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High School Earth and Space Sciences · 10–12 年级

2A Habitable Planet

Three rocky planets circle the Sun side by side. On the surface of Venus the air is so dense that it presses ninety-two times harder than on the Earth, and so hot — more than $460{}^{\circ}\mathrm{C}$ — that lead would melt. On Mars the thin air presses less than a hundredth as hard as ours, and the average temperature is far below freezing. Between them, the Earth has oceans, rain and life. The three planets formed from the same disc of gas and dust around the young Sun, out of much the same rock. Why such different fates? This chapter looks at what sets the temperature of a planet, what lets it keep an atmosphere, under what conditions water can be liquid, and where, around the Sun and around other stars, such conditions can be found.

Venus seen by the probe Mariner 10 in 1974, in colours close to what the eye would see: a planet entirely wrapped in cloud. Image: NASA/JPL-Caltech (public domain).

You already know

The Sun is a star; around it the four rocky planets Mercury, Venus, the Earth and Mars, then the four giant planets, travel on nearly circular orbits (the physics books and Book 1). Every body radiates energy as light, more and more of it as its temperature rises, mostly as infrared light at the temperatures of planets (the physics books). Greenhouse gases — water vapour, carbon dioxide, methane — absorb part of the infrared emitted by a surface and keep it warmer (the chemistry books). Water exists as ice, liquid water and water vapour; pressure is a force per unit area, measured in pascals (the physics books).

2.1 The temperature set by the star

The Sun sends its light equally in all directions. At a distance $d$ , its power $L$ (its luminosity) is spread over a sphere of area $4\pi d^2$ .

Proposition 2.1 (The inverse-square law of sunlight)

The power received per square metre facing the Sun, the solar irradiance, at a distance $d$ from the Sun is

S = \frac{L}{4\pi d^2}.

At the Earth’s distance, $1\,\mathrm{au}$ , $S_0 = 1361\,\mathrm{W}/\mathrm{m}^{2}$ ; at a distance $d$ measured in astronomical units, $S = S_0/d^2$ .

Proof. No light is lost in empty space, so the whole power $L$ crosses every sphere centred on the Sun; on a sphere of radius $d$ it is shared among $4\pi d^2$ square metres. Doubling $d$ multiplies the area by four and divides $S$ by four. ∎

The light that crosses one square at distance d is spread over four squares of the same size at distance 2d: the irradiance falls as the inverse square of the distance. — The light that crosses one square at distance $d$ is spread over four squares of the same size at distance $2d$ : the irradiance falls as the inverse square of the distance.

A planet absorbs the part of this light that it does not reflect, and it radiates energy to space as infrared light. If it absorbs more than it radiates, it warms up; warmer, it radiates more; it settles at the temperature where the two balance.

Definition 2.2 (Equilibrium temperature)

The equilibrium temperature of a planet is the temperature at which a body radiating like a perfect emitter would send to space exactly the power the planet absorbs from its star. The fraction of the incoming light that the planet reflects as a whole is its Bond albedo $A$ .

Proposition 2.3 (Computing the equilibrium temperature)

For a planet that spreads the heat it absorbs over its whole surface,

T_{\mathrm{eq}} = \left(\frac{S\,(1-A)}{4\sigma}\right)^{1/4}, \qquad \sigma = 5.67 \times 10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-4}.

At the Earth’s distance and with no reflection at all, this gives $278.3\,\mathrm{K}$ ; at a distance $d$ in astronomical units and with albedo $A$ , $T_{\mathrm{eq}} = 278.3\,\mathrm{K} \times (1-A)^{1/4}/\sqrt{d}$ .

Partial proof. The planet intercepts sunlight on a disc of area $\pi R^2$ and absorbs the power $(1-A)\,S\,\pi R^2$ . It radiates from its whole surface, $4\pi R^2$ . A perfect emitter at temperature $T$ radiates $\sigma T^4$ per square metre (a law proved in the physics books, admitted here). Equating $4\pi R^2 \sigma T^4 = (1-A)\,S\,\pi R^2$ gives the formula; the radius cancels. With $S = S_0/d^2$ , the fourth root of $1/d^2$ is $1/\sqrt d$ . ∎

Example 2.4 (The Earth)

With $S_0 = 1361\,\mathrm{W}/\mathrm{m}^{2}$ and $A = 0.294$ , $T_{\mathrm{eq}} = 278.3 \times 0.706^{1/4} \approx 255\,\mathrm{K}$ , that is $-18{}^{\circ}\mathrm{C}$ . (NASA’s fact sheet lists $254\,\mathrm{K}$ .) The average temperature of the Earth’s surface is $288\,\mathrm{K}$ , $15{}^{\circ}\mathrm{C}$ : the greenhouse effect of the atmosphere warms it by about $33\,\mathrm{K}$ . Without it the oceans would freeze.

