Exercises — Groundtrack analysis — swath, revisit

The subsatellite point is the spot on Earth's surface directly below the satellite — the point where the straight line from Earth's centre through the satellite pierces the surface. Since $i=82°\le 90°$, we use the top branch: ${\varphi }_{\max }=\min (82°,98°)=i=82°$. The track oscillates between $+82°$ and $-82°$ latitude and never reaches the poles.

Start from the parent's radian form and convert to degrees (as set up above): since $S={\omega }_{\oplus }T=\frac{2\pi }{T_{\oplus }}T$ radians, and one turn is both $2\pi$ rad and $360°$, the $2\pi$ cancels leaving $S=360°\times \frac{T}{T_{\oplus }}.$ Why convert: we want a human-readable angle in degrees, not radians. Substitute: $S=360°\times \frac{6000}{86164}=25.07°.$ Each equator crossing lands $\approx 25°$ west of the previous one.

The repeat condition is $NT=D{T}_{\oplus }$ with $N=15,D=1$ coprime. The track repeats after $D=1$ sidereal day (15 orbits). Using the closed-cycle spacing $\delta =360°/N$ defined above, the equatorial track spacing is $\delta =360°/15=24°$.

Step 1 (period). Orbital radius $a=R_E+h=6371+567=6938\text{km}$. $T=2\pi \sqrt{\frac{a^3}{\mu }}=2\pi \sqrt{\frac{6938^3}{398600}}=5760\text{s}(\approx 96\text{min}).$ Why Kepler here: period is not free — altitude fixes $a$, and $a$ fixes $T$. Step 2 (spacing). $S=360°\times \frac{5760}{86164}=24.07°.$ Step 3 (crossings/day). Orbits per day $=T_{\oplus }/T=86164/5760=14.96\approx 15$ crossings — matching $360°/24.07°=14.96$. Consistent.

We use the sine rule on the triangle Earth-centre → satellite → ground-edge (see the figure: the lavender edge is $R_E+h$, the dashed slate edge is $R_E$, the coral ray is the sensor line of sight, angle $\eta$ sits at the satellite and ${\lambda }_s$ at Earth's centre). We use the sine rule because we know two lengths ($R_E$, $R_E+h$) and one angle ($\eta$) and want the far angle — exactly what the sine rule delivers. Step 1. $\sin (\eta +{\lambda }_s)=\frac{R_E+h}{R_E}\sin \eta =\frac{7071}{6371}\sin 20°=1.1099\times 0.3420=\mathrm{0.3796.}$ Step 2. $\eta +{\lambda }_s=\arcsin (0.3796)=22.31°\Rightarrow {\lambda }_s=2.31°.$ Step 3. Convert to radians for the arc length ($\text{arc}=R{\theta }_{\text{rad}}$): $W=2R_E{\lambda }_s=2\times 6371\times (2.31\times \frac{\pi }{180})=2\times 6371\times 0.04032=514\text{km}.$

The gap between neighbouring same-day passes at the equator equals the spacing $S$ converted to arc length. One degree of equatorial arc is $\frac{2\pi R_E}{360°}=\frac{2\pi \times 6371}{360}=111.2\text{km/deg}.$ Gap $=24.07°\times 111.2=2677\text{km}$. Since swath $W=376\text{km}\ll 2677\text{km}$, one day's passes leave huge gaps — coverage requires the multi-day repeat pattern to interleave the tracks.

$\gcd (43,3)=1$, so the fraction is already lowest terms — good, else the true cycle would be shorter. (a) Orbits/day $=N/D=43/3=14.333$. (b) Repeat period $=D{T}_{\oplus }=3\times 86164=258492\text{s}$. (c) From $NT=D{T}_{\oplus }$, divide both sides by $N$ to isolate $T$ (we want $T$, so we undo the multiplication by $N$): $T=\frac{D{T}_{\oplus }}{N}=\frac{258492}{43}=6011.4\text{s}$. (d) $\delta =360°/N=360°/43=8.372°$. In km: $8.372°\times 111.2=931\text{km}$. Why $\delta$ uses $N$, not $D$: over the whole cycle the $N$ distinct tracks tile the $360°$ equator into $N$ equal strips.

