# End of the Earth V: There Goes the Sun

The Sun [photo courtesy of NASA]

People that know me well know that I have a lot in common with Robert Frost. We both were born in March and we both employ rural New England settings to explore complex social and philosophical themes in our poetry. We also like the same rap groups. In honor of my literary doppelganger, I will now, having already had the world end in fire, try my hand at ice. Let's try to answer the question: "If the sun blinks out of existence this instant, what is the temperature of the Earth as a function of time?" The Sun, in addition to being the King of Planets, is also what keeps us all warm and toasty and alive. What happens if we turn that off? Well, the Earth will cool by radiating its heat away into space. To see how long this would take, let's make some assumptions. Let's model the surface of the Earth as an ocean 1 km deep and let's pretend that all the heat is stored in this ocean. Let's take the ocean to be liquid water at T = 0 degrees Celsius. How long will it take this ocean to freeze into ice at 0 degrees Celsius? Well, the amount of energy released from the oceans as the water freezes is given by $$ Q = L_{w} M_{ocean} $$ where L is the "latent heat of fusion" and M is the mass of the water. The "latent heat of fusion" is a fancy way of saying "the amount of energy released per unit mass as water turns to ice at constant temperature." For water, we have that $$ L_{w} = 3.3 \times 10^5 \mbox{J/kg} $$ And for the mass of the ocean, it will be convenient later to write it as $$ M_{ocean} = 4\pi {R_{\oplus}}^2 \Delta R \rho $$ Alright, so now we've got enough to say how much heat energy we have, so how fast do we lose it? We can take the Earth to be a blackbody radiator, so the power lost in such a case is: $$ P =4\pi \sigma R_{\oplus}^2 T^4 $$ Since Power is just Energy per unit Time, we now have all we need to get the time for total freezing of all the oceans. We have: $$ t = \frac{Q}{P} = \frac{4\pi {R_{\oplus}}^2 \Delta R \rho L_{w}}{4\pi \sigma R_{\oplus}^2 T^4} $$ Simplifying the above expression a bit, we get $$ t =\frac{\Delta R \rho L_{w}}{\sigma T^4} $$ Now we can plug in some numbers, $$ t =\frac{\left(10^3 \mbox{ m}\right) \times \left(10^3 \frac{kg}{m^3}\right) \times \left(3.3 \times 10^5 \mbox{J/kg}\right)}{\left( 5.67 \times 10^{-8} J s^{-1} m^{-2} K^{-4}\right) \times \left( 273 K\right)^4} $$ where we have made sure to put our temperatures in Kelvin. Crunching the numbers with the calculator we "borrowed" from Nic three months ago gives: $$ t = 10^9 \mbox{ s} $$ Remembering that a year is very nearly $$ 1 \mbox{ year} = \pi \times 10^7 \mbox{ s}, $$ we find that the time for the oceans to freeze after the sun disappears is about 30 years. Hooray! Now this model was very simple. First of all, I assumed that the ocean temperatures were already at 0 degrees, but they are a bit warmer. If the oceans are about 300 K (ie 80 degrees in not-Yariv units), then we get another 30 years to cool down to freezing temperatures. Secondly, I have completely neglected the heat stored in the Earth. Will this change my answer by an embarrassingly large factor? Lastly, I have ignored all internal heating mechanisms (ie, radioactive decay) that will also heat up the Earth. But ignoring all that.... So is there a way for anyone to survive this? Well, for the most part it will mean the end of life on Earth. There could potentially be a few exceptions, like by geothermal vents and such. But for the most part, it's one quick cold spiral down to eternal nothingness. But what about a few people, could they survive for a bit even if the human race is doomed? I'm glad you asked! You see, I have this plan involving mine shafts. Hunkering down underground with a nuclear power plant and all the canned food food we can stomach should allow us to at least ride the rest of our lives. Details can be found here.

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