Equilibrium temperatures computed from each body’s distance and Bond albedo, beside the measured mean temperatures of their surfaces. Data: NASA planetary fact sheets.

Remark 2.5 (Reading the chart)

For Mercury and Mars, which have almost no atmosphere or a very thin one, the measured mean is close to the equilibrium temperature. The Moon’s mean lies below it: its days and nights each last about two weeks, and its surface swings from about $95\,\mathrm{K}$ to $390\,\mathrm{K}$ , which the simple formula, made for a planet that spreads its heat evenly, does not capture. Venus is the striking case: with a Bond albedo of 0.77 — its clouds reflect most of the sunlight — its equilibrium temperature is lower than the Earth’s, yet its surface is more than $500\,\mathrm{K}$ hotter than that: the greenhouse effect of its thick carbon-dioxide atmosphere.

2.2 Holding an atmosphere

A gas is made of molecules in constant motion, faster the hotter the gas. A molecule at the top of an atmosphere that moves upwards faster than the planet’s escape speed — the speed needed to leave its gravity for good (the physics books) — is lost to space. Over billions of years, a small planet with a weak gravity loses its air; a large one keeps it.

body	escape speed ( $\mathrm{km}/\mathrm{s}$ )	surface pressure (bar)	atmosphere
Moon	2.4	0	none
Mercury	4.3	0	none
Mars	5.0	0.006	thin, carbon dioxide
Venus	10.4	92	thick, carbon dioxide
Earth	11.2	1	nitrogen and oxygen

Remark 2.6 (Size is not the whole story)

The Moon and Mercury, small and (for Mercury) close to the hot Sun, have lost almost all their gases. Mars, with half the Earth’s escape speed, kept a thin atmosphere. Venus and the Earth have nearly the same size and escape speed, yet very different atmospheres: their histories — the fate of their water and of their carbon dioxide — diverged, as Chapter 26 will show for the Earth.

2.3 Liquid water

The phase diagram of water, with the mean surface conditions of Venus, the Earth and Mars (squares). The pressure axis is logarithmic: each tick is a hundred times the one below. Curves computed from the equations of the International Association for the Properties of Water and Steam; planetary values from NASA.

Proposition 2.7 (When water can be liquid)

Liquid water exists only at pressures above that of the triple point, $611.657\,\mathrm{Pa}$ ( $0.006\,\mathrm{bar}$ ), where ice, liquid and vapour coexist at $273.16\,\mathrm{K}$ , and at temperatures below the critical point, $647.096\,\mathrm{K}$ ( $220.64\,\mathrm{bar}$ ), beyond which liquid and vapour can no longer be told apart. At a given pressure above the triple point, water is liquid between its melting point, close to $273\,\mathrm{K}$ , and its boiling point, which falls as the pressure falls.

Example 2.8 (Three planets on one diagram)

The Earth’s surface, at $288\,\mathrm{K}$ and $1\,\mathrm{bar}$ , lies inside the liquid field: water there can be liquid, ice or vapour as the weather goes. Venus, at $737\,\mathrm{K}$ , lies far beyond the critical temperature: no liquid water is possible at all. Mars, at $6.36\,\mathrm{mbar}$ , lies just above the triple point: water there boils at about $0.5{}^{\circ}\mathrm{C}$ , so liquid water could survive only in a window of about half a degree, and the mean temperature, $210\,\mathrm{K}$ , is in the ice field.

Mount Sharp, inside Gale Crater on Mars, photographed by the rover Curiosity in 2012, in the colours recorded by its camera. Its lower layers were laid down in water, billions of years ago. Image: NASA/JPL-Caltech/MSSS (public domain).

2.4 The habitable zone

Definition 2.9 (Habitable zone)

The habitable zone of a star is the range of distances from the star at which a rocky planet with enough water and an atmosphere like the Earth’s could keep liquid water on its surface. Its inner edge is where the planet would lose its water to the heat; its outer edge, where even a strong greenhouse effect could no longer keep it from freezing.

Proposition 2.10 (The Sun’s habitable zone)

Detailed models of planetary atmospheres place the habitable zone of the Sun between $0.99\,\mathrm{au}$ and $1.70\,\mathrm{au}$ . The Earth ( $1\,\mathrm{au}$ ) lies just inside its inner edge; Mars ( $1.52\,\mathrm{au}$ ) lies inside it; Venus ( $0.72\,\mathrm{au}$ ) lies well inside the inner edge.

Proposition 2.11 (Habitable zones of other stars)

Around a star of luminosity $L$ , a planet receives the same irradiance as at a distance $d_\odot$ from the Sun when it lies at

d = d_\odot \sqrt{L/L_\odot}.

The edges of the habitable zone move in proportion, to a first approximation (the colour of the star’s light also matters a little).

Proof. Equal irradiance means $L/(4\pi d^2) = L_\odot/(4\pi d_\odot^2)$ , so $d^2 = d_\odot^2\,L/L_\odot$ . ∎

$Top: the habitable zone of the Sun and the four rocky planets. Bottom: the habitable zone of the faint red star Proxima Centauri, computed by scaling the Sun’s with the square root of the star’s luminosity, and its planet Proxima b. Note the change of scale: the whole bottom strip is 0.1\, au long.$

Top: the habitable zone of the Sun and the four rocky planets. Bottom: the habitable zone of the faint red star Proxima Centauri, computed by scaling the Sun’s with the square root of the star’s luminosity, and its planet Proxima b. Note the change of scale: the whole bottom strip is

0.1\,\mathrm{au}

long.

2.5 Planets of other stars

Definition 2.12 (Exoplanet)

An exoplanet is a planet that orbits a star other than the Sun.

Exoplanets are far too faint and too close to their stars to be seen directly, apart from a few giants. Most are found indirectly: either the planet passes in front of its star once per orbit and dims it very slightly, or its pull makes the star wobble towards and away from us, which shifts the colour of the star’s light back and forth. The second method gives only a minimum mass for the planet, because the tilt of the orbit is not known. These methods are studied in the Year 3 volume.

Figures as of 2026

On 15 October 2026 the NASA Exoplanet Archive listed 6 451 confirmed exoplanets.

Example 2.13 (Proxima b)

The nearest star to the Sun, Proxima Centauri, $1.30\,\mathrm{pc}$ away, is a small red star whose luminosity is only about $0.0015$ that of the Sun. In 2016 a planet was found orbiting it every 11.2 days, at $0.0485\,\mathrm{au}$ , with at least 1.06 times the mass of the Earth. At that distance it receives about 0.64 of the irradiance the Earth receives, and it lies inside its star’s habitable zone. Whether it has an atmosphere, or water, is not known.

Remark 2.14 (Habitable is not inhabited)

The habitable zone is a statement about distance and light only. A planet inside it may have no water, no atmosphere, or the wrong one — Mars lies inside the Sun’s zone and has no liquid water at its surface today. The Earth is habitable because several conditions hold at once: the right distance, a mass large enough to keep an atmosphere, abundant water, and a greenhouse effect neither too weak nor too strong.

2.6 Exercises

Exercise 2.1 ★

Define the equilibrium temperature of a planet. Why is the Earth’s mean surface temperature higher than its equilibrium temperature?

Solution

Solution of Exercise 2.1.

The temperature at which a perfect emitter would radiate to space exactly the power the planet absorbs from its star. The Earth’s surface is warmer ( $288\,\mathrm{K}$ against about $255\,\mathrm{K}$ ) because greenhouse gases absorb part of the infrared it emits and send part of it back down.

Exercise 2.2 ★

Using the bar chart of equilibrium and measured temperatures, name the body whose surface is warmed most by its atmosphere, and estimate by how many kelvins.

Solution

Solution of Exercise 2.2.

Venus: measured $737\,\mathrm{K}$ against an equilibrium temperature of about $227\,\mathrm{K}$ , a warming of about $510\,\mathrm{K}$ .

