"Here’s another question. Suppose Mars was at Earth’s orbit around the Sun, same atmosphere as it has now, what would its temperature be?"
The answer to this question will illustrate the Planet Surface Rotational Warming Phenomenon complete rightness!
The measured Mars' mean surface temperature at the actual Mars' distance from sun R.mars = 1,524AU is Tmean.1,524AU =210K
R = 1AU is the Earth's orbit distance from the sun in AU (astronomical units)
Let's apply the inverse square law
(1/R²) = (1/1,524²) = 1/2,32
Mars has 2,32 times less solar irradiation intensity than Earth has Let's calculate using the Stefan-Boltzmann emission law 4 power. When estimating the emission temperature the 4th root of the emission intensity should be applied.
Thus, by analogue, Mars at the earth's orbit being irradiated 2,32 times higher, should have (2,32)¹∕ ⁴ times higher the mean surface temperature, than on its actual orbit of R.mars=1,524AU, but with the same rotational spin...
Mars performs 1 rotation every 24,622 hours or 0,9747 rot/day.
(2,32)¹∕ ⁴= 1,23416
Tmean.mars.1AU = 1,23416*Tmean.mars.1,524AU =
= 1,23416*210K = 259,17K or rounded 259K
Mars at earth's orbit would have Tmean.mars =259K.
When comparing with the measured moon's, because Moon also orbits sun at earth's distance, and Moon having a lower than Mars Albedo (a.moon =0,11; a.mars =0,250), thus Moon at earth's orbit being more intensively irradiated Moon/Mars = (1-0,11)So /(1-0,250)So = 0,89/0,75 = 1,187 times Moon is more intensively irradiated, and the measured Tmean.moon =220K
It is a solid proof of the Planet Surface Rotational Warming Phenomenon. The higher Mars' average surface temperature at Earth's orbit than Moon's, the 259K vs 220K, and the huge and undisputable difference of 259K -220K =Δ39C can be explained only by the Mars' rotational spin being 28,783 times faster than that of Moon's.