# Tag Info

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Because stars are not confined. As @MariusMatutiae says the fusion in a star is maintained at equilibrium by the thermostat of pressure versus gravity. An even more apt appliance for analogy is a nuclear power plant. In nuclear fission power, control rods or other mechanisms adjust concentration so as to prevent explosion. The fissionable material is ...

4

A brief overview of stellar evolution can be depicted in the following image: (From here which says it is originally from an encyclopedia; click here for larger image). The heavier stars (top track) have very short life times (a few million years) because they run through hydrogen, helium, carbon+oxygen, ..., iron fusion in the core. Once a particular ...

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You are neglecting two important facts. The first one is that stars, toward the end of their lives, return to the interstellar medium (ISM) a lot of their initial mass, but now enriched with heavy elements produced by nuclear reactions inside the stars themselves. In this way, younger stars which form from the ISM begin their life with a larger fraction ...

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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 ...

9

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 ...

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The reason why a massive star does not immediately collapse to a black hole is radiation pressure. When a star is in that phase of its life called Main Sequence (MS), its luminosity depends approximately on its mass roughly as $M^4$. This means a star 10 times as massive as the Sun would be 10,000 times more luminous. This enormous luminosity is mostly ...

3

It wasn't a black hole because the density wasn't sufficiently high. The density was lower than what is needed for a black hole because the volume was larger. The volume was larger because the atoms (mostly hydrogen) were kept away from each other by the pressure produced by the fusion processes. Once the fusion processes stop, this source of repulsion ...

35

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 ...

9

None of these answers seems to explain correctly why the Sun differs from a nuclear bomb. The reason is that any star, including the Sun, acts as a thermostat. If the Sun were to produce more energy than it can radiate away, the energy thus freed would make it hotter; a hot gas expands, and simultaneously cools. Both factors (lower densities and lower ...

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Fusion, as it occurs within stars, is in fact very unlike what happens in a bomb. An "H-bomb" is actually a mixture of fission and fusion. The fission part works on a chain reaction: when a fissile nucleus absorbs a neutron, it vibrates madly and then splits into several components, in particular two or three neutrons. These extra neutrons go on breaking ...

1

Gyrochronology is semi-empirical in the sense that there is some justification for the temporal dependence that you mention. There is a line of argument for the $t^{1/2}$ dependence and it can be found on pp7-8 of this pedagogical review by Jerome Bouvier. http://arxiv.org/pdf/1307.2891v1.pdf The basic idea is of a spherically symmetric, ionised wind that ...

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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 ...

66

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 ...

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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 ...

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The interior of a star is a hot ionized gas at high pressure and temperature. High temperature means high average kinetic energy per particle, so all the nuclei of the atoms are whizzing around very fast (though for relatively short distance between collisions because the gas is so dense). The thing is that they are not all whizzing around at the same ...

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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 ...

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The answer is yes! And it has been used in this way many times. To pick a random, recent example by Vogt et al. (2014) - a 4-planet system orbiting a K0V star. http://arxiv.org/abs/1404.7462 The problem is akin to separating out multiple frequencies in a blended chord and is accomplished with fourier analysis techniques. A useful public resource ...

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