Foundations — Groundtrack analysis — swath, revisit

Before you can read the parent note Groundtrack analysis — swath, revisit, you need every symbol it uses to already feel obvious. This page builds them one at a time, from nothing.

0. The picture the whole topic lives in

Everything below is a label on one drawing: a spinning sphere (Earth), a tilted ring (the orbit), and a dot on the surface (the subsatellite point). Keep this picture in your head.

• The grey sphere is Earth.
• The yellow ring is the orbit — a circle whose plane is tilted.
• The pink dot on the surface is where the satellite's shadow lands (the subsatellite point).
• The blue arrow on the equator is Earth's spin direction (eastward).

1. Angles measured on a sphere: latitude $\varphi$ and longitude $\lambda$

Before any orbit, we need a way to name a spot on Earth.

The picture: stand at Earth's centre and point at a city. Two angles pin down that direction — how far up from the equator ($\varphi$), and how far around ($\lambda$). That is exactly a globe's grid lines.

Why the topic needs them: the groundtrack is a list of positions of the moving dot. A position on a sphere is precisely a $(\varphi ,\lambda )$ pair. When the parent says "plot $\varphi$ vs $\lambda$", it is drawing that dot's path.

2. Inclination $i$ — how much the orbit ring is tilted

The picture (figure above): take the yellow orbit ring and tip it. The angle it opens up from the flat equator is $i$. The higher you tip it, the closer to the poles the satellite flies.

Why the topic needs it: the highest latitude the dot ever reaches is set entirely by this tilt. Look at the ring: its topmost point sits exactly at latitude $i$ above the equator. That is why the parent states ${\varphi }_{\max }=i$.

3. Prograde vs retrograde — the sign of the tilt's effect

Why it matters here: you cannot separate "which way the ring is tipped" ($i$) from "which way you run around it". The two together decide whether ${\varphi }_{\max }=i$ or $180°-i$. The special retrograde orbits that keep a fixed Sun angle are Sun-Synchronous Orbits — the parent's "Landsat-like" example is one.

4. Time to go around once: the orbital period $T$

The picture: the satellite is a bead sliding once all the way around the yellow ring. $T$ is the stopwatch reading for one lap.

Why the topic needs it: the Earth keeps spinning during one lap. So $T$ is the clock that tells us "how long does Earth get to rotate between two equator crossings?" — the seed of the westward drift. How $T$ depends on altitude comes from Orbital Period & Kepler's Third Law: higher orbit → bigger loop → longer $T$.

5. How fast Earth spins: the rotation rate ${\omega }_{\oplus }$ and the sidereal day

The symbol $\oplus$ (a circle with a cross) is the standard astronomy shorthand for Earth. So ${\omega }_{\oplus }$ = "Earth's spin rate".

Why $86164$ s and not $86400$ s? There are two "days":

• The solar day (86400 s, the 24 h we live by) is the time for the Sun to return to overhead. But the Earth has also moved along its yearly orbit in that time, so it must spin a tiny bit extra to face the Sun again.
• The sidereal day (86164 s) is the time for Earth to spin exactly $360°$ relative to the distant stars — one true rotation, no extra.

The satellite's orbit is fixed relative to the stars, so to compare orbit and spin fairly we use the star-fixed (sidereal) rotation — the $86164$ s day. This is exactly the parent's most common warning. Full detail: Earth Rotation & Sidereal Time.

6. Putting spin and period together: the nodal spacing $S$

Why this exact product: in one lap the stopwatch reads $T$ seconds; in $T$ seconds Earth turns by (rate $\times$ time) $={\omega }_{\oplus }T$. The crossing appears that much farther west because the ground rotated east out from under it. This is just "distance = speed × time" wearing angle clothes.

Worked check (LEO): $T\approx 5760$ s ⇒ $S=360°\times 5760/86164=24.07°$. About $360/24\approx 15$ crossings tile the day.

7. From angle to ground distance: radians and arc length

We keep wanting kilometres on the ground, not just angles. Here is the one tool that converts.

Why radians and nothing else: the arc-length rule $R\theta$ is only true when $\theta$ is in radians — that is literally the definition of a radian (the angle whose arc equals the radius). Feeding degrees gives nonsense. So every "angle → km" step in the parent slips a $\pi /180$ in first.

The symbol $R_E$ is Earth's radius ($\approx 6371$ km); $h$ is the satellite's height above the surface. So the satellite sits at distance $R_E+h$ from Earth's centre.

8. Seeing sideways: the swath and its half-angle $\eta$

The parent gets ${\lambda }_s$ from the sine rule on the triangle (Earth-centre → satellite → swath edge). We do not re-derive it here — Remote Sensing Sensor Geometry does that — but you must recognise the three symbols $\eta$, ${\lambda }_s$, $W$ and that the final ground width is an arc-length: $W=2R_E{\lambda }_s$ (using §7).

9. Coming back: repeat cycle $N/D$, spacing $\delta$, and revisit

Why: after $D$ whole sidereal days Earth is back to the same orientation, and after $N$ whole orbits the satellite is back to the same spot on its ring. When both happen at once, the track overlays itself and the pattern repeats. In between, $N$ equally spaced strips fill the $360°$ around the equator, so each strip is $\delta =360°/N$ apart.

The slow drift of the orbit plane that makes these numbers land on nice integers is the $J_2$ effect in Nodal Regression & J2 Perturbation.

Worked check: $N=233,D=16$ ⇒ orbits/day $=14.5625$, and $\delta =360°/233=1.545°$, which at the equator is $6371\times 1.545\times \pi /180\approx 172$ km.

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