Worked examples — Groundtrack analysis — swath, revisit

This page is the drill-ground for Groundtrack analysis — swath, revisit. The parent gave you four formulas; here we push every one of them into every corner it can reach — small angles, big angles, straight-down looks, sideways looks, prograde and retrograde orbits, the exact-repeat trap, and a real-world word problem. Each example tells you which cell of the scenario matrix it fills.

Before any symbol appears, one reminder of the cast (all built in the parent note, but restated here so each worked example is self-contained):

Recall The full symbol list we will use (tap to re-read)

• $h$ ::: satellite altitude — the height of the satellite above Earth's surface (not above the centre), in km. The orbit radius from Earth's centre is then $a=R_E+h$.
• $R_E=6371$ km ::: Earth's mean radius (surface to centre).
• $T$ ::: orbital period, seconds per one lap around Earth.
• $\eta$ ::: sensor half-look-angle — half the total field-of-view cone, measured at the satellite from straight-down (nadir) to the outermost sight-line.
• ${\lambda }_s$ ::: the Earth-central half-angle — the angle at Earth's centre from the subsatellite point out to the ground edge the sensor can see.
• ${\varphi }_{\max }$ ::: the highest latitude the groundtrack ever reaches (its poleward turning point).
• $i$ ::: inclination — the tilt of the orbit plane relative to the equator.
• $S=360°\times \frac{T}{86164}$ ::: how far west the next equator crossing lands, in degrees.
• $W=2R_E{\lambda }_s$ with ${\lambda }_s=\arcsin \left(\frac{(R_E+h)\sin \eta }{R_E}\right)-\eta$ ::: swath width from the sensor geometry.
• Repeat: $NT=D{T}_{\text{Earth}}$, i.e. orbits/day $=N/D$; track gap $\delta =360°/N$ ::: when the pattern closes.

Throughout: $R_E=6371$ km, $T_{\text{Earth}}=86164$ s (the sidereal day — never the 86400 s solar day).

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The scenario matrix

Every problem this topic can throw is one of these cells. The examples below are labelled [C#].

#	Cell class	What makes it special	Example
C1	Small look-angle swath	$\eta$ tiny → ${\lambda }_s$ almost linear in $\eta$	Ex 1
C2	Large look-angle swath	$(R_E+h)\sin \eta$ approaches $R_E$ → arcsin near its limit	Ex 2
C3	Degenerate: nadir-only ($\eta =0$)	swath collapses to zero	Ex 3
C4	Limiting: grazing horizon	argument of arcsin hits $1$ → maximum possible swath	Ex 3
C5	Nodal spacing, low orbit	short $T$ → small westward $S$	Ex 4
C6	Nodal spacing, high orbit	long $T$ → large $S$ (altitude ↑ ⇒ $S$ ↑)	Ex 4
C7	Exact repeat $N/D$	integers, coprime, close the pattern	Ex 5
C8	Gap test $W$ vs $\delta$	does swath cover the track gap?	Ex 6
C9	Prograde vs retrograde ${\varphi }_{\max }$	$i\le 90°$ vs $i>90°$, altitude-independent	Ex 7
C10	Real-world word problem	disaster-monitoring revisit requirement	Ex 8
C11	Exam twist	"what altitude gives a 15-orbit-per-day repeat?"	Ex 9

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Example 1 — Small look-angle swath [C1]

Forecast: For a tiny cone the swath should be roughly proportional to $\eta$. If $15°$ gave $\approx 376$ km (parent note), guess that $5°$ gives roughly a third — around $120$ km. Let's see.

Look at the figure: the satellite (top), Earth's centre (bottom), and the ground edge form a triangle. The red ray is the sensor's outermost sight-line, tilted $\eta$ off straight-down. The arc it cuts on the surface is the half-swath.

