cosmictotem
Registered Senior Member
In post 41, you suggest 40,000 biospheres in the first circular orbit, at about 1AU from the sun. You have also suggested that they each house 100,000 people and be sized like there largest stadiums.
These "stadiums" will need to be enclosed in spherical mass shells, for protections from cosmic rays. For life time long protection these shells will need to be at least 30 meters (40 feet) thick. So my most important "remaining objection" is the one i've already asked at least 4 times now:
Where does all this mass come from?
Here is how to estimate it:
If R is the mean radius of the 40 foot thick spheres and we call pi = 3.14 then the surface Area of the spherical shells is A= 4x3.14R^2 and the Volume of mass in each is V = TxA, where T is the Thickness of the mass shell.
For good shielding, the average density of the mass should be more than that of water. You would want the outer most layers to be lead, to convert each high energy cosmic ray into a dozen or more "daughters" of lesser energy, which could be converted into more than 100 still lesser energy "grand-daughters." The material doing this could be only half as dense as lead. Finally, the two innermost layers should be even lower density. Innermost is mainly carbon as the 10,000 great, great grand daughters of each primary ray need to collide with modest weight atoms - best for "sucking" energy out of them with the struck atom's recoil. The layer next to the carbon layer should have about twice the density of water. The overall density of the shell, averaged, would be more than 3 times that of water, but I'll call it 3. If we work in feet (water is 64 pounds/ ft^3)
So the mass of your 40,000 bioshells is 40,000x3x64x4x3.14xT(R^2) =32,153,600T(R^2) pounds where both T & R are given in feet. If T = ~31 feet (probably to little, but makes for nice round number.) I. e. 1,000,000,000R^2 pounds. I think R must be at least 1000 (length of 10 foot ball fields is very small to cram 100,000 people inside) then here it the weight, of just the innermost of a set of rings between Earth and Mars: 1,000,000,000x1000x1000 pounds or in scientific notation: 1E15 pounds. That is about equal to taking the top layer half a mile deep from all the dry land parts of the earth. Doing that might reduce the dry land to only 80% of the present area.
So again, for the fifth time, where will you get all the needed mass, just for the cosmic ray shield? (and then more for the rest of the biosphere)?
Also, note that the area growing food for 100,000 people can't be inside the opaque (no sunlight) bioshield. Perhaps some fungus growing outside can survive bombardment by cosmic rays, but I doubt it. This is the 2nd problem I would raise.
Facts are very ugly enemies, arn't tthey.
Didn't I address this objection using potential advances in cellular maintanence and rejuvenation? In other words, injured cells could be periodically replaced with healthy cells generated in internal rooms which are smaller versions of your spheres with 40 foot thick walls. (See Aubrey de Grey for a rundown of coming advances in human cellular maintenance).
As for the food, well, if solar panels are used to collect light, the plants could be placed in opaque rooms fitted with LED lighting. But I see your point of having to enclose the plants in large rooms. But again, if we can master cell maintanence in humans, it's foreseeable we could do it for plants, raise them in smaller rooms, then bring them out later to the large exposed farms.
But, as our cells would be periodically replaced with healthy cells, it doesn't really matter if we are eating plants exposed to some radiation, does it? We'll just replace those sick cells with healthy ones generated in those much smaller protected rooms I suggested.
Additionally, haven't scientists proposed hydrated plastic as an alternative shielding?