Built on Facts

Nuke the Moon

Reader Scott writes in with a question:

Okay, let’s assume something knocked the moon out of its orbit and it is going to crash into the earth in an arbitrary amount of time. (7 days?)

You are humanity’s last hope. You have the entire nuclear salvos of the USA and Russia at your disposal, which are being fitted to be capable of being launched into outer space and to reach the moon at another arbitrary rate. Your goal is to knock the moon back onto its original orbit by striking it from the side in such a way that will redirect its net force back to a position that will result in the moon having its original distance from the earth and angular momentum.

Is this possible? If so, under what conditions?

i-4e5345dbfc97c4141fe477ac976b6794-bomb.png
Fig. 1: The continuation of astrophysics by other means.

(The answer is sort of long, so continued below)

The short answer is that if the moon’s going to hit us in 10 days, we’re toast. There just ain’t no way around it, and the world will be sterilized as thoroughly as if you put the entire planet in a giant autoclave. For smaller and more realistic dangers like asteroids, and given more warning time, we might have a chance. Let’s run the numbers, shall we?

First, let’s talk a little bit about how an explosion can move something. Newton’s laws require that an isolated system with no external forces must have a constant momentum. Fire a rifle and the forward momentum of the bullet is matched by the equal and opposite backwards recoil of the rifle. A nuclear explosion on the lunar surface is not so different – the momentum of the blast debris leaving the moon is matched by an equal and opposite recoil of the entire moon. Momentum is mass times velocity, so the momentum of the blast is very roughly the mass of the bomb times the velocity of the fragments. The moon’s recoil momentum in the opposite direction will have the same value, but the moon’s mass is huge. Huge mass means the corresponding recoil velocity is small.

How big do we need the recoil velocity to be to save us? That’s an extremely difficult question in general, but as a rough estimate we might say that we need enough velocity to move the moon (or asteroid) by a distance equal to the diameter of the earth in the time before impact in order for it to miss. Given the earth’s diameter and a week to move that distance, this corresponds to a velocity of about 20 meters per second. This doesn’t sound so bad. Given a lot more time – maybe 10 years for a very faraway asteroid detected early – we only need to give it a velocity of 0.04 meters per second. This is downright promising.

So how much impulse can we get from a bomb? This too is an extremely difficult question in general. We can’t get the velocity of the fragments directly from the energy because plenty of it will be lost in the form of low-momentum electromagnetic radiation and other inefficiencies. The blast isn’t perfectly directed away from the moon either. Nonetheless we can give ourselves the benefit of a doubt. Let’s say all available nuclear energy goes into the kinetic energy of the fission fragments and it’s all directed away from the moon. Using the usual classical equations for kinetic energy and momentum, and assigning the Greek letter rho = E/m for the released-energy-per-mass of the nuclei, we can derive an equation:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Savvy readers may object to my use of the classical rather than relativistic expressions here, but we’re ok. Typical nuclear fusion energies are in the ~5 MeV per nucleon range, which is well below the mass of a proton or neutron and thus mostly classical. For electrons we’d have no choice but to use the relativistic equation, but for the heavier nuclear particles the classical expressions are still quite accurate.

All right. Taking a wild guess that the innards of a fusion bomb are about 100 kg and using the energy density estimated above, a single bomb should net us an impulse of 3 x 109 Newton-seconds. Translated into less inside-baseball terms, this means each blast can give a 1 kilogram target a velocity of 3 x 109 meters per second, or a 3 x 109 kilogram target a velocity of 1 meter per second, or anything else with that ratio. Extrapolating to the 7.3 x 1022 kg mass of the moon, one blast will change the velocity of the moon by… 4 x 10-14 m/s. Yikes. Considering we need 10 m/s total, that’s quadrillions of bombs. As I said at the top of the post, it ain’t happening.

Asteroids are another story. The near-earth asteroid 99942 Apophis has a mass of only about 2 x 1010 kg. That’s only a dozen or so bombs, even with only a week’s notice. Now our estimate is very generous in terms of the effects of one bomb, but it’s certainly plausible that a few dozen or a few hundred bombs might well do the trick if we caught it reasonably early.

Let’s hope we never need to test this experimentally.