1. Ratio factor. $\frac{R_E+h}{R_E}=\frac{7071}{6371}=1.1099$. Why this step? The sine rule $\frac{\sin \eta }{R_E}=\frac{\sin (\eta +{\lambda }_s)}{R_E+h}$ rearranges to $\sin (\eta +{\lambda }_s)=\frac{R_E+h}{R_E}\sin \eta$; this factor is the multiplier.
2. Far angle sine. $\sin (\eta +{\lambda }_s)=1.1099\times \sin 5°=1.1099\times 0.08716=0.09674$. Why? This is the angle at the satellite between straight-down and the tilted ray, opened up by the altitude leverage.
3. Undo the sine. $\eta +{\lambda }_s=\arcsin (0.09674)=5.550°$, so ${\lambda }_s=5.550°-5°=0.550°$. Why arcsin? $\arcsin$ answers "which angle has this sine?" — it undoes the $\sin$ we built. Here the physical sum $\eta +{\lambda }_s$ is only $5.55°$, comfortably below $90°$, so the principal value arcsin returns (which always lands in $[-90°,90°]$) is exactly the angle we want — no branch ambiguity.
4. Arc length. $W=2R_E{\lambda }_s=2(6371)(0.550°\times \frac{\pi }{180})=2(6371)(0.009601)=122.3$ km. Why $\pi /180$? Arc length needs the angle in radians: $\text{arc}=R{\theta }_{\text{rad}}$.

Verify: $122$ km is about a third of the $376$ km at $15°$ — matching the near-linear forecast (small angles keep $\sin \eta \approx \eta$, so $W$ scales almost with $\eta$). Units: km × (dimensionless) = km. ✔

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Example 2 — Large look-angle swath [C2]

Forecast: Wide angles peer far toward the horizon, so the swath should be much larger than linear scaling suggests — expect a big jump, maybe thousands of km, because the geometry is bending.

1. Far angle sine. $\sin (\eta +{\lambda }_s)=1.1099\times \sin 60°=1.1099\times 0.86603=0.96114$. Why? Same sine-rule rearrangement — but note the result is now close to 1, meaning the ray is nearly tangent to Earth.
2. Undo the sine — and check the branch is valid. $\arcsin (0.96114)=73.98°$. So $\eta +{\lambda }_s=73.98°$, giving ${\lambda }_s=73.98°-60°=13.98°$. Why is the principal value the right one? Arcsin could in principle return either $73.98°$ or its supplement $180°-73.98°=106.02°$ (both have the same sine). We keep $73.98°$ because the physical angle $\eta +{\lambda }_s$ is the angle at the satellite inside the triangle, and it must stay below $90°$: the outermost ray always leans outward from nadir but never doubles back past the horizon. As long as the sensor still hits the ground (see Ex 3b), $\eta +{\lambda }_s<90°$, so the principal value arcsin gives — which lives in $[-90°,90°]$ — is automatically correct. Here $73.98°<90°$ confirms we are inside the valid domain.
3. Arc length. $W=2(6371)(13.98°\times \frac{\pi }{180})=2(6371)(0.24404)=3109$ km. Why? Radians again; the same $R\theta$ arc rule.

Verify: From $5°\to 122$ km and $15°\to 376$ km, doubling-ish look-angles gave near-proportional swaths; but $60°$ jumps to $\sim 3100$ km — the nonlinear blow-up the forecast expected, because $\sin (\eta +{\lambda }_s)$ is crowding against $1$. ✔

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Example 3 — Degenerate & limiting look-angles [C3][C4]

Forecast: (a) Looking exactly straight down sees a point, so $W=0$. (b) The biggest swath happens when the ray is tangent to Earth — the far angle hits $90°$; guess a swath of a few thousand km.

Case (a) — nadir, $\eta =0$:

1. Start from the general far-angle relation $\sin (\eta +{\lambda }_s)=\frac{R_E+h}{R_E}\sin \eta$. Substituting $\eta =0$ makes the argument on the left collapse: $\eta +{\lambda }_s=0+{\lambda }_s={\lambda }_s$, so the equation literally becomes $\sin ({\lambda }_s)=1.1099\times \sin 0°=0$. Why did the argument of $\sin$ change from $\eta +{\lambda }_s$ to just ${\lambda }_s$? It didn't change form — we simply set $\eta =0$, and $\eta +{\lambda }_s$ with $\eta =0$ is ${\lambda }_s$. So $\sin {\lambda }_s=0\Rightarrow {\lambda }_s=0$, and $W=2R_E(0)=0$ km. Physical why? Zero cone angle means the sensor's edge ray is the centre ray — no width. The formula degrades gracefully to zero.

