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If a nuclear weapon is detonated on the moon it would launch debris into space. The debris would then fall to earth like asteroids.

Would this impact wipe out life on Earth? In other words, are our present nuclear weapons capable of breaking off enough big chunks of the moon to kill life on Earth?

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    $\begingroup$ Well, if you just detonated it on the surface, it wouldn't do too much. In order to break off chunks, you'd have to detonate it below the surface, which is a difficult task. $\endgroup$
    – Phiteros
    Commented Oct 23, 2016 at 17:51
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    $\begingroup$ A popular SF scenario is the dust orbiting the earth creating a permanent winter en.wikipedia.org/wiki/La_Compagnie_des_glaces $\endgroup$
    – Antzi
    Commented Oct 24, 2016 at 2:06
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    $\begingroup$ On the moon: No. In the moon: probably yes. See Seveneves. $\endgroup$ Commented Oct 24, 2016 at 9:40
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    $\begingroup$ @MartinSchröder: That too is rather unlikely. The gravitational binding energy of the moon is immense. Not as high as the earth, of course, but we're still talking megatons of antimatter. Not a few grams of mass annihilated in an ordinary nuclear explosion. $\endgroup$
    – MSalters
    Commented Oct 24, 2016 at 11:48
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    $\begingroup$ "The debris would then fall to earth like asteroids." Unlikely. You'd need to accelerate the debris in the right direction by about 2.38km/s to escape the moon's gravity, plus 1km/s to "fall" on earth $\endgroup$
    – njzk2
    Commented Oct 24, 2016 at 17:47

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Probably not. Just to give you an idea, lunar rocks hit the Earth on a somewhat regular basis. The power required to have a rock hit Earth is equivalent to that of making a 450 m crater. This comes from a 30m asteroid, roughly. The Tunguska event was caused by a 60m rock, and had an explosive power equivalent to that of a large nuclear weapon, around 15 MT. And that was far too little to damage Earth, so it's really unlikely.

Let's assume you have a diabolical setup. Let's say you have a perfect cannon, such that the entire force of the nuke will be lifted to a perfect Earth intercept trajectory. The velocity leaving the Moon would be about 2.74 km/s. The velocity at Earth would be about 11.2 km/s, which is Earth Escape Velocity (Probably slightly less). Let's say you have the highest theoretical yield weapon that has been planned, 100 MT. Energy effectively scales as the square of the velocity, thus there would be about a 16 times multiplier on the force of your nuke, in absolutely perfect conditions. That would become a 1.6 GT yield weapon equivalent. That's roughly the equivalent of a 300m asteroid impact, which occurs on average every 73,000 years. It would cause some local damage, maybe even regional, but certainly wouldn't end life on Earth as we know it.

If the rock broke in to several chunks, it could again cause substantial damage, even more than indicated above, but would most likely be limited to a region, causing significant damage over the area of something like Australia, and might cause a global cooling event, but would most likely not be a major event.

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    $\begingroup$ Did you mean cannon instead of canon? $\endgroup$
    – user
    Commented Oct 24, 2016 at 8:04
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    $\begingroup$ Can you please detail your math? I am not clear why the velocity of a nuclear bomb would affect the energy created by nuclear fission. Certainly we can talk about the mass of the bomb and its velocity imparting additional force, but I'm pretty sure that the power unleashed during fission is irrespective of velocity of the mass being moved. Otherwise any given ICBM would have hundreds of MT of yield. $\endgroup$
    – GrinningX
    Commented Oct 24, 2016 at 13:26
  • $\begingroup$ It would not affect the energy created by fission. Basically, I'm imagining a rock that was launched from the Moon in a cannon which launches the rock somehow safely using the entire energy of the bomb (Worst case). The minimum energy would be enough to escape the Moon, at the 2.7 km/s listed. When said rock hits Earth, it will be moving at close to Earth escape velocity, as listed. Thus, the rock will have in effect picked up energy. $\endgroup$
    – PearsonArtPhoto
    Commented Oct 24, 2016 at 13:46
  • $\begingroup$ Ahhhh... thank you for the clarification. Perhaps I need coffee. Yes, that's an interesting point. $\endgroup$
    – GrinningX
    Commented Oct 24, 2016 at 13:59
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    $\begingroup$ Also, what is meant by "The power required to have a rock hit Earth is equivalent to that of making a 450 m crater"? The power required to have a rock hit Earth depends on the mass of the rock, and small pebbles definitely will not make a 450 m crater, nor would require much energy to be put in an Earth-intercept course. $\endgroup$ Commented Oct 24, 2016 at 18:44
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No, even a 100 megaton bomb would not endanger life on earth. A nuclear weapon on the surface of the moon might knock some small rocks loose, but mostly it would vaporize material. Under-surface weapons wouldn't do that much either, you only have to look at the history of underground nuclear explosions on earth to see that nuclear explosions underground pulverize the material above it rather than blowing solid chunks free. A large nuke optimally below the surface would throw up lot of small pieces, some of which would make it to the earth and then harmlessly burn up. The rest would fall back to the surface of the moon.

