The excellent readers of this blog have left numerous astute comments about the Nuke the Moon post, assessing the difficulty of knocking aside asteroids via nuclear explosions. The two most common themes are orbital mechanics and using the lunar mass itself in a sort of mass-driver configuration. Both are excellent points. The orbital mechanics are more trouble than I want to go through at the moment (though maybe later), but the lunar surface mass is easier pickings.

The equation I derived for impulse-per-nuke was this:

This is of course with dramatically simplifying assumptions, foremost among which is that all the energy of the blast is released as kinetic energy of the outbound fragments. But remember that rho was the energy density of the nuclear material. If we put the bomb under the surface then the energy density is decreasing because we’re adding inert lunar mass with no explosive energy in and of itself. But we’re also increasing the total mass m, so the equation is being pulled in two directions. But m increases faster than root m, so scrapping the energy density we have (for total energy E):

Here m is the entire ejected mass, bomb and lunar rocks combined. By increasing it without limit, we can increase the momentum transfer without limit. Well of course we really can’t – one hard and fast limit is that the kinetic energy per mass must still be enough to reach escape velocity. Any mass that ends up falling back down does us no good. Lunar escape velocity is about 2.37 km/s, so every kilogram we launch has to retain 2.8 million joules of energy. Using the same rough numbers from last time, each kilogram of nuclear fuel contains about 4.8 x 10^{14} joules. Dividing out, each kilogram of bomb can launch 172 million kilograms into space with sufficient velocity – again with the simplifying but somewhat unrealistic assumption that the energy transfer is totally efficient.

The total energy of a nuclear explosion is usually measured in kilotons or megatons of TNT equivalent. The units are not meant to be especially precise, but if the internet is to be trusted (and surely it is!) a megaton is about 4.2 x 10^{15} joules. The largest device the US ever tested was about 15 megatons. So plug that into our equation for E, and again using our very rough but probably order-of-magnitude-right-ish guess that the nuclear fuel has a mass of about 100 kilograms and including our multiplier above for the ejected lunar mass (deep breath), we get a momentum transfer per blast of…

4.65 x 10^{15} newton-seconds. That’s an improvement by a factor of 10,000ish from our last estimate of a purely surface blast. So instead of a quadrillion bombs to knock aside the moon, the simple expedient of burying them might bring that figure down into the mere trillions. Asteroids are of course much easier, but we have to be cautious. Blowing them apart doesn’t solve many problems; we’d rather not turn one big problem into a few hundred medium problems. We want to push it to the side gently.

Nukes aren’t the only option for gently pushing things in space, of course. There’s plenty of others from ion engines to space sails and beyond. We’ll save those for later.