Re: Dyson shell redux

From: Anders Sandberg (asa@nada.kth.se)
Date: Wed Feb 21 2001 - 04:11:24 MST


You would need a gas pressure counteracting the weight of one square
meter of the shell, or P=GmM/r^2. The ideal gas law (insert wild
handwaving here) PV=nRT gives us VGmM/(Rr^2)=nT or 4*pi*GmMr/(3R)=nT
when we factor in the volume.

Putting in numbers,
R=8.31451 m^2 kg/s^2 K mol
G=6.673e-11 m^3/kg s^2
M=1.989e30 kg
m=42 kg/m^2 (from my own FAQ, http://www.nada.kth.se/~asa/dysonFAQ.html#ENOUGH)
r=1.50e11 m
gives us nT=4.2e32.

Assuming we spread out all of the sun as hydrogen gas, we get
n=1.989e30/2.016e-3 =9.8e32 mols. Wow, same order of magnitude! So in
this case you could manage with a mere T=0.42 K gas. However, this is
of course unrealistic as the gas itself will start contracting if it
is cold.

If we accept T=1000 K as the maximum temperature our system can stand,
then we just need a thousandth of this mass and things look a little
better - but not much, since now we will have an active fusing star in
the middle heating the gas up! A more likely gas temperature would be
the sun's surface temperature 6000 K or solar wind temperature
150,000K - ouch! And in this case the gas will not be as thin as the
solar wind is now, it will really transfer heat to the
construction. You better have good cooling.

-- 
-----------------------------------------------------------------------
Anders Sandberg                                      Towards Ascension!
asa@nada.kth.se                            http://www.nada.kth.se/~asa/
GCS/M/S/O d++ -p+ c++++ !l u+ e++ m++ s+/+ n--- h+/* f+ g+ w++ t+ r+ !y



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