Realistic Example: Satellite Retrieval of XCO2

This example demonstrates how satellite measurements lead to realistic XCO2 values (~420 ppm), consistent with modern atmospheric observations.

Step 1: Total Air Column from Hydrostatic Balance

The total dry air column is given by:

$$N_{\rm air} =\frac{P_s}{m_{\rm air}g}$$

Using realistic values:

  • Surface pressure: Ps = 101325 Pa
  • Gravity: g = 9.81 m/s²
  • Mean molecular mass of air: mair = 4.8 × 10-26 kg/molecule
$$ N_{\rm air} ≈ 2.1 \times 10^{29} \text{molecules}/\text{m}^2 $$

This represents the total number of air molecules above 1 m² of Earth's surface.

How Mean Molecular Mass of Air is Calculated?

Dry air is mainly composed of:
Gas Fraction Molecular Mass (g/mol)
N₂ 78% 28
O₂ 21% 32
Ar 1% 40

The mean molecular mass is simply the weighted average:

$$M_{\rm air} = \sum (f_i \times M_i)$$ Where:
  • \(f_i \) = = mole fraction
  • \(f_i \) = molar mass
From the calculation above: $$M_{\rm air} = 28.96 \text{g/mol}$$

Convert to Mass per Molecule:

For hydrostatic balance we need kg per molecule, not per mole. $$m_{\rm air} = \frac{M_{\rm air}}{N_A}$$ where \(N_A= 6.022 \times 10^{23} \) molecules/mol. Therefore: $$m_{\rm air} = \frac{0.02896}{6.022 \times 10^{23}} \approx 4.8\times 10^{-26}$$ kg/molecule

Step 2: Use Realistic Atmospheric CO₂ Mixing Ratio

Current global average:

$$ XCO_2 \approx 420 ~\text{ppm} $$

Convert ppm to fraction:

$$ 420 \text{ppm} = 420 \times 10^{-6} = 4.2 \times 10^{-4} $$

Step 3: Compute Total CO₂ Column

$$ N_{{\rm CO}_2} = XCO_2 \times N_{\rm air} $$

Substituting values:

$$ N_{{\rm CO}_2} = (4.2 \times 10^{-4}) \times (2.1 \times 10^{29}) $$ $$ N_{{\rm CO}_2} \approx 8.8 \times 10^{25} ~~ \text{molecules}/\text{m}^2 $$

This is the realistic atmospheric CO₂ column sensed by satellites.

Step 4: Connect to Radiative Transfer

Optical depth is given by:

$$ \tau = \sigma \times N_{{\rm CO}_2} $$

Using an effective band-averaged absorption cross-section:

$$ \sigma \approx 2 \times 10^{-26}~ \text{m}^2/\text{molecule} $$ $$ \tau = (2 \times 10^{-26}) \times (8.8 \times 10^{25}) $$ $$ \tau \approx 1.76 $$

This optical depth is realistic for strong CO₂ absorption bands near 1.6 μm.

Step 5: Reverse Calculation (Satellite Retrieval Concept)

If a satellite measures:

$$ \tau \approx 1.8 $$

Then:

$$ N_{{\rm CO}_2} = \tau/ \sigma $$ $$ N_{{\rm CO}_2} = 1.8 / (2 \times 10^{-26}) $$ $$ N_{{\rm CO}_2} ≈ 9 \times 10^{25} \text{molecules}/\text{m}^2 $$

Now compute \(XCO_2\):

$$ XCO_2 = N_{{\rm CO}_2} / N_{\rm air} $$ $$ XCO_2 = (9 \times 10^{25} / (2.1 \times 10^{29}) $$ $$ XCO_2 ≈ 4.3 \times 10^{-4} = 430 ~\text{ppm} $$

Final Interpretation

The atmosphere contains approximately:

  • 1029 air molecules per m²
  • 1025 CO₂ molecules per m²

Even though CO₂ is only ~0.042% of the atmosphere, it significantly affects Earth's radiative balance.

Reference

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