Phases and the brightness of planets

Builds on: Sections 4.2, 6.1 and 6.2
Reading time: 40 min
Exercises: 4
Olympiad problems: 2

In 1610 Galileo turned his small telescope on Venus and watched it change over the months from a small round disc to a large thin crescent. In the old picture with Earth at the centre, Venus always stays between us and the Sun, so it could only ever show us a crescent or nothing. A full Venus could only mean that Venus goes round the Sun, and Galileo's drawings became one of the strongest arguments for the new picture. The same changes explain something we can check with our own eyes. Venus shines brightest as a crescent, about five weeks before or after it passes between us and the Sun, rather than when its whole face is lit. In this section we work out how much of a planet we see lit, and how bright that makes it.

6.3.1The phase of a planet

Before you readMake a guess first

How much of Mars's face do we ever see in shadow?

Half of every planet is lit by the Sun at all times. How much of that lit half we see depends on the angle at the planet between the directions to the Sun and to Earth, called the phase angle. At a phase angle of 0° we look at the planet from the Sun's direction and see the lit half face on, so the planet is full. At 90° we see half of the lit half, like the Moon at first quarter, and at 180° we look at the dark side.

The boundary between the lit and the dark part, the terminator, is a circle on the planet. Seen from the side it looks like half an ellipse, whose width depends on the phase angle. Adding up the lit area gives the fraction of the disc we see lit, called the phase.

ResultThe phase of a planet(6.6)

\begin{aligned} k & = \frac{1 + \cos i}{2} \\ \cos i & = \frac{r^{2} + Δ^{2} - R^{2}}{2 r Δ} \end{aligned}

$k$	the phase, the fraction of the planet's disc that we see lit
$i$	the phase angle, the angle at the planet between the Sun and Earth
$r$	the distance of the planet from the Sun
$Δ$	the distance of the planet from Earth
$R$	the distance of Earth from the Sun

The second line is the law of cosines in the triangle formed by the Sun, Earth and the planet. For the Moon the Sun is so far away that the lines from it to Earth and to the Moon are almost parallel, so the phase angle is close to

180 °

minus the elongation.

Show the derivation

Seen from Earth, the planet's disc has radius $ρ$ . The lit half is bounded on one side by the edge of the disc, a half circle of radius $ρ$ , and on the other by the terminator, which we see as half an ellipse with the long semi-axis $ρ$ and the short semi-axis $ρ | \cos i |$ . When $i$ is less than 90° the lit part is the half disc plus half of the ellipse, with area $\frac{1}{2} π ρ^{2} + \frac{1}{2} π ρ^{2} \cos i$ . When $i$ is more than 90° it is the half disc minus half of the ellipse, which gives the same formula because $\cos i$ is then negative. Dividing by the area of the disc, $π ρ^{2}$ , gives $k = (1 + \cos i) / 2$ .

An inferior planet goes through the full range of phases. It is full at superior conjunction, half lit at greatest elongation, where the right angle of the triangle is at the planet, and new at inferior conjunction. A superior planet's phase angle is largest at quadrature, where $\sin i = R / r$ . For Mars this is 41° on circular orbits, so Mars shows at least 88% of its disc lit (47° and 84% when Mars is near perihelion), and for Jupiter the largest phase angle is 11° and at least 99% of the disc is lit.

Quick checkThe Moon as a crescent

The Moon is 60° east of the Sun in the evening sky. What fraction of its disc do we see lit?

Figure 6.4The phases and brightness of Venus and Mars

Planet

Time

Watch

Elongation
Phase angle
Lit fraction of the disc
Distance from Earth
Apparent diameter
Magnitude in the simple model

The figure shows the planet as seen in a telescope at top left, with north up and its lit side towards the Sun, drawn to the same scale of angle for both planets. At top right are the Sun S, Earth E and the planet P on circular orbits seen from the north, with the phase angle i marked at P. The graph below gives the planet's magnitude in the simple model of (6.7), brighter upwards, through one synodic period. The dashed curve shows the magnitude the planet would have if it were fully lit, so the gap between the two curves is what the phase costs. The settings can be changed to , , or .

In the figure Venus's disc grows six times wider, from 9.7″ at superior conjunction to 60″ at inferior conjunction, because its distance from Earth falls from 1.72 au to 0.28 au. At the same time its lit fraction shrinks from all of it to nothing. Its brightness comes from the balance of the two.

6.3.2How bright a planet looks

Before you readMake a guess first

Mars was at opposition in July 2018, 0.39 au from Earth, and in January 2025, 0.64 au away. How many times brighter was it in 2018?

