Lesson plan · Expert · Adults · 45 minutes

Perturbed satellite orbits: J2 and Sun-synchronous orbits

Earth's equatorial bulge turns satellite orbit planes: the lesson derives the nodal rate from Kozai's theory, computes the ISS's 4.95 degrees a day of regression and the 98.19 degree inclination a Sun-synchronous orbit needs at 700 km, and checks both against real element sets from the viewer's live layer.

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Level

Expert

Ages

Adults

To present

45 minutes

Steps presented

Steps for pupils

Questions

Pupil tasks

The lesson

The class sees the live satellite layer at true scale, derives the nodal regression from Kozai's first-order theory, and computes it for the ISS from a real element set. It then finds the inclination a Sun-synchronous orbit needs at 700 km and checks that Landsat 9, in the same live layer, really advances its node a degree a day and crosses the equator at 10:12 local time.

What they take away

J2 turns an orbit's node at −(3/2) n J2 (R/a)² cos i / (1 − e²)²: −4.95 degrees a day for the ISS, and +0.9856 degrees a day, the Sun's mean rate, for a 700 km orbit inclined 98.19 degrees. Two CelesTrak element sets a week apart show both, and the viewer's SGP4 propagation carries the same rates.

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.

01
An oblate Earth
Presented and for pupils

Caption

Earth's equatorial bulge, measured by J2, makes satellite orbit planes turn. The node regresses at dΩ/dt = −(3/2) n J2 (R/a)² cos i / (1 − e²)².
Talking points
- Kozai (1959) writes Earth's potential as U = (GM/r)[1 + (A₂/r²)(1/3 − sin²δ) + …]. The usual form, (GM/r)[1 − J2 (R/r)² P₂(sin δ)] with P₂(x) = (3x² − 1)/2, matches it term for term when A₂ = (3/2) J2 R².
- Kozai's first-order secular results (p. 372, just before his eq. 14): Ω̄ = Ω₀ − (A₂/p²) n̄ t cos i and ω̄ = ω₀ + (A₂/p²) n̄ (2 − (5/2) sin² i) t, with p = a(1 − e²). With A₂ = (3/2) J2 R², dΩ/dt = −(3/2) n J2 (R/p)² cos i, which is the formula above. Prograde orbits (i < 90°) regress; retrograde orbits (i > 90°) advance.
- Constants (IERS Conventions 2010, Table 1.1): GM⊕ = 3.986004418 × 10¹⁴ m³ s⁻², equatorial radius a_E = 6,378,136.6 m, J2 = 1.0826359 × 10⁻³ (zero-tide values). NASA's fact sheet gives J2 = 1082.63 × 10⁻⁶ and an equatorial radius of 6,378.137 km.
- The viewer's satellite layer (about 16,600 satellites from CelesTrak) and its ISS are propagated with SGP4, which includes these secular J2 terms, so the orbit planes turn in the viewer as they do in space. This scene uses the live sky, close to the epochs of the element sets the viewer loads.
Sources 1 2 3 5Present from this step
02
The ISS's turning plane
Presented and for pupils

Caption

The ISS, inclined 51.63° and making 15.488 revolutions a day, regresses its node by about 4.95° a day. Two element sets a week apart agree with that to about 0.001° a day.
Talking points
- CelesTrak GP element set for the ISS (NORAD 25544), epoch 2026-10-08 20:29:05.961 UTC, as the viewer's satellite relay served it: mean motion 15.48782232 revolutions a day, eccentricity 0.0006778, inclination 51.6315°, right ascension of the ascending node 97.5838°.
- From the mean motion, a = (GM/n²)^(1/3) = 6,798.4 km, 420 km above the equatorial radius.
- Observed: the viewer's fallback snapshot holds the set with epoch 2026-10-01 02:39:00.159 UTC and node 135.8796°. Over the 7.7431 days between them the node moved −38.2958°, −4.9458° a day. Propagating the newer set one day with the viewer's own SGP4 code moves the orbit-averaged node by −4.948°.
- This scene follows the station on its real elements at a minute a second.
Ask the class

From n = 15.48782232 rev/day, e = 0.0006778 and i = 51.6315°, with GM = 3.986004418 × 10¹⁴ m³ s⁻², R = 6,378.1366 km and J2 = 1.0826359 × 10⁻³ (IERS), compute the ISS's nodal rate in degrees per day. Take a from n by Kepler's third law.

