Visual walkthrough — Groundtrack analysis — swath, revisit

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Step 1 — Two spinning things, one frame

WHAT. We have two motions that must be measured in the same reference frame — the frame fixed to the distant stars (the "inertial" frame). One motion is the satellite going around its orbit. The other is the Earth turning on its axis.

WHY the star-fixed frame. If we measured the satellite's orbit against the stars but the Earth's spin against the Sun, we'd be comparing apples to oranges — the two numbers would live in different clocks. Physics forces one choice: the orbit plane is fixed relative to the stars, so we must clock Earth's spin against the stars too. (This is exactly the sidereal-vs-solar-day subtlety of Earth Rotation & Sidereal Time.)

PICTURE. Look at the figure: the faint dotted circle of stars does not move. Against it, the blue satellite sweeps its orbit and the yellow arrow on Earth's equator turns. Both angles are read off the same star background.

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Step 2 — Name the rates: how fast each thing turns

WHAT. Give each rotation a symbol so we can do arithmetic.

• ${\omega }_{\oplus }$ (read "omega-Earth") $=$ how many radians Earth turns per second. One full turn is $2\pi$ radians, and Earth takes one sidereal day $T_{\oplus }=86164\text{s}$ to do it, so ${\omega }_{\oplus }=\underset{\text{full turn }÷\text{ time for it}}{\underbrace{\frac{2\pi }{T_{\oplus }}}}=\frac{2\pi }{86164\text{s}}.$ Here $2\pi$ is "one whole revolution measured in radians" and $86164\text{s}$ is "the seconds that revolution takes against the stars."

• $T$ (read "period") $=$ the seconds the satellite needs for one lap of its orbit. It comes from Orbital Period & Kepler's Third Law — bigger orbit, longer $T$.

WHY radians and not degrees here. Radians make "arc length $=$ radius $\times$ angle" true with no conversion factor. We will turn an angle into a distance on the ground soon, and radians are the currency that lets us do it cleanly.

PICTURE. The figure marks the same clock twice: the blue wedge is how far the satellite moves in a slice of time; the yellow wedge is how far Earth spins in that same slice. We are about to ask: after one whole satellite lap, how much did the yellow wedge grow?

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Step 3 — One lap later, the Earth has turned underneath

WHAT. Freeze the satellite at the moment it crosses the equator going up (the "ascending node"). Let it fly exactly one lap and cross the equator going up again. That took time $T$. During that same time $T$, the Earth turned by $\Delta ={\omega }_{\oplus }T=\underset{\text{rad per second}}{\underbrace{{\omega }_{\oplus }}}\times \underset{\text{seconds per lap}}{\underbrace{T}}=\text{radians Earth turned in one lap.}$ Every symbol here is now earned: ${\omega }_{\oplus }$ from Step 2, $T$ from Step 2.

WHY multiply. "Rate $\times$ time $=$ amount" — the same reason speed $\times$ time gives distance. Rate is radians-per-second, time is seconds, the seconds cancel and leave radians.

PICTURE. Two snapshots side by side. On the left, the green subsatellite point sits at some longitude on the equator. On the right (one lap later) the orbit plane is where it was against the stars, but the Earth's yellow reference mark has rotated eastward by the angle $\Delta$. The satellite crosses the equator at the same inertial place — but a different patch of ground is now underneath.

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Step 4 — Why the new crossing lands to the WEST

WHAT. The next crossing appears west of the previous one by the angle $\Delta$. We attach a sign to record direction: eastward-positive longitude, so a westward shift is negative: $\Delta {\lambda }_{\text{node}}=-{\omega }_{\oplus }T.$ Here $\lambda$ (lambda) is longitude, the "how far east" angle, and the subscript "node" reminds us we are measuring at the equator crossing.

WHY west and not east. Earth spins eastward. Picture standing at the first crossing longitude. While the satellite went around, the ground you were standing on rode eastward. So when the satellite returns to the same star-fixed spot, the ground that is now there had to travel from farther west to get under it — meaning the crossing is over a more-westerly longitude than before.

PICTURE. A single equator line. Red mark = first crossing. Green mark = second crossing, sitting to the left (west). The yellow arrow shows the ground's eastward drift; the red-to-green gap is that same amount, appearing as a westward jump of the track.

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Step 5 — Put a number on it (LEO worked example)

WHAT. Take a low orbit with $T\approx 5760\text{s}$ (about $96$ minutes, altitude near $700$ km). $S=360°\times \frac{5760}{86164}=24.07°\text{west per orbit.}$

WHY this is the useful form. $24.07°$ is a distance you can feel: about $24°$ of longitude, roughly $2670$ km at the equator. Fifteen such jumps ($15\times 24°=360°$) almost tile the whole globe in a day — which is why a LEO satellite gives you $\sim 15$ passes daily.

