Spectra & Composition

Section 7 of 8

Planetary Spectroscopy

Part 6: Spectroscopy Meets Climate — Why CO₂ Warms Planets

Everything you’ve learned about spectral absorption applies not only to stars but to planetary atmospheres. The same quantum physics that creates stellar absorption lines also creates the greenhouse effect — one of the most consequential applications of spectroscopy in all of science.

Planetary Energy Balance: Stefan-Boltzmann Applied to Planets

In Lecture 2, you learned the Stefan-Boltzmann law $L = 4\pi R^2 \sigma T^4$ . It applies to any blackbody — and, to first approximation, a planet is one. A planet doesn’t generate its own luminosity; it absorbs sunlight and re-radiates that energy as thermal (blackbody) radiation in the infrared. The Sun and Earth are both approximate blackbodies, but at very different temperatures: the Sun’s Planck curve peaks in visible light (~0.5 μm), Earth’s in the thermal infrared (~10 μm). These two blackbody spectra barely overlap — and that spectral separation is what makes the greenhouse effect possible.

At energy balance, the planet radiates exactly as much as it absorbs. The power absorbed is built from three factors: the stellar flux at the planet’s distance $L_\star / 4\pi d^2$ ; the intercepted cross-section $\pi R_p^2$ (the area of the shadow the planet casts, not its full surface); and the albedo correction $(1 - A)$ , where the albedo $A$ is the fraction of incoming light reflected away. So $\text{Absorbed} = (L_\star/4\pi d^2)\,\pi R_p^2\,(1 - A)$ . The power radiated, modeling the planet as a blackbody radiating from its full surface $4\pi R_p^2$ , is $\text{Radiated} = 4\pi R_p^2 \sigma T_{\text{eq}}^4$ . Setting these equal and solving for the equilibrium temperature gives:

This is $T_{\text{eq}}$ — the blackbody temperature the planet would have with no atmosphere. The planet’s radius $R_p$ cancels entirely; only $L_\star$ , $A$ , and $d$ remain.

Albedo

The fraction $A$ of incident light a surface or planet reflects rather than absorbs. $A = 0$ is perfect absorption; $A = 1$ is a perfect mirror. The Moon has $A \approx 0.12$ , Earth $\approx 0.30$ , Venus $\approx 0.77$ . Only the absorbed fraction $(1-A)$ heats the planet.

Equilibrium temperature

The temperature at which a planet radiates exactly as much energy as it absorbs from its star, assuming no atmosphere: $T_{\text{eq}} = (L_\star(1-A)/16\pi\sigma d^2)^{1/4}$ . The greenhouse effect raises the real surface temperature above this baseline.

Ratio Method: Planetary Temperature

The general formula contains five quantities, three of them constants or stellar properties. The ratio method evaluates the messy constants once for a reference case, then expresses every new problem as dimensionless ratios.

Reference case — a zero-albedo planet orbiting the Sun at 1 AU ( $L_\star = L_\odot$ , $A = 0$ , $d = 1\ \text{AU}$ ). Plugging in CGS values gives $T_{\text{ref}}^4 = 6.00 \times 10^9\ \text{K}^4$ , so $T_{\text{ref}} \approx 279\ \text{K}$ .

General formula as ratios — divide by the reference; $\sigma$ and $16\pi$ cancel:

$\boxed{T_{\text{eq}} = 279\ \text{K} \times \left(\frac{L_\star}{L_\odot}\right)^{1/4} (1-A)^{1/4} \left(\frac{d}{1\,\text{AU}}\right)^{-1/2}}$

Apply to Earth ( $L_\star/L_\odot = 1$ , $d = 1\ \text{AU}$ , $A \approx 0.3$ ): $T_{\text{eq}} = 279\ \text{K} \times (0.7)^{1/4} = 279 \times 0.915 = 255\ \text{K} = -18°\text{C}$ . That’s well below freezing — colder than Earth’s actual 288 K (+15°C). Something warms Earth by about 33 K beyond the bare equilibrium. That something is the greenhouse effect. Mars at 1.52 AU with $A \approx 0.25$ ? $T_{\text{eq}} \approx 279 \times (0.75)^{1/4} \times (1.52)^{-1/2} \approx 210\ \text{K}$ — no calculator needed.

The Greenhouse Effect: Absorption Lines in the Atmosphere

Earth’s surface radiates as an approximate blackbody at 288 K. Wien’s law gives the peak wavelength:

$\lambda_{\text{peak}} = \frac{b}{T} = \frac{2.898 \times 10^6\ \text{nm·K}}{288\ \text{K}} \approx 10{,}000\ \text{nm} = 10\ \mu\text{m}$

This is deep in the thermal infrared — far from the visible band where sunlight arrives. The figure below makes the separation vivid: the Sun’s incoming Planck curve (peaking near 0.5 μm) and Earth’s outgoing curve (peaking near 10 μm) barely overlap.

