Temperatures of planets

Builds on: Sections 4.1 and 4.6
Reading time: 30 min
Exercises: 3
Olympiad problems: 2

Venus is almost twice as far from the Sun as Mercury, yet its surface, at about 737 K, is hotter than any part of Mercury, whose day side reaches about 700 K while its night side cools to about 100 K. Without its air, Earth would be well below freezing everywhere. Every planet is warmed by the sunlight it absorbs and cooled by the heat it radiates into space, and the temperature at which the two balance can be found with the rules of Section 4.1 and Section 4.6. In this section we find that temperature, and then see why real planets such as Venus and Earth depart from it.

4.8.1The balance of energy

Before you readMake a guess first

Two planets orbit the Sun at the same distance and reflect the same fraction of sunlight, but one has ten times the radius of the other. Which one is warmer?

A planet of radius $R$ at distance $d$ from a star of luminosity $L$ receives the flux $L / (4 π d^{2})$ of (4.1). Seen from the star, the planet is a disc of area $π R^{2}$ , so that is the area that catches the light. A fraction $A$ of the light is reflected straight back into space. This fraction is the planet's albedo, 0.30 for Earth and 0.76 for Venus with its bright clouds. The rest is absorbed and heats the planet.

A warm planet radiates like a blackbody, mostly in the infrared. If it rotates quickly, so that its whole surface reaches nearly the same temperature $T$ , it radiates $σ T^{4}$ from every square metre of its surface, an area $4 π R^{2}$ , by (4.12). The temperature settles where the heat radiated equals the light absorbed,

4 π R^{2} σ T^{4} = (1 - A) \frac{L}{4 π d^{2}} π R^{2} .

The radius $R$ cancels, as the guess suggested. Writing the star's luminosity as $4 π R_{⋆}^{2} σ T_{⋆}^{4}$ gives a form that needs no constants at all.

ResultEquilibrium temperature of a planet(4.16)

\begin{aligned} T & = {(\frac{(1 - A) L}{16 π σ d^{2}})}^{1 / 4} \\ = T_{⋆} (1 - A)^{1 / 4} \sqrt{\frac{R_{⋆}}{2 d}} \end{aligned}

$T$	the equilibrium temperature of the planet
$A$	its albedo, the fraction of the light falling on it that it reflects
$L, R_{⋆}, T_{⋆}$	the luminosity, radius and effective temperature of the star
$d$	the distance of the planet from the star

This assumes a planet that rotates fast enough to have one temperature all over, radiates as a blackbody, and has no heat of its own. The temperature falls as

1 / \sqrt{d}

, so a planet four times as far from the star is half as warm.

Example 4.15The temperature of Jupiter

Jupiter is 5.20 au from the Sun and has albedo 0.34. Using $T_{⊙} = 5772$ K, $R_{⊙} = 6.957 \times 10^{8}$ m and 1 au $= 1.496 \times 10^{11}$ m, find its equilibrium temperature.

Solution

By (4.16),

\begin{aligned} T & = 5772 \times {0.66}^{1 / 4} \times \sqrt{\frac{6.957 \times 10^{8}}{2 \times 5.20 \times 1.496 \times 10^{11}}} \\ = 110 K . \end{aligned}

AnswerJupiter's equilibrium temperature is about 110 K. Measured in the infrared, Jupiter is warmer, at about 124 K, because it still gives out heat left over from its formation, about 60 per cent as much as it absorbs from the Sun.

Quick checkFour times as far

A blackbody at 1 au from the Sun, with albedo 0, has an equilibrium temperature of 278 K. What is its temperature at 4 au?

4.8.2Slow rotation and air

A planet that turns slowly, or that always keeps the same face to its star, cannot share its heat between day and night. In the simplest model its night side is cold and its day side radiates all the absorbed light from the half of its surface facing the star, an area $2 π R^{2}$ instead of $4 π R^{2}$ . The day side is then warmer than the equilibrium temperature by a factor of $2^{1 / 4} = 1.19$ . The point directly under the star is warmer still, since there the light falls straight down. Mercury, which turns once in 59 days, comes close to this case.

