# The Planet's Effective Temperature and the Planet's Mean Surface Temperature Equation: Tmean = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴

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Neptune Voyager 2 (29347980845) flatten crop

Triton moon mosaic Voyager 2 (large)

### Neptune’s Mean Temperature Calculation

Neptune’s Mean Temperature Equation Tmean.neptune is:

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

Neptune’s sidereal rotation period is16 h 6 min 36 sec, or 0,6713 day

N = 1/0,6713 rotations/per day

R = 30,33 AU, 1/R² = 1/30,33² = 0,001087 times lesser is the solar irradiation on Neptune than that on Earth.

So = 1.362 W/m² is Solar constant

Neptune’s albedo, aneptune = 0,290

Neptune is a gaseous planet, Neptune’s surface irradiation accepting factor Φneptune = 1

(Neptune has not surface to reflect the incident sunlight. Accepted by a Gaseous Hemisphere with radius r sunlight is S*Φ*π*r²(1-a), where Φ = 1)

Atmosphere composition 80% ± 3,2% H₂, 19% ± 3,2% He, 1,5% ± 0,5% CH₄.

Cp.H₂ = 3,1388 cal/gr. oC , H₂ specific heat at 175 K, Cp.He = 1,243 cal/gr. oC , He specific heat, Cp.CH₄ = 0,531 cal/gr. oC , CH₄ specific heat.

Cp.neptune = 80%*Cp.H₂ + 19%*Cp.He + 1%*Cp.CH₄ = 0,80*3,14 + 0,19*1,243 + 0,01*0,531 = 2,512 + 0,236 + 0,00531 = 2,753 cal/gr.oC

β = 150 days*gr*oC/rotation*cal – it is the Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law constant

σ = 5,67*10⁻⁸ W/m²K⁴, a Stefan-Boltzmann constant

So we have

Neptune’s mean temperature Tmean.neptune is:

Tmean.neptune = {1*(1-0,290)1.362*0,001087(W/m²) [150*(1/0,6713)*2,753]¹∕ ⁴ /4*5,67*10⁻⁸(W/m²K⁴) }¹∕ ⁴ =

Tmean.neptune = 69,31 K is the calculated.

And below is the measured by satellites

Tsat.mean.neptune = 72 K (at 1bar level)

Tsat.mean.neptune = 55 K (at 0,1 bar level).

Let's, for comparison reason, calculate Neptune’s Effective Temperature Incomplete Formula Te.neptune.incompl is:

Te.neptune.incompl = [ (1-a) So (1/R²) /4σ]¹∕ ⁴

Te.neptune.incompl = [(1-0,290)1.362*0,001087(W/m²) /4*5,67*10⁻⁸(W/m²K⁴) ]¹∕ ⁴ =

Te.neptune.incompl = 46,40 K is the calculated.

And below is the measured by satellites

Tsat.mean.neptune = 72 K (at 1bar level)

Tsat.mean.neptune = 55 K (at 0,1 bar level).

And comparing with the Mean Temperature Equation Tmean.neptune = 69,31 K we may conclude here that the without-atmosphere Mean Temperature Equation, when applied to Gaseous Giant (Ice Giant) Neptune gives us much closer to the measured by satellites result.