Wednesday, September 24, 2008

"Also, who the shooting back will do?", part I

As a kid I got Dragon for the game-able stuff, not the fiction. But one of my favorite pieces of fiction published in Dragon was "The Gun That Shot Too Straight" by Ralph Roberts. The story appeared in issue #94, February 1985. To sum up briefly, the story hinges on a young genius named Henry inventing an energy pistol that fires tiny balls of energy that disrupt molecular bonds. Henry's uncle, a defense contractor, is very interested in the device until they catch on the news that the test firing brought down a plane on the other side of the world. It seems the ever-expanding fireball punched all the way through the Earth, wrecking an Aussie military jet. The illo that opens the story depicts the event quite nicely:

Near the end of the story the uncle asks "I was wondering, if the energy balls continue travelling, how big will they get and what will they hit?" The young inventor doesn't have an answer. The last line of the story is the title of today's post. Over the years I've often thought about running some numbers on this scenario and today I'm going to finally do it. Please point out any math errors you spot in the comments.

Let's start with the easy one: the velocity of the fireball. We'll assume a constant velocity because we have no evidence of acceleration once the disruption ball exits the pistol aperture. According to the story it took approximately ten hours (36,000 seconds) to punch through the Earth, which has a diameter of roughly 12,742 kilometers. That's about 350 meters/second. Or roughly 390 yards/second, or about 1,200 feet/second, or about .2 mile/second if metric isn't your bag. Or 790 miles per hour. Not anywhere close to lightspeed, but not too shabby in general. By way of comparison that's about 40% faster than the muzzle velocity of a bullet fired out of a M1911 pistol.

Now let's look at the rate of expansion of the disrupting sphere. Again we're working with estimates. All we really know from the story is that the emitting aperture on the pistol is "a barely discernible pinhole" from the point of view of the old uncle. Let's call that a 1 millimeter diameter for our purposes. The fireball that brought down Captain Smythe's plane was described as "the size of a large beachball". I'm going to make the calculations as easy as possible by calling that 1 meter in diameter. So that's a thousandfold increase over the ten hour timeframe. For simplicity's sake we'll assume a constant growth of 1 meter per ten hours, or about .03 mm/second.

Tomorrow comes the fun part. We'll aim the fireball at various celestial objects and determine how long it takes to reach the target and how big the fireball is at the moment of impact. As I type this I'm not sure exactly how these scenarios are going to turn out, but I suspect that the relatively slow velocity is going to result in some pretty dang big fireballs once this baby starts hitting things in outer space.