Armageddon prevention

I was watching the movie Armageddon and it made me think of a few things. We would probably not need to send a crew to blow the asteroid in two, I think we could still send a warhead with enough power to blow it out of our orbit if we had enough time. This is just for my own interest but i do not know enough physics to do this. So i decided to ask the experts.

I want to know the best type of nuclear warhead to use to deflect this asteroid, What is the explosive power needed and how much time NASA will need to launch this warhead.

I made a little research on a few items we may need to figure this out so hopefully i got all the information. It would also be nice on the process you take to figure this out. I think it would a great conversation starter.

I think the asteroid was made of iron, so lets say this asteroid will be iron so the density of iron is 7500.0 $kg/m^3$, 10,000 $m$ in diameter and a velocity of 20,000 $m/s$

The largest nuclear warhead ever created clocks in at about 50,000 megatons, so how far would the asteroid need to be for NASA to have enough time to deflect the asteroid?

If the launch speed is 11 $km/s$, how long would it take for the warhead to reach the asteroid?

And so based on the velocity of the warhead, and the asteroid, and assuming NASA can launch the warhead as soon as the asteroid is detected, how far away from the warhead impact point can the asteroid be when detected and still be deflected and not crash into the earth?

I'm excited to share what I find from you guys with the physicists I know and see what they have to say. :)

• JFYI the more elegant approach would be to tug away the asteroid by using gravity - en.wikipedia.org/wiki/Gravity_tractor – Hubble07 Sep 18 '14 at 18:22
• This is not a homework assignment. It is for my own curiosity CuriousOne ;) That is why the homework tag was not originally added. Qmechanic added it when he edited my question. – user2856118 Sep 18 '14 at 21:12
• Hi user2856118. Welcome to Phys.SE. (For the record: CuriousOne's and my actions were independent.) If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Sep 18 '14 at 22:29
• I see, thank you for that explanation. That makes sense. – user2856118 Sep 18 '14 at 22:51

Forget Armageddon and Deep Impact. Movie physics is not real. It's a movie, after all.

Should some large dinosaur killer class asteroid or comet be on a trajectory that eventually impacts the Earth, our only hope is to detect that object decades in advance. A large number of options exist given adequate advance warning.

No options exist were we to detect that large body just a few weeks before impact. Blowing it up may well result in even more damage than had we just let it be. It's still going to hit us, but now we'll have lots of pieces covered with radioactive debris heading toward us. Do it just wrong and we'll lose our atmosphere.

Edit

We would need a huge amount of lead time to divert an extinction level asteroid. A hundred years might do it. Those once in a hundred million year or so events are not the key worry. We should be able to see those a hundred years in advance. It's the Tunguska-style events (50 meter diameter asteroid) that occur every few hundred year or so years (some say more, others less) and the Ch'ing-yang-style events (100 meter diameter asteroid) that occur every few thousand years (once again, some say more, others less) that scare people.

To put things in perspective, we never saw the Chelyabinsk meteor (< 20 meter diameter asteroid) of February, 2013 before it hit Chelyabinsk. Those things are very numerous, but not so damaging. Lots of people were injured, lots of buildings were damaged, lots of money was lost, but no one died. However, damage is roughly proportional to the cube of the linear size. A fifty meter asteroid would cause over twenty times the damage of that Chelyabinsk meteor; a one hundred meter asteroid, almost 200 times the damage. A Tunguska-style event over the center of a major metropolitan area would kill hundreds of thousands of people. The death toll would be in the tens of millions with a Ch'ing-yang-style event.

The problem with these relatively high frequency events is that even with advanced detection capabilities, we'd be lucky to have a decade of advanced warning. That rules out things such as the gravity tractor in WetSavannaAnimal aka Rod Vance's answer. That technology requires multiple decades using existing technology, and with but a decade's advanced warning, we wouldn't have time to do research.

We would need a few years to nail down the orbit so we could intercept it, a few years to prepare for the mission, a year or so to wait for the launch window to open, and a few more years for the vehicle to get there. Some of those things can be done in parallel, but even with that, a decade doesn't give much time.