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As a first approximation, ignoring many details: for any nuke detonated on Earth, our atmosphere has proved to be a impenetrable barrier: no debris made it into outer space. Detonating a nuke on the Moon will have the same effect: Earth's atmosphere stops the debris from arriving on Earth's surface. The debris will burn up as it enters the atmosphere.

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  • $\begingroup$ What? Moon has no atmosphere you know. Nothing to stop debris to leave except gravity $\endgroup$
    – Allahjane
    Commented Oct 23, 2016 at 18:43
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    $\begingroup$ Earth's atmosphere stops the debris from arriving on Earth's surface: the debris will burn up as it enters the atmosphere. $\endgroup$
    – Hobbes
    Commented Oct 23, 2016 at 19:26
  • $\begingroup$ An air-burst rock could potentially cause some damage. $\endgroup$
    – PearsonArtPhoto
    Commented Oct 23, 2016 at 22:44
  • $\begingroup$ @Hobbes while that may be true, it doesn't seem to follow from your original reasoning, which states that 'since no debris caused by Earth explosions reached space, no debris caused by space explosions can reach Earth'. Or am I missing something? $\endgroup$ Commented Oct 24, 2016 at 0:23
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    $\begingroup$ Nuclear debris may well have made it into space; en.wikipedia.org/wiki/… $\endgroup$
    – Richard
    Commented Oct 24, 2016 at 13:46
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If a nuclear weapon is donated on the moon it would launch debris into space. The debris would then fall to earth like asteroids.

This is incorrect, and reflects a common misconception about space.

I am not trying to ridicule you - this is a very common point of confusion shown by these XKCD submitters and the producers of GI Joe: Retaliation. But let's oversimplify it: Lunar astronauts are not dangling upside down like Australians. They do not hang from the underside of the moon by the soles of their shoes, where they could bend up towards their feet, grab a rock and stretch down towards the earth, and then drop the rock on the earth with catastrophic results. They are well planted on the moon, and chucking a rock, hitting a golf ball, blowing up a small bomb, or even firing a high-speed rifle (1000 m/s muzzle velocity) at the Earth high overhead would simply cause it to land on the moon.

Even when holding onto the ISS, throwing the rock down at the earth would (counterintuitively) not cause it to hit the Earth! It would simply change the shape of the orbit, rotating it about the Earth. Eventually, the rock would intersect your path again. Throwing it radially inward, you would need to either throw it incredibly fast so that the shape of the new orbit intersects the Earth before it gets 1/4 of the way around - which might take as little as 22 minutes. That projectile in GI Joe was not moving nearly fast enough to reach the Earth again in minutes - it appeared to be traveling prograde, meaning it would end up further from the Earth, but even going retrograde that launch speed would have required days or weeks to deorbit.

To get off the surface of the moon way, way up into low-Earth orbit requires about 5 km/s of Delta-V. From there it's downward onto the surface, requiring about 8 km/s. If you could direct the energy of your nuclear weapon into motion of a projectile - producing a big gun with a nuke instead of gunpowder - you could launch $m$ kg from the moon to low-Earth orbit with $K$ terajoules of energy according to:

$$ K = \frac{1}{2}mv^2 $$

where $m$ is in kg, $v$ is in meters (not kilometers) per second, and $K$ is in joules. 1 kg requires 12.5 MJ of energy.

