So a few years back I wrote Star Wars, Einstein and When Lucas Got It Right where I said, “Saying that because the Death Star could blow up Alderaan it could blow up Yavin is like saying something that can blow up Earth can blow up Jupiter.”
Naturally, that got me to thinking about how much harder it would be for the Death Star to blow up varying objects in the solar system.
I wasn’t sure how to calculate this so I asked on a math website – a website claiming to have people who were good at math and would help people who asked questions. (I didn’t mention it was about the Death Star; I put it in more generic terms.) But no one had an idea of how to do this mathematically.
Now with the new Star Wars movie having come out (which I haven’t seen yet - so no spoilers, please), it’s impossible not to think of Star Wars because it’s on TV and at the stores. And then I saw this thread (which I have not read) Death Star vs. Starkiller Base: both are unrealistic, the latter especially so (open spoilers) and that made me think about the power of the Death Star, which made me think of that piece I wrote about the power of the Death Star and wonder about the math I used (as — let me make this clear — I am not a math expert, I was just figuring it out the best way I could come up with).
So, as there are a lot of smart people on this site who clearly know math better than I do, I wanted to ask
[ol]
[li]How valid is my formula?[/li][li]What would have been a better way to calculate this?[/li][/ol]
Now, I put that piece up years ago; I’m not going to be changing the numbers. But if someone does come up with a better calculation, I’d probably do an edit of the piece giving the new information.
Point of order: At the end of Star Wars, the Death Star was not going to blow up Yavin – the rebels had their base on a moon of Yavin, and the Death Star was basically catching up to that moon on its orbit around Yavin.
The moon where the base was located was way smaller than Yavin, and quite possibly smaller than Alderaan even. So, if we accept that the Death Star was able to blow Alderaan, it shouldn’t have had any trouble blowing up the Yavin moon where the rebel base was located.
The end of Star Wars, though, is sometimes criticized, because the Death Star wastes a lot of time trying to clear Yavin to get a shot at the moon, instead of just blowing up Yavin itself. I think the OP is arguing that a laser designed to blow up (relatively) small, rocky planets would not necessarily be effective against gas giants.
I’m not really sure that volume and density matter to the equation if you’re already dealing with mass. In fact, since mass = volume * density, you’re sort of double-counting the same things. You add them up and then average, so it’s not exactly the same as double-counting, but if we’re talking about completely blowing a planet away, mass and gravity are all that really matters.
The results aren’t all that different if you look only at mass, though. Jupiter is 317x the Earth’s mass, and your calculation says it is 547x as hard to blow up. In the context of implausible super weapons, that seems close enough.
Assume the planet is a sphere to make the math easier…
Well, maybe. No specific mechanism is given for the Death Star weapon to cause a planet to “explode”. It has the appearance of some kind of really powerful laser or particle beam projector, but this really doesn’t make much sense; such a weapon would be expected to heat the facing surface and cause it to evaporate while the far side is protected from primary effects. Even enough energy to cause the near surface to turn to vapor with sufficient force to project material into an escape trajectory wouldn’t just burn through to the other side; realistically you’d expect to carve a crater or divot through the crust to the mantle; the resulting release of internal energy is probably enough to wreck havoc on the rest of the planet, but for the planet to explode uniformly like an overstuffed piñata would require some other mechanism which directly transfers momentum or globally changes the character of the chemical or gravitational bonds. Such effects may well vary with volume or density.
The minimum energy to cause a planet or other solid celestial body to “explode” and release debris into an escape trajectory as depicted would have to negate the gravitational binding energy, which can be derived as U = -(3Gm[SUP]2[/SUP])/(5*r), assuming the planet has a uniform density. For Earth, that would require the application of 2.242 * 10[SUP]20[/SUP] TJ of energy, which the Death Star appears to deliver in a pulse of less than one second, which is roughly equivalent to the average total luminal output of Earth’s sun over a 6.7 day period. On this basis the Death Star must be mostly extremely cold temperature reservoirs to avoid being vaporized by its own internal energy, or else has a means of storing and releasing energy which likely violates the laws of thermodynamics as we understand them.
Unless they use…midichlorians. I understand those things can patch up any plotholes or continuity errors.
I always assumed that the first Death Star had an enormous recharge time – days or weeks or more. So orbiting Yavin would have actually been faster than blowing it up and then blowing up the moon, since it takes it so damn long to recharge (or maybe rebuild!) the weapon between shots.
Why would they want a Death Star that can destroy a gas giant, anyway? The point of the Death Star is to terrorize planets into submission. All the SW universe’s life forms appear comfortable on terrestrial planets with atmospheres and gravity levels similar to Earth (the odd alien might be wearing a respiration device.) It appears everyone in the SW universe lives on a terrestrial planet, so that’s all the Death Star needs to be able to blow up.
It comes from the fact that you are integrating the potential energy of a mass of a spherical shell (S = 4⋅π⋅r[SUP]2[/SUP]) through the enclosed volume (V = 4/3⋅π⋅r[SUP]3[/SUP]) times the density of each. For a sphere of uniform density the constant of proportionality works out to exactly 3/5. This is a bit of a fudge because the Earth is neither strictly spherical nor of uniform density (the iron core is substantially more dense than the mantle) but I think it’s a pretty close order of magnitude estimate, assuming that whatever effects your weapon uses acts uniformly over the entire body simultaneously.
I worked out the integral in this post, if you’re interested. However, note that I did the computation in terms of radius and density as opposed to radius and mass. You can use the volume of a sphere formula to convert.
ISTM we don’t really need to offset 100% of the gravitational binding energy. Consider an explosion that shatters the planet into rocks that fly outwards for 1000 pre-explosion planet radii and then slow to a halt and all fall back into their collective gravity well to recondense.
I think it’s fair to say the planet has been comprehensively destroyed. Certainly the biosphere and all human scale constructions are done for.
So I think we can back your energy requirements down a bunch. e.g. in the Earth example we could do the job with 10[SUP]17[/SUP] TJ or maybe 10[SUP]18[/SUP] TJ tops. Depending on the breaks.
Not that that changes the big picture too much, but reducing the energy required by a couple / few orders of magnitude might help get this thing done on time and under budget. Which seemed to be an issue in the very first episode.
My problem with the death star has always been why is it necessary? It seems like it would be much easier to sterilize the surface of a planet than to destroy it entirely.
Honestly, once you can build a hyperdrive big enough to move a moon, you don’t even need a laser. Just park it in low orbit and let the tides wreck everything on the surface.