Death Star Explosions: The Foliage Factor

More calculatory bliss.

On a forum it was recently asked just how powerful the Death Star explosion over Yavin IV could've possibly been . . .that is, "what is the most energy released by the Death Star that Yavin IV could have survived?"

Well, that sounded like fun.

A beginning estimate would be to basically figure out a level of energy that would observably damage the moon on the surface. This is useful because of three factors.

1. As the Rebels take off to go fight the Death Star, the red gas giant is visible in the sky. This suggests that the Yavin base ought to have been within line of sight of the Death Star blast.

2. After the battle, we get two useful scenes.

A. We see a brief shot of the planet from a great distance as the surviving Rebels fly home.

B. We see a close shot of the jungle near the Rebel base with its impenetrably dense stone before the medal ceremony.

So, as a result, we know that the condition of the base and nearby jungle was basically the same before and after the Death Star was destroyed. With that fact in mind, we can estimate an energy intensity that would have produced observable effects, then rock the inverse square law back out to the range of the Death Star. Since the planet and jungle are not observably affected, this would serve as a back-of-the-envelope upper limit.

- Atmosphere

Now, in reality, there would be a number of other factors to consider, such as atmospheric attenuation (especially in the sense of what parts of the spectrum make it through), cloud cover more specifically, and so on. So while we could use surface effects and ignore the atmosphere, technically the actual energy would need to be greater. For sunlight on Earth, it is said that about 50 percent makes it to the surface to be absorbed, 30 percent gets reflected at the surface, and 20 percent is absorbed by the atmosphere. That 20 percent is what we need to zoom in on at the moment, because I presume the 30 percent figure is a general value including such things as oceans (what with their being 75 percent of the planet and all), which are not relevant to our inquiry.

Just in case, though, let's assume that a third of the radiative emission of an exploding Death Star would be absorbed by the atmosphere . . . that means we need to increase our relevant values below by 33 percent.

- Range

As for the range to the Death Star from Yavin, most estimates are in the 150,000km range, which I believe is true for both the SFJ pages and Wong's SDN. (Also, some nincompoop on SDN claimed I am the source for a very specific figure from Michael January's pages, but that's just a funny thing I've been wanting to share and is not relevant right now.)

I haven't personally studied the matter, but we'll roll with this figure for now as it is not really inconsistent with the ~77,000km range of the Alderaan shot.

- Damaging the Surface

For ease, we can use estimates from http://glasstone.blogspot.com/2006/04/ignition-of-fires-by-thermal-radiation.html . . . that for a nuclear explosion, the energy required to ignite assorted things like dry and dark rotted wood falls somewhere in the range of 15-25 calories per square centimeter (my own selected range from the wide assortment of figures he reports). Now, Yavin IV didn't exactly look dry, so we could crank this up further to 50 just to try to be sure, and to try to cover the 33% atmosphere absorption mentioned above.

I don't see where a precise time factor is provided, but we can presume for the purpose of this that the nuclear weapon and the Death Star blast would radiate for about the same amount of time.

Translating to joules per square centimeter, then, with a calorie being 4.184 joules, that's 209 J, which in square meters would be 2,090,000 . . . so,  two megajoules per square meter. (The total energy incident on Yavin would be interesting but not relevant here.)

- The Inverse Square Law

The inverse square law is a fancy term for the idea that the intensity of something per unit area at a range of, say, one meter (e.g. an intensity of 1 watt per square centimeter at one meter's range) will fall off in a geometric fashion. So, at two meters, you'll actually see the intensity drop to 1/4th watt per square centimeter, because while the distance has only doubled, the affected area will quadruple. At three meters it's 1/9th, et cetera. And note that while I used watts above, joules work just as well if you don't need a time factor . . . indeed, the inverse square law applies to all sorts of things, what with it being fairly basic geometry, applied.

The fun part for our purposes, though, is that you can also estimate the strength of the actual source itself if you know this intensity at a given range. This isn't really the inverse square law per se, but it is often discussed in the same breath.

So, in the case of 1 watt per square centimeter at one meter's range, we can work backward. It is 100 centimeters to the point where the intensity is 1 watt per square centimeter, meaning that if we want to work backwards to a 'zero point' all we need to do is get the total wattage for the whole sphere. That's just figuring the whole area of the sphere and multiplying the known wattage on one square centimeter accordingly. So, a 100cm sphere has an area of 4πr^2, or 4π(100cm)^2, or 125,663.7cm^2. Since we know the intensity is 1 watt per square centimeter and we know the whole sphere is 125,663.7 square centimeters, then the point source of energy must be outputting at a rate of 125,663.7 watts, or about 126 kilowatts. (That's a lot, but so is a watt per square centimeter at one meter's range. Per the basic geometry, you'd be looking at 10,000 watts per square centimeter at 1 centimeter range . . . yet the sun only hits us with 0.14 watts per square centimeter, and we only receive about 0.1 watts per square centimeter at the surface.)

- Rocking the Law

So with 150,000,000 meters as the range and two megajoules per square meter as the energy intensity at that range, the total energy of the Death Star blast would be 4π(150,000,000)^2 times two megajoules, or 282,743,338,823,081,391.5m^2 times two megajoules. That gives us about 565,500,000,000,000,000 megajoules, or 565,500 exajoules, or 5.655E23 . . . or, about 135,000 gigatons.

If, however, the estimated range is wrong and it was really, say, 500,000 kilometers (500,000,000 meters), then the final value would be much higher, in the range of 6.3E24 J, or about 1.5 million gigatons.

The sun puts out 4E26J every second, meaning that the Death Star explosion upper limit from the first  example would be equivalent to 1/700th the energy output of the sun at the range and with the energy on the surface specified.

Given a range of 500,000km, the comparison comes out rather better, with the Death Star explosion coming in at about 1/64th of the sun's energy every second.

However, given the way the Death Star II played out so very much closer to the surface of Endor, all of these figures seem ridiculously high.

- Addenda

I seemed to recall doing something like this before, so I went to look and found that this is a different technique (insofar as calculating via affecting foliage) than what I used in regards to getting an upper limit for the DS2 explosion in the Superlaser Effect section of my site. See #4 at http://www.st-v-sw.net/STSWsuperl-5.html and you'll see a technique involving evenly heating the atmosphere by fifty degrees Celsius. While I still don't think that was a completely terrible approach, I think the full methodology above is much superior.

Applying it, and using the 3500km value, we get 3E20 J as the upper limit, which is 1/1,333,333 of the sun's output in one second.

Even that is very much too high, given that we know exposed skin was unaffected by the DS2 blast . . . two megajoules per square meter in a short timespan like that would definitely provide a touch of sunburn.  Leia didn't even seem to notice the DS2 had gone up until several seconds later, so we can reasonably suppose that she was not burned, blinded, et cetera.

3E20 joules . . . about 73.6 gigatons . . . is thus, to my mind, a very solid upper limit value for the DS2 destruction, and probably applies with caveats to the first Death Star, to boot.

Also, just for bonus points, and to actually apply the main part of the inverse square law a little, if we assume the Falcon and friends were only about 500 kilometers from the second Death Star when it blew, then they would've been getting about fifty times more energy per square meter, or 100 megajoules per square meter.  The Falcon would therefore probably have had to deal with, say, 20,000 megajoules . . . 20 gigajoules . . . incident upon her, at worst case.  That's a bit less than a five ton bomb.

1 comment:

William R York as templerman said...

Wow! I am really dumb because I still am not sure what your answer is. Would Yavin-IV/Endor have survived the DS's explosion?