Reentry

Thinsats in orbit, even broken or obsolete ones, are valuable where they are. They can be used as ballast for future thinsats, and they thin out radiation. As long as they can be controlled or externally manipulated, they are useful. But we might lose control of them; then what?

Assume everything goes wrong, and all the aluminum foil thinsats eventually fall into the atmosphere. They will likely be tumbling, slowly or quickly, and will shatter into flakes. What will re-entry be like?

Assume the perpendicular from a flake can point in any direction. Either the front or the back is "shining" in the direction of motion, so a round flake has 2 \pi R^2 area, and that area can be "expressed" in 4 \pi steradians different directions. So we can assume that the average area facing forwards during tumble is half the total area. Another way of looking at that is projecting an area element onto the surface of a half sphere ( 2 \pi R^2 ) and projecting that onto a disk ( \pi R^2 ). We can compute the ballistic coefficient by doubling the thickness: If the coefficient of drag for a flat plate is around 1.0, a micron 0.2kg/m² aluminum thinsat with will have an average ballistic coefficient B of = 0.4 Kg/m². A typical satellite has a ballistic coefficient of around 100 Kg/m².

The drag acceleration is the gas density, times the velocity squared, divided by the ballistic coefficient, or a = \dot v = \rho { v^2 } / B . The gas density at the m288 6411 km altitude is around 5E-17 kg/m³, and the orbital velocity is 5583 m/s, so the acceleration at m288 is 1E-12 m/s². Assuming a slowly decaying circular orbit, the orbital radius will decay as dr / dt = ( 2 r / v ) \dot v or about 3 nanometers per second, or 10 centimeters per year. At these high altitudes, uncompensated light pressure will have a much larger effect, moving the orbit (perigee down, apogee up) many hundreds of kilometers per year. Moving perigee thousands of kilometers down will increase perigee drag greatly, which lowers apogee.

The orbital decay rate is an inverse exponential function of altitude. The density decreases and the scale height increases with altitude. It is easier to work the problem backwards in time, since the decay gets a lot faster as we get closer to the surface. So let's consider the various regimes in order, from near the ground to operating altitude.

Near the ground

In thick atmosphere near the ground, with a density around 1.2 kg/m³, tumbling thinsat fragments will fall at terminal velocity, ag = \rho { v^2 } / B or v = \sqrt{ B a_g / \rho } or about 1.8 meters per second. Faster than a falling leaf, but the largest fragments, weighing grams or less, will not carry very much energy, a few millijoules. Sharp corners of a fractured thinsat might get caught in a blinking eyelid, worst case. Or, slivers might drop much faster, point-first like an arrow - this needs study!

If a shard is 50 milligrams, and moving at 1.8 m/s, its kinetic energy is 8 0 microJoules . The kinetic energy of a 2mm ( 4mg ) raindrop moving at 7 m/s is about 200 microJoules. The shards are likely to be much smaller; we can expect thinsats to fracture at the slot antennas, and typical bits might be as small as the antenna size and spacing, perhaps 9 mm² and weighing 2 milligrams.

Higher up

At 50 km altitude, the air density is 1e-3 kg/m³, and gravity is reduced to 9.6 m/s². The vertical terminal velocity of a tumbling thinsat is 60 m/s .

At 100 km altitude, the air density is 4.6e-7 kg/m³, and gravity is reduced to 9.5 m/s². The vertical terminal velocity of a tumbling thinsat is 3000 m/s ; quite fast, but much slower than 7400 meters per second atmosphere-relative orbital velocity. An orbiting thinsat coming near the atmosphere at this height decelerates rapidly, losing horizontal speed and orbital lift, increasing vertical velocity while drag forces build up exponentially. Chances are that differential acceleration and pressure (higher towards the edges) will shatter a thinsat as it moves much faster than the speed of sound.

Simulation, from 500 km down

Above 150km, solar minimum and maximums change density greatly. Solar maximum will raises the atmosphere, so drag increases. Most thinsats will reenter during solar maximums. What will their last few hours in orbit look like?

Precise models of the atmosphere, like the NRLMSISE-00 model, are too complicated for simple calculations. For an approximate guess, we will use 1962 standard atmosphere density numbers, and interpolate exponentially between them.

We will not compute horizontal position, but we will compute horizontal velocity and deceleration. We will also compute vertical position, velocity, and acceleration. We will start with circular orbit velocity at 500 km altitude. Server sky thinsats normally operate above 6000 km altitude, and cannot maneuver properly below about 1000 km.

