Read a Ground Track
A ground track is the orbit drawn on a turning Earth. Compute how far west the next pass lands using nothing but the orbital period, and find out why the inclination sets how far north the track reaches.
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Overview
An orbit is a path in space. A ground track is the shadow of that path on the Earth below, the line traced by the point directly beneath the satellite. The satellite goes round; the Earth also turns underneath it. Everything interesting about a ground track comes from those two motions happening at once.
The Earth completes one turn relative to the stars in a sidereal day of 23 hours 56 minutes 4 seconds, that is 1436.07 minutes, so it turns 360° / 1436.07 min ≈ 0.2507° every minute. While the satellite takes one orbital period to come back round, the ground underneath has rotated eastward by that rate times the period. The result is that the next track lands to the west of the last one.
For a period near 92.9 minutes, the shift works out at about 23° of longitude per revolution. At the equator that is well over 2000 km. It is why a satellite that passes over you this evening will miss you on its very next orbit.
The north-south extent comes from somewhere else entirely. A ground track reaches its highest latitude at the orbital inclination, so a 51.6° orbit never crosses over anywhere further north than 51.6°N or further south than 51.6°S. A near-polar orbit at 98° reaches almost to both poles. Nothing about the period changes that.
One refinement worth knowing before you go further: Earth’s equatorial bulge also drags the orbit plane slowly round, which shifts the crossing point by a fraction of a degree per revolution on top of the rotation term. It is a small correction here, and it is the entire subject of the Sun-synchronous lesson.
At a glance
Learning objectives
- Define a ground track as the path of the sub-satellite point over a rotating Earth.
- Compute the westward shift per revolution from the orbital period and the sidereal day.
- Predict the highest latitude a ground track reaches from the orbital inclination alone.
- Explain why a satellite does not fly over the same place on consecutive passes.
Prerequisites
- Comfort with a two-step calculation and a calculator.
- Familiarity with latitude and longitude.
Required software
- A web browser, a calculator or spreadsheet.
Dataset version
OrbitalWiki live catalog. Record the dataset-release label from /datasets when available, or the exact access date for a live lookup.
Student materials
Student instructions
- 1Write down the length of a sidereal day in minutes (1436.07) and compute Earth’s rotation rate in degrees per minute. Show the division.
- 2Open a catalog record with orbital elements and record its NORAD ID, orbital period in minutes, inclination, element epoch, and your access date.
- 3Compute the westward shift per revolution: shift in degrees = 360 × (orbital period ÷ 1436.07). Show your working and round sensibly.
- 4State the highest north and south latitudes this ground track can reach, and say which single field you used to get them.
- 5Repeat steps 2 to 4 for a second record with a clearly different period, for example a navigation satellite with a period of several hours. Compare the two shifts.
- 6Answer in writing: if a satellite passes directly over your town, roughly how far west will the next pass be, and will it still be visible to you? Justify from your own numbers.
- 7Find a record whose orbital period is close to 1436 minutes. Explain what its ground track looks like and why.
Expected output
- Earth’s rotation rate computed as approximately 0.2507° per minute, with the division shown.
- A two-row table of NORAD ID, period, inclination, epoch, access date, computed shift, and maximum latitude.
- A correct statement that maximum latitude equals inclination, with no reference to period.
- A short written answer explaining that a longer period gives a larger westward shift, and that a period near one sidereal day gives a track that closes on itself.
Teacher materials, not student-facing
Teaching notes
- The two-source confusion is the thing to teach against: students routinely try to derive the latitude reach from the period, or the westward shift from the inclination. Insist that each answer names the single field it came from.
- Use the sidereal day (1436.07 min), not the 1440-minute solar day. The difference is small but it is the physically correct quantity, since the orbit plane is fixed relative to the stars, not to the Sun. A student who uses 1440 gets about 23.2° instead of 23.3°; accept it if the reasoning is stated, but explain the distinction.
- A geostationary record makes an excellent step 7: with a period of one sidereal day and near-zero inclination the ground track collapses towards a single point, which is exactly why those satellites appear to hang still.
- For students who finish early: an orbit that is not quite geosynchronous, or one with a few degrees of inclination, traces a small figure-of-eight. Ask them to predict its shape before showing them one.
- This lesson deliberately does not use a live map animation. The point is that the number falls out of two fields on a record.
Answer key
- Earth’s rotation rate in degrees per minute?
- 360 ÷ 1436.07 ≈ 0.2507°/min, using the sidereal day.
- Westward shift for a 92.9-minute orbit?
- 360 × (92.9 ÷ 1436.07) ≈ 23.3° of longitude per revolution.
- Highest latitude for a 51.6° inclination orbit?
- 51.6°N and 51.6°S. The maximum latitude equals the inclination.
- Which field sets the maximum latitude?
- Inclination, and only inclination. The period plays no part in it.
- What happens when the period approaches one sidereal day?
- The westward shift approaches 360°, that is, zero net shift. The track closes on itself, and at near-zero inclination it collapses towards a fixed point over the equator.
- Will the next pass still be visible from the same town?
- Usually not. A shift of roughly 23° of longitude is over 2000 km at the equator, far outside the visibility circle of a low-orbit pass.
- Why use the sidereal day rather than 24 hours?
- The orbit plane stays fixed with respect to the stars, so the relevant rotation is Earth’s rotation relative to the stars, which takes 23 h 56 min 4 s.
How to cite
Cite each record by NORAD ID with its element epoch and your access date. The sidereal day and the shift formula are standard results; cite the NASA orbit catalog for the orbit-type context.
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- NASA Earth Observatory: Catalog of Earth Satellite OrbitsRetrieved 2026-08-03Confirmed
- CelesTrak: Two-Line Element Set format (field definitions and column positions)Retrieved 2026-08-03Confirmed