[The title for this post is stolen from a tribute to the immortal IMAO essay of the same name]

Comments

  1. #1 The Science Pundit
    August 5, 2009

    Don’t forget that the moon (or an asteroid) is also spinning. Unless the blast is conducted at one of the poles (of the axis of rotation), you run the risk that some of that energy from the blast will go into changing the impending object’s angular momentum without really changing its course.

  2. #2 Uncle Al
    August 5, 2009

    If you hit or pulse thrust the moon hard enough to make a difference, you’ll shatter it. It’s going to shatter when it enters its Roche radius from the Earth anyway. The moon’s current orbital radius/Roche limit ratio is 41:1. Bingo at center-to-center distance of 5900 miles. Sum of Earth and moon equatorial radii is 5042 miles. Whole or fragmented won’t make much difference.

    The most effective tactic is to sign many strongly worded Enviro-whiner petitions about the presumed effects of terrestrial impact. The moon smashing into the Earth is carbon-neutral, so Al Gore will see no revenue from it.

  3. #3 Willy
    August 5, 2009

    Aren’t you only assuming the bomb’s mass in the equation? Isn’t where the bomb detonates important? If it’s burst at or above the surface (less than 100 meters) then your calculations are generally correct. However, should the bomb be detonated subsurface then there would be (hundreds/thousands of) tons of lunar debris ejected and that would increase the thrust on the moon (via conservation of momentum.)

    Still, whatever got the moon to disengage from its orbit probably would’ve already messed things up here on Earth.

  4. #4 anon
    August 5, 2009

    Once upon a time in Classical Mechanics, they told us that, when a cannonball explodes, the center of mass of the resulting debris field follows the same path that the cannonball was on before the explosion.

    If, then, the moon is approaching the earth at 25000 miles/day, and we land a shipload of nukes on it and blow it into gravel, we will have a moon-sized cloud of radioactive gravel, some of which will be coming our way even faster.

    How is this an improvement?

  5. #5 SimplyHarmonic
    August 5, 2009

    @anon,

    The exploding cannon ball probably explodes due to some internal mechanism, not because someone hit it with a bomb from the side. The idea is to blow the moon into radioactive gravel that has just enough sideways momentum that it won’t murder everything on Earth. Unfortunately, as Matt pointed out, we can’t.

    Actually, if we can hit the moon hard enough to make sufficiently small radioactive gravel pieces, most of it might burn up in the atmosphere. Although, that is likely to still be really bad news.

  6. #6 Matt Springer
    August 5, 2009

    #3, a subsurface blast will eject a lot more mass, but that mass will also be moving a lot more slowly than the near-lightspeed fission fragments of the nuclear fuel itself. We’d also expect to lose a lot more energy as disorganized heat. Nonetheless it could well be that we might be able to squeeze some improvement out by adding more inert mass. It won’t radically change the overall difficulty though.

  7. #7 Kobra
    August 5, 2009

    @2: Also, the change in the tides will be plenty catastrophic, even if we were to somehow correct the moon’s orbit.

  8. #8 Joshua Zelinsky
    August 5, 2009

    Did you not know that NASA is already planning on this? http://www.youtube.com/watch?v=Csj7vMKy4EI

  9. #9 Robert
    August 5, 2009

    You need a thermonuclear device optimized for the production of neutrons. Neutrons penetrate the surface and heat the outer shell of the asteroid enough to blow it off. Only material ejected fast enough to exceed the escape velocity counts; if stuff goes up and falls back to the surface there is no recoil. Kinetic and gamma ray effects of the blast are relatively minimal.

  10. #10 travc
    August 6, 2009

    For the more plausible deflecting an asteroid scenario, you point out very well why some sort of cannon or mass-driver makes a lot more sense than a bomb. Not so much to shoot the asteroid (though that plausibly could work in some cases), but much better to put it on body and shoot chunks at high speed in the opposite direction to the way we want the body to move. I appreciate that you used the term “recoil”, because that evokes the right metaphor (as well as being technically correct).

    Also helpful, the mass expelled to generate the recoil can be (somewhat) controlled so that it doesn’t hit the Earth either… not so much if a bomb is used. Of course, the line between a carefully placed subsurface bomb and a gun is a bit blurry. For the moon spiraling down that quickly though, we’d be totally screwed.


