# Tag Info

61

The symmetry of the Sun has got very little to do with any symmetry in its formation. The Sun has had plenty of time to reach an equilibrium between its self gravity and its internal pressure gradient. Any departure from symmetry would imply a difference in pressure in regions at a similar radius but different polar or azimuthal angles. The resultant ...

41

Actually, it doesn't have the same mass, it has significantly less mass than its precursor star. Something like 90% of the star is blown off in the supernova event (Type II) that causes the black holes. The Schwarzschild radius is the radius at which, if an object's mass where compressed to a sphere of that size, the escape velocity at the surface would be ...

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When you watch a pop-sci TV show, you need to take everything you see with a very healthy grain of salt. This is particularly the case if the show's host isn't a scientist, but even when a scientist is the host, you need to be suspicious. Stellar black holes do not turn into monsters that reach out and pluck objects from the heavens. From far away, a black ...

21

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 ...

16

We think that most neutron stars are produced in the cores of massive stars and result from the collapse of a core that is already at a mass of $\sim 1.1-1.2 M_{\odot}$ and so as a result there is a minimum observed mass for neutron stars of about $1.2M_{\odot}$. See for example Ozel et al. (2012; http://adsabs.harvard.edu/abs/2012ApJ...757...55O ). Update - ...

15

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 ...

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

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 formation/...

12

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 ...

11

What you are looking for is called the stellar mass function by astronomers. It is the distribution of masses for stars. There is a nice review of the definitions, measurements, and basic theory in Galactic Stellar and Substellar Initial Mass Function, Chabrier 2003, PASP 115 763. It discusses both the initial mass function (IMF) and the present-day mass ...

11

It actually goes the other way around: when a star collapses to form a black hole, its planets (if it has any) will become unbound and fly away to infinity. Simple reason: when the star explodes to form a compact object (neutron star or black hole), it releases most of its mass in the form of a SuperNova explosion, so that the central object around which ...

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 ...

9

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 ...

9

The question is dealt with in some detail in this article by John Baez. Although the article assumes only a basic understanding of physics it's probably a bit too much for the non-physicist so I'll summarise. As a gas cloud collapses the particles within it are confined to a smaller volume of space so the entropy associated with their position (call this $... 9 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 "... 8 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 ... 8 Observed neutron stars range from$1.0 \pm 0.1 M_{\odot}$to$2.7 \pm 0.2 M_{\odot}$according to table 1 of The Nuclear Equation of State and Neutron Star Masses, which lists dozens of examples. Keep in mind that the mass of the neutron star is typically substantially smaller than the mass of its progenitor star; late in the stellar life cycle a lot of ... 8 I think the colloquial term for that type of plot is "spaghetti diagram" because you have a bunch of lines running across it. It's really the mass fraction as a function of interior mass. From our stellar structure equations, we have that $$\frac{dm}{dr}=4\pi r^2\rho,$$ which is derived from the mass-continuity equation, so you can relate the radius,$r$(... 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 ... 7 Star are fighting against gravitational forces by pressure gradients due to fusion in the core (and the shells outwards). Once fusion stops, there is no pressure gradient and gravity wins the "battle." The classic picture of a massive star at the end of its life is (and obviously not to scale), But each star star started off with just hydrogen in the ... 7 Measuring the star formation history of the Universe is a key component in understandng the evolution of galaxies. It is closely related to the other uestion, recently asked by the same person: Are stars getting more metal-rich, less massive and shorter-lived with cosmic time? Although this question pertains to the Universe as a whole, an understanding ... 6 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$\rm{...

6

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 ...

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 ...

6

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.

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First, a star does not become a red giant when helium fusion begins, instead it becomes a red giant earlier when an inert degenerate core of helium forms and a shell of hydrogen begins fusion. When shell hydrogen fusion begins, the star expands to be a red giant. The core is degenerate (sustained from collapse by electron degeneracy pressure) and ...

6

I like to explain this using a figure from a talk by Marco Limongi some years ago. Based on a given set of models, the $x$-axis shows the initial mass of the models and the $y$-axis the final mass. The different coloured layers show the composition of the star at the moment of collapse. The mass ejected in the supernova is the difference between the curve ...

6

I) In this answer we will only discuss the equilibrium shape. Recall that when we discussed the shape of Earth in this Phys.SE post, the gravitational quadrupole moment was important. Unlike the Earth, from a surface perspective, it is a very good approximation to assume that all the mass of the Sun sits in the center, cf. below graph. Moreover, the ...

5

If you measure the large-distance strength of the gravitational acceleration $g\approx \frac{GM}{r^2}$ of a star / black hole with the assumption that your distance $r$ is much further out than the various mass parts, shock wave, and ejected material; then $g\approx \frac{GM}{r^2}$ is (within a percent or so) the same before and after the supernova. This is ...

5

Short answer: gravitational potential energy is converted into heat. Let's look at the Sun as an example. Its mass is $M_\odot = 2.0\times10^{30}\ \mathrm{kg}$ and its radius is $R_\odot = 7.0\times10^8\ \mathrm{m}$. If its density were uniform, its gravitational binding energy would be  U_{\odot,\,\text{uniform}} = -\frac{3GM_\odot^2}{5R_\odot} = -2.3\...

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