Diagram showing the electromagnetic spectrum with two Planck curves. Left panel (yellow background): incoming solar radiation peaking near 1 micron in the visible band. Right panel (blue background): outgoing terrestrial radiation peaking near 10 microns in the thermal infrared, with deep absorption notches carved by atmospheric greenhouse gases. The full EM spectrum from gamma rays to long waves is labeled across the top. — Figure 12Two blackbodies, two very different temperatures. The Sun (~5,800 K) radiates a Planck curve peaking in visible light (~0.5 um). Earth (~288 K) re-radiates a Planck curve peaking in thermal infrared (~10 um). Greenhouse gases absorb in the infrared window where Earth is trying to radiate — not where the Sun's energy arrives. The sharp absorption features in the outgoing spectrum are the molecular fingerprints of CO2, H2O, and O3.NOAA JetStream

For Earth to maintain balance, this outgoing infrared must escape to space. But greenhouse gases — CO₂, H₂O, CH₄, N₂O — have molecular absorption bands in the infrared, precisely where Earth’s blackbody curve radiates. These molecules absorb outgoing infrared photons and re-emit them in random directions — some back downward. The surface receives extra energy and warms until it radiates enough to compensate.

This is Kirchhoff’s law 3 applied to a planet: the atmosphere is the “cool gas” absorbing from the “hot continuum” (Earth’s surface thermal emission). The physics is identical to stellar absorption lines — only the relevant wavelengths differ (infrared, not visible).

Greenhouse effect

Warming of a planet’s surface caused by atmospheric gases absorbing outgoing infrared radiation and re-emitting part of it back downward. It is Kirchhoff’s third law applied to a planet: a cool atmosphere absorbing from the warm surface continuum.

A Tale of Three Planets

Venus, Earth, and Mars orbit the same star and formed from the same disk — yet their surface temperatures are wildly different. Spectroscopy tells us why:

Planet	Distance (AU)	Albedo	$T_{\text{eq}}$ (K)	Actual $T_{\text{surface}}$ (K)	Greenhouse warming (K)	Dominant greenhouse gas
Venus	0.72	0.77	227	735	+508	CO₂ (96.5% of atmosphere)
Earth	1.00	0.30	255	288	+33	H₂O + CO₂ (0.04%)
Mars	1.52	0.25	210	210	~0	CO₂ (95%, but very thin)

Bar chart comparing equilibrium temperature (teal) and actual surface temperature (gold) for Venus, Earth, and Mars. Venus shows the largest greenhouse gap of +508 K, Earth shows +33 K, and Mars shows approximately zero. Dominant atmospheric greenhouse gases are labeled beneath each planet. — Figure 13Venus is closer to the Sun yet has a lower equilibrium temperature than Earth because its high albedo reflects 77% of sunlight. The greenhouse warming (mauve arrows) is what separates the equilibrium temperature from the actual surface temperature — Venus's massive CO2 atmosphere adds 508 K, Earth's adds 33 K, and Mars's thin atmosphere adds essentially nothing.ASTR 201 (generated)

Venus is closer to the Sun than Earth, yet its equilibrium temperature is lower (227 K vs. 255 K). How? Albedo. Venus’s thick sulfuric-acid clouds reflect 77% of incoming sunlight before it can be absorbed. The $(1-A)^{1/4}$ factor means only the absorbed fraction matters: Venus absorbs just 23%, Earth 70%. The higher albedo more than compensates for the shorter distance — making the +508 K greenhouse warming even more staggering. Venus’s massive CO₂ atmosphere creates a runaway greenhouse: surface temperatures hot enough to melt lead (735 K = 462°C). Mars has a CO₂-dominated atmosphere too, but it’s so thin (0.6% of Earth’s surface pressure) that its greenhouse effect is negligible. Earth sits in between — a modest greenhouse that makes the planet habitable.

Worked Example 5Earth's Equilibrium Temperature

Problem

Calculate Earth’s equilibrium temperature assuming no atmosphere. Use $L_\odot = 3.828 \times 10^{33}\ \text{erg/s}$ , $d = 1.496 \times 10^{13}\ \text{cm}$ , and $A = 0.30$ .

StepNumerator

$L_\odot(1-A) = (3.828 \times 10^{33}\ \text{erg/s})(0.70) = 2.680 \times 10^{33}\ \text{erg/s}$

StepDenominator

$16\pi\sigma d^2 = 16\pi (5.67 \times 10^{-5})(1.496 \times 10^{13})^2 = 6.376 \times 10^{23}\ \text{erg s}^{-1}\text{K}^{-4}$

StepSolve for $T_{\text{eq}}$

$T_{\text{eq}}^4 = \frac{2.680 \times 10^{33}}{6.376 \times 10^{23}} = 4.203 \times 10^9\ \text{K}^4 \;\Rightarrow\; T_{\text{eq}} = (4.203 \times 10^9)^{1/4} \approx 255\ \text{K}$

Dimensional check

$\left(\dfrac{\text{erg/s}}{\text{erg cm}^{-2}\text{s}^{-1}\text{K}^{-4}\cdot\text{cm}^2}\right)^{1/4} = (\text{K}^4)^{1/4} = \text{K}$ ✓.

