What mechanism could cause Krypton to explode?

[Andy_L]

@Dr.Strangelove

In case anyone’s curious about the binding energy equation and the crazy dependence on r[sup]5[/sup] and weird constants… here’s the derivation.

First we figure out the gravitational energy for a small lump of matter. The force equation is simply:
F = G*m[sub]1[/sub]*m[sub]2[/sub]/r[sup]2[/sup]

To get energy, we integrate forcedistance, or:
E = ∫[r…inf] Gm[sub]1[/sub]m[sub]2[/sub]/r[sup]2[/sup] dr =
Gm[sub]1[/sub]m[sub]2[/sub]∫[r…inf] r[sup]-2[/sup] dr =
Gm[sub]1[/sub]m[sub]2[/sub]|[r…inf] -r[sup]-1[/sup] =
Gm[sub]1[/sub]*m[sub]2[/sub]/r

Easy enough. Now, we want to see how much energy it takes to take a layer at a time off of the planet. We integrate again over the radius, this time from the original radius down to zero, slicing off thin layers until there’s nothing left.

The mass of a thin layer is just the surface area times the thickness, times the density (p), or:
m[sub]2[/sub] = 4pir[sup]2[/sup]pdr

And the mass of the rest of the planet up to a radius r is just the volume times density, or:
m[sub]1[/sub] = 4/3pir[sup]3[/sup]*p

Plug into our first energy equation and stuff inside an integral:
E = ∫(r…0) Gm[sub]1[/sub]m[sub]2[/sub]/r =
G∫(r…0) (4/3pir[sup]3[/sup]p)(4pir[sup]2[/sup]pdr)/r =
G16/3pi[sup]2[/sup]p[sup]2[/sup]∫(r…0) (r[sup]3[/sup])(r[sup]2[/sup]dr)/r =
G16/3pi[sup]2[/sup]p[sup]2[/sup]∫(r…0) r[sup]4[/sup] dr =
G16/3pi[sup]2[/sup]p[sup]2[/sup]|(r…0) r[sup]5[/sup]/5 =
-G16/15*pi[sup]2[/sup]*p[sup]2[/sup]*r[sup]5[/sup]

And there you go. The r[sup]5[/sup] and p[sup]2[/sup] factors really make for a lot of energy.

It’s sometimes convenient (if you’re the type of person who does calculations of gravitational binding energy often enough to be concerned about the convenience of that calculation) to express this as -0.6GM[sup]2[/sup]/R, where M is the total mass of the planet, by the way

[kaylasdad99]

@Dr.Strangelove

To make a planet explode, you need something on the order of the gravitational binding energy of its matter. When objects fall in on each other, they release energy–and to pull them apart requires the same energy.

I don’t know about Krypton, but we can compute the binding energy for Earth. The equation is:
-16/15*pi[sup]2[/sup]*p[sup]2[/sup]GR[sup]5[/sup]

p is the density and R radius. For Earth, that’s p=5510 kg/m[sup]3[/sup] and R=6371000. That comes to:
E = 2.24e32 J

For comparison, a 1 megaton bomb is 4.2e15 J, so you need about 50 quadrillion of them to blow up Earth.

If you can get your hands on 1.24 trillion tons of antimatter, that would do the trick. But that’s a little hard to come by.

Yabbut how would you weigh it to know you have enough? It’s not like you could put antimatter on a scale.

[dzeiger]

Krypton Oil and Gas got really, really, aggressive with their fracking and wastewater injection wells.

[Dr.Strangelove]

@kaylasdad99

Yabbut how would you weigh it to know you have enough? It’s not like you could put antimatter on a scale.

You can, in fact. You need a containment vessel, of course, but if you know the mass of your vessel, you can weigh the whole thing and take the difference. Antimatter behaves largely the same way as normal matter, and (unless much of what we know is wrong) falls “down”.

It’s a bit like the “birds in a van” puzzle (i.e, how much does a 5000 pound van with 100 pounds of flying birds in the back weigh? It’s 5100 lbs.). The antimatter may not be touching the vessel, but whatever is levitating it (lasers, electric fields, etc.) will push down with the exact same force (on average) as the weight of the antimatter.

[Dr.Strangelove]

@Chronos

Eh, the topic’s been covered plenty of times before, and the binding energy of a planet is easy to look up. Munroe usually covers things a lot more bizarre than that.

Ahh, but do you read Munroe’s columns in Dr. Strangelove’s voice? It’s important for the full effect. (also, I did re-derive the equation)

I’m certainly not Randall, but I’ve always liked the idea behind his What If columns–that you can actually answer some pretty fun questions using roughly high-school physics. Even some basic unit analysis goes a long ways.

And yeah, I’m by no means the first to suggest the binding energy answer to this question. But hey, there’s always someone around in the 10,000.

[Lumpy]

How much energy would a mini-black hole release as Hawking radiation in the last hour of it’s life? Would that be enough to explode a good-sized planet?

