# Tag Info

83

The bottleneck in Solar fusion is getting two hydrogen nuclei, i.e. two protons, to fuse together. Protons collide all the time in the Sun's core, but there is no bound state of two protons because there aren't any neutrons to hold them together. Protons can only fuse if one of them undergoes beta plus decay to become a neutron at the moment of the ...

75

This is an answer that I made, as suggested by John Rennie, by cutting and pasting his answer and dmckee's and adding a little more material. There are four factors involved: Velocity distribution of the nuclei Small geometrical cross-section for head-on collisions of nuclei Quantum-mechanical tunneling probability For the p-p reaction, a weak-force ...

68

It has more to do with physiology of the eye rather than the spectrum of light produced by stars. Stars emit light over the full range of visible wavelengths. Hot stars emit more blue/violet light, cool stars emit more red light. The Sun is relatively neutral in that regard, so does not have a strong colouration, but many other stars in the sky have ...

62

There is no alignment between the Sun or the Solar System's net angular momentum and the "spin axis" of the Galaxy. Think for a moment about whether the line of the ecliptic (which marks the "equatorial line" of the Solar System) and the Milky Way (which roughly marks the plane of the Galaxy) are lined up? If this were so, then you would always see the ...

57

The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons). You can do the calculation: If you're 10 light-years away from ...

47

One thing to keep in mind is that objects that are bound gravitationally actually revolve around each other around a point called a barycenter. The fact that the earth looks like its revolving around the sun is because the sun is much more massive and its radius is large enough that it encompasses the barycenter. This is a similar situation with the Earth ...

45

The division is conventionally made at the boundary between where stars end their lives as white dwarf stars and where more massive stars will end their lives in core collapse supernovae. The boundary is set both empirically, by observations of white dwarfs in star clusters, where their initial masses can be estimated, and also using theoretical models. ...

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 ...

40

If a dense, spherical star were made of uniformly charged matter, there'd be an attractive gravitational force and a repulsive electrical force. These would balance for a very small net charge: $$dF = \frac1{r^2}\left( - GM_\text{inside} dm + \frac1{4\pi\epsilon_0}Q_\text{inside} dq \right)$$ which balances if $$\frac{dq}{dm} = \frac{Q_\text{inside}}{M_\... 40 Describing the sun as an average star is probably more of a reaction against the idea that there is something unique about it. Obviously there is for us, since it is the star that we happen to be in orbit around, and much closer to than any other star, and hence historically the sun has been considered rather unique. But over the centuries we've discovered ... 36 The answer kind of depends on how old you are. At a very introductory level, say, maybe middle school or younger, it's "okay" to refer to Jupiter as a failed star to get the idea across that a gas giant planet is sort of similar to a star in composition. But around middle school and above (where "middle school" refers to around 6-8 grade, or age ~12-14), I ... 36 Anything the mass of a star is going to get hot like a star and fuse hydrogen like a star. In other words it will be a star not a planet! While it's technically possible to have a rocky planet the mass of a star, in practice when stellar systems form there aren't enough metals available to build such a large object. Large objects are invariably built from ... 36 The answer lies in something called the virial theorem. You are correct, a cloud that is in equilibrium will have a relationship between the temperature and pressure in its interior and the gravitational "weight" pressing inwards. This relationship is encapsulated in the virial theorem, which says (ignoring complications like rotation and magnetic fields) ... 32 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 ... 30 Although I agree with all three of the above answers let me present a slightly different perspective on the problem. It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light ... 30 The conditions at the core of the Sun are very different from those in a thermonuclear bomb. The first thermonuclear bomb used deuterium as the secondary. The Sun has to create deuterium before getting to this stage. It's the creation of deuterium that's the bottleneck in the fusion that occurs inside the Sun. Later bombs used lithium deuteride, which is ... 28 Not quite like in the photo above, which shows more than what the naked eye can see, but yes, absolutely! Our galaxy (well, the chunk of it visible from these parts) is a naked-eye object. The fact that your question even exists shows how much time is now spent by people under light-polluted skies. It will not be visible from the city, however. You need to ... 27 The premise that the sun has the same conditions all throughout is incorrect. For the most part the conditions (Temperature and Pressure) necessary for nuclear fusion to occur are only found within a small region in the core. For example, when hydrogen fusion occurs and creates helium, since that helium is heavier it tend to coalesce as the core. In ... 26 For visible stars, the answer is no. In Newtonian physics, a star that would pull something travelling at light speed back to itself, i.e. a star for which the escape velocity were c, was called a dark star and seems to have been first postulated by the Rev. John Mitchell in a paper to the Royal Society in London in 1783. The great Simon Pierre de Laplace ... 26 Why shouldn't the orbits of stars be Keplerian? The answer is simple. Keplerian orbits are predicated on a single central point mass. That assumption fails to some extent even in a solar system. It fails massively in a galaxy. A galaxy is not a point mass. 25 The estimates I've read are similar to yours: 200 to 400 billion stars. Counting the stars in the galaxy is inherently difficult because, well, we can't see all of them. We don't really count the stars, though. That would take ages: instead we measure the orbit of the stars we can see. By doing this, we find the angular velocity of the stars and can ... 22 A lot (to put it mildly) of elements are created in stars and supernovae. These elements then travel through space until they fall to Earth (or, to be exact, some microscopic portion of them reach us). Earth itself wouldn't exist if stars hadn't generated elements which then clumped into dust, into minerals, and so on until a big ball of matter started to ... 22 To address your last point, there are several stars of which we have been able to resolve images i.e. see the star as more than just a featureless point. There is a list of these stars on Wikipedia (I love that they put the Sun at the top of the list - true but pedantic :-). The farthest away of the stars in the list is Epsilon Aurigae at about 2000 light ... 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 ... 21 According to Opacity of an Ionized Gas, "light from regions [of the sun] where the pressure is greater than 0.01 atm. is cut off completely, so that all we see comes from a spherical shell of rarefied gas". There is no real surface of the Sun. Instead, the density and pressure of gas/plasma progressively increase from an infinitesimal value far from the ... 20 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 ... 19 I'll add a theoretical limit to the actual record put forward by John Rennie. To image an object as more than a featureless "point source", it must be resolved by the telescope. The angular resolution \theta of a telescope is:$$\theta\sim1.22\frac{\lambda}{D_{\rm aperture}} $\lambda$ is the wavelength of light, $D_{\rm aperture}$ is the diameter of ...

19

The heavier-than-iron elements are not formed during stellar fusion, but they are formed during supernovae. Then the oldest stars cannot have these heavier elements, but new generations, formed from 'recycled' material of other stars that went supernova can. See Stellar populations . There are heavier that iron elements on Earth, the Earth was formed from ...

18

It is a myth that heavier elements than iron are not produced in stars, slow-neutron-capture-process is a nucleosynthesis process that occurs at relatively low neutron density and intermediate temperature conditions in large stars. For details of what elements are produced and about the process itself, see S-process.

Only top voted, non community-wiki answers of a minimum length are eligible