INTERACTIVE COMPANION · METHODS · EARTH'S ENERGY BUDGET

The Layer Bench

A live model accompanying “A Greenhouse Model You Can Check by Hand”

Two benches sit below. The first stacks perfectly absorbing layers above a planet and prints the temperature of every level, beside the value the analytic formula demands. The second keeps one layer, lets you set how absorbing it is, and then shows you what that convenient grey assumption is costing. Every constant here is the one in the analysis file, and the defaults reproduce the article's headline numbers exactly.

Model 1. Stack the layers

The rules are short enough to hold in your head. Sunlight passes through every layer untouched and is absorbed at the ground. Each layer absorbs everything in the infrared and therefore emits everything, radiating σT⁴ upward and the same amount downward. Space above the top layer sends nothing back. Balance the energy at every level and you get a chain of linear equations that the code below solves twice, once by stepping down the chain and once by the closed form, so you can watch the two agree.

F + x₁ = x₀ xᵢ₋₁ + xᵢ₊₁ = 2xᵢ xₘ₊₁ = 0 → Tₛ = Tₑ(N+1)^1/4

Number of layers N 1 Start at 1, which is the version almost every textbook draws, and read the surface temperature: 303.50 K. The real planet is 288.00 K. Now drag to 0 and you get 255.21 K, which is 32.79 K too cold. There is no setting between them, and that gap is the whole subject of the article. Push on to 10 and watch the steps shrink: each new layer adds less than the one before, because the temperature goes as the fourth root. Solar constant S 1361 W/m² The measured value is 1361 with an uncertainty of about 0.5, so this slider's whole range is far wider than anything real. It is here to show you how little the answer cares: moving S by a full standard deviation changes the inferred emissivity in Model 2 by 0.00045, which is nothing. Planetary albedo α 0.293 0.293 is the CERES satellite value. Most textbooks round to 0.30, which moves the no-atmosphere temperature by 0.63 K, more than either value's distance from the famous 255. Try 0.15, which is roughly what Earth's albedo would be with the clouds removed along with the air, and watch the greenhouse effect shrink from 33 K to 21 K without a single new measurement. Arrow animation The arrows drift to suggest which way the radiation is going. Turn it off if it distracts. It turns itself off if your system asks for reduced motion, and it stops whenever this tab is in the background.

surface: – effective: – vs observed 288 K: – layers Earth needs: –

–

Model 2. Let the layer be imperfect, then count the cost

One layer, absorbing a fraction ε of the infrared rather than all of it. The layer's own balance loses the ε entirely, so a thin veil and an opaque slab sit at the same temperature and differ only in how much of the ground they can see. What is left is one formula:

σTₛ⁴ = F / (1 − ε/2) → ε* = 2[1 − (Tₑ/Tₛ)⁴] = 0.7667

Underneath the plot sits the comparison the article is really about. The two-band model splits the infrared at the edges of the atmospheric window, gives the window an optical depth you can set and the rest of the spectrum whatever depth is needed to reach the same surface temperature, and computes the band fractions by summing the exact blackbody series. On the present day the two models agree exactly. Then you multiply the absorber and they stop agreeing.

Layer emissivity ε 0.767 The marked value is 0.7667, which is what it takes to reproduce 288.00 K. Slide to 1.000 and you are back at Model 1's single perfect layer, 303.50 K, fifteen and a half degrees too hot. Slide to 0 and the atmosphere has vanished. The curve is steepest at the right-hand end, which is why one percentage point of emissivity is worth 0.58 K here and less further left. Window optical depth τ_w 0.30 How opaque the 8 to 12 µm window is. The two-band model refuses to work below about 0.08, because with a perfectly clear window a single layer cannot reach 288 K however opaque you make the rest of the spectrum. Try 0.10 and watch the grey error factor jump towards 8. Try 0.90 and watch it fall towards 1. The direction of the conclusion survives the whole range; its size does not. Multiply the absorber by 2.0× This is the experiment. At 1.0 both models sit on 288.00 K by construction. At 2.0, which is one doubling, grey gives +11.51 K and two-band gives +5.44 K. Keep going and grey runs into its own ceiling of 303.50 K, the perfect absorber again, while the two-band model keeps climbing on the window alone. What the multiplier scales Carbon dioxide works in the band and barely touches the window, so “absorbing band only” is the more realistic of the two. Choose it and watch the two-band curve flatten completely: a hundredfold increase buys 2.64 K and then stops, because a saturated band cannot be made to radiate from any higher. Grey has no ceiling of that kind at all, which is the single most important thing it hides.

grey T_s: – two-band T_s: – window escape: – grey overstatement: –

–

τ grey, calibrated: –
τ_b, calibrated: –
window fraction at T_s: –
effective absorptivity A: –
total escape to space: –
observed total escape: 22.0 W/m² (Costa & Shine 2012)

What to try

Set Model 1 to N = 1 and read 303.50 K, then to N = 0 and read 255.21 K. Those two numbers bracket the planet and neither is the planet. That is the whole problem in ten seconds.

In Model 2, put the multiplier back to 1.0 and confirm that both models read 288.00 K and that the effective absorptivity A equals ε* to six decimals. Splitting the spectrum changed nothing at all about the present day, which is exactly why the grey assumption survives in textbooks. Then move the multiplier to 2.0 and watch the two numbers separate by six degrees.

Last, set the scaling to “absorbing band only” and push the multiplier to the top. The two-band curve goes flat. Nothing you do to that band matters any more, because it is already opaque and radiating from its own top. Grey cannot represent that, and never will, because a single optical depth can always be made larger.