If we're going to have significant numbers of people spending significant amounts of time in space, we're going to need supplies for them. I've written previously about how we can provide for farms in space for their food: but we'll also need water. Water's heavy (or, at least, the amounts of it we're prone to using come to quite a lot of mass) so it'd be good to avoid lifting it out of a gravity well.
Thankfully, there's plenty of water out in space: most comets and many asteroids have substantial water content, albeit usually in the form of ice and mixed up in varying proportions with dirt. Furthermore, we have strong reasons for wanting to develop competence in moving asteroids and comets around: if we find one whose orbit involves a significant risk of hitting Earth, we'll want to adjust that orbit so that we're confident it'll miss. In practice, the only way to be sure we're going to be competent at making such course corrections is to do similar to asteroids we know aren't in any danger of hitting us, so as to get some practice. When our lives depend on a celestial mechanic doing a job right, it'll be reassuring to know the mechanic has done this before, more often successfully than not. Once we've got celestial mechanics who know how to control orbits reliably, it shouldn't be too big a problem to bring some asteroids into orbit around the Earth and set to work on mining them for their minerals, for use as raw materials for our activities in space. One of those minerals is water.
Of course, such a mining operation shall tend to yield a lot of water from each asteroid mined, but we won't (at least to begin with) be doing this very frequently, so we'll need a reservoir in which to store our water. That needs to be reasonably close to Earth and the Moon (since that's where most of our activity likely shall be); yet adequately clear of their gravity wells and in a stable orbit. It so happens that there's a pair of natural candidates for this job: the L4 and L5 Lagrange points of the Moon's orbit, a sixth of a turn ahead of and behind it. This is slightly complicated by the fact that the Sun's gravitation perturbs these Lagrange points, disrupting the stability they would normally have; and we may need to clear out some dust that's loitering there. All the same, there's a fair chance we can stabilise its orbit by judicious sporadic interventions. Besides, that instability should mean that it's relatively easy to move water to and from such a point.
So let's suppose we have a reservoir at a Lagrange point on the Moon's orbit. Naturally, it should be named after Joseph-Louis Lagrange; and the French for lake is lac, so let's call it lac Lagrange. Its water doesn't have to be pure enough to drink – it can be further processed later – but it'll make sense to get it about as pure as sea water. If it's liquid, it'll be evaporating from its surface, have an atmosphere of water vapour (we could add nitrogen too, if we ever get it big enough to hold any down) and be slowly losing that to space. The evaporation shall tend to cause it to cool and I suspect it would freeze (based on asteroids whose orbits cross that of Earth tending to be solid), at least on its surface. Since ice makes a good thermal insulator, it shouldn't take much effort to keep its interior liquid (albeit cold) for ease of pumping out to water-collecting visitors.
However, it'd be more fun to keep it warm by using some big mirrors (e.g. made of metalised Mylar) to reflect infra-red light onto it. You'd have to be a little careful on the side of it away from the Sun, since it'll focus the Sun's light and produce a hot-spot there which may be a bit dangerous. However, you'd be able to go scuba-diving (you have to take your oxygen with you anyway, after all) in it. You could float boats on the surface, although sailing wouldn't be very practical – I doubt solar wind and radiation pressure are strong enough to overcome water-hull friction; and you'd have trouble getting it to hold onto an atmosphere thick enough to provide a decent wind; if it were big enough to do that, it'd be disrupting the Moon's orbit and its gravity well would be big enough to negate the whole point of having it there ! Still, it might usefully get big enough to support a modest amount of gravity, which may make it practical to give it enough of an atmosphere that we might be able to get a live population of algae going and stock it with fish.
As long as it's liquid, it's a spherical body of uniform density, ρ ≈ 1 tonne per cubic metre. Let its surface radius be R. The gravitational field strength at radius r from its centre, for r ≤ R, is G.M/r/r = 4.G.π.ρ.r/3. Let Q = 4.G.π.ρ/3 ≈ 3.622/hour/hour or 0.27947e-6/s/s or 8.54 g per light second, where g ≈ 9.81 m/s/s is standard gravity. The gravitational potential per unit mass at its surface is −G.M/R = −Q.R.R and we're keen to keep this reasonably small (for ease of taking water away) but not too small (since we want the water to stay there until we're ready to take it away); escape velocity would be R.√Q, wherein √Q is 0.52865e-3/second or 1.9031/hour. Gravitational potential per unit mass, φ, at radius r≤R would satisfy dφ/dr = Q.r, implying φ = −(3.R.R −r.r).Q/2.
Presume that pressure at its surface is zero (for comparison, on Earth, the pressure at ocean surface is equal to what 10 metres of water adds; so treating it as zero would only add about a 10 metre inaccuracy to my calculations); let the pressure at radius r be P(r), so we have P(r) = 0 for r ≥ R. For r ≤ R, dP/dr = −ρ.Q.r so P = (R.R −r.r).ρ.Q/2. Pressure at the centre would thus be ρ/2 times gravitational potential per unit mass at the surface; or, equally, a third of the gravitational potential per unit volume at the centre.
| R | surface gravity Q.R / g | escape velocity R.√Q . s/m | −φ(R) . kg/J | −φ(0) . kg/J | P(0) / kPa |
|---|---|---|---|---|---|
| 10 km | 0.2850e-3 | 5.2865 | 27.947 | 41.92 | 13.973 |
| 27 km | 0.769e-3 | 14.273 | 203.73 | 305.6 | 101.86 |
| 1 light ms ≈ 300 km | 8.54e-3 | 158.48 | 25.117e3 | 37.676e3 | 12.558e3 |
| Earth's radius ≈ 6375 km | 0.182 | 3370 | 11.4e6 | 17.03e6 | 5.68e6 |
I illustrate the 27 km case because its P(0) is approximately the atmospheric pressure at ocean surface on Earth. The water-ball as big as the Earth has roughly the same surface gravity as the Moon (which is c. 3.34 times as dense as water; Earth is about 5.5 times as dense as water); but 14.8 times the mass (so it'd be rather disruptive to the Earth-Moon system's stability) and four times the gravitational potential per unit mass at surface (so twice the escape velocity). The pressure at its centre would be 56 thousand atmospheres; I'm not sure how water behaves at such pressures but you almost certainly don't want to go swimming there – that's equivalent to being at the bottom of an ocean about 580 km deep at the Earth's surface (the real ocean doesn't go deeper than about 11 km).
Written by Eddy.