Experiencing Technical Difficulties (resonant) wrote,
Experiencing Technical Difficulties
resonant

Terraforming Venus

rfmcdpei mentioned that in one of Kim Stanley Robinson's books, the planet Venus was shaded with orbital mirrors so that its atmosphere cooled, thus terraforming it.

Venus is hot, about 735 K (460 C, 860 F). It has a diameter of about 12 100 kilometers.

Let's assume that we put a mirror at the Lagrange point between Venus and the Sun, sized so that it prevents all sunlight reaching the surface.

http://en.wikipedia.org/wiki/Lagrangian_point

Let's assume that the entire surface of Venus then radiates heat into the coldness of outer space with 100% efficiency.

So, we have a surface area of around 4.6 x 10^14 square metres, radiating heat at 735 K to outer space at around 4 K. Using the Stefan-Boltzman constant, that means Venus is now cooling at 7.6 x 10^13 kJ/s.

Let's assume that there is little heat transfer between the hot rocks of the surface and the hot atmosphere. Let's assume the atmosphere is effectively 100% carbon dioxide, with a specific heat of 0.98 kJ/kgK, and a mass of 4.8 x 10^20 kg.

Therefore, to cool one degree K, the atmosphere needs to lose 4.7 x 10^20 kJ. At a rate of 7.6 x 10^13 kJ/s, that will take 6.2 x 10^6 seconds, or about two years.

To cool the atmosphere of Venus to Earth-normal temperatures at this rate, it would take a minimum of 900 years. The rate of cooling would drop as the temperature decreased (hot bodies lose heat faster), and it's overly optimistic to assume Venus would act like a perfect black-body radiator. We'd also have to cool down the surface rocks to comfortable temperatures. Removing the carbon from the atmosphere and burying the resulting graphite would also give produce heat which would need to be removed. So, without some other means of cooling, we'd need millennia to terraform Venus.


[EDIT: Mindstalk pointed out that I got my math wrong, and the cooling rate is closer to 7.6 x 10^15 kJ/s)
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