# If Earth was converted into energy by antimatter-matter collision would the energy “destroy” the observable universe?

If half the earth was made antimatter and contacted the other half all at once, would the resulting energy be enough to "destroy" the observable universe?

I did some math based on the mass of the earth being around 5*10^24kg and an antimatter explosion given a mass of 1kg being about 43 megatons, and a nuclear explosion with a 4 megaton load leaving a fireball with a 1 mile radius. So that would be 1kg of antimatter and matter = about 10 mile radius and 5*10^24kg of antimatter = 5*10^25 miles radius where as the observable universe of 13.8 billion light years can be converted to 8*10^22 miles < 5*10^25 miles. Is this accurate?

• Explosion radius does not scale in a linear way with explosive ..force/power(?). At the very best the radius is the cubed root of the raw explosive power & AFAIR it might be the 4th root. – Andrew Thompson Apr 8 '16 at 5:57
• @AndrewThompson I believe the word you were looking for is explosive yield. – user Jul 21 '16 at 9:13
• @MichaelKjörling Yes. The 'explosive yield' was the term I was after. :) – Andrew Thompson Jul 21 '16 at 10:57

Assuming the earth has a mass of 5 x $10^{24}$ kg, the energy released would be $$E = mc^2 = 5\times 10^{24} \times 9\times 10^{16}= 4.5\times10^{41}\text{ J}$$ The diameter of the earth is about 13,000 km, so the resulting radiation pulse will last about $$t = \frac{13\times 10^3}{3\times10^8} = 4.3 \times 10^{-5} \text{ seconds}$$ At what distance will the energy density be lowered to 1 J/$m^2$? That's easy $$A = 4\pi r^2$$ so $$r = \sqrt{\frac{A}{4\pi}} = \sqrt{\frac{4.3\times 10^{41}}{4\pi}} = 1.9\times 10^{20} \text{ m}$$ Which equals 6 x $10^{11}$ light seconds, or only about 19,000 light years. Of course, it will look like God's own lightbulb, with a peak intensity of roughly a quarter of a MW per square meter at this range, but the duration is too short to do much damage.

So the answer is no, it won't destroy the universe. It won't even do any damage outside of our galaxy, and not even most of the galaxy, at that. The local neighborhood will get pretty well trashed, though. At 19 light years the energy produced is in the neighborhood of a megajoule per square meter, which ought to be pretty tough on anybody who's looking in our direction.

• nicely formatted. I have been working on a similar answer, but yours seems to be structured much better. You could add where the formulas you're using come from though :) – Doomed Mind Apr 8 '16 at 14:47
• Does the anti-matter contribute to m? I just don't actually know, but if it does, there's a missing factor of 2. – Scott Whitlock Apr 8 '16 at 16:12
• @ScottWhitlock - the OP says the earth is half matter, half AM, so I ran the numbers with the total equal to the "real" earth. And even if that's not what was intended, it only produces an error of cube root of 2, or 25%. – WhatRoughBeast Apr 8 '16 at 16:45
• @DoomedMind - You need a source for E = mc^2? For the area of a sphere? Huh? – WhatRoughBeast Apr 8 '16 at 17:53
• And if this scenario is not catastrophic enough for you, consider an earth made of protons and a moon made of electrons: what-if.xkcd.com/140 – Shufflepants Sep 16 '16 at 17:58

I'd like to borrow WhatRoughBeast's number:

$$E = mc^2 = 5\times 10^{24} \times 9\times 10^{16}= 4.5\times10^{41}\text{ J}$$

Now consider a fantastical unit, known as the foe: $10^{51}$ ergs ("foe" being a shorthand abbreviation for "Ten to the Fifty One Ergs"). That's about $10^{44}J$

This is a unit used by astronomers when talking about the energy emitted from supernovae. Supernovae typically range in the 1-2 foe range.

Thus your antimatter-earth explosion, while certainly violent, would register as an explosion roughly one tenth of a percent of the size of a supernova.

Space is big. Really big.

• Right on, although a supernova takes a great deal longer than the hypothetical perfect mixing of matter/antimatter which I tacitly assumed. So, with a longer duration the peak power of a supernova will be a good deal less. – WhatRoughBeast Apr 10 '16 at 15:17