The code is here, and the compressed results are here. Lower orbits are faster, and drag on a satellite puts it in a lower orbit, but speeds it up. On the other hand, we drop out of orbit and into a steep fall in only a few minutes, so we are no longer in a near-circular orbit.

The drag at 500 km is significant, given the very low ballistic parameter. After an hour, the altitude drops to 496 kilometers. After 103 minutes, the thinsat passes below 400 km, the upper International Space Station altitude, 8 minutes later it passes below 330 km, the lower ISS altitude. Two hours after starting at 500 km, the thinsat drops to 165 kilometers. Things happen fast after that.

Most of the fragmentation occurs during a 4 minute period, 2 hours after passing 500 km altitude, and the flakes reach the ground about 5 hours after passing that altitude. From upper ISS altitude down to 96 kilometers altitude takes 25 minutes - thinsat orbits decay very fast. Even if a thinsat impacts the outer meteorite shield of a satellite, they will vaporize before they penetrate it.

Fragmentation of aluminum substrates

The distribution of aerodynamic forces on a thinsat flake determine how small the flakes will be, and I'm no aerodynamicist. WAG follows.

Assume that the aerodynamic forces are zero at the center and increase linearly towards the edges of a square plate. The average 175 m/s² shown above corresponds to a maximum 350 m/s² when the plate is face on to the stream, and acceleration forces of 700 m/s² at the edge and 0 in the middle; perhaps only very briefly. How big a plate can survive this without fracturing in the middle?

redo this! The slot antennas create natural high-stress fracture points

If the plate has a width of W , then the average lever arm of the torque produced force is 1/3 W . The total torque per meter in the center is a \rho t W^2 / 6 where t is the thickness and \rho is the acceleration. If Y is the tensile yield strength (and there are no through-holes or other stress concentrators, and the tumbling centrifugal force is not extreme), the torque on the center-line of the plate at yield is Y t^2 / 6 . Equilibrating, the width of the plate is given by W = sqrt { Y t / a \rho } . For pure aluminum, Y ~ = ~ 10 MPa, \rho ~ = ~ 2700 kg/m³, and at max Q, the average acceleration is a = 160 m/s² and the thickness is 8E-5 meters. This results in W = 1.2cm. Vibration, stress risers, and centrifugal forces reduce this, so flakes are unlikely to be much larger than 1 cm², less than 20 mg.

Environmental hazards

Will a disintegrating thinsat throw off a lot of very small dust grains, which could put a lot of dust in the atmosphere? 100 terawatts of thinsats, at 1 gram per watt, is 10 billion kilograms of thinsat, about 3% of the mass of a large dust storm. Since the orbits may take thousands of years to decay, the global deposition rate might be on the order of 5 million kilograms per year, or about 1 milligram per square kilometer per hour, about the same rate as meteors enter the earth's atmosphere. This is less populous and energetic than the meteors that pass near the earth and pose a risk to space assets. Unlike micrometeorites, we can expect most of the thinsats will be larger flakes. Since meteorites don't noticeably damage eyes and lungs, compared to terrestrial dust kicked up by wind, the portion that is reentering dust will probably have no discernible effect. Thinsats will likely burn and turn into fine alumina powder.

Mount Saint Helens erupted in 1980 and spewed about 10 trillion kilograms of rock dust (2.8 cubic kilometers) over the Pacific Northwest in a few days. Where I lived, it produced a blanket of sandy-powdery ash about half a millimeter thick, or about 2 kilograms per square meter. The ash was vexing, but the power grid stayed up, and semiconductor fabs in the area kept their class 10 clean rooms operating. Besides some clogged automobile air filters, scratched windshields, and clogged sewers, life went on everywhere but the immediate vicinity of the volcano (scalded and buried by a collapsing mountain and pyroclastic flow).

The first thinsats may have micron thick Indium Phosphide solar cells, and a few other exotic materials in thin layers on top, but they are mostly silicon and aluminum. We will probably run out of the exotic materials before producing 10 billion kilograms of server satellite, and use clever molecular arrangements of more abundant materials like silicon and aluminum and carbon to produce thinsats, eventually from lunar materials. Most of the oxides (like indium oxide) that would be produced by reentry will be biologically inert.

While there is no guarantee that all manufacturers of thinsats will use biologically inert materials, there is no good reason to incorporate toxins in large quantities. The thinsats can be assayed before launch - the ITU or other international licensing bodies may discourage the use of toxics. Thus, even in the worst case, the main environmental cost of abandoning server sky to long term reentry will be the return of polluting behavior it no longer replaces.