[ExI] Life must be everywhere!

Anders Sandberg anders at aleph.se
Sun Apr 15 08:23:21 UTC 2012

(Summary: Panspermia is fun! So many different factors interacting, so 
much cool physics, mathematics and biology! Introduce your family to the 
joys of trying to estimate whether it works or not today! Oh, and big 
rocks are likely *much* more viable vehicles for spores than small ones. )

On 2012-04-15 08:13, Kelly Anderson wrote:
> 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.

You can check some of the physics in Collins, Melosh and Marcus' 
*excellent* paper describing their calculator:

The real issue for us is sequential breakups, since we are interested in 
the conditions that produce pieces that have cores that are not too hot; 
whether they get deposited on the ground with a bang or just float down 
as dust doesn't matter.

>> 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?

We have to start somewhere with the calculations. We can be fairly 
confident that the interior of a major impact is going to be a fireball, 
so a thin molten layer is likely not a bad approximation.

In fact, for doing the calculation properly the thickness of the molten 
layer matters a lot. Obviously the initial amount of hot rock vs cold 
rock matters for estimating the end temperatures. But then there is 
this: the surface is radiating away P = 4*pi*R^2sigma*epsilon*T^4 Watts 
of heat. This will decrease its temperature as P/(4*pi*R^2*t*K) = 
sigma*epsilon*T^4/(t*K) Kelvin/second (ignoring heating from the 
inside), where t is the thickness of the IR-optically transparent outer 
layer and K is the thermal capacity. A thick layer of melt means that it 
will cool quicker since it can move heat out faster (also, it might shed 
droplets, a far more potent cooling mechanism than radiation). However, 
I don't know how much t should be in this estimate - any thermal 
physicist or hot material scientist around?

>> 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!

Me too! Unfortunately my dreams tonight were about running a restaurant 
at the Cote d'Azure (involving noneuclidean geometry and some minor plot 
for world domination), so I will still have to do the calculations while 

But the basic point you made is right: big boulders will not be fried as 
easily as pebbles. They also have the benefit that they are less likely 
to be ablated to nothing when passing through a terrestrial atmosphere 
on impact. On the downside, and this could be major, there will be fewer 
of them. I suspect the size distribution is a power-law, with far more 
small pieces than bigger ones. So even if the probability of life 
surviving in small pebbles is much lower the sheer number of pebbles 
might outweigh the boulders in terms of viable cells lofted into space. 
More calculations are needed...

>> 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.

I did find this excellent little page:

Running the numbers likely for a small pebble suggests that it does 
indeed take more than a minute for it to cool down below temperatures 
that would denaturate life. A bit of a caveat: this assumes the pebble 
is all a uniform temperature and conducts heat instantly, so it will be 
somewhat wrong. But it still shows that the timescale is not in favor of 


The real problem to solve is this:

Solve the spherical heat equation
r^2 dT/dt = alpha*d/dr(r^2 dT/dr) + r^2 q'/rho c_p
where the thermal diffusivity = k/rho c_p is alpha, k is the thermal 
conductivity, rho is the density, c_p is the specific heat capacity and 
q' is the rate of addition/removal of heat.

Boundary conditions:
r<R, t>0
T(r,0) = T0 (say 300 K) for r < R-thickness (where thickness might be 1 mm)
R(r,0) = T1 (say 1200 K) for R < thickness < r < R
q' = 0 for r<R
q' = - sigma*pi*T(R,t)^4 / c_p rho thickness

What conditions on R, T1 and thickness will guarantee that T(0,t) always 
remain under a critical denaturation temperature T_d(say 500 K)?


A quick mathematician's look at the problem and borrowing a bit from 
suggests that core temperatures will change roughly like Tx - 
(Tx-T0)exp(-alpha*lambda^2*t), where Tx is some hot temperature and 
lambda is the smallest eigenvalue, ~pi/R. So if that is true, the time 
to serious heating of the core will scale roughly as R^2. But again, 
this ignores the radiative cooling issue and the shape of the Fourier 

> 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. :-)

Yup, DNA degradation is likely the big problem. To be honest, I think 
launch due to big impacts is less of a problem: if it is possible at 
all, the amount of mass launched and the number of cells are going to be 
large. But then they will drift around for potentially millions of years.

In most papers I have read the idea is that bacteria on the surface of 
grains are going to get killed quickly, but bacteria inside are fairly 
safe. Of course, that might be biased because the papers are typically 
written by pro-panspermia people.

I guess the relation to look at is something like this: radiation will 
penetrate into the material with some decay length lambda, so over time 
T cells at depth d will have received a dose of D_0*exp(-lambda*d)*T. So 
a cell at the core will survive until its dose reaches D_max, at time 
T_max = D_max/(D_0*exp(-lambda*R)). D_max for Deinococcus radiodurans is 
5,000 Gray, although I am uncertain that relates to spores or the living 
bacterium. Lambda depends on the type of radiation; UV mostly affects 
the surface (a few hundred microns in basalt, 
), while cosmic rays go deeper (64 cm for typical rocks on Rarth, 
according to 
). D_0 in the solar system is on the order of 0.1 Gray per year
(I am here ignoring the difference between Grays and Sievert dose 
equivalents - Bad Anders! Bad Anders! - but I don't think I know how to 
estimate it properly for bacterial spores...)
So, putting it all together: T_max = 5000/(0.1*exp(-R*0.64)) years. For 
R=0.01 pebbles T_max is 50,000 years. For R=0.1 rocks T_max is 53,000 
years. For R=1 m boulders T_max is 95,000 years. For big R=10 m boulders 
T_max is 30 million years.

Conclusion: bigger is better. Especially since damage and cell death is 
a random (Poisson) process: more cells means that some are more likely 
to be lucky - expect to find some viable spores beyond a few multiples 
of T_max. Since the number of cells likely scales as some power of 
volume (why a power? fractal distribution!) and the earlier discussion 
shows that surface heating is less likely to kill everything in big 
boulders, big is good.

Of course, big is also likely to be rare, as the earlier power law 
argument pointed out. This might be another reason why relatively low 
velocity impactors on light icy moons produce more viable spores than 
hits on terrestrials. KT impactors are rare, hot and tend to spray 
short-lived vehicles for spores, while small impacts on tiny moons are 
common, cold and can launch big chunks.

OK, My work here is done. Time to run off and give a lecture. Wheeeee!

(As you can tell, I got good morning coffee today :-) )

Anders Sandberg
Future of Humanity Institute
Oxford University

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