Mercury / Mars satellite measured mean temperatures 340 K and 210 K comparison

These ( Tmean, R, N, and albedo ) planets' parameters are all satellites measured. These planets' parameters are all observations.

Planet….Mercury….Moon….Mars

Tsat.mean.340 K….220 K…210 K

R…......0,387 AU..1 AU..1,525 AU

1/R²…..6.6769….....1….…0,430

N…1 /175,938..1 /29,531..0,9747

a......0,068......0,11......0,250

1-a….0,932……0,89…….0,75

coeff...1,1371.............0,7524

Comparison coefficient calculation

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

Mercury:

Tsat.mean = 340 K

[ (1-a)*(1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,932*6,6769*(1/175,938)¹∕ ⁴ ] ¹∕ ⁴ = 1,1433

Mars:

Tsat.mean = 210 K

[ (1-a)*(1/R²)*(N)¹∕ ⁴ ]¹∕ ⁴ =

= [ 0,75*0,430*(0,9747)¹∕ ⁴ ] ¹∕ ⁴ = 0,7524

Let's compare

Mercury coeff. / Mars coeff. =

= 1,1433 /0,7524 = 1,5195

And

Tmean.mercury /Tmean.mars =

= 340 K /210 K = 1,6190

The Mercury's comparison coefficient (1,1433) is calculated for Mercury's Semi-major axis which is 0,387 AU. But half of the time, Mercury comes closer to the sun at its Perihelion of 0,307 AU. The fact Mercury's orbit has high eccentricity e = 0,205 partly explains the difference between the calculated (1,5195) and the measured (1,6190).

Conclusion:

Everything is based on observations in

the comparison coefficient 

[ (1-a) (1/R²) (N)¹∕ ⁴ ]¹∕ ⁴

And

It is the confirmation that the planet's axial spin (rotations per day) "N" should be considered in the Tmean planet's surface mean temperature equation in the fourth root:

Tmean.planet = [ Φ (1-a) So (1/R²) (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴.