Reentry

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

Assume the perpendicular from a round fragment can point in any direction. Either the front or the back is "shining" in the direction of motion, so there is 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. We can compute the ballistic coefficient by doubling the thickness: If the coefficient of drag for a flat plate is around 1.0. a 50 micron thick glass thinsat will have a ballistic coefficient B of 2650*2*5E-5 = 0.265 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 5 nanometers per second, or 15 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 down will increase peregee 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 1.5 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 1/3 of a thinsat, 1g or so, and moving at 1.5 m/s, its kinetic energy is 1mJ . The energy of a 2mm ( 4mg ) raindrop moving at 7 m/s is about 0.2 mJ. So this large might sting more than a raindrop, perhaps, or less because it is more spread out.

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 50 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 2300 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 - the size of the remaining shards are likely to be far smaller than a third of a thinsat as above.

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. N The code is here, and the results are here. NOTE: I may not be properly accounting for conservation of angular momentum. Lower orbits are faster, and drag on a satellite puts it in a lower orbit, but speeds it up.

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.

|| 6641 || 330 || 7111 || 0.3 || Lower ISS altitude ||

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.

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 server satellite, at 1 gram per watt, is 10 billion kilograms of thinsat. 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, about the same as meteoric rate, or about 1 milligram per square kilometer per hour. Unlike micrometeorites, we can expect most of the server satellites will be larger flakes. Since meteorites don't noticably damage eyes and lungs, compared to terrestrial dust kicked up by wind, the portion that is reentering dust will probably have no discernable effect. And if server satellites are made from carbon, rather than silicon, chances are they will literally burn and turn into CO2.

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