[Space, Spaceships] Armor needed for Aerobraking/Re-entry

Anthony · 02-27-2017, 02:39 PM

As far as aerobraking being hard, though, what matters isn't really the delta-V involved, it's the total energy you're dissipating. A vehicle aerobraking is dissipating energy at a rate of force * velocity.

Let's say a spaceship's approach velocity towards Earth started at 3 km/s, which is a kinetic energy of 4.5MJ/kg. At the top of atmosphere, it has gained another 62.5 MJ/kg from potential energy, so its total energy is 67 MJ/kg for a velocity of 11.57 km/s. To be captured, its energy needs to drop below escape energy (62.5 MJ/kg), so we need to dissipate 4.5 MJ/kg and about 400 meters per second.

Now, the big trick here is that atmospheric density varies substantially with altitude, and our altitude during this pass will vary a lot. You can reasonably approximate this pass as if we were falling 'up', so at 10s away from apogee you have gained 500 meters; as the scale height of the atmosphere is about 8 km, we can figure our total time in atmosphere as if it were about 80 seconds (it's actually longer than that but a lot of it is in very thin atmosphere), during which time we need to shed 400 m/s and 4.5 MJ/kg. 400m/s in 80s isn't hard (it means a peak deceleration of about half a G), the challenge is the 4.5MJ/kg. If we assume we have a 10 ton probe that is a 4 meter diameter sphere (33 cubic meter volume), it has a surface area of 50 m^2, so we're looking at 45000 MJ/50 m^2 = 900 MJ/m^2 (it's actually uneven heating, max is about 4x that).

It takes about 7.5 MJ to ablate a kilogram of steel, so if we just ablated away it would be 120 kg/m^2 (about 15mm), or a total of 6 tons of steel gone. Since we're only a 10 ton craft and we still have another 625 GJ of orbital energy we want to get rid of, that's not really a viable option. Thus, we have to figure out how to resist it without ablation.

Our peak energy flow rate is 900/80 = 11.25MW/m^2. An idealized blackbody sheds heat at a rate of 5.67e-8 W*m^-2*K^-4, which we can solve for temperature, T(K) = (1.125e+7/5.67e-8)^0.25, or 3750K (and as noted, it's not going to be evenly spread, so it actually reaches 5300K on the leading edge). That's higher than any realistic material can withstand, but it's not dramatically higher; it we can find some trick that causes heat to be radiated outwards without ever striking the craft, we should be okay.

At that point, the surface of the craft isn't ablating -- and we no longer need armor at all, we just need a framework that can hold our shielding in place without buckling, plus insulation. This roughly describes the way the space shuttle's tiles work -- while they would likely stop attacks for a little while, they aren't really armor as such.

02-27-2017, 02:39 PM	#23
Anthony Join Date: Feb 2005 Location: Berkeley, CA	Re: [Space, Spaceships] Armor needed for Aerobraking/Re-entry As far as aerobraking being hard, though, what matters isn't really the delta-V involved, it's the total energy you're dissipating. A vehicle aerobraking is dissipating energy at a rate of force * velocity. Let's say a spaceship's approach velocity towards Earth started at 3 km/s, which is a kinetic energy of 4.5MJ/kg. At the top of atmosphere, it has gained another 62.5 MJ/kg from potential energy, so its total energy is 67 MJ/kg for a velocity of 11.57 km/s. To be captured, its energy needs to drop below escape energy (62.5 MJ/kg), so we need to dissipate 4.5 MJ/kg and about 400 meters per second. Now, the big trick here is that atmospheric density varies substantially with altitude, and our altitude during this pass will vary a lot. You can reasonably approximate this pass as if we were falling 'up', so at 10s away from apogee you have gained 500 meters; as the scale height of the atmosphere is about 8 km, we can figure our total time in atmosphere as if it were about 80 seconds (it's actually longer than that but a lot of it is in very thin atmosphere), during which time we need to shed 400 m/s and 4.5 MJ/kg. 400m/s in 80s isn't hard (it means a peak deceleration of about half a G), the challenge is the 4.5MJ/kg. If we assume we have a 10 ton probe that is a 4 meter diameter sphere (33 cubic meter volume), it has a surface area of 50 m^2, so we're looking at 45000 MJ/50 m^2 = 900 MJ/m^2 (it's actually uneven heating, max is about 4x that). It takes about 7.5 MJ to ablate a kilogram of steel, so if we just ablated away it would be 120 kg/m^2 (about 15mm), or a total of 6 tons of steel gone. Since we're only a 10 ton craft and we still have another 625 GJ of orbital energy we want to get rid of, that's not really a viable option. Thus, we have to figure out how to resist it without ablation. Our peak energy flow rate is 900/80 = 11.25MW/m^2. An idealized blackbody sheds heat at a rate of 5.67e-8 Wm^-2K^-4, which we can solve for temperature, T(K) = (1.125e+7/5.67e-8)^0.25, or 3750K (and as noted, it's not going to be evenly spread, so it actually reaches 5300K on the leading edge). That's higher than any realistic material can withstand, but it's not dramatically higher; it we can find some trick that causes heat to be radiated outwards without ever striking the craft, we should be okay. At that point, the surface of the craft isn't ablating -- and we no longer need armor at all, we just need a framework that can hold our shielding in place without buckling, plus insulation. This roughly describes the way the space shuttle's tiles work -- while they would likely stop attacks for a little while, they aren't really armor as such. __________________ My GURPS site and Blog.