And on reentry, you'd loose a lot of that energy. It would slow down and ablate in the atmosphere. You're going up by 5 km/s to LEO, then down from LEO by 5 km/s to get back to your original energy level, and then you've got an additional 3 km/s of energy which needs to make up for the losses of hoisting the nuke to the moon and all that atmospheric drag.

You're probably better off redirecting an asteroid from further out. Rather than being deep in the Moon's gravity well, you just start way above that of Earth. Speaking of gravity wells, that reminds me of another xkcd source which appears to show the Moon's gravity well as being much shallower than I interpreted it as based on these Delta-V numbers, which cite some inaccuracy. Check my work, please!

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  • $\begingroup$ moon's escape velocity is 2.38km/s and moon's orbital speed is only 1km/s. If one accelerates backward to 2.38km/s to escape the moon's influence, then forward 1.38 to compensate the overshoot, wouldn't they end up with orbital speed ~0, and literally fall on the earth? $\endgroup$
    – njzk2
    Commented Oct 24, 2016 at 17:55
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With no atmosphere on the moon, a nuclear detonation would not have a blast effect like on Earth.

The above answers about how any moon rocks hitting Earth from a nuclear explosion there all said the effect wouldn't be very much, but it would be especially little without atmosphere to result in a shock wave to cause rocks to go anywhere.

So, extreme no.

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  • $\begingroup$ Uhm, you don't need an "atmosphere" to explosively vaporise surrounding rock through thermal energy. $\endgroup$
    – JimmyB
    Commented Oct 24, 2016 at 8:21
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    $\begingroup$ @JimmyB Actually, we have plenty of data from underground nuclear tests on Earth. You'll get melting, a small bit of vaporising (which creates a cavity the surrounding rock will tend to fall into, rather than an effect similar to a steam explosion), and a whole lot of pulverising. Unless you design a nuclear-bomb-powered-cannon (a "nuclear shaped charge", if you will) you're not going to get any significant "rock through outwards at huge velocities" effect. There's a reason high explosives are used for mining (and nukes were considered) - they pulverize the rock without throwing it around. $\endgroup$
    – Luaan
    Commented Oct 24, 2016 at 9:24
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Well, theoretically at least, if you were to destroy a sizable portion of the moon, (which might take quite a few nuclear weapons) you could change its centre of mass, destabilising its orbit and eventually causing it to careen into the earth as an extinction event on steroids. If you're not too worried about crunching numbers it could be plausible enough (although you may need the entire nuclear arsenal of earth to pull it off)

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    $\begingroup$ Compute the energy to give escape speed to 1% of lunar mass, you will need much more energy than all nuclear weapons in the world. Instead if you detonates all those weapons on Earth you get nuclear Winter $\endgroup$ Commented Oct 24, 2016 at 9:31
  • $\begingroup$ Still not the worst use for earth's nuclear arsenal. $\endgroup$ Commented Oct 24, 2016 at 10:36
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    $\begingroup$ Changing the center of mass has no effect on orbit stability. $\endgroup$
    – jkavalik
    Commented Oct 24, 2016 at 13:01
  • $\begingroup$ Aren't orbits dependent on the speed of rotation counteracting the pull of gravity on the body? Gravity functions roughly the same as magnetic attraction, so while the Earth would still have the same mass it would have less Lunar mass to attract gravitationally, and the lowered lunar mass would also have less of a gravitational pull on the earth. That would break the equilibrium that an orbit would require $\endgroup$ Commented Oct 25, 2016 at 1:31
  • $\begingroup$ No. Chaaos kitty the center of mass of moon debrils system stays in the came place. For the moon to destabilize this much would mean debrils leave really really far away. Or it will just collapse back together. $\endgroup$
    – joojaa
    Commented Oct 25, 2016 at 2:07

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