<--
.. title: End of the Earth VII: The Big Freeze
.. date: 2012-04-22 19:34:00
.. tags:
.. category: old
.. slug: end-of-the-earth-vii-the-big-freeze
.. author: Jesse
.. has_math: true
-->
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[![image](http://1.bp.blogspot.com/-c8vJR4CVwZc/T5R7_62SLuI/AAAAAAAAAHU/POCT5Fhx-CQ/s320/Space_Scene_Frozen_Earth_WP_BG_by_PimArt.jpg)](http://1.bp.blogspot.com/-c8vJR4CVwZc/T5R7_62SLuI/AAAAAAAAAHU/POCT5Fhx-CQ/s1600/Space_Scene_Frozen_Earth_WP_BG_by_PimArt.jpg)
http://tinyurl.com/7rdj996
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It is traditional here at The Virtuosi to
[plot](http://thevirtuosi.blogspot.com/2010/04/end-of-earth-physics-i.html)
[the](http://thevirtuosi.blogspot.com/2010/04/end-of-earth-ii-blaze-of-glory.html)
[destruction](http://thevirtuosi.blogspot.com/2010/04/end-of-earth-physics-iii-asteroids.html)
[of](http://thevirtuosi.blogspot.com/2011/04/end-of-earth-iv-shocking-destruction.html)
[the](http://thevirtuosi.blogspot.com/2011/04/end-of-earth-v-there-goes-sun.html)
[earth](http://thevirtuosi.blogspot.com/2011/04/end-of-earth-vi-nanobot-destruction.html).
We also are making secret plans for our volcano lair and death ray.
However, since it is earth day, we will only share with you the plans
for the total doom of the earth, not the cybernetically enhanced guard
dogs we're building for our [moon
base](http://thevirtuosi.blogspot.com/2012/04/earth-day-2012-escape-to-moon.html).
The plan I reveal today is elegant in its simplicity. I intend to alter
the orbit of the earth enough to cause the earth to freeze, thus ending
life as we know it. According to the internet at large, the average
surface temperature of the earth is \~15 C. This average surface
temperature is directly related to the power output of the sun. More
precisely, it is directly related to the radiated power from the sun
that the earth absorbs. Assuming that the earth's temperature is not
changing (true enough for our purposes), the then power radiated by the
earth must be equal to the power absorbed from the sun. More precisely
$$ P_{rad,earth}=P_{abs,sun}$$ Now, the radiated power goes as
$$P_{rad}=\epsilon \sigma A_{earth} T^4 $$ where A_earth is the
surface area of the earth, T is the temperature of the earth, and
epsilon and sigma are constants. I'll be conservative and say that I
want to cool the temperature of the earth down to 0 C. The ratio of the
power the earth will emit is
$$\frac{P_{new}}{P_{old}}=\frac{T_{new}^4}{T_{old}^4} \approx
.81$$ Note that the temperature ratio must be done in Kelvin. The power
radiated by the sun (or any star) drops off as the inverse square of the
distance from the sun to the point of interest: $$P_{sun} \sim
\frac{1}{r^2} $$ To reduce the power the earth receives from the sun
to 81% of the current value would require
$$\frac{P_{sun,new}}{P_{sun,old}}=\frac{r_{old}^2}{r_{new}^2}=.81
$$ This tells us that the new earth-sun distance must be larger than the
old (a good sanity check). In fact, it gives $$r_{new}=1.11 r_{old} $$
So I'll need to move the earth by 11% of the current distance from the
earth to the sun. No small task! The earth is in a circular orbit (or
close enough). To change to a circular orbit of larger radius requires
two applications of thrust at opposite points in the orbit It turns out
that the required boost in speed (the ratio of the speeds just before
and after applying thrust) for the first boost of an object changing
orbits is given by
$$\frac{v_{f}}{v_{i}}=\sqrt{\frac{2R_{f}}{R_i+R_f}}=1.026$$ To
move from the transfer orbit to the final circular orbit requires
$$\frac{v_{f}}{v_{i}}=\sqrt{\frac{R_{i}+R_f}{2R_i}}=1.027$$ Note
that despite the fact that we boost the velocity at both points, the
velocity of the final orbit is less than that of the initial. Now, how
could we apply that much thrust? Well, the change in momentum for the
earth from each stage is roughly (ignoring the slight velocity increase
of the transfer orbit) $$\Delta p = .03M_E v_E $$ The mass of the
earth is \~6*10^24 kg, the orbital velocity is \~30 km/s, so $$\Delta
p = 5\cdot 10^{27} kg*m/s$$ A solid rocket booster (the booster
rocket used for shuttle launches, when those still happened) can apply
about 12 MN of force for 75 s (thank you wikipedia). That's a net
momentum change of \~900 *10^9 kg*m/s (900 billion!). So we would
only need $$\frac{2*5\cdot 10^{27}}{9\cdot 10^{11}}=12 \cdot
10^{15}$$ That's right, only 12 million billion booster rockets! With
those I can freeze the earth. I assure you that this plan is proceeding
on schedule, and will be ready shortly after we have constructed our
volcano lair.