[Dr.Strangelove]

@Lumpy

How much energy would a mini-black hole release as Hawking radiation in the last hour of it’s life? Would that be enough to explode a good-sized planet?

Nope. Just a little one.

Wikipedia gives this as the evaporation time for a black hole of mass M:
t[sub]ev[/sub] = 5120piG[sup]2[/sup]*M[sup]3[/sup]/hbar/c[sup]4[/sup]

For t[sub]ev[/sub] = 3600, I get M = 3500000 kg. Using good old E = mc[sup]2[/sup], that’s 3.15e23 joules, or 75,000,000 megatons.

Plugging back into the binding energy calculation, using a density similar to Earth, I get a radius of 108 km. Just barely enough to blow up Phoebe (assuming all the energy were used efficiently, which wouldn’t happen).

[Ranger_Jeff]

Maybe some Geology grad student at Krypton State just got really drunk or his girlfriend dumped him for a theoretical physics student or someone on the football team and he and programmed Krypton’s version of Deep Thought to divide by zero in order to get even?

[Grim_Render]

@Anne_Neville

A bit unlikely- most astronomical instruments aren’t developed by lone wolf inventors, but by a group of scientists.

Jor-El was a member of the ruling Science Council of Krypton and probably had a research staff and budget at least equal to a major university if not minor country.

@Anne_Neville

Well, except for the fact that the smallest black hole we’ve found so far is 3.8 times the mass of the Sun. It’s about 24 km in diameter, so it would fit inside a planet, but it’s not going to whiz around inside the planet- you wouldn’t have a 3.8 solar mass object in a tight orbit around the center of a planet. You’d need two black holes smaller than the mass of the planet. We haven’t found even one black hole this size, let alone two that collided with the same planet. If there are black holes this size, then you have to have a planet capture two of them in very small orbits. Unlikely. I suppose if the Kryptonians had some way of manufacturing small black holes it could happen.

While black holes are easier to detect the larger they are, there is nothing in the theory that cuts them off at any size. Short-lived Quantum Black Holes are fairly accepted by physicists today. And yes, theoretically manufacturable.

@Chronos

Difficulty there: While black hole collisions can release insane amounts of energy, it’s almost all in the form of gravitational waves, which will mostly just pass right through the planet.

Now this I don’t know a lot about, but would not the gravity waves be very strong at the point of collision (inside the planet!), and tear the planet a new one due to tidal stresses? Which incidentally would look a lot like it exploded to an outside observer.

[Chronos]

What gravitational waves did get absorbed would do a lot of hurt, that close. It just wouldn’t be a literal world of hurt (unless you went way overkill with the mass of the black holes), since no matter how you arrange it, most of the energy is just going to escape.

[Leo_Bloom]

@kaylasdad99

Yabbut how would you weigh it to know you have enough?..

“Yabbut”– Great. Did you just coin that?

[Anne_Neville]

@Ethilrist

What’s the current theory about the formation of the asteroid belt? A common theme in SF is that it’s a former planet.

It’s not a former planet. What we think happened is, Jupiter was close enough to the asteroid belt that its gravity prevented the material there from coalescing into a larger body.

If the asteroid belt were a former planet, it would be a very small one. We think the total mass of the material in the asteroid belt is about 4% of the mass of the Earth’s moon.

@kaylasdad99

Yabbut how would you weigh it [antimatter] to know you have enough?

Very carefully.

Actually, assuming that our current understanding of physics is right and antimatter would interact gravitationally the same way normal matter does, you could figure out its mass by orbiting a satellite around it.

[SimonMoon5]

@Anne_Neville

But then how did kryptonite get spewed across the universe?

Come to think of it, the existence of kryptonite on Earth means the event that shattered Krypton had to be even more energetic than what would be required to shatter a planet. The explosion not only shattered a planet, it propelled at least some of the pieces outside of that planet’s home solar system.

In the pre-Crisis universe, this was explained by the existence of space-warps which spread kryptonite through out the galaxy (especially near Earth). Possibly, baby Kal-El’s ship opened a warp that brought kryptonite to Earth, but I may be conflating the pre- and post-Crisis histories.

In the post-Crisis universe, there was only one chunk of kryptonite that traveled to Earth, as the exploding planet sent out a chunk of rock that attached itself to baby Kal-El’s ship.

In the new52 universe… eh, I don’t know, I haven’t been reading it.

[dtilque]

@Anne_Neville

It’s not a former planet. What we think happened is, Jupiter was close enough to the asteroid belt that its gravity prevented the material there from coalescing into a larger body.

To be more specific, the current best idea (as far as I know) is called the Grand Tack Hypothesis. It postulates that Jupiter formed about where the asteroid belt is now, moved inward for a while, then “turned around” and moved outward. During both phases of movement, it scattered some planetesimals which ended up being the asteroids.