An atmosphere changes the balance in another way. Air lets most visible sunlight through to the ground, but gases such as carbon dioxide and water vapour absorb much of the infrared that the warm ground gives out, and send part of it back down. The ground is therefore warmer than the equilibrium temperature, an effect called the greenhouse effect. Earth's equilibrium temperature is 255 K, about −18 °C, while its mean surface temperature is about 288 K. Venus, with a thick atmosphere of carbon dioxide, has an equilibrium temperature of only about 229 K, which its greenhouse effect raises by more than 500 K.

Example 4.16A planet that always faces its starfinish it

A planet at 1.523 au from the Sun has albedo 0.25 and an equilibrium temperature of 210 K if it rotates fast. What would the mean temperature of its day side be if it always kept the same face to the Sun?

The start

The day side radiates all the absorbed light from half the surface, so $2 π R^{2} σ T^{' 4} = 4 π R^{2} σ T^{4}$ .

Your stepNow find

T^{'}

Common errorsSection 4.8

A planet catches sunlight on a disc of area $π R^{2}$ but radiates from its whole surface, $4 π R^{2}$ . Using the same area for both gives a temperature $\sqrt{2}$ times too high.
The absorbed fraction is $1 - A$ , not $A$ . A planet of albedo 0.9 reflects most of the light and absorbs only a tenth.
The equilibrium temperature of a planet is not its surface temperature. An atmosphere can make the surface much warmer, and heat from inside can make the whole planet warmer.

ExercisesSection 4.83 questions

4.8.1
A blackbody at 1 au from the Sun has an equilibrium temperature of 278 K. What is the equilibrium temperature of a blackbody at 9 au?
K
4.8.2
Venus is 0.723 au from the Sun and has albedo 0.76. A blackbody of albedo 0 at 1 au has equilibrium temperature 278.3 K. What is the equilibrium temperature of Venus?
K
4.8.3
A planet's albedo rises from 0.2 to 0.6, with nothing else changed. What happens to its equilibrium temperature?

Olympiad problemsPast IOAA problems, restated2 problems

2007A planet and its surface temperature10 points

Adapted from IOAA 2007 (Chiang Mai), theory problem Q2. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

A fast-rotating planet of radius $R$ and albedo $a$ orbits a star of luminosity $L$ at distance $D$ . Assume that all the energy the planet absorbs is radiated again as a blackbody.

(a) What is the flux of starlight at the planet? (1.5 points)

(b) At what rate does the planet absorb energy? (1.5 points)

(d) What is the mean blackbody temperature of the planet's surface? (2 points)

(e) If the planet always kept one face towards the star, what would be the mean temperature of that face? (2 points)

(f) Find the temperature of part (d) in kelvin for $a = 0.25$ , $D = 1.523$ au and $L = 3.826 \times 10^{26}$ W. (1 point)

I solved it without the solution

AnswerThe flux is $L / (4 π D^{2})$ , the planet absorbs $(1 - a) L R^{2} / (4 D^{2})$ and reflects $a L R^{2} / (4 D^{2})$ , its temperature is ${((1 - a) L / (16 π σ D^{2}))}^{1 / 4}$ , which is 210 K for the numbers given, and a face kept towards the star would be $2^{1 / 4}$ times warmer.

2021The temperature of the Earth10 points

Adapted from IOAA 2021 (Bogotá, held online), theory problem Q2. We have restated it in our own words and kept the original data, adding the Sun's temperature and radius and the astronomical unit. The solution was last checked on 20 October 2026.

For at least the last few million years, Earth has been roughly in balance with the sunlight it receives. Take the Sun's effective temperature as 5772 K, its radius as $6.957 \times 10^{8}$ m and 1 au as $1.496 \times 10^{11}$ m.

(a) Treating Earth as a perfect blackbody, find its equilibrium temperature in degrees Celsius. (4 points)

(b) Earth's albedo is about 30%. Find its temperature, in degrees Celsius, taking the albedo into account. (2 points)

(c) Earth radiates the energy it absorbs from its surface, but the atmosphere absorbs 58% of that radiation and sends it back to the ground, which warms the ground further. Treat this return as happening only once, and do not treat the atmosphere as a separate blackbody. Find the temperature of the ground, in degrees Celsius. (4 points)

I solved it without the solution

AnswerEarth would be at about 5 °C as a blackbody and −19 °C with its albedo, and the returned heat raises the ground to about 12 °C.