Answer-4.95 degrees per day. Answers from -4.98 to -4.92 are marked right.

Whyn = 15.48782232 × 2π / 86,400 = 1.126306 × 10⁻³ rad/s; a = (GM/n²)^(1/3) = 6,798.42 km; (R/a)² = 0.880179; cos i = 0.620717; (1 − e²)² = 0.999999. dΩ/dt = −1.5 × 1.126306 × 10⁻³ × 1.0826359 × 10⁻³ × 0.880179 × 0.620717 = −9.993 × 10⁻⁷ rad/s = −4.947 degrees per day. Leaving out cos i gives −7.97; using R/a unsquared gives −5.27.

Sources 1 2 5 6 7Present from this step
03
Your turn: the plane against the Sun
Pupils only

Caption

Follow the ISS for a few orbits at ten minutes a second, then click the Earth to stand back. While the node regresses, the Sun moves the other way.
Talking points
- The mean Sun moves 360° in a tropical year, 365.242 days (NASA): +0.9856° a day. Against it the ISS's node turns −4.947 − 0.986 = −5.933° a day.
- At ten minutes a second one ISS orbit, 93.0 minutes, takes about 9 seconds.
Pupil question

The ISS's node moves −4.947° a day and the mean Sun +0.9856° a day (360° in 365.242 days). How many days does the ISS's orbit plane take to turn once relative to the mean Sun?

Answer60.7 days. Answers from 60.2 to 61.2 are marked right.

WhyRelative rate = −4.947 − 360/365.242 = −5.933 degrees per day, so one turn takes 360 / 5.933 = 60.7 days.

Pupil task

Click the Earth to focus on it.

HintEarth fills much of the view behind the station; click anywhere on it.

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

Sources 3 5 6Open this step as a pupil
04
Sun-synchronous
Presented and for pupils

Caption

Make the node advance 0.9856° a day, once round in a year, and the orbit keeps its angle to the Sun: it crosses the equator at the same local solar time every day. For a circular orbit 700 km up that needs i = 98.19°.
Talking points
- NASA Earth Observatory: a Sun-synchronous orbit means that "whenever and wherever the satellite crosses the equator, the local solar time on the ground is always the same", and it "keeps the angle of sunlight on the surface of the Earth as consistent as possible, though the angle will change from season to season".
- Required rate: 360° per tropical year of 365.242 days (NASA) = 0.98565° a day = 1.99106 × 10⁻⁷ rad/s. Using the sidereal year (365.25636 days, JPL) changes the answer by 0.0003°.
- dΩ/dt > 0 needs cos i < 0, so Sun-synchronous orbits are retrograde and near-polar.
- The view is square on to the Sun line, the terminator edge-on, with the live satellite layer.
Ask the class

Find the inclination of a circular Sun-synchronous orbit 700 km above the equatorial radius, in degrees: the node must advance 360° in 365.242 days. Use R = 6,378.137 km, J2 = 1.08263 × 10⁻³ and GM = 3.986004418 × 10¹⁴ m³ s⁻².

Answer98.19 degrees. Answers from 98.16 to 98.22 are marked right.

WhyWith R = 6,378.137 km and J2 = 1.08263 × 10⁻³ (NASA) and GM = 3.986004418 × 10¹⁴ m³ s⁻² (IERS): a = 7,078.137 km; n = √(GM/a³) = 1.060206 × 10⁻³ rad/s; (R/a)² = 0.811988. cos i = −1.99106 × 10⁻⁷ / (1.5 × 1.060206 × 10⁻³ × 1.08263 × 10⁻³ × 0.811988) = −0.14242, so i = 98.19°. The IERS radius and J2 give 98.188° too. Using R/a unsquared gives 97.37°.