PICTURE. The equator drawn as a strip $0°$ to $360°$. Successive crossings marked at $0°,-24°,-48°,\dots$ marching leftward, each labelled with its orbit number. You can see them wrap around and interleave.

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Step 6 — Edge case: what if the jump divides the circle evenly?

WHAT. Sometimes after a whole number of orbits $N$, and a whole number of days $D$, the satellite lands exactly back on a previous track. This happens when $\underset{\text{time for }N\text{ laps}}{\underbrace{NT}}=\underset{\text{time for }D\text{ Earth turns}}{\underbrace{D{T}_{\oplus }}}⟹\frac{N}{D}=\frac{T_{\oplus }}{T}=\text{orbits per day},$ with $N$ and $D$ sharing no common factor (coprime).

WHY this closes the pattern. After $D$ sidereal days the Earth is back in exactly the same orientation; after $N$ orbits the satellite is back at exactly the same orbital position. Both true at once means the green dot retraces its old trail — a repeat groundtrack. The tracks then sit at even spacing $\delta =360°/N$ around the equator. Sun-synchronous survey satellites are deliberately tuned this way — see Sun-Synchronous Orbits and the drift bookkeeping of Nodal Regression & J2 Perturbation.

PICTURE. Same equator strip. When the jumps are a rational fraction of $360°$, the marks fall onto a finite comb of evenly spaced teeth (here $N$ teeth), instead of drifting forever. The figure shows the teeth filling in until the pattern locks.

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Step 7 — Degenerate cases: check the boundaries

WHAT & WHY. A formula you trust must behave sensibly at its extremes.

• $T\to 0$ (impossibly fast orbit): $S={\omega }_{\oplus }T\to 0$. The Earth barely turns during a lap, so the track hardly drifts — correct.
• $T=T_{\oplus }=86164\text{s}$ (geosynchronous): $S=360°$, a full turn per orbit. The track lands back on itself every single orbit — the groundtrack becomes a fixed figure-8 (or a point for a circular equatorial orbit). Correct.
• $T>T_{\oplus }$: $S>360°$; the jump wraps past a full circle, so the effective shift is $S\mathrm{mod}360°$ — the track can even appear to drift east.
• Inclination $i$ from Inclination & Orbital Elements only fixes latitude reach, never $S$: the maximum latitude is ${\varphi }_{\max }=i$ (or $180°-i$ for retrograde), a purely geometric fact, independent of $T$ and therefore of $S$.

PICTURE. Three little equator strips stacked: $T$ tiny (dense clustered dots), $T=T_{\oplus }$ (all dots stacked on one longitude), $T$ large (dots overshooting and wrapping). One glance shows the whole family.

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The one-picture summary

WHAT this compresses. Star-fixed frame $\Rightarrow$ measure both spins there $\Rightarrow$ Earth turns ${\omega }_{\oplus }T$ per lap $\Rightarrow$ track jumps that much west $\Rightarrow$ if the jump is a rational slice of the circle, the track repeats after $N/D$ orbits-per-day. The whole chain in one drawing.

Recall Feynman retelling — the whole walkthrough in plain words

Imagine a runner circling a spinning merry-go-round. To compare the runner's laps with the merry-go-round's spin fairly, you must watch both from the outside (against the walls of the room — our "stars"). Each time the runner completes a full loop, the merry-go-round has turned a little, so the runner passes over a different seat — and always one that's shifted the same way (westward, opposite to the spin) by the same angle, because "how much it turned" is just its spin-rate times how long a loop takes. If that shift happens to be a neat fraction of a full circle, then after a whole number of loops the runner comes right back over an old seat: the trail repeats. That repeat spacing, together with how wide the runner can glance sideways (the swath), decides how long anyone waits to be looked at again.

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Quick self-check

Recall

Why must Earth's spin be clocked against the stars, not the Sun? ::: Because the orbit plane is fixed relative to the stars; both motions must be compared in the same inertial frame, so we use the sidereal day $86164\text{s}$. In one orbit of period $T$, how far west does the track jump? ::: $S={\omega }_{\oplus }T=360°\times T/86164\text{s}$. What is $S$ for a $96$-minute LEO? ::: About $24.07°$. When does the groundtrack exactly repeat? ::: When $NT=D{T}_{\oplus }$ with $N,D$ coprime integers, i.e. orbits per day $=N/D$. What happens to $S$ at geosynchronous period? ::: $S=360°$ — the track lands on itself each orbit.