How to calculate the time taken for the sun to cool down?

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Calculating the time it would take for the sun to cool down if nuclear energy production ceased involves determining the total energy of the sun and its luminosity. The total energy can be approximated using the formula (3/2)NkT, while the luminosity indicates the rate of energy loss through radiation. The discussion highlights the complexity of the sun's structure and the challenges in applying theoretical models like Wien's law and the Stefan-Boltzmann law. A simplified approach might involve assuming a uniform temperature for the sun, despite the temperature gradient between its core and surface. Ultimately, the cooling process is intricate and cannot be easily modeled with basic equations due to the sun's dynamic nature.
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We're asked how long would it take for the sun to cool down if nuclear energy production suddenly came to a hault

This is the question asked in one of my lecture courses given that the radius of the sun is ~ 10^9m, mass ~ 10^30 kg, luminosity ~ 10^26, and G = 6.7x10^-11

Am i correct in assuming we just calculate the total energy of the sun using (3/2)NkT then using the luminosity calc the time?
 
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In that example, the only way for the sun to lose energy is to radiate it. Given the starting energy, you could approximate the frequency of the radiation at a given absolute temperature using Wien's law (qg) and from the frequency estimate the energy lost at that temperature. And then...
 
Oh i see,

I've learned Wien law before but in this particular course we haven't covered it yet, so i guess there must be another way of doing it.
 
Hmm, somewhat tricky, since your lecturer seems to be wanting you to follow a path that is quite unreal*, so you have to choose (guess?) the method she is fishing for! Perhaps it's 'just' "what is the instantaneous rate of cooling of a blackbody, of mass x, radius y, and temperature z?"

*a real Sun will undergo a very complicated set of changes if its internal fire were extinguished ... for a start, without that fire, the internal pressure which stabilised it against gravitational contraction would diminish rather quickly ... next, energy is transported from the core to the surface of stars principally by two mechanisms - radiative and convective; the balance between the two in real stars composed of varying mixtures of real matter isn't something to be solved with pencil and paper; ...
 
While it is impossible for fusion to cease in the core of the sun [given the amount of hydrogen and gravitational forces], I will suspend disbelief for a moment. The pauli exclusion limit would kick in and the sun still ends up as a white dwarf.
 
The Stefan-Boltzmann law gives the power radiated by a blackbody of temperature T:

P = \frac{dE}{dt} = \sigma A T^4

If you knew how much thermal energy was in the Sun at a specific temperature, you could setup a differential equation to calculute the time it would take. The Sun's structure is quite complicated, and this relationship would be very difficult to find.

Perhaps you could get a reasonable approximation by just assuming a sphere of gas at uniform temperature (despite the fact that such a sphere is not in hydrostatic equilibrium), but I'm not even sure how you'd find the average temperature of the Sun!

- Warren
 
I know that's what was puzzeling me :-/ I didn't know what to do about the temp gradient
 
If you think the approach outlined by chroot is what your lecturer is looking for, then you are free to choose any 'reasonable' temperature (if you get it wrong, you may lose some points, but since the main thrust is how the Sun cools, not what it's 'average temperature' is, that shouldn't hurt too much). Clearly, the Sun's 'average temperature' is greater than that of its photosphere and less than that of its core ... how to pick something in between?

Of course, the Sun is *currently* radiating as if it were a spherical blackbody (radius x, temperature y) - well, 'sorta' - if you replaced it with a spherical blackbody of the same radius but a temperature of 10y (say), it would 'cool' an awful lot more quickly than it is now! :-p :rolleyes:
 
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