Sources 1 2 3 4 9Present from this step
05
Landsat 9 in the live layer
Presented and for pupils

Caption

Landsat 9, one of the points in the live layer, flies at 98.22° and 14.571 revolutions a day. Its node advanced 7.38° in 7.49 days, 0.986° a day, and its southbound equator crossing falls at 10:12 local mean time.
Talking points
- NASA: Landsat 9 flies a near-polar, Sun-synchronous orbit at 705 km, inclination 98.2°, period 99 minutes (about 14.5 orbits a day), with an equatorial crossing time of nominally 10:12 AM (± 5 min) local time at the descending node.
- CelesTrak element set via the viewer's relay (NORAD 49260), epoch 2026-10-08 15:20:19.761 UTC: mean motion 14.57108674 revolutions a day, eccentricity 0.0001379, inclination 98.2172°, node 350.375°. The viewer's snapshot set, epoch 2026-10-01 03:41:55.457 UTC, has node 342.9913°.
- Rates: observed 7.3837° in 7.4850 days = 0.9865° a day; Kozai's first-order formula +0.9879° a day (a = 7,080.7 km); one day of the viewer's SGP4 +0.9863° a day; the Sun-synchronous target 0.98565° a day.
- Local time of the node: the mean Sun's right ascension at the epoch, from the IERS Earth rotation angle plus the IERS expression for Greenwich mean sidereal time less (UT − 12 h) × 15°, is 197.272°. The ascending node is (350.375 − 197.272)/15 = 10.207 hours east of it, crossed at 22:12 local mean time; the descending node, on the opposite side of the Earth, at 10:12. The 1 October set gives 10:12 too.
Ask the class

With Landsat 9's node at 350.375° and the mean Sun's right ascension 197.272° at the same instant, at what local mean solar time does it cross the equator going south? Give decimal hours.

Answer10.21 hours. Answers from 10.16 to 10.26 are marked right.

Why(350.375 − 197.272) / 15 = 10.207 hours: the ascending node is crossed at 12 + 10.207 = 22.207 h local mean time, and the descending node, 180° away, at 22.207 − 12 = 10.207 h, 10:12. NASA gives 10:12 AM ± 5 minutes.

Sources 1 2 5 6 7 8Present from this step
06
Measure the rate
Pupils only

Caption

From the two Landsat 9 element sets, measure its nodal rate and compare it with the Sun-synchronous target of 0.98565° a day.
Talking points
- Epochs 2026-10-01 03:41:55.457 and 2026-10-08 15:20:19.761 UTC: 7.4850 days apart. Nodes 342.9913° and 350.375°.
- The measured 0.9865° a day is 0.08 per cent above the target, and Kozai's first-order rate 0.23 per cent above it. That formula keeps J2 to first order only; Kozai's paper goes on to second-order terms and the third and fourth harmonics.
Pupil question

Landsat 9's node was 342.9913° at 2026-10-01 03:41:55.457 UTC and 350.375° at 2026-10-08 15:20:19.761 UTC. What nodal rate does that give, in degrees per day?

Answer0.9865 degrees per day. Answers from 0.9845 to 0.9885 are marked right.

Why(350.375 − 342.9913) / 7.4850 = 7.3837 / 7.4850 = 0.9865 degrees per day, within 0.1 per cent of the 0.98565 a Sun-synchronous orbit needs.

Sources 3 5 6 7Open this step as a pupil
07
The critical inclination
Presented and for pupils

Caption

J2 also turns the line of apsides, at (3/4) n J2 (R/p)² (4 − 5 sin² i). That rate vanishes at one inclination below 90°, where, to first order, an eccentric orbit's perigee stays put.
Talking points
- From Kozai's first-order secular result for ω (p. 372), dω/dt = (A₂/p²) n (2 − (5/2) sin² i) = (3/4) n J2 (R/p)² (4 − 5 sin² i).
- For the ISS (sin² i = 0.61471) the perigee advances 3.69° a day.
- For a Sun-synchronous orbit near 98° the bracket is negative and the perigee regresses.
Ask the class

At what inclination below 90° is dω/dt zero?

Answer63.43 degrees. Answers from 63.41 to 63.45 are marked right.

Why4 − 5 sin² i = 0 gives sin² i = 0.8, sin i = 0.894427, i = 63.435°; the retrograde solution is 116.565°.

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Sources

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Perturbed satellite orbits: J2 and Sun-synchronous orbits

Present to the class

Pupil link

Print

The lesson

What they take away

Step by step

An oblate Earth

The ISS's turning plane

Your turn: the plane against the Sun

Sun-synchronous

Landsat 9 in the live layer

Measure the rate

The critical inclination

Sources

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Perturbed satellite orbits: J2 and Sun-synchronous orbits

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