[ExI] Life must be everywhere!
Kelly Anderson
kellycoinguy at gmail.com
Sun Apr 15 06:13:51 UTC 2012
On Sat, Apr 14, 2012 at 4:26 PM, Anders Sandberg <anders at aleph.se> wrote:
> The survival of small impactor is a bit complicated. If I remember the
> literature right very small ones vaporize far up, likely burning all spores
> (they heat up in the very thin high atmosphere, where they are not strongly
> slowed). Larger impactors have a hot crust but reach denser lower
> atmosphere, where they reach terminal velocity and are even cooled by the
> surrounding air. Even larger impactors are subject to enough force that they
> can split apart, with fragments of different sizes and temperatures, some of
> which hit the ground and have fairly low temperatures. Even bigger "pancake"
> and explode - and then the biggest ones hit the ground.
It is even more complicated than that when you figure in iron vs rock,
but I would assume for purposes of this discussion we're only
interested in rock.
>> Undoubtedly, but we have recently learned that they can withstand
>> 20,000 Gs of shock, which is rather amazing!
>
> Bacteria have been grown under 100,000 Gs. It is not the shock that is the
> problem. The problem is any temperature above (say) 200 degrees C.
Ok, good enough.
> A simple problem: a spherical granite pebble of radius R starts out with a
> core temperature ~300 K and a surface that is molten, ~1500 K. How long will
> it take for the core to become 500 K hot, and is this time shorter than the
> time it takes to cool the surface in a space environment down to around 300
> K?
And why would we assume that the entire surface is molten? And how
thick would that molten layer be? Wouldn't it make a difference if a
lot of the surface was melted vs just a little bit?
> It is too late in my evening for me to try to solve the spherical heat
> equation for the initial value or with a Stefan law boundary condition...
> but it might be fun when I am awake.
>
> But let's assume the temperature gradient is linear. Then the heat flux is
> F=k*delta T/R Watts, where k is the thermal conductivity of granite (1.7-4.0
> W/mK) and delta T is the temperature difference between core and surface.
> The core will heat up as F/CM, where C is the heat capacity of granite, 790
> J/kg K and M is the mass of the core. Let's make it equal to a fourth of the
> volume, giving a mass M=pi*rho*R^3/3, with rho the granite density 2691
> kg/m^3. Putting it all together, the initial heat flux will make the core
> temperature go up by 3*k*delta T/(C pi*rho*R^2) = (0.00107/R^2) Kelvin/s. If
> R=1e-2, a pebble, the temperature rise will be 10.7 degrees per second. So
> unless the pebble has cooled significantly in the first 20 seconds the core
> will be denaturated.
You're amazing Anders. If I'm reading right though, there is a bigger
size that works better... I'm sure you'll figure out how big it has to
be to work in your sleep, and will just awaken with the right answer
on your pillow.. :-) I love hanging around with smart people!
> Now, how fast is radiative cooling of ejecta? I am too sleepy to solve that
> differential equation right now. But I bet it is slower than 20 seconds.
No doubt.
One of the most interesting factoids dancing around in my Ken Jennings
type mind is that water bears can survive in outer space. If a water
bear can do that, then why not a bacteria? It could just go into a
state of suspended animation... no need to reproduce, or keep extra
junk around. One worry is the degradation of DNA in the radiation of
space over thousands of years. That one is probably worthy of some
math by someone smarter than me. :-)
-Kelly
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