    Anyone want to do the math regarding what it would take to stabilize the Moon’s orbit. It is drifting away after all, and that seems like something we could plausibly change.

  11. #11 Robert Waldmann
    August 6, 2009

    Seems that you never watched Armageddon (Bruce Willis nukes and asteroid).

    The calculation is get momentum from kinetic energy using
    mv = 0.5mv^2 so the larger is m the better. You assume only the bomb gets blasted away.

    Nah if the bomb penetrates before detonating it can push lots of moon rock much bigger M much smaller v much more momentum.

    Can bombs penetrate before blowing up ? Sounds tough, but some of ours can (Russky bombs would be pretty much useless funny how no one mentioned that during the cold war).

    More importantly you make a crater with small bomb 1, then send bomb2 into the crater. Whole lot of moon is gonna leave. It won’t go in the right direction and you have to make sure that it is well above escape velocity (in the calculation you assume zero gravity).

  12. #12 John Whitesell
    August 6, 2009

    While a momentum calculation is probably best for an accurate estimate, I think you need a conservation of energy equation to get a ballpark figure. This is because going with momentum alone, you are likely to get either unrealistically high or low estimates about the efficiency with which the bomb’s energy is translated into kinetic energy. As Robert Waldmann points out, you wouldn’t be detonating the bombs on the surface, a deeper explosion would result in more mass to expell and be much more feasible, something which your momentum balance allowed you to overlook.

    A back of the envelope type energy calculation reveals that we need about 2.5*10^25 Joules of kinetic energy to give the m/s change in velocity you want. We can estimate that there are about 20,000 nukes in the world and the average yeild is somewhere around half a megaton. A megaton is thermodynamic work, not total energy, so work done should be a sizable fraction of yeild. So the nukes total energy is around 8*10^19. That means we are still screwed, but assuming around 50% of yeild becomes kinetic energy, we only need billions not quintillions of nukes to achieve your 20 m/s.

  13. #13 Matt Springer
    August 6, 2009

    Removing more of the Moon’s surface in the blast should be equivalent to raising m and lowering rho by the same factor. The improvement in impulse should thus go as the square root of added mass. The limiting factor is the need to not let the velocity of the ejected mass be less than the lunar escape velocity. That should be a fairly easily tractable calculation, and I think I’ll probably write an update exploring that possibility.

  14. #14 Robert
    August 6, 2009

    What about the orbital mechanics? If the moon is approaching, that means it’s orbital speed is increasing. To raise it’s orbit, wouldn’t you need to decrease the orbital speed? So exactly which direction do you need to aim the recoil?

  15. #15 anon
    August 6, 2009

    Another problem is that any impact large enough to knock the moon loose from its orbit has certainly also created a lot of fragments of various sizes; we will have to deal with hundreds of big moon chunks, not just one intact moon.

  16. #16 Robert
    August 6, 2009

    You don’t necessarily need an impact to get the moon on a collision course with the earth. A close passage by a massive object could do it. But I’m wondering what kind of orbit or trajectory the moon would have, and what exactly would it take to make the moon safe again. It seems to me that if you simply push the moon away from the earth, and look at it in a rotating reference frame, it would be like throwing a ball in the air and the moon will eventually continue its fall toward the earth.

  17. #17 foo
    August 6, 2009

    I hate to add to this kind of speculative discussion, but here goes: you are ignoring orbital mechanics here. It actually helps a bit. If you make the minimum assumption of what “hit the earth” and “miss the earth” mean, you could interpret it to mean: the moon is now on an orbit that grazes the surface of the earth (R_perigee = 6000km). Assume the moon is still at apogee. We want to boost the orbit enough at apogee to make perigee large enough to be comfy. Someone above suggests keeping it above the Roche limit (where tides rip it apart) and implies that value is ~10,000 km. Let’s make a minimal, but slightly more conservative requirement that R_perigee = 2R_earth = 12,000km. The velocity at apogee is sqrt((mu/a)*(R_per/R_ap)). mu = 400,000 km^3 sec^-2, a (semi-major axis) is about 200,000km, R_ap in both cases is about 400,000km, so what matters is R_per. For 6000km, v = 17.3cm/s for 12,000km, v = 24.5cm/s. Thus, we “only” need to boost the moon’s velocity by 7.2cm/s. Not that this is easy (and we have to get the nukes there in zero time…)

  18. #18 Robert
    August 6, 2009

    Here’s another way to look at it. The moon has an orbital velocity of 1,000 m/sec. In order for the moon to fall into the earth, its orbital velocity has to be somehow reduced to near zero without destroying it.

