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⇤ ← Revision 1 as of 2012-02-27 20:14:58
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| 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: A 50 micron thick glass thinsat will have a ballistic coefficient of 2650*2*5E-5 = 0.265 Kg/m^2^. | 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/m^2^. A typical satellite has a ballistic coefficient of around 100 Kg/m^2^. 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/m^3^, and the orbital velocity is 5583 m/s, so the acceleration at m288 is 1E-12 m/s^2^. 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/m^3^, 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 points might get caught in a blinking eyelid, worst case. |
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, 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 points might get caught in a blinking eyelid, worst case.
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