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In a follow-up to this question,

What would it be like to watch an average space rock / ice comet at 1-20 Pluto distances from a supernova of $10^{44}$ joules?

At what proximity would they become vapor/lava/above $100^\circ C$? How fast would they deflect into space? Would it be like smashing them with a cosmic hammer with a sudden force? Would the force rise over minutes and hours like a rail gun?

$$ E'=\frac{\pi R^2}{4\pi r^2}E\approx 10^{-16}E. $$ Chiral Anomaly's formula linked above tells me that a melon sized boulder of 8kg at 1ly from the supernova will get $10^{26}$ joules of energy, and $10^{21}$ joules at 3ly.

There's also a lot of matter ejected nearly at c, At 10 pluto distances a melon sized rock would take about 1-3kg of that matter at 300AU it's $118kg^{m3}$ ( $6*10^{30}kg$ / $4*10^{22}km^2$ )..

What is mysterious is the thermal energy accrued by the rock, and gamma rays/neutrinos which traverse through it.

I'm wrongly imagining that supernovas throw spheres of millions of local comets outwards like a game of cosmic billiards, including ice comets with DNA life from nearby planets.

bandybabboon
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1 Answers1

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A typical supernova emits an energy of $10^{44}$ joules. According to Supernova - Light curves a typical supernova explosion has a duration of $100$ days, i.e. $10^7$ seconds. This results in a power of $$P=\frac{10^{44}\text{ J}}{10^7\text{ s}}=10^{37}\text{ W}$$

Let's consider a distance $10$ times the distance to Pluto. $$r=10 \cdot 6 \cdot 10^9\text{ km}=6\cdot 10^{13}\text{ m}$$

At this distance from the supernova you have a radiation intensity $$I=\frac{P}{4\pi r^2} =\frac{10^{37}\text{ W}}{4\pi\cdot (6\cdot 10^{13}\text{ m})^2} = 2\cdot 10^8\text{ W/m}^2$$

We can use the Stefan-Boltzmann law ($I=\sigma T^4$) to roughly calculate the temperature $T$ a body aquires by absorbing this radiation intensity. $$T=\sqrt[4]\frac{I}{\sigma} =\sqrt[4]\frac{2\cdot 10^8\text{ W/m}^2}{5.7\cdot 10^{-8}\text{ W/m}^2\text{K}^4} =8000\text{ K}$$ This temperature is well above the boiling point of rocks and metals. So everything will evaporate there.

Thomas Fritsch
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