Sunday, April 13, 2014

Asteroid deflection

Interesting article below. (sorry to make this a post, but it was too large to be allowed as a comment.) However, the writer of that linked article needs to brush up on his basic math a bit. :).

He describes the asteroid as "three to five times larger" than the dino-killer, yet also lists the size ranges for both as: 6 miles wide vs. 23 to 36 miles wide. 

What's the actual difference in size for the small end estimate of 23 miles?
For Chixcalub, at 6 miles wide it has a volume of 113 cubic miles.

But for the big one, at 23 miles wide, it's 6370 cubic miles, so it's 56 times larger.

The good news is that the bigger they are, the easier they are to see. So, we'd probably have plenty of advance warning of a big asteroid that size.

The bad news; it's a minimum 6370 cubic miles of rock. That's a lot of mass.

Now, if we assume we see it a decade before it's going to hit, and it'll take us 5 to build something and get it out to it, that's 5 years from impact.
How much detla/v do we need to impart? Also fairly easy; worst case is it's heading for the center of earth's disk, so that means we need to move it by the radius of the earth plus a couple hundred miles for margin and to clear the atmosphere. so, say 4000 miles.    And we've got 5 years. The math is easy; 4000 (miles needed to move) divided by 1820 days (days in 5 years) is 2.1978 miles per day. that works out to 8 feet per minute: impart that much delta/V in almost any direction, and you'll move the trajectory at earth by enough to miss.

But, how much force would it take to get a mass like that moving at 8 feet per minute? It's weight dependent. So, how much does a rock 23 miles across weigh, anyway? That's not hard to figure out. If in density is close to a mid-range rock, like crushed river rock (asteroids are thought to be aggregate, so river rock works well for a density substitute) then we're looking at about, roughly, two tons per cubic yard. That makes this super easy; 
5,451,776,000 cubic yards in a cubic mil3, so each cubic mile weights. rounding a bit, 11 billion tons. And we're dealing with 6370 cubic miles, so, 70,070 billion tons, round it to 70,000 billion, and that means 70 trillion tons. 

So, to get 1 ton moving 8 feet per minute, what kind of force do we need? One foot pound for one second would impart a delta/v on one pound of (rounded) 32 feet per second. Divide that by 2000 (pounds in a ton)and multiply by 60 to get feet per minute, and we get .96 so, rounding again, we can say one foot pound for one second  imparts a delta/v of one foot per minute to a ton. So, we need a foot pound for eight seconds to impart the needed 8 feet per minute. But, as we're dealing with 70 trillion tons, we need 70 trillion foot pounds for eight seconds. Or, to break it into minutes, 9.33 trillion foot-pounds for a minute.

To put that in perspective, the Saturn 5 first stage fired for about 2 minutes, and produced about 7.6 million foot-pounds of force. So, how many Saturn 5 first stages mounted on the asteroid would we need?
613,815 Saturn5 first stages.

Kind of hard to do, eh?
The good news is there's an easier way: use the asteroid as its own reaction mass.

50 megatons has a potential of 1,527,159,999,997,874 foot pounds. 1527 trillion. We need to impart 70 trillion, which is about 4.6% How do we do it? Standoff detonation. You won't get a shockwave as you would in atmosphere, so the majority of the energy is imparted as radiation, all across the spectrum. A large chunk is heat (photons).

When this full spectrum radiative pulse hits the asteroid, it would vaporize (up to a few inches in depth) the surface of the hemisphere facing the blast, flashing it to plasma. This plasma would violently expand away from the asteroid. What you'd have would be, in effect, rocket thrust across almost half of the asteroid's surface, which the greatest force in the area directly facing the nuclear blast. It's a lossy process, so you might need to do it twice.

One enormous benefit; you don't need your bomb carrier to match course and speed with the asteroid. A high relative velocity would be fine, so that saves a huge amount of spacecraft delta/v. Secondly, the rather primitive, over-engineered tsar-bomba weighed 30 tons (and we can't test it in time, so we'd need to use a design we know works, and in bomb engineering, that means heavy, so let's assume 30 tons). You'd need less than a TLI equivalent delta/v to send it on an asteroid intercept course, but it's certainly feasible to send a 35 ton spacecraft through 4kps from LEO (about what you'd need)

It's a hell of a lot more feasible to think of doing that twice than to think of sending over half a million Saturn5 first stages through at least twice that Delta/v to get them to the asteroid and install them there.

And that's why, for asteroid (Anything more than a few hundred yards in diameter) deflection, I think all the non-nuclear options are bogus; they require a course and velocity match with the asteroid (far more Delta/v) and require a lot of upmass. They simply aren't practical unless you have many decades of lead time, and even then, against a large one, they're rather useless.

Asteroids are indeed a threat, but due to their nature, we'd likely see a big one coming in time to do something about it. What really worries me are comets, which unless previously seen tend to be discovered about a year out from Earth's orbit. Some of them are as large, or larger, than the asteroid mentioned above. Hale-Bopp, for example, had a nucleus about 60 miles across. While it's basically a dirty snowball, at that size, it'd easily mass as much as the large asteroid we're talking about, and make as big an impact. And we wouldn't have enough lead time to do anything about it.

The dinosaurs would still be alive if they had a space program. Shouldn't we have one?

1 comment:

ken_anthony said...

Why sorry? This blog is yours as much as mine. Make posts as often as you like.