Aero wrote:A small asteroid, discovered yesterday, will pass close to Earth tonight.
My problem to you is just this. If this asteroid actually was headed for a bulls eye strike on earth, what would it take to deflect it? Here are some questions to ponder:
1. If it actually struck Earth, what would its impact velocity really be? Certainly more than the quoted 18,000 mph, as Earth's escape velocity is like 25,000 mph. (I would guess impact velocity to be more like 40,000 mph.)
The velocities don't add, but the energies do. The specific kinetic energy of the asteroid is (8100 m/s)^2, while the specific kinetic energy it would get from falling to the Earth is at most (11200 m/s)^2. Add that together, and you get a specific kinetic energy of 190000000 m^2/s^2, for a impact speed of the square root of that, or about 14km/s (31,000 mph).
2. If it was a nickel iron asteroid, what would be its mass? I guess, assuming it is spherical, that it masses something over 2000 metric tons.
Nickel-iron asteroids are about 7-8tonnes/m^3, so a 7m diameter spherical asteroid would be... about 1250-1400tonnes. Call it 1300 tonnes.
But a spherical rock 7m in diameter would be the exception, not the rule. Most asteroids are irregularly shaped, and therefore smaller than they would be for the same diameter than if they were spherical. Two common asteroid shapes are "cigar-like" and "dumbbell". If it were dumbbell shaped, we could approximate it as two spheres each 3.5m in diameter, and that would cut it's mass by a factor of 4, down to about 350tonne.
3. What would be its energy release on impact? I'm guessing it would be quite a bit more than the 4 kilotons quoted in the news article, more nearly 100 kilotons is my guess. What would be its radius of destruction?
For 1300 tonnes at 14km/s, I get 60 kton tnt.
For 1300 tonnes at 8.1km/s (their quoted speed), I get 20kton tnt.
That's for nickel-iron. The vast majority of meteorites are not nickel-iron, and have a density of about half that, and thus about half the energy (so about 10-30kton tnt). Add in the irregular shape, and you lose another factor of 4, so we're talking 2.5-7.5tkon tnt.
But that's the total energy dumped into the Earth. It doesn't account for energy dissipated in the atmosphere burning up the asteroid, etc. It wouldn't surprise me in the least if a "typical" asteroid of that size had a low-altitude airburst in the kton range, at most. Remember that Little Boy (Hiroshima) was about 10-15kton tnt.
Now for the fun part! With total freedom to use anything we have seriously discussed on this forum, what would it take to divert this hypothetical asteroid? That is, we have two days notice, so our system obviously has to be in place, docked somewhere, but not on ready alert though its crew should be stationed nearby. What will this system look like? I'm guessing that it is not a gravity tractor.
Probably not a gravity tractor, but let's see what we can do... I'm going to relax the conditions a bit, and assume we can get to this asteroid with 2 days to spare. At that point, it would be 1.4GM away. Let's say we wish to divert it so that it misses the center of the earth by 20Mm (about 3 Earth radiuses, or two Earth radiuses above the surface). The delta-V required would be 115m/s.
The rocket equation says the mass ratio (initial/final) of a rocket needs to be equal to e^(delta-V/exhaust-V). For the Shuttle engines, the exhaust-V is about 4400m/s, so the mass ratio is 1.03. To move a 1300tonne rock, you'd only need 39 tonnes of fuel. An SSME could produce that much delta-V in under 2 minutes, and add only 3.2tonnes. Not bad, a Saturn V could lift that much payload (but not really with enough to spare for matching speed with the rock).
With a VASIMR, and an exhaust-V of 100km/s, the mass ratio is even lower, so only 1.5 tonnes of fuel would be needed. If we spread that over 2 days, that's less than 1000N of thrust, which could be handled by 200 VF-200 engines, each at 300kg, which would unfortunately weigh in at 60tonne. The chemical rocket is the better choice.