Case (b) — grazing the horizon:

1. Tangency condition. The ray is tangent to Earth when the angle at the ground edge is $90°$, i.e. $\eta +{\lambda }_s=90°$. Then $\sin (\eta +{\lambda }_s)=1$, its maximum. Why $=1$? $\sin$ can never exceed $1$; the sine-rule right side $\frac{(R_E+h)\sin \eta }{R_E}$ is capped there. Push $\eta$ past this and the arcsin has no answer — physically the ray misses Earth entirely (looks into space). This is also the exact boundary of the valid-branch discussion in Ex 2: $\eta +{\lambda }_s$ reaches $90°$ precisely at the horizon and never exceeds it.
2. Solve for ${\eta }_{\max }$. $\frac{(R_E+h)\sin {\eta }_{\max }}{R_E}=1\Rightarrow \sin {\eta }_{\max }=\frac{R_E}{R_E+h}=\frac{6371}{7071}=0.90101$, so ${\eta }_{\max }=\arcsin (0.90101)=64.24°$. Why? This is the largest half-look-angle that still touches ground — beyond it, no swath.
3. Half-swath. ${\lambda }_s=90°-{\eta }_{\max }=90°-64.24°=25.76°$. Why this step? With $\eta +{\lambda }_s$ pinned at $90°$ by tangency, the half-swath ground angle is simply whatever is left after subtracting ${\eta }_{\max }$ — that leftover is the arc from the subpoint out to the horizon.
4. Swath. $W=2(6371)(25.76°\times \frac{\pi }{180})=2(6371)(0.44961)=5729$ km. Why this step? Same arc-length rule $W=2R_E{\lambda }_s$ as every other case; we double ${\lambda }_s$ (both sides of the subpoint) and convert to radians so $R\theta$ gives a length in km.

Verify: (a) gives exactly $0$ — the sensible degenerate limit. (b) The maximum swath $5729$ km is a real "edge of visible Earth" figure; larger than the $60°$ case ($3109$ km) as required, and its half-angle $25.76°$ is the biggest ground reach possible from 700 km. ✔

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Example 4 — Nodal spacing: low vs high orbit, and the sign convention [C5][C6]

Forecast: More time per lap = more Earth-spin per lap = bigger westward jump. So (b) $>$ (a).

Sign convention first. The parent writes $\Delta {\lambda }_{\text{node}}=-{\omega }_{\oplus }T$: the minus sign means the next crossing lands west of the last, because Earth spins eastward under the orbit. We report the magnitude $S={\omega }_{\oplus }T=360°\times T/86164$ and remember "west" as its direction. This drift comes purely from Earth's rotation, so it happens for every orbit regardless of the satellite's direction of travel — see the retrograde note below.

1. Low orbit. $S_a=360°\times \frac{5400}{86164}=360°\times 0.062671=22.56°$ west. Why this step? $S=360°\times T/86164$: in one period the Earth turns this fraction of a full turn, and that is how far west the next crossing lands.
2. High orbit. $S_b=360°\times \frac{12000}{86164}=360°\times 0.139269=50.14°$ west. Why this step? We reuse the exact same formula but with the longer period; because $T$ is bigger, Earth spins through a larger fraction of a full turn during one lap, so the next crossing lands even farther west.
3. Compare. $S_b/S_a=50.14/22.56=2.222$ — the ratio equals $12000/5400$ exactly, because $S$ is linear in $T$. Why? $360°/86164$ is a constant multiplier, so $S\propto T$; and Kepler makes $T$ grow with altitude, so altitude ↑ ⇒ $S$ ↑. Forecast confirmed.

Verify: $S_b-S_a=27.58°$, both positive (westward), both under $360°$ (a single lap can't wrap Earth twice at these periods). ✔

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Example 5 — Exact repeat pattern [C7]

Forecast: $43/3\approx 14.3$ orbits/day → period a bit over 100 min. With 43 strips wrapping the globe, spacing near $8°$.