Result

$255\ \text{K} = -18°\text{C}$ . Without an atmosphere, Earth would be a frozen world. The observed average of 288 K (+15°C) means the greenhouse effect warms Earth by 33 K — the difference between a habitable planet and an ice ball, supplied by the infrared absorption bands of water vapor, CO₂, and other trace gases.

Predict First

Think about Kirchhoff's third law applied to a planet.

Earth's no-atmosphere equilibrium temperature is 255 K, but the observed average is 288 K. What must the atmosphere be doing to account for the 33 K difference, and which molecules could be responsible?

The 15-micron band below is the key.

Why CO₂ Matters: The 15-Micron Band

Carbon dioxide has a particularly strong absorption band centered at $\lambda \approx 15\ \mu\text{m}$ (15,000 nm). This sits right near the peak of Earth’s outgoing thermal radiation. When CO₂ concentration increases, this band deepens and broadens — blocking more outgoing infrared and forcing the surface to warm until the planet radiates enough through the remaining transparent windows.

Graph showing reflectance vs wavelength (400-2400 nm) for snow (high, flat), vegetation (red edge jump at 700 nm), dry soil (rising toward IR), and water (absorbs in IR). Each material has distinct spectral signature. — Figure 14Different materials reflect different wavelengths. Snow reflects broadly; water absorbs IR; vegetation has a sharp 'red edge' at 700 nm. Spectra can identify surface composition — even on exoplanets.JWST/STScI

The physics has been understood since John Tyndall’s experiments (1861) and Svante Arrhenius’s calculations (1896): CO₂ absorbs infrared at specific wavelengths set by its molecular energy levels — the same quantum mechanics that produces hydrogen Balmer lines. Adding CO₂ to the atmosphere is, spectroscopically, equivalent to adding more absorbing atoms to a stellar atmosphere: the absorption features deepen.

Deep Dive: Enrichment: Why N₂ and O₂ Are NOT Greenhouse Gases

Earth’s atmosphere is 78% N₂ and 21% O₂. Why don’t these dominant gases trap heat? The answer is spectroscopic. Infrared absorption requires a molecule to change its electric dipole moment when it vibrates — no change in dipole, no absorption.

N₂ and O₂ are homonuclear diatomics — both atoms identical. Their only vibrational mode (stretching) doesn’t change the dipole moment because the charge distribution stays symmetric. They are essentially invisible to infrared. H₂O and CH₄ are asymmetric with permanent dipole moments; their vibrations modulate the dipole strongly. CO₂ is the interesting case: a symmetric linear molecule (O=C=O) with no permanent dipole, but it has a bending mode that breaks the symmetry and creates a transient oscillating dipole — that bending mode produces the strong 15 μm band (and an asymmetric stretch absorbs near 4.3 μm).

This is why trace gases (CO₂ at only 0.04%!) can dominate the greenhouse effect: they interact with infrared light while the dominant gases do not. It’s the same quantum selection rules that determine which atomic transitions are “allowed.” The universe is consistent.

Deep Dive: Enrichment: Why Is the Sky Blue (and Sunsets Red)?

The same physics that governs molecular infrared absorption explains the color of the sky. Earth’s atmosphere is mostly N₂ and O₂ molecules, $\sim 10^{-4}\ \mu\text{m}$ in size — roughly 5,000 times smaller than visible wavelengths. When a scatterer is much smaller than the wavelength, we are in the Rayleigh scattering regime, with cross-section $\sigma_{\text{Rayleigh}} \propto \lambda^{-4}$ .

Figure 15Rayleigh scattering explains both the blue sky (scattered short wavelengths) and red sunsets (transmitted long wavelengths). Same physics, different viewing geometry.NotebookLM

That $\lambda^{-4}$ is steep: blue light (450 nm) scatters about $(650/450)^4 \approx 4.4$ times more efficiently than red (650 nm). Blue photons scatter out of the direct beam in all directions, filling the sky dome with blue; red photons continue straight through. Sunsets are red because at the horizon sunlight traverses ~10 times more atmosphere — most blue scatters away before reaching you, leaving red and orange. Rayleigh vs. Mie: interstellar dust grains ( $\sim 0.01$ – $1\ \mu\text{m}$ ) are comparable to visible wavelengths, so they follow Mie scattering ( $\sigma \propto \lambda^{-1.3}$ roughly) — gentler reddening. The connection: Rayleigh scattering, molecular absorption, and stellar absorption lines are all the electromagnetic interaction between photons and matter, governed by quantum mechanics and the relative scales of wavelength and structure size.

Part 6 takeaway: the greenhouse effect is Kirchhoff’s third law applied to a planet — atmospheric molecules absorb infrared from the warm surface at specific wavelengths set by quantum mechanics. CO₂’s strong 15 μm band sits squarely in Earth’s thermal emission window. Increasing CO₂ deepens this absorption, traps more outgoing radiation, and warms the surface. The same spectroscopy that tells us what stars are made of tells us why our planet is warming.