To make matters worse, we wouldn't have a decade. We would have at most eight years. Even with nuclear weapons, the weapon would have to explode about two years prior to impact to change that impact into a miss. The delta V needed to change an impact into a miss is a highly non-linear function of time before impact. If the action occurs but a few weeks prior to impact, the energy needed to turn that impact into a miss would exceed the yearly energy consumption by all of humanity. The curve is nearly exponential with short time before impact. The knee in the curve is around 400 or 500 days. That means we would need to take final action a couple of years or so prior to impact.

The most reliable way to use a nuclear weapon to divert an asteroid is to make it explode at a short standoff distance from the asteroid. This wastes over half of the energy from the explosion, but it is something we know how to do. It is "Technology Readiness Level 9", or TRL 9. The explosion bathes that side of the asteroid with gamma rays, X-rays, and high energy nuclides. That near instantaneous pulse of energy vaporizes a small layer of one side of the asteroid. It's this secondary explosion rather than the primary explosion of the bomb that changes the asteroid's trajectory.

A nice side effect of performing this standoff explosion a couple of years prior to impact is that even if the asteroid is a rubble pile and is blown apart, it will reform as a rubble pile within a year or so. The primary explosion doesn't do much to the asteroid; that explosion occurs in the vacuum of space. The secondary explosion ejects that thin layer of vaporized asteroid, but the kick given to the rest is less than escape velocity.

Alternatives to a nuclear standoff explosion

There are a couple of other approaches involving nuclear devices, an explosion at the surface and an explosion below the surface. The surface level explosion would impart more energy to the asteroid, a subsurface explosion, even more. There are multiple challenges with both approaches that reduce the readiness level well below TRL 9.

Another approach is kinetic impactors. We've done that before; for example, we intentionally dropped an expired satellite into a crater near the Moon's South Pole to determine if those polar lunar craters contained ice. They do. The problem with kinetic impactors is that they don't have near the impact of a nuclear bomb, and rubble piles and rotation present even more challenges. We would need to use a number of impactors working in concert with one another, once again reducing the readiness to well below TRL 9.

The gravity tractors mentioned by WetSavanna fall into the broad class of slow push / slow pull approaches, with the gravity tractor being at the top with regard to readiness. There have been a number of papers published on gravity tractors, with a wide range of time spans needed to generate the necessary delta V. The papers that pay close attention to technology, to control theory, to the very lumpy gravity fields of small mass (< 1 km diameter) asteroids, and to the quirky rotational behavior of these asteroids all say that the rockets must fire for many years to generate the requisite delta V. The papers that say that only a few weeks of thrusting is needed are written by people who ignore technology issues, who don't know control theory, who don't know the nasty consequences of those lumpy gravity fields, and who don't know the nasty consequences of the polhode rolling without slipping on the herpolhode lying in the invariable plane.

There are a number of other approaches as well. Most of them have some critical piece that is somewhere between TRL 1 to TRL 3. Technologies oftentimes take decades make it out of that low TRL quagmire. That's not good when millions of lives might be lost within a decade.

References

The best reference to date is the 2006-2007 study performed by NASA. It's bit outdated, but it does cover the main concepts quite nicely. NASA's 2007 Report to Congress is a bit brief and shorts on the science and engineering a bit. The 2006 Near-Earth Object Survey and Deflection Study preliminary study report contains a lot of substance. Section 6 (pages 72-119) cover deflection alternatives.

Another widely used reference that summarizes available options is Brent Barbee's Master's thesis: Barbee, B (2005), Mission Planning for the Mitigation of Hazardous Near Earth Objects. As is the case with the NASA report, nuclear standoff is the clear winner, both in terms of delivered delta V and not requiring any technological breakthroughs.

For a set of more recent references, look to the Asteroid Deflection Research Center at Iowa State University. A summary page at nasa.gov describes their most recent research. Details can be found at the Asteroid Deflection Research Center web site.