Exercise 2.3 ★

Mars orbits at $1.52\,\mathrm{au}$ from the Sun. Using the inverse-square law with $S_0 = 1361\,\mathrm{W}/\mathrm{m}^{2}$ , compute the solar irradiance at Mars.

Solution

Solution of Exercise 2.3.

$S = 1361/1.52^2 \approx 589\,\mathrm{W}/\mathrm{m}^{2}$ (the fact sheet gives $586\,\mathrm{W}/\mathrm{m}^{2}$ , with the exact mean distance).

Exercise 2.4 ★

Using the phase diagram of water, give the state of water at $1\,\mathrm{bar}$ and $330\,\mathrm{K}$ ; at $0.001\,\mathrm{bar}$ and $250\,\mathrm{K}$ ; at $100\,\mathrm{bar}$ and $700\,\mathrm{K}$ .

Solution

Solution of Exercise 2.4.

$1\,\mathrm{bar}$ , $330\,\mathrm{K}$ : liquid (below the boiling point, about $373\,\mathrm{K}$ ). $0.001\,\mathrm{bar}$ , $250\,\mathrm{K}$ : ice (the pressure lies above the ice–vapour curve, which is at about $0.0008\,\mathrm{bar}$ at $250\,\mathrm{K}$ ). $100\,\mathrm{bar}$ , $700\,\mathrm{K}$ : vapour — the temperature is above the critical temperature, so no liquid can exist.

Exercise 2.5 ★

What is an exoplanet? Describe in one sentence each the two main ways in which exoplanets are detected.

Solution

Solution of Exercise 2.5.

A planet orbiting a star other than the Sun. Transits: the planet passes in front of its star and dims it slightly at every orbit. Wobble: the planet’s pull moves the star back and forth along our line of sight, which shifts the colour of its light periodically.

Exercise 2.6 ★★

Compute the Earth’s equilibrium temperature with $S_0 = 1361\,\mathrm{W}/\mathrm{m}^{2}$ and $A = 0.294$ , then with $A = 0$ . Convert both to degrees Celsius.

Solution

Solution of Exercise 2.6.

$T = (1361 \times 0.706/(4 \times 5.67\times10^{-8}))^{1/4} \approx 255\,\mathrm{K} \approx -18{}^{\circ}\mathrm{C}$ . With $A = 0$ : $\approx 278\,\mathrm{K} \approx 5{}^{\circ}\mathrm{C}$ .

Exercise 2.7 ★★

A star is four times as luminous as the Sun. Between which distances does its habitable zone lie? Would the Earth, placed at $1\,\mathrm{au}$ from it, be inside?

Solution

Solution of Exercise 2.7.

$\sqrt 4 = 2$ : from $2 \times 0.99 = 1.98\,\mathrm{au}$ to $2 \times 1.70 = 3.40\,\mathrm{au}$ . At $1\,\mathrm{au}$ the planet would receive four times the Earth’s irradiance: far inside the inner edge, too hot.

Exercise 2.8 ★★

Venus orbits at $0.723\,\mathrm{au}$ . Compute the irradiance it receives, then its equilibrium temperature with its Bond albedo of 0.77. Compare with its surface temperature of $737\,\mathrm{K}$ and explain the difference.

Solution

Solution of Exercise 2.8.

$S = 1361/0.723^2 \approx 2604\,\mathrm{W}/\mathrm{m}^{2}$ ; $T_{\mathrm{eq}} = (2604 \times 0.23/(4\sigma))^{1/4} \approx 227\,\mathrm{K}$ . The surface is about $510\,\mathrm{K}$ hotter: its thick carbon-dioxide atmosphere ( $92\,\mathrm{bar}$ ) produces an enormous greenhouse effect.

Exercise 2.9 ★★

On Mars the mean surface pressure is $6.36\,\mathrm{mbar}$ , just above the triple point of water, and water boils there at about $0.5{}^{\circ}\mathrm{C}$ . Explain why liquid water cannot last on the surface of Mars today, even on a summer afternoon when the ground reaches $10{}^{\circ}\mathrm{C}$ .

Solution

Solution of Exercise 2.9.

At $6.36\,\mathrm{mbar}$ water is liquid only between about $0{}^{\circ}\mathrm{C}$ and $0.5{}^{\circ}\mathrm{C}$ . At $10{}^{\circ}\mathrm{C}$ the pressure is below the boiling curve: water boils away into vapour; at night it freezes.

