Built on Facts

Nuke the Moon. Again!

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:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

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):

i-88516ca505d84f969b7ab15a965c3fb3-2.png

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 1014 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 1015 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…

i-a3c7180761ba4dcbb2588bfc8bccaaf4-1.jpg
Fig 1: Castle Bravo. Kaboom!

4.65 x 1015 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.

Comments

  1. #1 Uncle Al
    August 7, 2009

    The largest yield thermonuke was the USSR RDS-220 “Tsar Bomba” at 50-57 MT without a depleted uranium tamper or device casing. Had it been a uranium burner it could have done 100+ MT (~4.6×10^17 J). It massed 27,000 kg rather than 100 kg.

    Tidal power generators subtract energy from the moon’s orbit. Humanity could easily subtract scores of terawatt-hrs/year of energy. Enviro-whiner alas, the moon would spiral outward not inward as lunar laser ranging folks published papers.

  2. #2 travc
    August 7, 2009

    Al, IIRC, the moon is already spiraling outwards. No doubt in part due to lots energy being transformed into ‘work’ like cyclically raising massive amounts of water ;)

  3. #3 Matt Springer
    August 7, 2009

    My guess of the 100kg figure is just for the fusion fuel itself. The the rest of the device is dead weight. In fact that blind guess might be pretty good; dividing 15 MT by the energy-per-fuel estimate I think we get something like 131 kg.

  4. #4 D. C. Sessions
    August 7, 2009

    Blowing them apart doesn’t solve many problems; we’d rather not turn one big problem into a few hundred medium problems.

    Yes and no. Done early enough, reducing a somewhat-cohesive pile of gravel to a cloud of gravel would cause most of it to miss the Earth entirely. Even close in, it would reduce the mass of the impacting particles enough that most would not reach the surface and the thermal bloom from the atmospheric disintegration would be spread over enough area to at least reduce the incendiary effect.

    As is so often the case, timing is everything.

  5. #5 Robert
    August 7, 2009

    If you just want the impulse due to debris, ignoring everything else:

    Figure 20% of the fusion device energy goes into kinetic energy of debris. Say the bomb weighs 1000 kg. You can calculate the RMS velocity of debris based on joules of energy in the bomb. It turns out that the optimitum height for a debris impulse to a spherical object is to detonate at a height of 0.4 to 0.5R (R=radius)and about 15% of the debris will hit the object.

  6. #6 Joel
    August 19, 2009

    So my 12 year old asked me last night, no kiddding, “How many nuclear bombs would it take to move Mars into Earths orbit so that we could terraform it?”

    Will these equations work? I am thinking that some form of continuous ion engine and millions of years….

  7. #7 someone else
    May 24, 2010

    Theoretically, directed nuclear bombs could exist. Orion (the project from the ’60) was supposed to use directed nuclear bombs for propulsion.

    And a project called Casaba-Howizer was supposed to deal with the use of such technology as a weapon. It is still classified.

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