## Tuesday, March 20, 2001

### Talk about astronomical chances …

It seems that Taco Bell is going to give everyone in the US a free taco if the Mir hits a 40'x40' (40 feet by 40 feet, or 12 meters by 12 meters for the Imperially challenged) target in the South Pacific. I even heard a rumor that they took out insurance*just in case*they have to pay out.

So what are the odds of Mir (or any piece thereof) hitting a 40'x40' area?
Well, the area of a sphere is 4πr²—to make things easier let's
just say it's 12r² and use 4,000 miles (6,437 km) as r. That's
12*(4*10^{3} * 4*10^{3}) which is 19.2*10^{7},
normalize it to 1.9*10^{8} and oh, let's say 2*10^{8} or 20
million square miles (you gotta love exponents—makes the math easier to
work with). The target is 1,600 square feet (1.6*10^{3}) and there
are 5,280 feet per mile, so that makes some 28 million square feet per mile
(2.8*10^{7}). So multiply 2*10^{8} by 2.8*10^{7} and
you end up with a surface area of 5.6*10^{15} square feet for the
Earth. Divide that by the target area and you get 3.5*10^{12} which
means if my math is sound, one chance in 3.5 trillion.

Well, actually, not quite. That's the entire Earth and since most objects orbit around the equator, the likelyhood of the Mir crashing in any arbitrary location, say, the South Pole, is not likely. So let's restrict ourselves to an area between 30° North to 30° South. To make it easy (again) we can treat this as a cylinder, so the surface area is 2πrh, and plugging the numbers in and rounding 2π to 6, we end up with an area that is approximately 10 million square miles, which increases the oods to one in 1.75 trillion.

I don't think Taco Bell has much to worry about.