    In order to get the moon back to its normal orbit, you’d have to accelerate its mass of 7 x 10^22 kg from a velocity of near zero to 1,000 m/sec, again without destroying it.

    So the needed kinetic energy is about 0.5 x (7 x 10^22) x (1000)^2 joules = 3.5 x 10^28 joules.

    If a 1 megaton nuclear bomb produces 4 x 10^15 joules, you’d need energy equivalent to 9 trillion bombs.

  19. #19 johno
    October 3, 2009

    no no no no i see the earth surrounded by dust and the earths tides and weather in kaos with 90% fatalities…/.this is not a thing to play with you dont really know what will happen anyway .. they say that im a seer please dont do this i was shocked and horrified when i herd this besause i saw this a long time ago.johno

  20. #20 rcroftensen
    October 6, 2009

    Your a moron we need the moon to keep the waves in the ocean and we should just send rockets up and push it away from the earth – it is now very close to the earth.

  21. #21 NJ
    October 9, 2009

    no no no no i see the earth surrounded by dust and the earths tides and weather in kaos

    But Maxwell Smart will save us!!!!!!

  22. #22 Keith M Ellis
    October 10, 2009

    Your a moron we need the moon to keep the waves in the ocean…

    There is something classic about “your a moron”. Also, tidal forces are not what creates “waves in the ocean”.

    This question, as well as the general phenomenon of the “nuke the Moon” discussion on the Internet the last couple of days, demonstrates a less moronic, yet still disturbing level of widespread scientific ignorance. It reminds me of talk about nuclear war “blowing up the Earth” during the Cold War era and related discussions about using nukes to trigger geological events (which is certainly possible, but in a “straw that breaks a camel’s back” rather than brute-force sort of a way). People greatly overestimate the energy released in nuclear weapons relative to the geological or astronomical context. But there’s a sort of popular atavistic irrationality associated with all nuclear technology, in general.

  23. #23 Mitchell Kidder
    October 14, 2010

    If we were no nuke the moon, would nutons 3rd law work. Theres no air in space so every force has an equal and opposite force. And also nukes make big boom wouldnt it blow up the moon. I mean the moon is smaller than the earth. Also there would still be pieces of the moon so it would come at the earth.

  24. #24 wqert
    February 2, 2011

    the moon is fake, placed there to distract ppl from the internet

  25. #25 altin ├žilek
    February 2, 2011

    no no no no i see the earth surrounded by dust and the earths tides and weather in kaos with 90% fatalities…/.this is not a thing to play with you dont really know what will happen anyway .. they say that im a seer please dont do this i was shocked and horrified when i herd this besause i saw this a long time ago.johno

  26. #26 adsense hack
    February 5, 2011

    Your a moron we need the moon to keep the waves in the ocean and we should just send rockets up and push it away from the earth – it is now very close to the earth.

  27. #27 Steve
    September 14, 2011

    Destroying the moon to stop it from hitting us would still destroy Earth as the moon is responsible for Earth maintaining an axis of 23 degrees. Any disruption to that would see chaotic and destructive weather patterns which would ultimately obliterate the planet.

    Mars is a good example of this, it has only 2 small moons with tiny amounts of gravity but the axis of the planet shifts dramatically and frequently.

    I can’t see how humanity could survive with current technology as the only way to “reset” the moons orbit around Earth would involve some kind of elaborate tractor beam or something of that ilk. The moon is actually moving away from the Earth at the rate of about 1 inch a year so getting it to its original orbit would need some incredible tech.

    I reckon Picard would have the problem solved in 40 mins – he always did! :)

    Scary problem – hope we never have to find out.