Okay, I have this villain who's able to create supernova-like explosions from his body that release a total energy of 1.67x10^39 J, and the heroes stop him from destroying Earth by portalling him away in the nick of time. The explosion takes a second to occur, and spreads out in an omnidirectional sphere, so it loses intensity with the square of distance from the epicenter. I know that he cannot be in the Solar System, lest catastrophe strikes, as if he was anywhere on Earth, the planet would become a superheated cloud of irradiated particles. From the Moon, the Moon will be incinerated beyond recognition, and the Earth will be destroyed, being hit by an energy equivalent to 1.44x10^36 J. The amount of energy needed to destroy Earth is 2.24x10^32 J, and the maximum distance from Earth he needs to be to destroy it is 30.7 million km, or .206 AU. To sterilize the planet, it'll take 6x10^26 J, so the villain will have to be 18.8 billion km, or 125.9 AU.

The explosion is as bright as the Sun from Earth at a distance of 117 lightyears. I ask, how far from any inhabited planet will he have to be to not cause mass destruction?

  • 1
    $\begingroup$ Well, 117 light years is a good place to start. $\endgroup$ Apr 11, 2021 at 23:51
  • $\begingroup$ Yeah, it could cause eye damage in a century or so, but no major destruction, due to its brevity. However, I want to know how close the explosion can be to an inhabited planet before there's possibly-fatal effects like asteroid showers, disrupted weather patterns, etc. $\endgroup$ Apr 12, 2021 at 0:43
  • $\begingroup$ Or you could come at the question from the opposite direction - decide how far away to teleport him and then ask how powerful the explosion can be without a major negative impact on Earth. (Given that alternative, I vote for teleporting him to the other side of some celestial body, like Jupiter, to use it as a shield.) $\endgroup$
    – Jedediah
    Apr 12, 2021 at 16:14
  • $\begingroup$ Jedediah, what if Jupiter's scattered remains reach Earth? What if the absence of its gravity endangers Earth, or its orbit is changed permanently? $\endgroup$ Apr 12, 2021 at 20:43
  • $\begingroup$ Is the energy gamma radiation, thermal or other and in what amounts?. A lethal dose of radiation is about 700 joules but a lethal does of heat is about 5.58*10^6 joules. $\endgroup$
    – user64888
    Apr 13, 2021 at 21:36

2 Answers 2


stop earth from being destroyed

We start with 1.67x10^39 J, and we just need to ensure that the cross section of the explosion with Earth doesn't exceed 2.24x10^32 J.

First, we know that the distribution of the energy at a certain distance is even over the surface area of a sphere at the given radius. So the energy density of the explosion at a distance R is E/4PiR^2.

The cross-section of earth is the projection of the sphere on another sphere, which is just a circle. This cross-section is Pi*R^2 with a radius of 6371 km. Don't worry about some parts of the earth being closer, the energy intercepted at a closer point is the same as if it would have been encountered at further away.

The amount of energy from the explosion at a distance r is then the energy density at a distance r times the cross section of the earth.

So we are solving for 1.67x10^39/4PiR^2 * Pi6371000^2 = 2.24x10^32. Where these are equal we will be at the minimum safe distance. This simplifies to 1.694610^52/R^2 = 2.24x10^32 which further simplifies to sqrt(1.694610^52/2.24x10^32) = R. Which gets a minimum distance of 8697803 kilometers or 22.6 times as far from earth as the moon is. The earth will be barely not destroyed. the energy density is 1.756610^18 Joules per meter. with a head down human cross section being 0.18 meters square. A human facing the event would get 3.1610^17 joules of energy. the heating the would endure if the radiation is all heat is. 3.1610^17 = 62 kg * 3.6 kj/kg * heat change or 1.415*10^12. So they would be thoroughly dead. But earth would not be destroyed.

safe for humans value

All the energy is heat

For a human safe value we need to work backwards. Assuming all the energy is heat we need to find how much energy we need to cause heat stroke. The average temperature of the earth is 15 Celsius, the maximum survivable temp most humans can endure is 40 Celsius. Note that you still get heat stroke and organ failure, but it is survivable in theory. The atmosphere will absorb about half of the radiation, so we can double the energy we need. 62 kg * 3.6 kj/kg * 25 = 5580000 Joules. Maximum energy density is 25580000/0.18 or 62000000 joules per meter. solving 1.67x10^39/4PiR^2 = 62000000 simplifies to 1.67x10^39/4Pi62000000 = R^2 which is sqrt(2.1410^30) = R so the final distance is 1.4610^15 meters or 1.4610^12 kilometers or 0.154 light years. That is 324 times further from the sun than Neptune is. That isn't far enough to be in a different star, in fact it won't even be outside of the Oort cloud.