• I have forgotten the movies. Those variables are not from the movie, I was just giving them as for the research i did. The movie did make me think of this. But the warhead will not go into the center of the asteroid. I want to shot it from the earth to the asteroid, and i want to know how far would the asteroid need to be for NASA to have enough time to deflect the asteroid, and how long would it take for the warhead to reach the asteroid. – user2856118 Sep 19 '14 at 0:42

One way to tackle the problem is with usage of an antimatter based bomb. It can yield enough destructive power in order to vaporize an asteroid. Before anyone starts to comment about how much we have produced antimatter in history or how much energy it would take in order to accumulate enough antimatter for the given purpose I want to remind that pushing this option is the silver bullet so to speak.

Antimatter bomb might seem to be unviable option today but scientific understanding tend to improve in huge steps. So, who knows, we might have the option in future.

• True: Tsar Bomba in its full configuration could have released about $4{\rm kg}$ of energy. One gigaton weapon is about $40{\rm kg}$ of energy. So I'd imagine that 1000 times that (if we could throw 20 tonnes of antimatter into the asteroid) and we're beginning to talk of serious destructive power given the calcs in my answer. You might have problems like "predetonation", where the exploding mass shoves itself away from the asteroid before full matter-anitmatter combining is complete. – WetSavannaAnimal Sep 19 '14 at 3:41

I'd like to add a little to David Hammen's answer. As he says, the centre of mass of the asteroid-nuclear weapon system is going to pass through the Earth no matter what happens. This is true simply by the principle of conservation of momentum (given that there are no outside forces on the rocket-weapon system) (see footnote).

Let's first address the idea of "nuking the threat". I guess most people have in mind that a fearsome bomb is going to vaporise the whole thing. Nothing like this would happen even in the most favourable conditions. As you say, the biggest bomb ever built was the Tsar Bomba, with a likely yield of a hundred megatons if fully laden (it was stripped of its natural uranium tamper at the last minute to prevent nuclear fallout, which means actually delivered "only" 57 megatons). From scale factors here (TNT Equivlent Wiki), I reckon 100MT TNT to be $4.2\times10^{17}{\rm J}$ (incidentally equal to roughly the amount of sunlight falling on the Earth in about four seconds!). Suppose the asteroid is 10km across, i.e. roughly $1000{\rm km}^3=10^{12}{\rm m}^3$, equating to about $5\times10^{15}{\rm kg}$ if it is made of rock. Optimistically, the asteroid's specific heat capacity is around about $10^3{\rm J\,kg^{-1} K^{-1}}$, so even if we could transfer all of the weapon's energy to the asteroid (there is no way we shall do this), we have enough energy to raise the whole thing's temperature by about 0.1K (=°C)! The Tsar Bomba was built in 1961 at its reason for being is best summarised by Khrushchev's promise to show the West Kuzma's Mother, i.e. it was part of the threat and bluff upholding Mutually Assured Destruction. Its actual military use would have been next to nought (it weighed 27 tonnes and was too heavy to take anywhere): there is only so much destruction you can wreak with a nuclear weapon before the excess energy from making the bomb bigger simply dissipates into outer space and the Tsar Bomba was well past this point. It's a pretty fair bet that nuclear weapons research since then has concentrated on "much smaller" weapons ("only" about 1MT) since the only driver for this research has been military applications. So the concept of a gigaton weapon almost certainly overbounds what we could do today on short notice. Even such a weapon will only raise the asteroid by 1° C! So no vaporisation is going to happen. That's aside from the fact something the weight of the Tsar Bomba is going to take something approaching the Saturn V launch vehicle to deliver it. Saturn V could put 47tonnes into trans-lunar injection.

The only good a bomb is going to do you is to shatter the asteroid so that the shards mostly miss the Earth. You can think of splitting the asteroid up into equal portions and giving them all equal sideways (i.e. normal to the Earth-asteroid line) velocity components so that their centre of mass still passes through the Earth but the shards miss the Earth. More likely, you would try to use the weapon to ablate (cleave and blast off) a smallish fragment so that fragment and main body have equal magnitude but opposite sign sideways momentum so that they both (or at least the main body) miss the Earth and its atmosphere. Even a glancing blow that bounces shards off the atmosphere could be dangerous; without interaction with the atmosphere, the asteroid's kinetic energy must be enough to escape Earth's gravitational field. Interaction with the atmosphere will dissipate some of this energy and allow Earth to gavitationally capture the shard, which means it will come back and pose the same threat at a later date.

