Timeline for Why does the Sun's (or other stars') nuclear reaction not use up all its "fuel" immediately?
Current License: CC BY-SA 3.0
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| Aug 18, 2014 at 9:14 | history | edited | MariusMatutiae | CC BY-SA 3.0 |
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| Aug 14, 2014 at 10:06 | comment | added | MariusMatutiae | @Kyle What you say is true, at least at Solar temperatures, but I fail to see how this is relevant. Stability is due to the fact that the dynamical timescale (about 10 minutes for the Sun) is much shorter than the energy removal timescale, aka Kelvin-Helmholtz timescale, about 30 million years for the Sun. This means: if excess energy is produced, it will be used to power an expansion before it manages to be carried away. Thus extra energy production is used exactly to quench the conditions that brought it about. That's all. | |
| Aug 14, 2014 at 1:13 | vote | accept | user.3898215 | ||
| Aug 14, 2014 at 1:13 | |||||
| Aug 13, 2014 at 21:55 | comment | added | Kyle Oman | Might be worth pointing out that the energy production for the two most common reaction chains in stars (p-p chain and CNO cycle) go as $T^4$ and $T^{20}$ respectively, which leads to very stable equilibrium states indeed. | |
| Aug 13, 2014 at 21:54 | comment | added | user4552 | Good point, +1. I would say that this is orthogonal to the other answers, which start from assumed conditions of pressure and temperature at the sun's core. However, the fact that the sun is in a stable equilibrium isn't the only difference between the sun and an H-bomb. The other differences include the need for the weak interaction and quantum-mechanical tunnelling. | |
| Aug 13, 2014 at 21:46 | history | answered | MariusMatutiae | CC BY-SA 3.0 |