A planet shines by reflected sunlight, and two inverse-square laws act on that light. The sunlight reaching the planet falls off as $1 / r^{2}$ with its distance from the Sun, and the light it sends back falls off as $1 / Δ^{2}$ with its distance from Earth. On top of that, we only receive light from the part of the disc that is lit. The simplest model takes the light we receive as proportional to the phase $k$ . With the magnitude scale of Section 4.2, the factors turn into terms added together.

To compare planets we need a fixed standard. The absolute magnitude $H$ of a planet is the magnitude it would have if it were 1 au from the Sun and 1 au from Earth, and fully lit. These three conditions cannot all hold at once for a real observer, but they give each planet one number that sets how bright it is.

ResultThe magnitude of a planet(6.7)

m = H + 5 \log_{10} (\frac{r Δ}{1 {au}^{2}}) - 2.5 \log_{10} k

$m$	the apparent magnitude of the planet
$H$	its absolute magnitude
$r, Δ$	its distances from the Sun and from Earth
$k$	its phase, from (6.6)

The model holds well at full phase. At other phases it is only a rough guide, because a surface does not send equal light from every lit point. Near the terminator the sunlight strikes at a glancing angle and spreads thinly, so a real crescent is fainter than the model says. The Moon at first quarter is only about a tenth as bright as the full Moon, while the model predicts a half.

Example 6.5Mars at two oppositions

At the opposition of 27 July 2018 Mars was 1.40 au from the Sun and 0.386 au from Earth. At the opposition of 16 January 2025 it was 1.63 au from the Sun and 0.644 au from Earth. With $H = - 1.60$ , find its magnitude on each date.

Solution

At opposition the phase angle is close to zero, so $k = 1$ and the last term of (6.7) drops out. In 2018, $m = - 1.60 + 5 \log_{10} (1.40 \times 0.386) = - 2.9$ . In 2025, $m = - 1.60 + 5 \log_{10} (1.63 \times 0.644) = - 1.5$ . The difference is 1.4 magnitudes, a factor of $10^{0.4 \times 1.4} = 3.8$ in brightness.

AnswerThe model gives magnitude −2.9 in 2018 and −1.5 in 2025, close to the −2.8 and −1.4 that observers saw. In 2018 it outshone Jupiter, which was then at −2.2, and in 2025 it was about as bright as Sirius.

For Venus the model shows why the crescent wins. Going towards inferior conjunction, the lit fraction falls but the distance falls faster for a while. The brightness, proportional to $k / Δ^{2}$ because $r$ stays fixed, is greatest about 36 days before and after inferior conjunction, at an elongation of 40°. Venus is then 27% lit and 39″ across. This moment is called greatest brilliancy, and its timing matches what we see. The model makes Venus about 1.6 magnitudes brighter there than at superior conjunction, but the real difference is smaller, and the real Venus at greatest brilliancy is about 0.7 magnitude fainter than the model's −5.5.

A more careful law treats the planet as a perfectly matte sphere, each lit point of which looks equally bright from every direction. Adding up the light from the whole lit part then gives a brightness at phase angle $i$ of

q (i) = \frac{(π - i) \cos i + \sin i}{π}

times its brightness at full phase, with $i$ in radians. At $i = 90 °$ this is $1 / π$ , so a matte sphere at quarter phase is 1.24 magnitudes fainter than when full, against 0.75 magnitude in the simple model. Some olympiad problems use this law. For Venus it does worse than the simple model, because it puts the brightest moment 51 days from inferior conjunction, at an elongation of 45°, while Venus's thick clouds send more light forwards at a crescent phase than a matte surface would.

6.3.3Size and albedo

The fraction of the sunlight falling on a body that it reflects is its (Section 4.8). A large dark planet and a small bright one can have the same absolute magnitude, so $H$ tells us the product of the area and the reflectivity. For this purpose we use the geometric albedo $p$ , the brightness of the body at full phase compared with that of a flat white disc of the same size facing the Sun. Comparing the planet at 1 au from both the Sun and Earth with the Sun seen from 1 au gives the link.

ResultAbsolute magnitude, size and albedo(6.8)

p R_{p}^{2} = (1 au)^{2} \times 10^{- 0.4 (H - m_{⊙})}

$p$	the geometric albedo
$R_{p}$	the radius of the body
$m_{⊙}$	the apparent magnitude of the Sun seen from 1 au, −26.74 in the V band

The sunlight caught by the body is proportional to

R_{p}^{2}

, and

p

sets how strongly it sends that light back towards us at full phase, so

H

fixes only the product

p R_{p}^{2}

. A measurement of either the size or the albedo then gives the other. In Section 6.6 we use this to find the sizes of asteroids.