In all of these possibly workable scenarios, I think you can see that we do need to burrow into the asteroid so that we can use the relatively small energy of the weapon to split the asteroid apart. Whether this is workable I do not know: it has now become a purely mining engineering problem. I should imagine this would all be fraught with great uncertainty: since we would know little about any flaws or faults inside the asteroid it would be anyone's guess as to which way it's going to split if at all. As in David Hammen's answer, all kinds of dangers can arise from even small shards hitting the Earth's atmosphere if they tally up to a significant fraction of the asteroid. The only mining engineering research humans have done that I am aware of that would be vaguely relevant is Project Plowshare.

I think you can see that even the nuclear weapon option is only workable if painstakingly planned and undertaken a long time ahead of impact. In contrast to all this uncertainty, the other methods, such as the Gravity Tractor are much more certain and much better behaved. The Gravity Tractor works somewhat like the Gravity Slingshot in reverse: you can think of the tractor and asteroid as being tethered by "rubber bands" as in Terry Bollinger's excellent description of the slingshot. Effectively, the rocket's thrust bears on the rocket-asteroid system: the effect, although small, is extremely well understood and certain to deflect the asteroid given enough time.

I've not seen Armageddon and Deep Impact made little impact on me: I've almost forgotten it.

Footnote: As David Hammen points out, all systems are orbitting the Sun. So, over longer timescales (probably more than the few weeks you seem to imply), the Sun needs to be included in the system to apply momentum conservation. In the short term, we need to be thinking of the Earth as one system, and the weapon-asteroid as another: both are freefalling and their centres of mass are on a collision course. Not much is going to change that in a few weeks, no matter what shape or collection of shapes the asteroid may end up in.

• Re As he says, the centre of mass of the asteroid-nuclear weapon system is going to pass through the Earth no matter what happens. I did not say that. It is not true. A gravity tractor is a nice approach with many decades of advance warning. With a decade or less, the nuclear option is just about the only option. With a few years or less, there is no option. – David Hammen Sep 19 '14 at 2:29
• @DavidHammen How so? Even in the gravity tractor system it is true if you take account of the propulsion gasses in a gravity tractor, i.e. the whole system, if there is no third body involved. The only way it cannot be true is if you include some other body in the system, i.e. deflecting the body into the influence of e.g. the moon. Is this what you mean? If so I can word to be a bit clearer. I also corrected a typo: I meant to say that the gravity tractor is much more CERTAIN (rather than uncertain) because it is much less dependent on the unknown structure inside the asteroid. – WetSavannaAnimal Sep 19 '14 at 2:50
• @DavidHammen Actually David you sound pretty sure about your timeframes: it sounds as though you've done some careful reading about what justifies $<10yr\Rightarrow \text{only nuclear}$ and so forth, so it might be good to cite your references in your answer. I think my answer shows that this is a highly technical question needing a great deal of research to answer it fully that is likely not withing the scope of Physics SE, i.e. the weapon option seems mainly like a mining engineering problem to me. – WetSavannaAnimal Sep 19 '14 at 2:59
• Even in the gravity tractor system it is true if you take account of the propulsion gasses in a gravity tractor, i.e. the whole system, if there is no third body involved. While that would be true in empty space, but not when things are orbiting. A simple counterexample: Consider a pair of objects orbiting the Sun as one. The push against each other to separate, with the $\Delta V$ directed along the orbital velocity vector. At some point in time, the objects will be on opposite sides of their orbits: The center of mass will be close to or inside the Sun. – David Hammen Sep 19 '14 at 3:11
• @DavidHammen "The center of mass will be close to or inside the Sun": as it always was in this three body problem, which is why I emphasised third body. I think the timescales are a bit vague in the OPs question, but they seem to imply doing something whilst the asteroid is weeks off. I've added a footnote to the answer. We need to be thinking of the Earth as one system, and the weapon-asteroid as the other: both are freefalling and their centres of mass are on a collision course. Not much is going to change that over the course of a few weeks. Also, do you have any references - I guess ... – WetSavannaAnimal Sep 19 '14 at 3:25