Physicist: A commenter from the original post was kind enough to point out a massively bone-headed mistake. My bad. In the calculation to figure out how long it would take the Sun to cool I had used Fourier’s law of thermal conduction. What I should have used was the Stefan–Boltzmann law of thermal radiation. So that calcu-estimation would have been perfectly valid if the Sun were floating in an expanse of very cold, and opaque, material and not valid if the Sun were floating in a whole lot of nothing. Sadly, the universe has taken effectively none of my constructive suggestions, and the Sun persists with being in space. After a revisit (included in the answer gravy below) it looks like we’d still have plenty of time to leave the planet, but longer time scale things would be done (continental drift, presidential campaigns, etc.). Life has existed on Earth for about 4 billion years, but interesting, multicellular life only exploded onto the scene about half a billion years ago. Assuming that this is because the Sun wasn’t doing it’s job at the time (the Sun gets brighter over time, and was dimmer in the past), that puts a floor of about 95% of its present output necessary for complex life. If fusion stopped, we’d be there in a little over a million years.
Science is based on the little-acknowledged axiom that not a single one of us is particularly smart, but that there’s a good chance that somebody out there is busy not making mistakes today (and who tomorrow may rely on you). To that end, what follows is a more direct calculation, for those of you interested in the details.
Answer Gravy:
The Stefan-Bolzman law says that the flow of heat out of the Sun should be 
Plugging this into the Stefan–Boltzmann law:
Fans of calculus can tell you that the solution of this is:
![\begin{array}{ll}\dot{T} = -\left(2\times 10^{-26}\right)T^4\\\Rightarrow \frac{dT}{dt} = -\left(2\times 10^{-26}\right)T^4\\\Rightarrow T^{-4}dT = -\left(2\times 10^{-26}\right)dt\\\Rightarrow -\frac{1}{3}T^{-3} = -\left(2\times 10^{-26}\right)t +k\\\Rightarrow T^{-3} = \left(6\times 10^{-26}\right)t +k\\\Rightarrow T = \left[\left(6\times 10^{-26}\right)t +k\right]^{-\frac{1}{3}}\end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7D%5Cdot%7BT%7D+%3D+-%5Cleft%282%5Ctimes+10%5E%7B-26%7D%5Cright%29T%5E4%5C%5C%5CRightarrow+%5Cfrac%7BdT%7D%7Bdt%7D+%3D+-%5Cleft%282%5Ctimes+10%5E%7B-26%7D%5Cright%29T%5E4%5C%5C%5CRightarrow+T%5E%7B-4%7DdT+%3D+-%5Cleft%282%5Ctimes+10%5E%7B-26%7D%5Cright%29dt%5C%5C%5CRightarrow+-%5Cfrac%7B1%7D%7B3%7DT%5E%7B-3%7D+%3D+-%5Cleft%282%5Ctimes+10%5E%7B-26%7D%5Cright%29t+%2Bk%5C%5C%5CRightarrow+T%5E%7B-3%7D+%3D+%5Cleft%286%5Ctimes+10%5E%7B-26%7D%5Cright%29t+%2Bk%5C%5C%5CRightarrow+T+%3D+%5Cleft%5B%5Cleft%286%5Ctimes+10%5E%7B-26%7D%5Cright%29t+%2Bk%5Cright%5D%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{ll}\dot{T} = -\left(2\times 10^{-26}\right)T^4\\\Rightarrow \frac{dT}{dt} = -\left(2\times 10^{-26}\right)T^4\\\Rightarrow T^{-4}dT = -\left(2\times 10^{-26}\right)dt\\\Rightarrow -\frac{1}{3}T^{-3} = -\left(2\times 10^{-26}\right)t +k\\\Rightarrow T^{-3} = \left(6\times 10^{-26}\right)t +k\\\Rightarrow T = \left[\left(6\times 10^{-26}\right)t +k\right]^{-\frac{1}{3}}\end{array}")
Here k is an “integration constant”, and it’s set so that the temperature right now (at t=0) is what it should be: 5878. Also, t is in seconds in σ. To fix that, just multiply t by 31,500,000,000 (the number of second in a thousand years) to change it to millennia.
The surface temperature in °K vs. time in thousands of years.
So all together, if fusion were to stop today, the Sun’s surface temperature should be about ![T = \left[\left(6.3\times 10^{-19}\right)t +5\times 10^{-12}\right]^{-\frac{1}{3}}](https://s0.wp.com/latex.php?latex=T+%3D+%5Cleft%5B%5Cleft%286.3%5Ctimes+10%5E%7B-19%7D%5Cright%29t+%2B5%5Ctimes+10%5E%7B-12%7D%5Cright%5D%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D&bg=ffffff&fg=000000&s=0 "T = \left[\left(6.3\times 10^{-19}\right)t +5\times 10^{-12}\right]^{-\frac{1}{3}}")
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