1. Coprime check. $\gcd (43,3)=1$ ✔ (43 is prime, 3 doesn't divide it). Why? If they shared a factor the pattern would close sooner, so the "fundamental" cycle needs coprime $N,D$.
2. Period. $NT=D{T}_{\text{Earth}}\Rightarrow T=\frac{D{T}_{\text{Earth}}}{N}=\frac{3\times 86164}{43}=\frac{258492}{43}=6011.4$ s $=100.2$ min. Why? After $D$ sidereal days Earth is re-oriented AND after $N$ laps the satellite is back — both true at once means the track overlays itself.
3. Orbits per day. $N/D=43/3=14.333$. Why this step? The fraction $N/D$ is the daily lap-rate: it tells you how many crossings are laid down each day, and its being non-integer ($14.333$, not a whole number) is exactly why the tracks interleave over $D$ days instead of retracing the same line every single day.
4. Track spacing. $\delta =360°/N=360°/43=8.372°$. In km at the equator: $8.372°\times \frac{\pi }{180}\times 6371=0.14612\times 6371=930.9$ km. Why $360°/N$? The $N$ crossings divide the equator ($360°$) into $N$ equal strips.

Verify: $43\times 6011.4=258490\approx 3\times 86164=258492$ s (rounding) ✔. Orbits/day $14.33$ gives period $86164/14.333=6011$ s — consistent. ✔

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Example 6 — Gap test: does the swath cover the track gap? [C8]

Forecast: $376$ km swath vs $931$ km gap — clearly not enough in one cycle. Guess we need roughly $931/376\approx 3$ cycles.

1. Coverage condition. Gap-free needs $W\ge \delta$. Here $376<930.9$, so NO — one cycle leaves bands unseen. Why? Each pass paints a $376$ km stripe; neighbours sit $931$ km apart, leaving $931-376=555$ km blind between them.
2. How many sub-cycles. Number of interleaved passes needed $=\lceil \delta /W\rceil =\lceil 930.9/376\rceil =\lceil 2.476\rceil =3$. Why $\lceil \rceil$ (round up)? You can't have a fraction of a stripe; $2$ stripes ($752$ km) still leave a gap, so you need $3$.
3. Effective revisit. If one repeat cycle is $D=3$ days, full gap-free coverage takes about $3\times 3=9$ days (three offset cycles). Why multiply by the cycle length? Each of the $3$ interleaved passes belongs to a separate $3$-day repeat cycle (the orbit is slightly nudged so the tracks fall between the previous ones), so filling the gap costs $3$ full cycles — and each cycle takes $D=3$ days, giving $3\times 3=9$ days total.

Verify: $3$ stripes give $3\times 376=1128$ km $>930.9$ km — overshoots the gap (with overlap), so $3$ suffices while $2$ ($752$ km) fails. ✔

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Example 7 — Prograde vs retrograde max latitude [C9]

Recall from the symbol list: ${\varphi }_{\max }$ is the highest latitude the groundtrack ever reaches — the poleward turning point of the sinusoid on the map. Its rule (parent note) is ${\varphi }_{\max }=i$ for $i\le 90°$ and ${\varphi }_{\max }=180°-i$ for $i>90°$.

Forecast: Max latitude is set purely by inclination, so A reaches $55°$, and B — being past $90°$ — uses $180°-i$, reaching $82°$. Altitude is irrelevant.

1. Satellite A. $i=55°\le 90°$, so ${\varphi }_{\max }=i=55°$ N and S. Why? A great circle tilted $55°$ to the equator peaks at latitude $55°$ — that's the geometric top of the tilted ring.
2. Satellite B. $i=98°>90°$, so ${\varphi }_{\max }=180°-i=180°-98°=82°$ N and S. Why the $180-i$ rule? A retrograde orbit ($i>90°$) tips over the top; its plane's tilt from the equator is measured as $180°-i$. So $98°$ reaches only to $82°$, not $98°$ (latitude never exceeds $90°$).
3. Altitude? Neither computation used $h$. ${\varphi }_{\max }$ is purely geometric — same at 300 km or 3000 km.

Verify: A: $55°\le 90°$ ✔. B: $180-98=82°\le 90°$ ✔ (a valid latitude). Both altitude-independent — a common exam trap avoided. ✔

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Example 8 — Real-world word problem [C10]

Forecast: $29$ strips → $\delta \approx 12°\approx 1370$ km, way bigger than the $376$ km swath — so a single 2-day cycle probably leaves gaps and fails the requirement.

