# Consequences of a huge electromagnetic burst

This is a sci-fi setting.

We have a planet with carbon-based life forms. This poor planet was exposed once to $$10^{32} J$$ of electromagnetic radiation of wide spectrum (from visible to gamma rays) released in one burst. Source of that radiation is omnidirectional. It is located in $$220\times10^6$$ kilometers from the planet.

From my calculations this amount of energy is usually emitted by the sun within 46 earth days.

What irreversible consequences we can expect for the life and global climate of the planet?

• What is the duration of the burst? And can you focus on just one question? – L.Dutch - Reinstate Monica Mar 4 '20 at 6:23
• @L.Dutch-ReinstateMonica duration of the burst should not be more than a couple of hours. – Diligent Key Presser Mar 4 '20 at 7:13
• Hang on. Is it the total omnidirectional burst has the 1e+32 J (and only whatever fell on Earth contribute) or is that one the total energy of that was delivered to Earth? – Adrian Colomitchi Mar 4 '20 at 7:42
• @AdrianColomitchi 1e+32 J is a total omnidirectional burst energy. – Diligent Key Presser Mar 4 '20 at 7:50

Ummm... let's see the amount of energy received by an Earth-like planet (with a radius $$R_⊕: 6371 km$$) at $$R_b = 220 \times 10^6$$km away from the blast with a total energy of $$E_b = 10^{32}J$$ - the blast point is far enough to consider the cross-section area of the planet as the "interception area"

Assuming an isotropic (omnidirectional and uniform) blast:

$$E_⊕ = \frac {\pi R_⊕^2}{4 \pi R_b^2} E_b = \frac{1}{4} (\frac{R_⊕}{R_b})^2 E_b = 2.1\cdot10^{-10} E_b = 2.1\cdot10^{22}J$$ with the $$4 \pi R_b^2$$ being the total area of a sphere with $$R_b$$ radius

Let's try to make sense of that $$E_⊕ = 2.1\cdot10^{22}J$$ amount of energy:

$$(1)$$ How much water can we boil with that energy from 20C?

The total energy required to boil a ton of water ($$1m^3$$) starting from 20C:

• specific heat capacity of water $$4.186\frac{J}{g \cdot K} = 4.186\cdot10^6\frac{J}{ton \cdot K}$$ - at 80K temp difference between 20C and 100C we get $$3.348\cdot10^8\frac{j}{ton}$$;

• latent vaporization heat of water (the heat required to bring 100C liquid water to water vapors): $$2260\frac{J}{g} = 2.260\cdot10^{12}\frac{J}{ton}$$

The total energy required to boil $$1m^3$$ of water from 20C = $$2.594\cdot10^{12}\frac{J}{ton}$$.

So we should be able to boil $$2.1\cdot10^{21}/2.594\cdot10^{12} = 8.0956\cdot10^{9}$$ tons of water. The mind still boggles down trying to make sense of it.

The Pacific Ocean area is $$= 161.8\cdot10^{12}m^2$$ (161.8 million km²), boiling $$8.0956\cdot10^{9} m^3$$ of its water will cause the water level to drop $$0.05mm$$.

Err... what? Not that impressive? Barely visible you say?
This will also create $$10^{13} m^3$$ sauna-quality steam. Bleah, large numbers again!) Let's say we'd use this steam to fill $$1.5m \times 1.5m \times 2m$$ personal saunas - then we'd have enough to fill 2222 billions of such sauna, or almost 300 saunas for everybody on Earth.

$$(2)$$ Hang on! About that claim of the amount of energy is usually emitted by the sun within 46 earth days. Can we check that and its consequences?

Solar constant on top of the atmosphere is $$1.36\frac{kW}{m^2}$$ at normal incidence. Which means the total energy that the Earth receives in 1 second from Sun is $$1.36\cdot10^3\frac{J}{m^2 \cdot s} \times\pi\cdot{(6371\cdot10^{3}m)}^{2}\times1s = 1.734\cdot10^{17}J$$

So the $$E_⊕ = 2.1\cdot10^{22}J$$ is received under normal circumstances (no solar flares or antimatter blasts) in $$121107 s = 33 h = 1.4 days$$.

Yeah, nah, mate. No need to worry that much, the Sun-Earth distance is 148.38 million km, this blast is defined as happening at about 150% longer distance and the intensity of EM radiation goes down with the square of the distance.

$$(3)$$ So, should I worry about orbital changes?

Earth's mass at $$5.972\cdot10^{24}kg$$ and an orbital speed of $$29.78\cdot10^3\frac{m}{s}$$ gives Earth's orbital kinetic energy of $$2.637\cdot10^{33}J$$.
$$2.1\cdot10^{22}J$$ is so puny by comparison, I'm not even gonna calculate the variation of radius of the Earth's orbit. No worries, mate.

$$(4)$$ What about the atmosphere? Will we get roasted or what?