Exercise 2.10 ★★

The escape speed is $2.4\,\mathrm{km}/\mathrm{s}$ on the Moon and $11.2\,\mathrm{km}/\mathrm{s}$ on the Earth. Compute their ratio. Explain why the Earth has kept an atmosphere and the Moon has not.

Solution

Solution of Exercise 2.10.

$11.2/2.4 \approx 4.7$ . A gas molecule must move almost five times faster to escape the Earth than the Moon; over billions of years the Moon’s weak gravity let its gases escape, while the Earth’s kept most of them.

Exercise 2.11 ★★

Suppose the Earth were covered with ice and its Bond albedo rose to 0.50. Compute its new equilibrium temperature. Explain why such a state, once reached, would tend to last.

Solution

Solution of Exercise 2.11.

$T = (1361 \times 0.50/(4\sigma))^{1/4} \approx 234\,\mathrm{K}$ , about $-39{}^{\circ}\mathrm{C}$ . The ice would reflect so much sunlight that the planet would stay cold enough to keep it: a reinforcing loop.

Exercise 2.12 ★★★

Show that $T_{\mathrm{eq}} = 278.3\,\mathrm{K} \times (1-A)^{1/4}/\sqrt{d}$ , with $d$ in astronomical units, follows from the formula of the chapter. By what factor does the equilibrium temperature change when the distance is multiplied by four?

Solution

Solution of Exercise 2.12.

$T_{\mathrm{eq}} = (S_0(1-A)/(4\sigma d^2))^{1/4} = (S_0/(4\sigma))^{1/4} (1-A)^{1/4} (d^2)^{-1/4}$ ; $(1361/(4\sigma))^{1/4} \approx 278.3\,\mathrm{K}$ and $(d^2)^{1/4} = \sqrt d$ . Four times farther: $T$ divided by $\sqrt 4 = 2$ .

Exercise 2.13 ★★★

At what distance from the Sun would a planet of albedo 0.294 have an equilibrium temperature of $273\,\mathrm{K}$ ? Compare with the inner edge of the habitable zone, and explain why the two are different.

Solution

Solution of Exercise 2.13.

$\sqrt d = 278.3 \times 0.706^{1/4}/273 \approx 0.934$ , so $d \approx 0.873\,\mathrm{au}$ , inside the inner edge ( $0.99\,\mathrm{au}$ ). The equilibrium temperature ignores the greenhouse effect: a real planet with an atmosphere is warmer than $T_{\mathrm{eq}}$ , so it overheats and loses its water already farther out, at about $0.99\,\mathrm{au}$ . The edges come from models of the whole atmosphere, not from $T_{\mathrm{eq}}$ alone.

Exercise 2.14 ★★★

On a planet whose surface pressure is $1000\,\mathrm{Pa}$ , water boils at about $7{}^{\circ}\mathrm{C}$ . Between which temperatures can water be liquid there? What would happen to a lake on that planet on a day at $15{}^{\circ}\mathrm{C}$ ?

Solution

Solution of Exercise 2.14.

Between the melting point, about $0{}^{\circ}\mathrm{C}$ , and the boiling point, about $7{}^{\circ}\mathrm{C}$ . At $15{}^{\circ}\mathrm{C}$ the lake would boil.

Exercise 2.15 ★★★

A planet orbits at $1.2\,\mathrm{au}$ from a star half as luminous as the Sun. Compute the edges of the star’s habitable zone and say whether the planet is inside. How sure can one be of the answer?

Solution

Solution of Exercise 2.15.

$\sqrt{0.5} \approx 0.707$ : from $0.707 \times 0.99 \approx 0.70\,\mathrm{au}$ to $0.707 \times 1.70 \approx 1.20\,\mathrm{au}$ . The planet is right at the outer edge. The edges are model results and the scaling is approximate, so it is impossible to say whether it is inside or outside.

2.7 Problem: Could Proxima b Hold an Ocean?

Problem 2.1

Weekend problem — from the light of a faint red star to the temperature of its nearest planet

A student reads the data sheet of the planet Proxima b. Its star has a luminosity $L = 0.00151\,L_\odot$ ; the planet orbits at $d = 0.0485\,\mathrm{au}$ in 11.18 days and has at least 1.055 times the mass of the Earth. Use $S_0 = 1361\,\mathrm{W}/\mathrm{m}^{2}$ , $\sigma = 5.67 \times 10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-4}$ , the Sun’s habitable zone $0.99\text{ to }1.70\,\mathrm{au}$ , and the Earth’s escape speed $11.2\,\mathrm{km}/\mathrm{s}$ .