1. Coprime & period. $\gcd (29,2)=1$ ✔. $T=\frac{2\times 86164}{29}=\frac{172328}{29}=5942.3$ s. Why this step? Same repeat condition $NT=D{T}_{\text{Earth}}$ as Example 5.
2. Track spacing. $\delta =360°/29=12.414°$; in km: $12.414°\times \frac{\pi }{180}\times 6371=0.21667\times 6371=1380.4$ km. Why convert this way? The gap is naturally an angle ($360°$ split into $N$ strips), but we compare it against a swath measured in km, so we turn the angle into a ground distance with the arc-length rule $\text{arc}=R_E{\theta }_{\text{rad}}$ — and $\theta$ must be in radians, hence the $\pi /180$.
3. Gap test. $W=376$ km $<1380.4$ km ⇒ gaps remain within one 2-day cycle. Passes needed to fill: $\lceil 1380.4/376\rceil =\lceil 3.67\rceil =4$. Why? $3$ stripes $=1128$ km still short of $1380$ km; need $4$.
4. Verdict. Full coverage needs $\sim 4$ interleaved 2-day cycles $\approx 8$ days $\gg 2$ days. Requirement NOT met. Fix hint: widen the swath (bigger $\eta$) or choose a denser repeat (larger $N$) so $\delta$ shrinks below $W$.

Verify: $4\times 376=1504$ km $>1380.4$ km ✔ (4 fills, 3 doesn't). $8\text{days}>2\text{days}$ ⇒ fails, matching forecast. ✔

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Example 9 — Exam twist: find the altitude for 15 orbits/day [C11]

Forecast: $15$ orbits/day → period $\approx 86164/15\approx 5744$ s $\approx 96$ min → a Low Earth Orbit, altitude maybe $500$–$600$ km.

1. Required period. $NT=D{T}_{\text{Earth}}$ with $N=15,D=1$: $T=86164/15=5744.3$ s. Why this step? $15$ equal laps must fit exactly one sidereal day for the track to repeat daily, so we solve the repeat condition for the period each lap is allowed to take.
2. Invert Kepler for $a$. From $T=2\pi \sqrt{a^3/\mu }$: $a^3=\frac{\mu T^2}{4{\pi }^2}$. Why this step? We know $T$ (from step 1) and want the orbit radius $a$, so we algebraically solve the period law for $a$ — square both sides and rearrange. See Orbital Period & Kepler's Third Law.
3. Plug in. $a^3=\frac{3.986\times 10^5\times (5744.3{)}^2}{4{\pi }^2}=\frac{3.986\times 10^5\times 3.29970\times 10^7}{39.478}=\frac{1.31527\times 10^{13}}{39.478}=3.3316\times 10^{11}{\text{km}}^3$. Why this step? Numerically evaluating the right side gives $a^3$; all inputs are known constants, so this is just arithmetic.
4. Cube root. $a=(3.3316\times 10^{11}{)}^{1/3}=6931.6$ km. Why cube root? It undoes the $a^3$; $a$ is a length so we need the real cube root.
5. Altitude. $h=a-R_E=6931.6-6371=560.6$ km. Why this step? $a$ is measured from Earth's centre, but altitude $h$ is height above the surface, so we subtract Earth's radius $R_E$.

Verify: Recompute $T$ from $a=6931.6$: $T=2\pi \sqrt{6931.6^3/3.986\times 10^5}=2\pi \sqrt{3.3312\times 10^{11}/3.986\times 10^5}=2\pi \sqrt{8.3573\times 10^5}=2\pi (914.2)=5744$ s ✔ — exactly $86164/15$. Altitude $\approx 561$ km is a sensible LEO, matching the forecast. ✔

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Recall One-line summary of the whole matrix

Which formula owns each cell? ::: Swath $W=2R_E{\lambda }_s$ owns C1–C4 (look-angle extremes); nodal spacing $S=360°T/86164$ owns C5–C6; repeat $N/D$ + $\delta =360°/N$ owns C7, C11; gap test $W\ge \delta$ owns C8, C10; ${\varphi }_{\max }=i$ or $180°-i$ owns C9.

See also: Orbital Period & Kepler's Third Law, Sun-Synchronous Orbits, Inclination & Orbital Elements, Remote Sensing Sensor Geometry, Earth Rotation & Sidereal Time, Nodal Regression & J2 Perturbation.