Elongation, retrograde motion and visibility

Builds on: Sections 1.4, 1.6 and 6.1
Reading time: 35 min
Exercises: 4
Olympiad problems: 2

On 7 December 2024 Mars stopped moving among the stars of Cancer. For the next 79 days it crept westwards, backwards compared with its usual eastward drift, past its opposition on 16 January 2025 and into Gemini. On 24 February it stopped again, 19° west of where it had turned, and then resumed its journey east. Astronomers of the ancient world built whole systems of circles upon circles to explain these loops. In this section we see that they come from looking at one moving planet from another, and we work out when they happen and how long they last. We then turn to a more practical question for anyone hoping to see Mercury or Venus, which is how high a planet near the Sun stands when the sky gets dark enough to see it.

6.2.1Elongation through the synodic period

Before you readMake a guess first

Mars is at opposition, 180° from the Sun. Thirty days later, roughly how far from the Sun is it in our sky?

In Section 6.1 we named the configurations and found how often they repeat. To know where a planet is in our sky on any day between them, we need its as the two planets move. With circular orbits this is a triangle problem. The Sun, Earth and the planet form a triangle, and the angle at the Sun between Earth and the planet, which we call $θ$ , grows steadily, by 360° in each synodic period. The side from Earth to the planet then follows from the law of cosines, and the elongation is the angle at Earth.

ResultElongation from the angle at the Sun(6.3)

\begin{aligned} Δ^{2} & = a^{2} + a_{\oplus}^{2} - 2 a a_{\oplus} \cos θ \\ \cos ε & = \frac{a_{\oplus} - a \cos θ}{Δ} \end{aligned}

$ε$	the elongation, the angle at Earth between the Sun and the planet
$Δ$	the distance from Earth to the planet
$θ$	the angle at the Sun between Earth and the planet, $360 ° \times t / S$ at the time $t$ after an opposition or an inferior conjunction
$a, a_{\oplus}$	the radii of the orbits of the planet and of Earth

This holds for circular orbits in one plane. The sine rule gives the same angle as

\sin ε = (a / Δ) \sin θ

, but the cosine form tells us at once whether the elongation is more or less than 90°.

Example 6.3Mars two months after opposition

How far from the Sun is Mars in our sky 60 days after an opposition? Take circular orbits of radius 1 au and 1.524 au, and the synodic period of 780 days.

Solution

In 60 days the angle at the Sun grows to $θ = 360 ° \times 60 / 780 = 27.7 °$ . By (6.3),

Δ^{2} = {1.524}^{2} + 1 - 2 \times 1.524 \cos 27.7 °,

so $Δ = 0.79$ au. Then $\cos ε = (1 - 1.524 \cos 27.7 °) / 0.79$ , which is negative, and $ε = 116 °$ . The sine rule alone would give $\sin ε = 1.524 \sin 27.7 ° / 0.79$ , whose smaller solution is 63.7°, and we would have to notice that the true angle is $180 ° - 63.7 ° = 116 °$ .

AnswerTwo months after opposition Mars is still 116° east of the Sun, high in the evening sky. It has moved 64° from the point opposite the Sun while the triangle turned by less than 28°.

Quick checkVenus after an inferior conjunction

Venus's synodic period is 583.9 days and its orbit has a radius of 0.723 au. What is its elongation 100 days after an inferior conjunction?

6.2.2Retrograde motion

Before you readMake a guess first

Earth overtakes Mars on the inside track around the time of opposition. Which way does Mars seem to move against the distant stars while this happens?

Most of the time a planet drifts eastwards along the ecliptic from night to night, in the same direction as the Sun and the Moon. This is called direct motion. For a few weeks or months around each opposition, a superior planet moves westwards instead. This is retrograde motion, and the two moments when the planet turns round and seems to stand still are its stationary points.

The planet always moves forwards round the Sun, and only our line of sight to it swings backwards. Near opposition, Earth on its inner orbit moves faster than the planet and in nearly the same direction, so we overtake it. Our line of sight then swings backwards across the stars, just as the car we overtake seems to slip back against the hills. Away from opposition our motion is mostly along the line of sight or against the planet's motion, and the planet's own eastward motion wins. An inferior planet shows the same effect the other way round. Near inferior conjunction Venus or Mercury overtakes Earth on the inside, and we see it slide westwards from the evening sky into the morning sky.

Figure 6.2Overtaking and retrograde motion

Planet

Time

Watch

Motion against the stars
Elongation
Backward motion in this model
Share of the synodic period spent going backwards

The figure shows the Sun S, Earth E and a planet P on circular orbits seen from the north, both moving anticlockwise. The heavy line is our line of sight to the planet and the faint lines show where it pointed at earlier times. For Jupiter and Saturn, Earth's orbit is drawn larger than to scale so that its motion can be seen, and the arrow marks where the line of sight leaves the panel on its way to the planet. The strip on the right shows the planet's position along the ecliptic from side to side, with east on the left as in the sky when we face south from the northern hemisphere, and time increasing downwards. It has the same scale in degrees for every planet. The dots are equal steps of time, and the heavy stretch between the stationary points 1 and 2 is the retrograde motion. The planet can be set to , , , or .

In the strip the curve runs straight down at the stationary points, where the planet's position along the ecliptic briefly stops changing. Switching from Mars to Saturn changes the share of the synodic period that the planet spends going backwards, from 9% to 36%, while its loop shrinks from 16° to 7°. The formula below explains why.

The line of sight stops turning when the motion of the planet relative to Earth points straight along it. Working this out for circular orbits, where each planet's speed is proportional to $1 / \sqrt{a}$ , gives the angle at the Sun at which this happens.

Show the derivation

We turn the picture so that Earth is at $(a_{\oplus}, 0)$ moving with speed $v_{\oplus}$ along $y$ , and the planet is at $(a \cos θ, a \sin θ)$ moving with speed $v$ at right angles to its radius. The planet's position relative to Earth is $(a \cos θ - a_{\oplus}, a \sin θ)$ and its velocity relative to Earth is $(- v \sin θ, v \cos θ - v_{\oplus})$ . The line of sight stops turning when these two vectors are parallel, which means that their cross product is zero,

(a \cos θ - a_{\oplus}) (v \cos θ - v_{\oplus}) + a v \sin^{2} θ = 0.

Multiplying out and using $\sin^{2} θ + \cos^{2} θ = 1$ ,

\cos θ = \frac{a v + a_{\oplus} v_{\oplus}}{a v_{\oplus} + a_{\oplus} v} .

For circular orbits $v = \sqrt{G M_{⊙} / a}$ , so $a v$ is proportional to $\sqrt{a}$ . Dividing above and below by $\sqrt{G M_{⊙}}$ and simplifying with $a^{3 / 2} + a_{\oplus}^{3 / 2} = (\sqrt{a} + \sqrt{a_{\oplus}}) (a - \sqrt{a a_{\oplus}} + a_{\oplus})$ gives the result.

ResultThe stationary points for circular orbits(6.4)

\cos θ_{s} = \frac{\sqrt{a a_{\oplus}}}{a + a_{\oplus} - \sqrt{a a_{\oplus}}}

$θ_{s}$	the angle at the Sun between Earth and the planet at a stationary point, measured from the line-up at opposition or inferior conjunction
$a, a_{\oplus}$	the radii of the orbits of the planet and of Earth

The formula is the same if we swap

a

and

a_{\oplus}

, so it serves inferior and superior planets alike. The retrograde motion runs from

θ_{s}

before the line-up to

θ_{s}

after it, and so it lasts a fraction

2 θ_{s} / 360 °

of the synodic period.

For Mars, $θ_{s} = 16.8 °$ . The angle at the Sun grows by 360° in 780 days, so Mars stops 36 days before opposition and again 36 days after it, and moves backwards for 73 days. The arc it covers works out to 16°. The real loop of 2024 and 2025 lasted 79 days and covered 19°. Mars's orbit is quite eccentric, and at that opposition Mars was near the far end of its orbit, moving slowly, so Earth spent longer overtaking it. For Venus the formula gives 42 days of retrograde motion, and for Mercury 23 days, both close to the real values.

The further out the planet, the more slowly it moves and the more of the synodic period it spends going backwards. Jupiter turns at $θ_{s} = 54.4 °$ and goes backwards for 121 days over about 10°, and Saturn for 138 days over about 7°, which is 36% of its synodic period against 9% for Mars.

Quick checkJupiter's retrograde motion

Jupiter's orbit has a radius of 5.203 au and its synodic period is 398.9 days. For how many days does Jupiter move backwards around each opposition?

days

Real retrograde paths are loops or zigzags on the sky, not just a backtrack along one line. Each orbit is tilted a little to Earth's, so the planet is slightly north or south of the ecliptic and its distance from the ecliptic changes as Earth passes it. In 2024 and 2025 Mars was 2.8° north of the ecliptic when it first stopped, 4.3° north at opposition and 3.7° north when it stopped again, so it traced an open loop. When the planet crosses the plane of Earth's orbit near opposition, the loop flattens into a Z.

6.2.3Seeing a planet near the Sun

Before you readMake a guess first

Mercury reached 18° from the Sun on 8 March 2025 and 24° on 29 October 2025, both times in the evening sky. On which date did it stand higher above the horizon at sunset in Ankara?

A planet close to the Sun can only be seen in twilight, after sunset or before sunrise, and only if it is high enough to clear the haze near the horizon. Its elongation is measured along the sky, but what we need is its altitude. The two are linked by the angle at which the ecliptic meets the horizon at the point where the Sun sets.

The ecliptic is tilted by 23.44° to the celestial equator, and the equator meets the horizon at the angle $90 ° - | φ |$ at the latitude $φ$ (Section 1.4). Where the two great circles cross, the ecliptic adds to or takes away from that angle. When the March equinox point is setting, the ecliptic stands as steeply as it can over the western horizon in the northern hemisphere, and when the September equinox point is setting it lies as flat as it can. The Sun is at the March equinox point in March, so spring evenings and autumn mornings are the best times to see Mercury and Venus from the north. In the southern hemisphere the seasons swap.

ResultThe ecliptic at the horizon(6.5)

\begin{aligned} ψ & = 90 ° - | φ | \pm 23.44 ° \\ \sin h & = \sin ε \sin ψ \end{aligned}

$ψ$	the angle between the ecliptic and the horizon where the Sun sets or rises, at the two extremes of the year
$φ$	the observer's latitude
$h$	the altitude of a planet on the ecliptic when the Sun is on the horizon
$ε$	the planet's elongation

The plus sign holds for spring evenings and autumn mornings, and the minus sign for autumn evenings and spring mornings. On other dates the angle lies between the two values. The second line comes from the right-angled spherical triangle formed by the Sun, the planet and the point of the horizon below the planet, and it ignores the planet's small distance from the ecliptic. If

ψ

comes out more than 90°, the ecliptic leans past the upright towards the celestial pole above the horizon, and the acute angle is

180 ° - ψ

From Ankara, at latitude 39.9° N, the angle is 73.5° at its steepest and 26.6° at its flattest. A planet 20° from the Sun then stands 19.1° high at sunset in March but only 8.8° high in September, so in September the planet sinks into the haze near the horizon soon after sunset.

Figure 6.3A planet in the twilight

Sky

DateLatitudeElongation

Angle of the ecliptic to the horizon
Altitude of the planet
Along the horizon

The figure shows the sky above the horizon facing the Sun S, which sits on the horizon. The dotted line is the ecliptic, which leaves the horizon at the angle ψ, and the planet P lies on it at the chosen elongation, with a dashed line down to the horizon to show its altitude. The settings can be changed to , or . The figure keeps the planet on the ecliptic.

The figure gives 17.4° for Mercury on 8 March 2025 and 11.6° on 29 October 2025. The real altitudes at sunset were 17.9° and 8.8°. On 29 October Mercury was also 2.9° south of the ecliptic, which pushed it even lower in the autumn sky.

The first morning on which a planet or star can be seen rising before the Sun, after weeks hidden in the Sun's glare, is called its heliacal rising. The date depends on the planet's brightness and on the angle $ψ$ , so the same planet appears sooner after conjunction when the ecliptic stands steeply. The ancient Egyptians timed their calendar by the heliacal rising of Sirius, which came shortly before the yearly flood of the Nile.

Example 6.4Venus from Sydney and from Ankarafinish it

Venus reached its greatest eastern elongation of 45.9° on 15 August 2026. Taking the extreme values of $ψ$ , compare its altitude at sunset from Sydney (latitude 33.9° S) and from Ankara (latitude 39.9° N).

The start

In the southern hemisphere August evenings are late-winter evenings, close to the steepest case, so for Sydney $ψ = 90 ° - 33.9 ° + 23.44 ° = 79.6 °$ . By (6.5), $\sin h = \sin 45.9 ° \sin 79.6 °$ , which gives $h = 45 °$ .

Your stepNow find the altitude from Ankara, where August evenings are close to the flattest case.

Common errorsSection 6.2

A planet in retrograde motion is not moving backwards round the Sun. Every planet always goes round the Sun in the same direction, and only the direction of our line of sight to it turns backwards.
The stationary points do not come at quadrature. For Mars they come about 36 days either side of opposition, while quadrature comes 106 days after it.
The elongation is the angle at Earth, not the angle at the Sun. The angle at the Sun grows steadily with time, but the elongation changes fastest when the planet is closest to Earth, near opposition or inferior conjunction, and slowest near superior conjunction.
A large elongation does not guarantee a good view. The altitude at sunset depends just as much on the angle of the ecliptic to the horizon, which changes with the season and the latitude.
The sine rule gives two possible angles for the elongation. When the planet is more than 90° from the Sun, the correct value is 180° minus the angle that a calculator returns.

ExercisesSection 6.24 questions

6.2.1
At its flattest, at what angle does the ecliptic meet the western horizon at sunset in London, at latitude 51.5° N?
°
6.2.2
Find the elongation of Mars 45 days after an opposition, with circular orbits of radius 1 au and 1.524 au and a synodic period of 780 days.
°
6.2.3
Where is Venus in our sky while it moves in retrograde?
6.2.4
From Hanoi, at latitude 21.0° N, how does the angle of the ecliptic to the western horizon at sunset compare between its two extremes?

Olympiad problemsPast IOAA problems, restated2 problems

2009Mars's distance in the manner of Copernicus10 points

Adapted from IOAA 2009 (Tehran), theory problem 13. We have restated it in our own words and kept the original data. The worked example of 6.1 runs the same geometry the other way, from the radius to the time. The solution was last checked on 23 October 2026.

Imagine living in the time of Copernicus, before Kepler's laws were known, but accepting that the planets go round the Sun. Mars's orbital period is 687 days, and Mars is seen at quadrature 106 days after an opposition. Find the distance of Mars from the Sun in astronomical units, taking both orbits as circles.

I solved it without the solution

AnswerMars is 1.52 au from the Sun, close to its real mean distance of 1.524 au.

2022The Sun moving backwards in Mercury's sky20 points

Adapted from IOAA 2022 (Kutaisi), theory problem 9. We have restated it in our own words. The original gives no orbital elements, so we add a = 0.387 au and e = 0.206, the values the official solution uses. The official solution sets v/r equal to the spin rate, which counts the small outward part of Mercury's velocity, and gets 8.28 days. The angular speed of the Sun round Mercury is the sideways speed divided by r, which gives 8.2 days. For the real Mercury the answer is 8.1 days. The solution was last checked on 23 October 2026.

Mercury's orbit is unusually eccentric, with a semi-major axis of 0.387 au and an eccentricity of 0.206, and Mercury spins once on its axis, measured against the stars, in two thirds of its orbital period. As a result, the Sun seen from Mercury moves backwards across the sky for a short time near perihelion. Find the total time in each orbit for which the Sun moves backwards, in Earth days.

I solved it without the solution

AnswerThe Sun moves backwards in Mercury's sky for about 8.2 Earth days in each orbit, centred on perihelion.