution is actually quite fast, depending what code you use. Certainly, it isn't like hydrodynamics or N-body simulations like those used in galaxy ...


7

Stars Indeed, the most readily apparent observables for stars are (1) their apparent luminosities, and (2) their spectra (or even just colors if you can only do photometry). The age has to be inferred, and this is where modelling comes into play. The Vogt-Russell "theorem" is the assumption that the initial mass and chemical composition of a star uniquely ...


7

I'll take a swing at this, but bear in mind that you probably won't get definitive answers because you're asking about two active and difficult areas of research (pop III star formation and re-ionization). I'll answer the particular questions, but I'm hoping you get a feel for the fact that we don't have clear-cut answers yet. During what range of years ...


6

Carroll and Ostlie, and Shu are both excellent introductory texts which have good discussions of star formation. The former is a little more quantitative, the latter qualitative. Also the online notes of Mark Krumholz are fantastic if you have some background in physics. The wikipedia page is also not bad for concepts. Star Formation The most basic ...


5

The condition for creation of a black hole is: $$ \text{gravitational potential} \le -\frac{ c^2 }{ 2 } $$ I won't go into the details of how to calculate the potential. But for the center of a star, suffice it to say that it's slightly more complicated than $-GM/r$. You can see that this makes no reference to the gravitational field itself. It comes ...


5

The answer depends upon the mass of the star. For stars of less than 2 solar masses, electron degeneracy pressure stops the collapse. For more massive stars, helium fusion begins which stops collapse, without a degenerate state being reached.


5

Red giants and asymptotic giants have some close similarities, and one actually evolves into the other. Both have an extended envelope of relatively cool, non-burning material (mostly $\rm{H}$, $\rm{He}$). They also each have a core of dense, non-burning material; in the case of the red giant this is mostly $\rm{He}$, while for the asymptotic giant it's ...


5

Assuming that by "shining" you simply mean "emitting light", the answer is that it starts slowly and gradually. Normally, when we think of a star shining, it is a hydrogen-burning star on the Main Sequence, where any star will spend the majority of its life time. But stars have significant light emission well before hydrogen fusion sets in and they settle ...


4

You could look at tools like EZ-web or interpolation formulae like those from Hurley, Pols and Tout 2000 to infer how much time a given star (say O-type) spend in a given state compared to the time spend in the Main Sequence. For example an initialy $10M_\odot$ star would spend around 25 Myrs on the main sequence and only 3 Myrs being a red giant as you can ...


4

I had hopes that someone who knew this subject well would answer as it's been about 20 years since I had the relevant course, but I guess I'll give it a try. What follows may be out of date in some ways, as I am going on my memory of a course I took in 1993 and on the basis of the text we used, Schwarzchild's 1958 Structure and Evolution of the Stars. I ...


3

Summary: There is no explosion at the birth of a star. It is a gradual process. The star first heats from the potential energy used by matter collapse, and starts radiating like any hot object. When the core temperature reaches some 10 millions debrees K, the fusion reaction starts. The energy liberated stops the collapse, and takes over the heating of the ...


3

To amplify a bit bit on Jerry's answer. Because of its small surface area, and large thermal mass (typically about a half the mass of the sun) the cooling time of white dwarves is billions of years. As he says they do cool, however the universe isn't old enough to have created condensed red dwarves. The stars currently called red dwarfs, are main sequence ...


3

White dwarves used to be the interior of a star, which was the hottest part of the star. They shine white because they are still very hot from this past part of their history. As they age, they will cool, and as they cool, they will lose temperature, and their blackbody profile will shift to redder and redder colors, and eventually into the infared and ...


2

It's a really complicated relationship that depends on the metallicity of the star. There is a paper that does show this though: See New grids of stellar models from 0.8 to 120 solar masses at Z = 0.020 and Z = 0.001 Here are the Geneva Grids: http://obswww.unige.ch/~mowlavi/evol/stev_database.html An extensive and homogenous database of stellar ...


2

Last week I wrote an answer to: Please clarify how entropy increases when matter gravitationally coalesces I illustrate how a collapsing body will increase its entropy. This also corresponds to an increased temperature. The imploded white dwarf has its internal energy compressed into a very small volume. Hence they initially shine at a very high ...


2

Well, we don't claim to know the status of the star at the present time. If the star is close enough (within our galaxy, or local neighbourhood), we measure the distance in light years, or parsecs. If they're further away than that, it's easier to quantify their distance in terms of a redshift. When we say that a star has some properties (size, temperature ...


2

Your question is somewhat poorly worded. You say, "... the star has size 2 times the mass of sun". I can't understand this phrase for sure. Still, they are completely different phases of very different stars. Wiki links on Neutron Star and Red Giant are much good for this question on Stellar evolution. Once the main sequence stage is complete, a star whose ...


2

There are two questions here. The second, What causes this if its temperature is expected to fall? How gas can expand if the temperature falls? is, IMO, answered by dmckee's answer. (You should probably also read it just for some background on the next bit.) The first question, What causes the dimensions of a star increase when its hydrogen fuel ...


2

Is it just after they have finished core H burning and the core contracts creating high temperatures which result in core He burning...? It is after the core finishes H burning, but He burning is not required. Hydrogen shell burning is sufficient to make it a red giant. Helium burning would make it a Horizontal Branch Star. See good explanation here: ...


2

Since every particle attracts all other particles, there is a net force directed towards the center of the star (or any object), for any particle not at the center. Therefore, the particles will move towards the center (collapse), unless some opposing force prevents it. In the case of a star, the kinetic energy of the particles creates the opposing force, ...


1

$$\frac{dP} {dr} = -\frac{GM\rho}{r^2} $$Where $P$, is electron degeneracy pressure, $ G $ is a constant, $M$ is the mass enclosed in a shell of radius $r$ with density $\rho$. In short, electron degeneracy pressure ( left hand side ) generated by electron degeneracy is opposite and equal to gravity ( Right hand side ). Where electron degeneracy is a ...


1

I think this question is more easy to conceptualize in stars than in planets, so this answer is going to reflect that. Stars stay the size they are due to hydrostatic equilibrium: the force due to gravity is matched by the pressure gradient, $$ \frac{dP}{dr} = -\frac{Gm\rho}{r^2} $$ If we consider a parcel of matter somewhere in your hypothetical star at a ...


1

You could look at tools like EZ-web or interpolation formulae like those from Hurley, Pols and Tout 2000 to do so based on the initial mass and metallicity of your star. Here is a picture of the evolution of an initial 10 solar masses star and solar metallicity computed using Hurley's formulae (accessible as a fortran library, see this page for a ...


1

Indeed, what we infer about stars from the light we see at the Earth is "old news". However for almost all practical purposes in stellar astrophysics this doesn't matter. The phases of a star's life last millions if not billions of years and most of the individual stars that are studied are within say 30 thousand light years of the Earth. The example you ...


1

The problem with answering your question is that red giant stars don't have a well defined edge. It's a bit like Earth's atmosphere. The density gets gradually lower as you move away from the centre but it's hard to draw a line and say "this the edge". So when you say the Sun engulfs the Earth, it depends on how deep into the Sun the Earth gets and how long ...


1

Temperature equates to the speed the nuclei are travelling at. Since Helium nuclei need to collide with greater energy to fuse this can only occur at a higher temperature. As a gas gets hotter it expands and so becomes less dense, reducing the amount of energy generated by fusion. With a mixture of Hydrogen and Helium the energy generated by Hydrogen fusion ...



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