# 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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