For the Moon, with $H = 0.25$ and a radius of 1737.4 km, (6.8) gives $p R_{p}^{2} = 3.6 \times 10^{5} {km}^{2}$ and $p = 0.12$ . The Moon's soil is about as dark as worn tarmac, and the full Moon only looks brilliant against the night sky.

Example 6.6Jupiter at opposition and at conjunctionfinish it

Jupiter has $H = - 9.40$ and its orbit has a radius of 5.2 au. Find its magnitude at opposition and at conjunction, taking it as fully lit both times.

The start

At opposition Jupiter is $5.2 - 1 = 4.2$ au from Earth, so by (6.7) with $k = 1$ , $m = - 9.40 + 5 \log_{10} (5.2 \times 4.2) = - 2.7$ .

Your stepNow find the magnitude at conjunction, when Jupiter is on the far side of the Sun.

Common errorsSection 6.3

The phase angle is the angle at the planet, between the Sun and Earth. The elongation is the angle at Earth, and the two add up to nearly 180° only for the Moon, where the Sun is so far away.
The distance from the Sun enters the magnitude as well as the distance from Earth. Leaving out $r$ in (6.7) treats the sunlight reaching the planet as the same everywhere in the Solar System.
The magnitude depends on $5 \log_{10}$ of the distances, because the flux goes as the inverse square of each. Using $2.5 \log_{10}$ halves the effect of distance.
The absolute magnitude of a planet is defined at 1 au from both the Sun and Earth. It is a different standard from the absolute magnitude of a star, which is defined at 10 pc.
The simple model with the lit fraction $k$ is reliable only near full phase. For Mercury at greatest elongation it gives magnitude −2.1, about two magnitudes brighter than Mercury really looks, because a rough, airless surface darkens quickly away from full phase.

ExercisesSection 6.34 questions

6.3.1
Venus has an absolute magnitude of −4.38 and a radius of 6051.8 km. What is its geometric albedo?
6.3.2
The dwarf planet Ceres has an absolute magnitude of 3.34 and a nearly circular orbit of radius 2.77 au. What is its magnitude at opposition?
6.3.3
In which configuration does Venus show its largest disc?
6.3.4
Which of these planets can never appear as a crescent from Earth?

Olympiad problemsPast IOAA problems, restated2 problems

2014The Moon at full and at first quarter

Adapted from IOAA 2014 (Suceava), theory problem P14. We have restated it in our own words and kept the original data. The original does not state the law of brightness with phase, and the official solution uses the matte-sphere law, normalised to 2/3 at full phase, which we state. The official solution rounds the factor 2/(3π) to 0.2 and then slips in the arithmetic, getting −10.5 at first quarter instead of −11.0. The official answer at full Moon is −12.25, from a distance of 0.00256 au. With $H$ defined at full phase, as in this section, the full Moon comes out at −12.70, close to the real −12.74. The real Moon at first quarter is near −10, fainter than a matte sphere, because of the shadows of its rough surface. The book gives no marks for this problem. The solution was last checked on 23 October 2026.

The absolute magnitude of the Moon is $H = 0.25$ . Find the Moon's apparent magnitude at full Moon and at first quarter. Take the distance from the Moon to Earth as 385 000 km and the distances from the Sun to Earth and to the Moon as 1 au. Assume the Moon scatters light as a matte sphere, so that the light it sends towards an observer at phase angle $i$ is proportional to $\frac{2}{3} [(1 - i / π) \cos i + (\sin i) / π]$ , with $i$ in radians, Take $H$ as the magnitude the Moon would have at 1 au from both the Sun and Earth if this expression were equal to 1.

I solved it without the solution

AnswerThe full Moon has magnitude −12.26 and the first-quarter Moon −11.0 in this model.

2012A perfectly white Moon10 points

Adapted from IOAA 2012 (Rio de Janeiro), theory problem S4. We have restated it in our own words. The original gives no data, so we add the full Moon's magnitude of −12.74 and its albedo of 0.14, the values the official solution uses. This albedo is a little higher than the geometric albedo of 0.12 found in this section from $H$. The solution was last checked on 23 October 2026.

What would the visual magnitude of the full Moon be if its albedo were 1? Take the full Moon's magnitude as −12.74 and its albedo as 0.14.

I solved it without the solution

AnswerA perfectly white full Moon would shine at magnitude −14.9, seven times brighter than the real one.