Albedo Effect

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

εσTa4 = 1/4So (1-A) = H = k4εσTs4                  Equation 1

Where  ε = emissivity = 0.97, σ  = 5.67E-8 watts/m2-K4, Ta = equivalent radiative temperature,
So = solar constant – sunlight incident on earth’s atmosphere = 1367 w/m2, A = albedo,
H = net sunlight incident on earth’s surface = earth’s heat radiated to space = 235.47 w/m2 for basline
Ts = temperature of earth’s surface  = 288.2K for baseline, k = Ta / Ts

(From the first and last terms of Equation 1, k4 = Ta4/Ts4, 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/m2

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

Computing Ta for no clouds:

Ta = [So(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/m2 (calculated from the two right hand elements of Equation 1.)

Compute k4 for 0.31 Albedo (increase of 0.1):

Ta = kTs = 0.8876 Ts

Using εσTs4 = So (1-A)/4

Then Ta4 = So (1-A)/(4εσ)

and since Ta = k Ts

k4Ts4 = So (1-A)/(4εσ)

Differentiate,

4k4T3 δTs = – SoδA/(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 = – SoδA/(16ϵσk4T3)

= -1367/(16*0.97*5.67E-8*(0.8876)4*288.23)*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 δTs.

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