# 1000 kilos of matter and 1000 kilos of antimatter...

Okay, forgive my nerdish broodings, but this has bugged me for a while. A Star Trek: Voyager episode - I forget the name… it’s the one where they come across an intelligent Cardassian missile/ship that Bel’anna programmed - anyway, the episode featured this highly-advanced missile that, after a malfunction, was heading towards a random alien planet, intent on causing massive amounts of destruction.

(Manny, Chronos, Jill… I SWEAR this post ends in a GQ.)

It’s mentioned that the warhead of the missile consists of 1000 kilograms of matter and 1000 kilograms of anti-matter. We all know that when antimatter touches matter, the entire mass is converted into energy, resulting in a big boom (which makes one wonder why they bother adding 1000 kilos of regular matter… oh well). One of the characters, upon hearing of the warhead’s stats, comments, “That’s enough to destroy a small moon.”

Well, okay. That sounds impressive… until you think of really dinky moons like Phobos or Deimos orbitting Mars. And then there are probably some truly HUGE moons running around out there in the cosmos.

So, seeing as how I don’t know the maths needed to figure such a ratio out, how much energy would this fictional (and plot-device-ridden) missile release, and how big of a moon would that destroy?

That would be a BIG explosion, though not sure HOW big. I heard somewhere years ago that if you converted one gram of matter completely to energy it would release more energy than the most powerful fusion bomb…anyway, I’m sure someone will come along who will be able to at least come up with a figure for the amount of energy, though it will be harder to figure out what kind of damage that would do.

Badtz Maru: I think you might be looking for E=MC[sup]2[/sup]

M=0.001 Kg x C[sup]2[/sup] = 299,792.458 * 299,792.458

= 89875517.87 (2dp) Joules

SPOOFE Bo Diddly: I can’t add anything yet but I will keep on searching…hang tight

Foundations of Astronomy (Sixth Edition) Michael A Seeds

So by E=MC[sup]2[/sup]

We get 1000 x 299,792.458 x 299,792.458 = 8.99 x 10[sup]13[/sup] joules

1 Tonne of TNT releases 4 x 10[sup]9[/sup] joules
1 Tonne of gasoline releases 48 x 10[sup]9[/sup] joules

[Carl Sagan’s Voice] The boundless potential for an entire Cartooniverse FILLED with Phobos-like moons expounds blithely upon the frontal lobe of the Cosmos, caressing each cerebral fold with it’s billions and billions of moon units. [/Carl Sagan’s Voice]

Cartooniverse

p.s. Spoofe? I love you man, ahhhh the mind that can come up with a rigorously difficult physics thread based on Star Trek. Nothing nerdish about it >sniff<. Takes a keen mind, a superior intellect and the ability to do THIS ( Vulcan Greeting inserted here ) on command.

Using E = mc[sup]2[/sup]:

E = (1000 kg)(2.9979 x 10[sup]8[/sup] m/s)[sup]2[/sup]

E = 8.99 x 10[sup]19[/sup] J

With the energy of 1 ton of TNT corresponding to 4.18 x 10[sup]9[/sup] J, the 1000 kg antimatter weapon has a yield of 21,500 megatons of TNT.

The largest thermonuclear (hydrogen) bombs were about 100 megatons…

Now hopefully, I didn’t make a math error as well.

[You know, I wonder if I should be using the total initial mass of 2000 kg (1000 kg of antimatter and 1000 kg matter). If so, double my numbers above.]

This is all assuming that you can get every single gram of antimatter to slap into a gram of matter, rather than having the leading front of the lump o’ antimatter go ‘boom’ and send the rest of the lump back the way it came.

Still. 21.5 gigatons. That’d leave a mark.

I think we should be using the total mass of matter and antimatter, which would give us (from robby’s calculations) 43000 megatons of TNT. To provide a sense of scale, that’s roughly 5.3 x 10[sup]-10[/sup] lunar masses or (using a small moon) 0.36% of the mass of Phobos.

Even a small moon is a pretty freakin’ big object; I’m not sure you could break it up with that amount of explosives applied at a single location on the surface, although you could certainly blow an impressive hole in it. <scrounges around the web some more>

Here’s a site with an interesting table of crater sizes vs. energy release for meteorite impacts (which seems reasonably analogous to this scenario). From this, we might estimate estimate that our missile would create a crater around 65-70 miles in diameter, assuming surface detonation. That’s over twice the diameter of Phobos on its long axis–a large portion of the energy would no doubt be lost. If the missile burrows deep before detonation, of course, all bets are off.

Summary: Maybe that missile, properly deployed, could destroy Phobos (the moon, that is–Phobos the poster may be immune to lame plot devices). As usual on Voyager, handwaving would have been better–“The missile has a matter/antimatter warhead massive enough to destroy a small moon…it will wipe out every lifeform on the planet.” would be a workable substitute.

How, exactly, would you contain antimatter?

The standard approach to antimatter containment is to use a magnetic field (“magnetic bottle”) to suspend it in an evacuated chamber. If you forget to pay the electric bill, you’re toast.

It occurs to me that magnetic containment of 1000 kg of antimatter would be a pretty power-intensive operation–how much fuel did the missile have? How long was it running unattended? If it ran out, it would go boom in a big way, smarty-pants AI or no.

A coupla notes on this discussion:

1.) When you get matter-antimatter reaction it doesn’t all go into energy. When they do electron-positron annhilation studies they typically see a lot of particle-antiparticle * creation* as well. You lose some of your energy to “incomplete yield” as you make a bunch of partcles you didn’t want.

2.) Back around the time the first Star Wars movie was released (circa 1977) The American Journal of Physics ran a calculation on just how much energy the Death Star would’ve NEEDED to destroy Alderaan. AJP is intended for teaching Physics, so they LOVE this kinda stuff. I don’t recall the result, but you can look it up. It was a lotta Joules – enough that you’d want to make it by annhilation. (Side note: at least THree different articles on the Tunguska explosion have posited that it was due to matter-antimatter or “contraterrene” matter interaction. The Tunguska event was only on the order of kilotons of explosive yield.)

3.) Hi, Opal! (I’ve wanted to do that for a while. Note that I have used it properly.)

I haven’t studied physics since secondary school and I’ve had a bit to smoke tonight, so correct me if I’m wrong, but surely the above calculations are wrong.
If you’re using E=mcc then m=2000 as there are 1000 kilos of matter and 1000 kilos of anti-matter.
I’m not going to do the sums as I couldn’t be bothered but isn’t that right?

I know, I know.

As I was going to bed I realised it took C in kilometres per second rather than metres per second.

I was praying that no-one would notice till the morning, whence I could correct it. But good job none the less