Part I — The light of the star.

Compute the irradiance at the planet as a fraction of the Earth’s, $L/d^2$ (with $L$ in solar luminosities and $d$ in au).
Express it in $\mathrm{W}/\mathrm{m}^{2}$ .
Compare it with the irradiance at Venus ( $2601\,\mathrm{W}/\mathrm{m}^{2}$ ) and at Mars ( $586\,\mathrm{W}/\mathrm{m}^{2}$ ).
How many of the planet’s years fit into one Earth year?

Part II — Temperatures without an atmosphere.

Compute the equilibrium temperature of a perfectly dark planet ( $A = 0$ ) at that irradiance.
Compute it for a planet as bright as Venus ( $A = 0.77$ ).
Which of these two temperatures is above the freezing point of water? What do the two results show about the role of clouds and ice?

Part III — The habitable zone.

Compute $\sqrt{L/L_\odot}$ .
Compute the inner edge of the star’s habitable zone.
Compute its outer edge.
Is Proxima b inside the habitable zone?
At what distance from the Sun would a planet receive the same irradiance as Proxima b? Where would that be in the Sun’s habitable zone?

Part IV — An atmosphere and water?

If the planet had the Earth’s mean density, its radius would be $1.055^{1/3}$ times the Earth’s. Compute this factor.
The escape speed is proportional to $\sqrt{M/R}$ . Estimate the planet’s escape speed. Could it keep an atmosphere?
Why is the mass known only as a minimum?
On the phase diagram, what is the lowest surface pressure that would allow liquid water at all?
Compute the equilibrium temperature of Proxima b with the Earth’s albedo, 0.294.
By how many kelvins would its greenhouse effect then have to warm it to bring its surface to the Earth’s mean temperature, $288\,\mathrm{K}$ ?

Solution

Solution of Problem 2.1.

1. $0.00151/0.0485^2 \approx 0.642$ .

2. $0.642 \times 1361 \approx 874\,\mathrm{W}/\mathrm{m}^{2}$ .

3. Between the two: about three times less than Venus, one and a half times more than Mars.

4. $365.25/11.18 \approx 32.7$ .

5. $(874/(4 \times 5.67\times10^{-8}))^{1/4} \approx 249\,\mathrm{K}$ .

6. $(874 \times 0.23/(4\sigma))^{1/4} \approx 173\,\mathrm{K}$ .

7. Neither: $249\,\mathrm{K}$ is still below $273\,\mathrm{K}$ . A bright cover of clouds or ice lowers the equilibrium temperature by more than $75\,\mathrm{K}$ : what covers a planet matters as much as its distance.

8. $\sqrt{0.00151} \approx 0.0389$ .

9. $0.99 \times 0.0389 \approx 0.0385\,\mathrm{au}$ .

10. $1.70 \times 0.0389 \approx 0.066\,\mathrm{au}$ .

11. Yes: $0.0385 < 0.0485 < 0.066$ .

12. $d_\odot = 0.0485/0.0389 \approx 1.25\,\mathrm{au}$ : in the inner half of the Sun’s habitable zone, between the Earth and Mars.

13. $1.055^{1/3} \approx 1.018$ .

14. $11.2 \times \sqrt{1.055/1.018} \approx 11.4\,\mathrm{km}/\mathrm{s}$ , about the Earth’s: its gravity could hold an atmosphere (whether it has one depends on its history).

15. The wobble of the star reveals only the part of its motion along our line of sight; since the tilt of the orbit is unknown, the mass found is the true mass only if the orbit is seen edge-on, and a minimum otherwise.

16. The pressure of the triple point, $611.657\,\mathrm{Pa}$ , about $0.006\,\mathrm{bar}$ .

17. $(874 \times 0.706/(4\sigma))^{1/4} \approx 228\,\mathrm{K}$ .

18. $288 - 228 = 60\,\mathrm{K}$ : Proxima b would need a greenhouse warming of about $60\,\mathrm{K}$ , almost twice the Earth’s $33\,\mathrm{K}$ .

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