0.154 light years

All the energy is gamma or x-ray radiation

Gamma/X-ray radiation is much more dangerous than heat, but it is also absorbed more by the atmosphere. About 700 joules will kill an average person. luckily, air will stop the radiation. Air halves the amount of radiation received for every 90 meters, so values similar to 1*10^-30 are about how much you need to multiply incoming radiation to account for the atmosphere.

Saved by the atmosphere, at least until it starts raining radioactive air.

All energy is Radio

Radio can pass through the atmosphere. 100 mW/cm^2 is a clear hazard according to this study. that study doesn't give a clear lethal dose, but 1000 times that might be a lethal dose in a super nova burst (taking into account the smaller time period). So we need about 1000000 joules. sqrt(1.67x10^39/4Pi1000000) = R gets a distance of 1.1510^16 meters or 1.1510^13 km or 1.22 light years. that is still in the Oort cloud, but a bit further.

1.22 light years

I have had trouble finding sources on lethal doses of UV and Visible light.

  • $\begingroup$ Thanks. I'd say that 1.22 ly is a good minimum distance, seeing your calculations. $\endgroup$ Apr 14, 2021 at 3:54
  • $\begingroup$ I live in Brazil, and there some days where the temperature exceeds 40 C. Although very discomfortable, it isn't really live threating. A few times I got temperatures exceeding 50 C that although being extremely discomfortable and even inducing mental confusion, I am still alive and I am surely not the only one. Also, some people that work close to ovens, furnaces or grills frequently are inside an air mass that can be over 60 C, and many have to work for long hours daily with that. So, I think that you need to redefine a bit that part about such temperature. $\endgroup$ Apr 14, 2021 at 4:36

I would teleport the villain deep into intergalactic space, hundreds of thousands or millions of light years from the nearest galaxy. Thus the probability of being near enough to a planet to destroy or damage it would be very low. and if the heroes have enough time and sufficient data they might select a location which they know is at least a few hundred light years from the nearest star in intergalactic space.

Unless the villain is suicidal, they somehow have the ability to survive the supernova explosions they create. So transporting them as far away as possible from any stars or planets will keep them from making trouble for anyone for as long as possible.

If the villian isn't teleported while in a spaceship or space suit, and can't survivie without breathing, teleporting them into intergalactic space will kill them, ending their threat forever.

If the villain can survive, they may be stuck without means of transportation in intergalactic space and thus never get near enough to anyone to ever be able to harm anyone.

If the villain can survive and also travel in space, they might be so far away that they can't find their way back. The villain might not be able to recognize which galaxy they came from and so head toward it. And maybe the villain only hates Earth people and doesn't have any desire to destroy other worlds.

If the villain can find their way back, it might take them a long time to do so, and everyone who would have died in their supernova explosion will live for the length of time it takes to the villain to return. And possibly the heroes might be able to defend Earth again when the villain does return and save everyone again.

If the heroes are capable of extremely accorate teleportation, they could teleport the villain inside the event horizon of a massive black hole. According to current physics, nothing can get out of a black hole, so any explosion within it should have no effect on the outside universe. Unfortunately, the ability of the heroes to teleport someone many light years and into a black hole is impossible according to modern physics, so if they can do that maybe some effects of the explosion can leave a black hole and devastate nearby star systems.

If the heroes don't have time to plot a spcific destination for the villain, teleporting him at least a hundred million light years in any random direction should be safe enough. Any random location in the universe should be at least a few thousand light years from the nearest galaxy, and probably tens or hundreds of light years from the nearest isolated stars in intergalactic space, and thus from the nearest potentially life bearing planets.

  • $\begingroup$ Problem is, he can teleport countless lightyears, and survive being at the epicenter of the explosion, even though unleashing it uses up a vast majority of his internal energy reserve, and forces him to take a rest, lest he risk death. He can also survive the vacuum of space, but he does get so angry he tries to kill the protagonist, but she kills him. Yes, it will look, but I'm looking at the minimum distance, and I'm sure that "well into extragalactic space" is much larger than the minimum. Great thinking though. $\endgroup$ Apr 12, 2021 at 19:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .