Our Solar System

Lesson plan · Expert · Adults · 40 minutes

The Laplace resonance

Io, Europa and Ganymede keep n₁ − 3n₂ + 2n₃ = 0 with their Laplace angle librating about 180 degrees: the lesson checks both against NASA periods and JPL mean elements, shows why a triple conjunction is impossible, and measures how sensitive the relation is to rounding the periods.

Present to the class

Opens the lesson in the viewer on its cover, ready for the projector. Step through it with the arrow keys or a clicker.

Present to the class

Pupil link

Pupils work through the lesson on their own devices, with questions to answer and tasks to do. To see their answers, set it to a class.

Open it as a pupil

Print

The lesson plan with every answer for you, or the questions alone on one sheet for pupils.

Level
Ages
Adults
To present
40 minutes
Steps presented
5
Steps for pupils
7
Questions
6
Pupil tasks
1

The lesson

The class watches Io, Europa and Ganymede run their 1:2:4 rhythm, verifies the Laplace relation from NASA's periods and the 180-degree Laplace angle from JPL's mean elements, and works out why the three can never line up together. It ends with what the viewer's own moons, on NASA's periods from random starting points, can and cannot show, and how far rounded periods would drift from the real lock.

What they take away

The mean motions satisfy n₁ − 3n₂ + 2n₃ = 0 to the precision of the published periods, the Laplace angle λ₁ − 3λ₂ + 2λ₃ sits at 180 degrees, so a triple conjunction cannot happen, and the whole pattern turns with the 487-day period set by ν = n₁ − 2n₂. The viewer runs the moons on NASA's periods from random starting angles, so its Laplace angle stays close to whatever value the page load gave it.

Step by step

The caption is what the class reads on screen; the talking points are in your notes drawer (N) when you present. Steps marked Pupils only appear only in the pupil lesson.

  1. 01

    One, two, four

    Presented and for pupils

    Caption

    Io, Europa and Ganymede orbit Jupiter in 1.769, 3.551 and 7.155 days, very nearly 1:2:4. The clock runs at six hours a second; the drawn distances are not to scale.

    Talking points

    • NASA Jovian Satellite Fact Sheet, orbital periods: Io 1.769138, Europa 3.551181, Ganymede 7.154553, Callisto 16.689017 days; semi-major axes 421.8, 671.1, 1070.4 and 1882.7 thousand km.
    • The ratios are close to, and not exactly, 2: P₂/P₁ = 2.00729 and P₃/P₂ = 2.01470, so P₃/P₁ = 4.04409. The departures matter: they are what the Laplace relation ties together.
    • In the viewer each moon turns at 360°/P with P from its data, the fact sheet's 1.769138, 3.551181 and 7.154553 days, starting from a random angle picked when the page loads. The periods are real; where each moon is on a given date is not.
    • This is the artistic scale at spread 1: the moons are drawn at 18, 24 and 32 units, ratios 1 : 1.33 : 1.78, against the real 1 : 1.59 : 2.54.

    Sources 1Present from this step

  2. 02

    The Laplace relation

    Presented and for pupils

    Caption

    Numbering Io, Europa and Ganymede 1, 2 and 3, their mean motions obey n₁ − 3n₂ + 2n₃ = 0, and the Laplace angle λ₁ − 3λ₂ + 2λ₃ librates about 180° with a small amplitude.

    Talking points

    • Peale and Lee (2002): the 2:1 resonance variables θ₁ = λ₁ − 2λ₂ + ϖ₁ and θ₃ = λ₂ − 2λ₃ + ϖ₂ librate about 0° and θ₂ = λ₁ − 2λ₂ + ϖ₂ about 180°, all with small amplitude. Since θ₂ − θ₃ = λ₁ − 3λ₂ + 2λ₃, that angle stays near 180° and n₁ − 3n₂ + 2n₃ = 0. "This last libration condition is called the Laplace relation because Laplace first understood its stability."
    • Peale and Lee on the geometry: conjunctions of Io and Europa occur with Io near its periapse and Europa near its apoapse; conjunctions of Europa and Ganymede occur with Europa near its periapse, and Ganymede can be anywhere in its orbit then because θ₄ = λ₂ − 2λ₃ + ϖ₃ circulates.
    • Origin: Peale and Lee contrast assembly by differential tidal expansion of the orbits (tides raised on Jupiter, mainly by Io, with libration damped by dissipation in the satellites) with their primordial assembly by migration in a circumjovian disk. The resonances sustain the tidal heating behind Io's volcanism.
    • Units: n = 360°/P in degrees per day, P from the fact sheet.

    Ask the class

    From the fact sheet periods (1.769138, 3.551181 and 7.154553 days), compute n₁ − 3n₂ + 2n₃ in degrees per day.

    Answer0 degrees per day. Answers from -0.0005 to 0.0005 are marked right.

    Whyn₁ = 360/1.769138 = 203.48893, n₂ = 101.37473, n₃ = 50.31761 degrees per day. 203.48893 − 304.12418 + 100.63522 = −0.00003 degrees per day, zero within the rounding of the periods (each to 10⁻⁶ day, worth up to ±0.0001 here). For scale, n₁ − 2n₂ alone is 0.7395 degrees per day.

    Sources 1 3Present from this step

  3. 03

    180 degrees from JPL's elements

    Presented and for pupils

    Caption

    JPL's mean elements at J2000 put the Laplace angle at 180.0°, and the three 2:1 variables at the centres Peale and Lee name.

    Talking points

    • JPL Planetary Satellite Mean Elements (ephemeris JUP365, Laplace-plane frame, epoch 2000-01-01.5 TDB), in degrees, node, argument of periapsis ω, mean anomaly M: Io 0.0, 49.1, 330.9; Europa 184.0, 45.0, 345.4; Ganymede 58.5, 198.3, 324.8. The mean longitude is λ = node + ω + M.
    • Each moon's elements are referred to its own Laplace plane, with the node counted from that plane's node on the ICRF equator. The three poles (R.A., Dec.) are 268.1°, 64.5°; 268.1°, 64.5°; 268.2°, 64.6°, so the origins agree to about a tenth of a degree.
    • JPL: mean elements are "simply the elements of a precessing ellipse which has been fit in a least squares sense to the numerically integrated orbit", "not intended for ephemeris computation". Each λ is the sum of three angles given to 0.1°, so good to 0.15°, and the coefficients of λ₁ − 3λ₂ + 2λ₃ add up to 6 in size: the Laplace angle from these elements is good to 0.9° at worst.
    • The same elements give θ₁ = 0.3°, θ₂ = 180.2° and θ₃ = 0.2° (with ϖ = node + ω), the libration centres of the previous step.
    • Libration of the Laplace angle: Lari (2018) finds a period of 2047.85 days in his semi-analytical model and quotes 2059.62 days from Lainey and colleagues' frequency analysis. Lari and Rossi (2026, Icarus) give the free libration, from the JUP365 and NOE-5 ephemerides, a period of 2060 days and an amplitude of 0.061°.

    Ask the class

    With λ = node + ω + M from JPL's elements (Io 0.0, 49.1, 330.9; Europa 184.0, 45.0, 345.4; Ganymede 58.5, 198.3, 324.8 degrees), compute λ₁ − 3λ₂ + 2λ₃, reduced to 0 to 360 degrees.

    Answer180 degrees. Answers from 179.5 to 180.5 are marked right.

    Whyλ₁ = 380.0 → 20.0°, λ₂ = 574.4 → 214.4°, λ₃ = 581.6 → 221.6°. 20.0 − 643.2 + 443.2 = −180.0, which is 180.0° in the range 0 to 360.

    Sources 2 3 4 5Present from this step

  4. 04

    No triple line-up

    Presented and for pupils

    Caption

    If Io, Europa and Ganymede were ever all in conjunction, their Laplace angle would be 0°. It stays near 180°, so the three never line up together on one side of Jupiter.

    Talking points

    • With λ₁ = λ₂ = λ₃ = λ, λ₁ − 3λ₂ + 2λ₃ = λ(1 − 3 + 2) = 0. Peale and Lee: the angle librates about 180° with small amplitude, so that configuration never occurs.
    • Rearranged, the Laplace relation reads n₁ − n₂ = 2(n₂ − n₃): Europa-Ganymede conjunctions come exactly half as often as Io-Europa ones. From the fact sheet periods the synodic periods are 3.525464 and 7.050927 days, a ratio of 1.999999.
    • In the viewer the three moons start from random angles, so its Laplace angle rests at whatever value the page load gave it (step 6). Three moons within ε of each other need that angle within 3ε of 0°, so when it happens to start near 0° the viewer shows close line-ups that the real system never makes.
    • NASA lists the inclinations as 0.04°, 0.47° and 0.18°, so the three orbits are close to coplanar.

    Ask the class

    Why can Io, Europa and Ganymede never all be in conjunction at the same moment?

    1. A triple conjunction makes λ₁ − 3λ₂ + 2λ₃ equal 0°, and that angle stays within a fraction of a degree of 180°Right answer
    2. Ganymede's orbit is inclined too steeply to Io's for them to line up
    3. Their periods are exactly 1:2:4, so they line up only once every 7.155 days
    4. Tides on Jupiter keep pushing Io ahead of the other two

    WhyEqual mean longitudes give λ(1 − 3 + 2) = 0, and the Laplace angle librates about 180° with a small amplitude. The periods are not exactly 1:2:4, and the inclinations are all under half a degree.

    Sources 1 3Present from this step

  5. 05

    The 487-day clock

    Pupils only

    Caption

    The small excess ν = n₁ − 2n₂ = n₂ − 2n₃ sets the slow clock of the system: the conjunction lines, and Io's and Europa's periapses with them, go round once every 360°/ν.

    Talking points

    • From the fact sheet periods, n₁ − 2n₂ = 0.73948 and n₂ − 2n₃ = 0.73951 degrees per day. Lari (2018) gives 486.89 days for the circulation related to ν = n₁ − 2n₂ = n₂ − 2n₃, and 486.81 days from Lainey and colleagues' frequency analysis.
    • θ₁ = λ₁ − 2λ₂ + ϖ₁ librating about 0° means ϖ̇₁ = 2n₂ − n₁ = −ν: Io's periapse regresses at ν. So does Europa's (from θ₂). Io's periapse-to-periapse period is then 360°/(n₁ + ν) = 1.762732 days, and Europa's 360°/(n₂ + ν) = 3.525464 days, which is exactly the Io-Europa synodic period, since n₂ + ν = n₁ − n₂. That is why the conjunctions always fall at the same point of Europa's orbit.
    • A trap in JPL's mean-element table: the period column, which the table's key calls the sidereal period, lists Io 1.762732 and Europa 3.525463 days, the periapse-to-periapse values above. Ganymede's 7.155588 and Callisto's 16.690440 also equal 360° over the mean anomaly's rate built from JPL's own apsidal and nodal periods. For mean motions use the fact sheet's orbital periods.
    • The clock in this view runs at a day a second.

    Pupil question

    With the fact sheet periods (Io 1.769138, Europa 3.551181 days), compute 360°/ν with ν = n₁ − 2n₂, in days.

    Answer486.8 days. Answers from 486.5 to 487.1 are marked right.

    Whyν = 203.48893 − 2 × 101.37473 = 0.73948 degrees per day, and 360 / 0.73948 = 486.83 days (486.81 from n₂ − 2n₃). Lari (2018) gives 486.89 days, inside the tolerance.

    Sources 1 2 3 4Open this step as a pupil

  6. 06

    What the viewer holds, and what it does not

    Presented and for pupils

    Caption

    The viewer gives each moon NASA's period and a random starting angle, so its Laplace angle stays close to an arbitrary starting value. Here the clock runs at a week a second.

    Talking points

    • In src/main.js each moon's angle is its starting angle plus (360°/P) × JD, with P parsed from the orbitalPeriod strings in src/data/celestialData.js: Io 1.769138, Europa 3.551181, Ganymede 7.154553 days, the fact sheet's values. createMoon picks the starting angle with Math.random(), so it differs every time the page loads.
    • With those periods n₁ − 3n₂ + 2n₃ is −0.00003 degrees per day (step 2), so the viewer's Laplace angle moves about a degree in a century of simulated time, from wherever it happened to start.
    • The relation is sensitive to the third decimal place of a day. Rounded to two decimals (1.77, 3.55 and 7.15 days), the same periods would turn the Laplace angle once round in about 2,640 days, 7.2 years.
    • What the viewer is good for here: the 1:2:4 rhythm (Io makes 4.044 laps per Ganymede lap), the real periods and the moons in Jupiter's equatorial plane. What it does not have: real positions on a date, the 180° lock, or the small eccentricities (NASA: 0.004, 0.009, 0.001), since it draws each orbit as a circle.
    • At a week a second and 60 frames a second, Io moves about 24° between frames, so the rhythm strobes; the earlier steps run slower.

    Ask the class

    Round the periods to two decimals (1.77, 3.55 and 7.15 days) and compute n₁ − 3n₂ + 2n₃ in degrees per day.

    Answer-0.136 degrees per day. Answers from -0.138 to -0.134 are marked right.

    Why360/1.77 = 203.3898, 360/3.55 = 101.4085, 360/7.15 = 50.3497. 203.3898 − 304.2254 + 100.6993 = −0.1362 degrees per day, a full turn of the Laplace angle every 360/0.1362 = 2,643 days. The fact sheet periods give −0.00003, which is why the viewer carries all six decimals.

    Sources 1 3Present from this step

  7. 07

    Your turn: count the laps

    Pupils only

    Caption

    Watch Ganymede make one lap of Jupiter and count Io's laps in the same time, then pause the clock.

    Talking points

    • At six hours a second Ganymede's lap takes 7.154553 / 0.25 = 28.6 seconds and Io's 7.1 seconds.
    • The viewer runs the real ratio, 7.154553 / 1.769138 = 4.044, so Io finishes its fourth lap just before Ganymede finishes its first.

    Pupil question

    How many laps does Io make while Ganymede makes one?

    Answer4 laps. Answers from 3.9 to 4.1 are marked right.

    Why7.154553 / 1.769138 = 4.044 with NASA's periods, which the viewer uses: four laps and a little.

    Pupil task

    Count Io's laps during one lap of Ganymede, then pause the clock.

    HintStart counting as Ganymede passes a label or an edge of the view, and stop when it comes back there.

    The viewer checks the task as the pupil works and says when it is done.

    Sources 1Open this step as a pupil

Presenter keys

→ PageDown Space
Next step
← PageUp
Back
R
Reveal the answer
N
Your notes
P
Play or pause the clock
L
Labels on or off
B .
Black screen
F
Full screen
Esc
Leave the lesson

Sources

Every fact in this lesson comes from these sources, and each step lists the ones it uses.

  1. Jovian Satellite Fact Sheet, NASA NSSDCA
  2. Planetary Satellite Mean Elements (JUP365, Laplace-plane elements at 2000-01-01.5 TDB), JPL Solar System Dynamics
  3. Peale, S. J. and Lee, M. H. (2002), A Primordial Origin of the Laplace Relation Among the Galilean Satellites, Science 298, 593, doi:10.1126/science.1076557 (arXiv:astro-ph/0210589)
  4. Lari, G. (2018), A semi-analytical model of the Galilean satellites' dynamics, Celestial Mechanics and Dynamical Astronomy 130, 50, doi:10.1007/s10569-018-9846-4 (arXiv:1802.07878)
  5. Lari, G. and Rossi, M. (2026), The recent crossing of the 7:3 resonance between Ganymede and Callisto, Icarus 460, 117247, doi:10.1016/j.icarus.2026.117247 (arXiv:2607.03505)

More Expert lessons

Our Solar System · 3dsolarsystem.online/teachers/lessons/laplace-resonance/

The Laplace resonance

NameDate
  1. From the fact sheet periods (1.769138, 3.551181 and 7.154553 days), compute n₁ − 3n₂ + 2n₃ in degrees per day.

    degrees per day

  2. With λ = node + ω + M from JPL's elements (Io 0.0, 49.1, 330.9; Europa 184.0, 45.0, 345.4; Ganymede 58.5, 198.3, 324.8 degrees), compute λ₁ − 3λ₂ + 2λ₃, reduced to 0 to 360 degrees.

    degrees

  3. Why can Io, Europa and Ganymede never all be in conjunction at the same moment?

    • A. A triple conjunction makes λ₁ − 3λ₂ + 2λ₃ equal 0°, and that angle stays within a fraction of a degree of 180°
    • B. Ganymede's orbit is inclined too steeply to Io's for them to line up
    • C. Their periods are exactly 1:2:4, so they line up only once every 7.155 days
    • D. Tides on Jupiter keep pushing Io ahead of the other two
  4. With the fact sheet periods (Io 1.769138, Europa 3.551181 days), compute 360°/ν with ν = n₁ − 2n₂, in days.

    days

  5. Round the periods to two decimals (1.77, 3.55 and 7.15 days) and compute n₁ − 3n₂ + 2n₃ in degrees per day.

    degrees per day

  6. How many laps does Io make while Ganymede makes one?

    laps

In the viewer