I’m wondering if Venus could have lost carbon in the form of methane evaporating into space from the upper atmosphere. methane CH4 has a molecular weight of just 16, half O2’s 32 and about 1/3 of CO2’s 44. Lighter molecules travel faster at the same temperature and thus are more likely to reach escape velocity. CO2+2(H2O) -> CH4 + 2O2. If this is the case, wonder what happened to the oxygen?

I’m attempting to do that very calculation including sedimentary rocks on Earth and I’m only getting a value of 18 bar for Earth, not 60. The estimates for the amount of limestone/dolomite in sedimentary rocks seems to vary widely depending upon the source of the information so it’s almost impossible to get an accurate number.

It all depends on the insolation which in turn depends primarily on the distance from the star, albedo of the planet, and luminosity of the star. In the case of Earth with a 243-day retrograde rotation the sun would appear to move in the sky 2.47 degrees per 24 hours (west to east). If you consider the subsolar point the sun will be within 20 degrees of zenith for 389 hours (16.2 days). The area near the subsolar point is going to get very hot with no chance to cool off at night.

Let’s compute how much energy that point receives. Assume the solar constant is 1366 watts/m^2 and the albedo is 30%. Assume the average elevation angle is 80 degrees. Then average solar power received over 389 hours is 1366 * 0.7 * sin(80) = 924 watts/m^2. Assume that the same energy is re-radiated back into space assuming a blackbody emitter. Then temperature of emission is (924 / 5.67E-8)^0.25 = 357 deg K = 84 deg C. This is only 16 deg C less than the boiling point of water. The evaporation of water will be tremendous causing huge quantities of water vapor to be released into the atmosphere. The temperature only has to rise 16 deg C before the oceans really start to boil in earnest. So a slow rotation is not a good thing for a terrestrial planet.