Ok, first some facts...
thickness of translucent aluminum foil in visible light: .013mm (possibly smaller)
radius of the earth:6378.1 km
length of the earth's center up to the karmen line(definition of space): 6378.1 km + 100 km = 6478.1km
area of the karmen line(surface area of the recognized boundary of space): 4(pi)r^2 = 4(pi)(6478.1 km)^2 = 527357094 km^2
volume of aluminum needed to cover space boundary: 527357094 km^2 * .016 mm = 527357094 km^2 * .000000016 km = 8.437713504 km^3
density of aluminum: 2.70 g/cm^3 = 2,700,000,000,000 kilograms/km^3
weight required to cover earth's surface: 2,700,000,000,000 kilograms/km^3 * 8.437713504 km^3 = 22,781,826,460,800 kilograms
weight of the atmosphere:5,000,000,000,000,000,000 kilograms
weight proportion of this aluminum foil shield to atmosphere:22,781,826,460,800 kilograms / 5,000,000,000,000,000,000 kilograms = 0.000455636529216%
or 1 part in 220,000.
atomic weight of aluminum: 26.982 g/mol
atomic weight of air: 28.97 g/ mol
Therefore, by weight ratio, 23 trillion kilograms of mass or only 0.000455636529216% of the atmosphere's weight could block out the entire visible
spectrum of light with huge effects to our climate. Since air and aluminum have similar molar masses, this is roughly equivalent to the proportion of
atoms in the whole system (air atoms vs. aluminum atoms). So, we are talking about 5 ppm (parts per million) blocking out a significant portion of
As you can see in the image above, visible light makes up a good portion of the energy that reaches the earth's surface(the red part). 5 parts per
million could block all that out. So, the OP's error is in making energy effects proportional to the atomic ratios in the atmosphere. It states that
you can multiply the proportion of the atmosphere's molecular count of CO2 by solar irradiance to come up with its max effect on the energy
environment. Obviously, this is not the case.
[edit on 15-2-2010 by ncb1397]