8.2 Solar Constant and Planetary Energy Balance

Convert solar luminosity to Earth-received intensity, justify the S/4 factor, and derive a no-atmosphere equilibrium temperature model.

Estimated time: 32 minutes

From Solar Luminosity to the Solar Constant

If the Sun emits total power $P_{\text{sun}}$ approximately isotropically, that power spreads across a sphere of radius $d$ (the Earth-Sun distance). Dividing by the sphere area gives the intensity at Earth's orbit. This intensity is the solar constant $S$ , defined at the top of the atmosphere.

$S = \frac{P_{sun}}{4\pi d^2}$

Using present-day values gives $S$ close to $1360$ to $1400\,\mathrm{W\,m^{-2}}$ , depending on averaging convention.

At this stage, keep geometric meaning explicit: S is not the globally averaged surface intensity. It is the beam intensity on a surface perpendicular to sunlight at Earth orbit distance. The global-average surface-relevant input is lower for geometric reasons even before albedo is applied.

Why Earth Receives S/4 on Average

Earth intercepts sunlight over a disk area $\pi R^2$ , but that absorbed energy is distributed over the full planetary surface area $4\pi R^2$ . The ratio of these areas gives the famous factor $1/4$ in globally averaged incoming intensity.

$I_{in,avg} = \frac{S}{4},\qquad I_{absorbed} = (1-\alpha)\frac{S}{4}$

Alpha removes the reflected fraction, leaving absorbed shortwave intensity.

This is one of the most important conceptual checkpoints in the chapter. Students often remember the formula but forget why the factor appears. If you can explain disk-versus-sphere geometry in words, you can rebuild the expression even if you forget it under exam pressure.

With $\alpha \approx 0.30$ and $S \approx 1360\,\mathrm{W\,m^{-2}}$ , the absorbed global-average flux is around $238\,\mathrm{W\,m^{-2}}$ . That number acts as the incoming side of first-pass equilibrium models.

No-Atmosphere Equilibrium Temperature

In the simplest model, Earth has no infrared-absorbing atmosphere and radiates like an effective black or gray body directly to space. Equilibrium sets absorbed solar intensity equal to emitted thermal intensity.

$(1-\alpha)\frac{S}{4}=\epsilon\sigma T_{eq}^4,\qquad T_{eq}=\left(\frac{(1-\alpha)S}{4\epsilon\sigma}\right)^{1/4}$

For $\epsilon$ close to $1$ and $\alpha \approx 0.30$ , this gives roughly $255$ to $256\,\mathrm{K}$ , below observed surface average.

That result is not a failure; it is a diagnostic clue. The model is intentionally missing atmospheric infrared absorption and reradiation. The temperature gap points directly to the greenhouse contribution needed to reconcile observed mean surface temperature.

Simulation: Planetary Equilibrium Without Atmospheric IR Trapping

Use planetary mode to vary solar constant, albedo, and surface emissivity, then track the resulting equilibrium temperature and top-of-atmosphere balance.

Greenhouse Energy Balance Lab

Solar constant S (W m^-2): 1361

Albedo alpha: 0.30

Surface emissivity epsilon_s: 1.00

Surface temperature

-18.6 deg C

254.6 K

Atmosphere temperature

not modeled

planetary mode

Absorbed solar

238.2 W m^-2

(1 - alpha)S/4

Outgoing to space

238.2 W m^-2

imbalance 0.000 W m^-2

Flux diagram (global average intensities)

Temperature sensitivity to albedo

Test Yourself

Take $S = 1360\,\mathrm{W\,m^{-2}}$ and $\alpha = 0.30$ . Enter the absorbed global-average solar intensity in $\mathrm{W\,m^{-2}}$ .

Hint: Compute $(1 - \alpha)S/4$ .

Your numerical answer (W m^-2)

8.1 Radiation from Real Bodies

8.3 Greenhouse Effect and Photon Absorption