Okay, so what would a supernova explosion do to a planet?
Well, let's say we have a supernova, a star that goes supernova.
These put out a power that can be up to about 10 to the 44 joules.
Well, that's the energy so that's the total power integrated over the time.
The supernova will be bright for a few weeks so if you add up the power of all that time, that's the total energy you get out.
Let's say we have a planet, some distance D, let's say about 10 to the 11 metres, a bit less than an astronomical unit,
and let's say that it's a rocky planet, so it's on a radius of about, well it's got 6000 kilometres,
that's a bit bigger than the Earth so 10,000 kilometres so it's 10 to the four, five, six, seven, about 10 to the seven metres in radius.
There's no point in trying to calculate it totally precisely,
we're just after a rough figure to see whether it's even feasible that a planet could survive.
If it's up by many orders of magnitude, then the odd factor of two doesn't matter.
If it comes up as being close then we've got to go back and look at the calculation in more detail.
So, we're going to be two steps to the calculation, the first step is "what fraction of this power actually hits the planet?"
The second is "what would that energy do to the planet?"
So, let's look at the fraction that's hitting the planet.
Let's draw an imaginary sphere of radius D around the supernova.
Most of the radiation will just go straight through and out into space, but the bit that lands here will hit.
So, you can imagine replacing the planet with a disk of radius R facing the star, which therefore has an area of pi R squared.
The total surface area of this whole sphere is four pi D squared.
The fraction of the energy, the total energy that hits, the fraction of energy is going to be this area here divided by the total area,
so pi R squared over four pi D squared.
If you call the total energy "E", energy absorbed by a planet is going to be roughly E,
the pi's cancel here, times E over four, R over D squared. So, how much is that?
So, if we assume that the total energy is about 10 to the 44 joules, so that's going to be about 10 to the 44 over 4,
and we've got about 10 to the seven metre radius planet about 10 to the 11 metres away, that's all squared,
so that's 10 to the minus four, so when you square it, it's 10 to the minus eight, take eight off that, so that's about equal to a quarter.
10 to the 44 minus eight orders of magnitude is going to be about 10 to the 36.
So, that is a lot of energy, but planets are big.
The mass of the Earth is about six by 10 to the 24 kilograms, let's assume it's a bit bigger so three or four times bigger,
like the pulsar planets, so that gives us a mass of the planet about, let's say, two by 10 to the 25 kilograms.
So, we compare this with that, we get an energy per unit mass, totals one over four times 10 to the 36 all over two by 10 to the 25,
so that's about one eighth times 10 to the 11, which is pretty much 10 to the 10 joules per kilogram.
So, that's a lot of joules for every kilogram on the planet. Now this, of course, may not be totally realistic.
A lot of the energy of the supernova is in neutrinos, that might go straight through the planet.
Some of it might be in the blast wave which might curve in some fewer dynamical way around the planet,
so it could well be that the energy that actually impacts the planet is less than this, but it's going to be something like that.
Now, let's ask, if you did get 10 to the 10 joules added to every kilogram of a planet, what's it going to do?
Well, let's say we've got a kilogram of rock and you add large amounts of energy to it, whats going to happen?
Well, the first thing is it'll warm up. So, let's warm it up to the melting point.
The melting point is going to be about a thousand degrees, depends on the exact type of rock, it might be 2000,
starting at a temperature somewhere between naught and 100, so we've got to increase the temperature by about roughly 1000 kelvin.
The energy needed to do that is the change in temperature times the specific heat capacity.
Specific heat capacity for rocks is about 1000 joules per kilogram,
so to bring the rock up to the melting point, you're going to need about 1000 times 1000, so about 10 to the six joules.
So, that's got it up to the melting point, now to melt it we need the latent heat of fusion.
This is the energy needed to convert a kilogram of a solid at the melting point temperature
to a kilogram of liquid at the melting point temperature.
It's basically going into breaking the bonds. Now, I don't know what that is for rock,
but let's assume that that's about the same as water which is about roughly 300 kilojoules, 300,000 joules per kilogram.
That's to turn ice into water, and it's going to be roughly the same for rock.
It might be five times smaller or five times bigger, but about the same.
That five times factor doesn't really matter here.
So, to convert out kilogram, we're going to need about, that's about three times 10 to the five joules, so less than that.
Then we've got to raise it up by another temperature difference to bring it up to the boiling point.
Let's say it boils, I have no idea what temperature rock boils at, I've never seen rock boiling,
but let's say it boils at 2000 degrees, probably not going to be too far off.
So, we're going to have to raise the temperature by another thousand,
let's assume that the heat capacity remains the same if it's liquid, so that's another 10 to the six joules,
and then we've got to supply the latent heat of vaporisation to turn a kilogram of liquid into a kilogram of gas.
Just like it takes the energy to turn water at 100 degrees into steam in a kettle.
Now, once again, I don't know the value of that for a rock, I haven't tried boiling any rock recently, curiously enough,
but for water it's about two by 10 to the six joules per kilogram. So, we'll assume that it's about the same.
So, add these all up and we've got about five by 10 to the six joules needed
to turn a kilogram of rock at a normal temperature into a kilogram of rock vapour.
Now, remember, the energy available is 10 to the 10. All we need is a few by 10 to the six,
so that's telling us that we have far, far lot's of energy so the energy supplied is much greater than energy needed to vaporise rock.
If it was only like two or three times as big, you might worry about maybe we've left something out,
maybe we needed to look up more precisely these values, try melting some rock in the lab,
you have to worry about whether the neutrinos are being absorbed by the radiation or flow around it,
maybe as it starts to vaporise that will give it a very silvery clouds of vaporised rock that reflects much of the radiation off.
All of those things might bring the energy input down by maybe a factor of 10, maybe a factor of 100, but that's nothing like enough.
It would need to be brought down by 2000 times to save the planet.
So, it looks like planets are DOOMED!