A planet’s * albedo* is its ability to reflect incident sunlight back to space.

**The intent of this parametric study is to determine the effect of increasing earth’s albedo from estimated present value of 0.30 to a value of 0.31**

**The result is the annual average temperature would be lowered by 0.01C**

**COMPUTATION OF SENSIVITY OF EARTH’S AVERAGE TEMPERATURE TO ITS ALBEDO.**

CALCULATIONS

Heat Balance Equation

εσT_{a}^{4} = 1/4So (1-A) = H = k^{4}εσT_{s}^{4} Equation 1

Where ε = emissivity = 0.97, σ = 5.67E-8 watts/m^{2}-K^{4}, *T*_{a}* = equivalent radiative temperature,*

S_{o} = solar constant – sunlight incident on earth’s atmosphere = 1367 w/m^{2}, A = albedo,

H = net sunlight incident on earth’s surface = earth’s heat radiated to space = 235.47 w/m^{2} for basline

T_{s }= temperature of earth’s surface = 288.2K for baseline, k = T_{a} / _{Ts }

(From the first and last terms of Equation 1, k^{4} = Ta^{4}/Ts^{4}, or k =Ta/Ts )

Rearranging the middle components of Eqn. 1, and assuming A = 0.30:

H = 1367 * (1 – 0.3) / 4 = 239.2 w/m^{2}

This is the H computed by ModTran for the no clouds condition.

Computing Ta for no clouds:

Ta = [S_{o}(1-A)/4εσ]^{ ¼} =[1367*(1-0.3)/(4*.97*5.67E-8)]^{1/}4 = 256.8 K

For Ts = 288.2 K, k = 256.8K / 288.2K = 0.8911, the same as for no-clouds from ModTran

If we use H from ModTran for the cloud condition Stratus-Stratocumulus 0.66 to 2 km top, then k would be that from the ModTran calcs = 0.8876, where H was 235.47 watts/m^{2} (calculated from the two right hand elements of Equation 1.)

**Compute k ^{4} for 0.31 Albedo (increase of 0.1):**

T_{a} = kTs = 0.8876 Ts

Using εσTs^{4 }= So (1-A)/4

Then T_{a}^{4} = So (1-A)/(4εσ)

and since T_{a} = k T_{s}

k^{4}Ts^{4} = So (1-A)/(4εσ)

Differentiate,

4k^{4}T^{3} δT_{s} = – S_{o}δA/(4εσ)

Rearrange, and assume Ts approx = Ta, the difference being about 1 degree out of 288, is represented as a constant T in the denominator below.

δTs = – S_{o}δA/(16ϵσk^{4}T^{3})

= -1367/(16*0.97*5.67E-8*(0.8876)^{4}*288.2^{3})*δA = -104.7δA

δA = 0.31 – 0.30 = 0.01

**δTs = –**

**1**

**.05C**

**, that is temperature decreases as albedo increases by 0.01K, or by 3.3%.**

Because differentials are finite, the smaller the δA, the more accurate the δT_{s.}

czorba3@gmail.comPost authorSorry, I don’t have any.

Don’t want any.

CMW