In terms of temperature, the absolute worse that can happen is to drop the entire energy in the atmosphere as heat and see how much the temperature would raise. Even more, we are going assume an isochoric process (constant volume) which will correspond to a heating fast enough that the atmosphere doesn't get to expand (and lose energy by mechanical work), yet slow enough to be a quasy-equilibrium process (it's never gonna happen this way, some parts of the atmosphere will heat faster and probably be blown away from Earth, but bear with me).

• Isochoric specific heat capacity of air at normal pressure is about $$1.07\frac{kJ}{kg\cdot K}$$ and stays pretty much constant up to 10atm.

• Total mass of Earth's atmosphere is $$5.15\cdot10^{18}kg$$. We'll take half of that, because only half of it it will be exposed to the burst short enough to keep the same volume assumption (it means twice as much energy for the atmosphere taking the blast full on)

The entire $$2.1\cdot10^{22}J$$ will (isochorically) heat half of the atmosphere by $$\Delta T = \frac{2.1\cdot10^{22}J}{\frac{5.15\cdot10^{18}kg}{2}\cdot1.07\cdot10^3\frac{J}{kg\cdot K}} = 7.6218K$$

So, nope, the planet isn't gonna get roasted (still some may get seriously Sun-burnt)

• Wow, cool! Great job, man! What about all the steam, will this amount affect the climate? – Diligent Key Presser Mar 7 '20 at 13:12
• All the considerations there are not realistic, they assume the energy is transferred to different planet components in full and try to assess the maximal effect. In a more realistic scenario, depending on the power, very likely the upper atmosphere will take the brunt of the hit, produce a lot of X/gamma fluorescence and deflect at least 45-60% into space - a good amount of ionosphere will be gone in the process. The rest of the burst is still likely to scald (rather than carbonize it) the vegetation and may start fires in dry areas (e.g California or Australia). – Adrian Colomitchi Mar 9 '20 at 23:13
• The immediate effect on civilization: anything relying on long wires are gone no matter the 'sunny/night' side of the burst (byebye electrical power) The oceans won't heat enough to boil, but expect some impressive hurricanes/typhoons in the next quarter. The amount of water vapors may get a peak during the event but very likely will subside to 'nothing impressive' over 2-3 days. – Adrian Colomitchi Mar 9 '20 at 23:23

Your planet is a little nearer to your sun than earth.
We receive $$1 \times 10^3 \frac{W}{m^2}$$ at sea level.

The burst will deliver 46 days worth of power in a moment.
$$1000 \frac{W}{m^2} \times 46 days \times 24 \frac{hours}{day} \times 60 \frac{minutes}{hour} \times 60 \frac{sec}{minute}$$

This equates to ~$$4 \times 10^9 \frac{W}{m^2}$$ over its entire surface at sea level.

This is enough energy to raise the temperature of the column of air from the upper atmosphere to sea level to 300,000 C. Which equates to a mean molecular velocity of $$1 x 10^5 \frac{m}{s}$$ ( 100 x escape velocity )

On the side facing the star, its atmosphere will be burned away and lost to space. Its oceans converted to plasma. I would imagine that most of the crust will vaporize too since silicates vaporize at 2000 C.

Your previously oblate sphere of a planet will look something like a deflated basketball. It’s moment of inertia will change suddenly and the rotation will be a strange wobble.

The ejecta from the massive energy burst will most certainly push the planet out of its orbit. Its angular velocity will remain the same or increase and its mass will be reduced by a significant fraction. The planet might settle in a higher and colder orbit or it might leave the solar system completely or become like Nostradamus’s Wormwood and orbit the sun once every 10 thousand years.

The planet will be wracked by incredible storms as the remaining atmosphere equilibrate and fill in the void left by the incident impulse.

The planet will drift and become like an icy planet like those past Saturn’s orbit.

• Please don't forget your sunblocker at home. – Erik Mar 4 '20 at 6:45
• I am not sure it is correct to neglect dynamic effects and the system inertia here. – L.Dutch - Reinstate Monica Mar 4 '20 at 7:00
• @L.Dutch-ReinstateMonica, I’ve added the dynamic effects on the atmosphere. I think I am account for the planet’s changes in inertia – EDL Mar 4 '20 at 7:18
• Any chance for some back-of-a-napkin calculations to support "atmosphere will be burned away, and its oceans converted to plasma" and the rest. It is a very plausible assertion indeed (given the disparity between the orders of magnitude), but I'd feel better with some numeric estimations. – Adrian Colomitchi Mar 4 '20 at 7:36
• Ummm... The total energy is already specified, one would only need to see how the energy distributes over all the "degrees of freedom" that the planet as a system has. Why should one consider the power and the make considerations based on the power? – Adrian Colomitchi Mar 4 '20 at 7:45