October 7, 2012

Generation Starship Material Selection

The space environment is, contrary to popular belief, not that empty.  It is full of various forms of radiation, many of them quite dangerous.  To keep a crew happy and healthy for generations this necessitates a thick outer hull composed of the right material.  Since the Starseed proposal also requires in-situ (or on-site) construction of new vehicles at their destination the material in question must also be one commonly found in, or easily fabricated from, rocky bodies in deep space.  Further we will desire the material to be the lightest possible choice that can fulfill these prerequisites.  Keeping the structure mass down will reduce the fuel requirements to complete the mission.
To begin with we'll need an understanding of the forms of radiation encountered in deep space.  Together with understanding of how radiation shielding works this will give us an idea how thick the hull must be.  Then we can consider the materials that might be available for use.  Finally by calculating the thickness needed, and from that the areal density of the hull, for each material we'll know which of our options will be the lightest.  This option, then, is the one used going forward.

In the previous post on Hazards of Interstellar Travel, we looked at galactic cosmic rays and the dangers they can pose to interstellar travelers.  Since the interactions of these particles with living matter is currently poorly understood we determined that an appropriate safety concern to so block all particles up to a given energy above which the total likelihood of interaction with any particle is less than 3% over a human lifetime.  This cutoff energy is about 15 GeV for ionized hydrogen nuclei, by far the most common GCR.  Thus the vehicle skin must be thick enough to block such a particle and must be able to contain spallation products produced by the reaction.  In order to achieve this the skin will necessarily be extremely thick, and so the material must be a one that is commonly found in the galaxy.
Recall from before the Bethe-Bloch equation:

Inputting the appropriate constants and using the Bloch simplification for ionization potential gives:

Consider a few points that this shows us.  First the larger the ion atomic number the easier it is to slow, so we have another verification that a hydrogen nucleus is our worst case shielding scenario.  Second the slowing power of a material is directly proportional to its density, which is logical since higher density means more material and thus a greater chance of interaction.  Normally we would want to avoid high density materials, but in this case we are well served by using them.  Third the ratio of atomic number to atomic weight, in theory the higher this is the better.  However in practice this ratio typically comes to about 2 for most stable elements, and the difference is not appreciable when compared to the effect of density.
From this we might think that they heaviest element is best, and in principle it is.  However material strength, nuclear stability, and availability must also be taken into account.  Many of the heaviest elements are also radioactive, which disqualifies them as potential radiation shields.
Consider the curve of binding energy as described in Chapter One.  This curve tells us the stability of various nuclei.  At its peak is iron; elements below iron can, in principle, be fused to produce meaningful energy and elements above it can be fissioned.  Anything heavier than iron, when fused produces less energy than was required.  Thus stars are capable of fusing elements until they produce iron, while heavier elements are only produced in the last stages of dying stars.  This fact makes iron one of the most common elements in the universe, after those produced by the Big Bang and those found in early stage stars.  It is because of this that we'll select iron in place of some heavier element like platinum.
Iron's high density, at 7,814 kg/m3, is also a great benefit as is its high strength.  In order to completely stop a 10 GeV proton some 7.34 m of iron are necessary.  With a reasonable factor of safety this calls for an approximately 10m thick hull.  For this thickness to cover the whole vehicle requires 80 million cubic meters, or 625 billion kg of iron.  While this is achievable, some alternate system allowing a lighter hull would be preferable.  Since the cosmic rays in question are charged particles many authors have suggested that electrical or magnetic fields could be used to divert or slow incoming particles without the need for thick mass shields.  The utility of such a system given the size of a generation ship is clear.  Iron is still the material of choice, but an electomagnetic shield should also be used to produce a minimum mass habitat.

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