Gap-free coverage needs $W\ge \delta =931\text{km}$.

• Sensor A: $W=800<931$ → shortfall $=931-800=131\text{km}$ uncovered strip between adjacent tracks. Not gap-free.
• Sensor B: $W=600<931$ → shortfall $=331\text{km}$. Worse. Neither covers the equator gap-free in 3 days; either the swath must grow past 931 km or the mission must accept a longer repeat cycle (larger $N$ ⇒ smaller $\delta$).

Qualitative: larger $a$ ⇒ larger $T$ (Kepler) ⇒ Earth rotates more per orbit ⇒ larger $S={\omega }_{\oplus }T$. Altitude and spacing move together. Numerical:

• $a_1=6938\text{km}$: $T_1=5760\text{s}$, $S_1=24.07°$ (from L2.1).
• $a_2=6371+1200=7571\text{km}$: $T_2=2\pi \sqrt{7571^3/398600}=6558\text{s}$, so $S_2=360°\times 6558/86164=27.40°$. Indeed $S_2>S_1$. ✔

Step 1 (period from repeat condition). Why: the repeat condition $NT=D{T}_{\oplus }$ fixes how long one orbit must last so that after $N$ orbits exactly $D$ days have passed. Solve for $T$ by dividing by $N$: $T=\frac{86164}{14}=6154.57\text{s}$. Step 2 (invert Kepler for $a$). Why invert: we know $T$ and want $a$, so we algebraically undo $T=2\pi \sqrt{a^3/\mu }$: square both sides ($T^2=4{\pi }^2{a}^3/\mu$) to remove the square root, multiply by $\mu$ and divide by $4{\pi }^2$ to isolate $a^3$, then take the cube root to isolate $a$: $a={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}={\left(\frac{398600\times 6154.57^2}{4{\pi }^2}\right)}^{1/3}=7270\text{km}.$ Step 3 (altitude). Why subtract $R_E$: $a$ is measured from Earth's centre, but altitude is height above the surface, so remove one Earth radius: $h=a-R_E=7270-6371=899\text{km}$. A circular orbit at $\approx 899\text{km}$ closes its track every single sidereal day after 14 revolutions.

Step 1 (required swath). Why $\delta$ and not $S$: coverage is decided by the closed-cycle spacing. $\delta =360°/14=25.714°$. In km: $25.714°\times 111.2=2860\text{km}$. So we need $W\ge 2860\text{km}$ — a wide-swath imager. Step 2 (back out ${\lambda }_s$). Why divide by $2R_E$: the swath relation $W=2R_E{\lambda }_s$ is linear in ${\lambda }_s$, so isolate it by dividing: ${\lambda }_s=\frac{W}{2R_E}=\frac{2860}{2\times 6371}=0.2245\text{rad}=12.86°.$ Step 3 (back out $\eta$). Invert the swath relation $\sin (\eta +{\lambda }_s)=\frac{R_E+h}{R_E}\sin \eta$ with $a/R_E=7270/6371=1.1411$: $\sin (\eta +12.86°)=1.1411\sin \eta .$ Expand with the angle-sum identity: $\sin \eta \cos 12.86°+\cos \eta \sin 12.86°=1.1411\sin \eta$. Divide every term by $\cos \eta$ (this is why $\tan$ appears — it collapses the two unknowns $\sin \eta ,\cos \eta$ into one quantity $\tan \eta$): $\tan \eta =\frac{\sin 12.86°}{1.1411-\cos 12.86°}=\frac{0.2225}{1.1411-0.9749}=\frac{0.2225}{0.1662}=1.339\Rightarrow \eta =53.2°.$ A $53°$ half-angle is very wide (grazing near the horizon), showing 1-day gap-free coverage from LEO is physically demanding.

(a) Why divide by $N$: the repeat condition $NT=D{T}_{\oplus }$ says $N$ orbits must exactly fill $D$ sidereal days, so one orbit lasts $T=\frac{D{T}_{\oplus }}{N}=\frac{16\times 86164}{233}=5916.6\text{s}$. (b) Why invert Kepler (square, isolate, cube-root): period comes from $a$ through $T=2\pi \sqrt{a^3/\mu }$; running it backward gives $a={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}={\left(\frac{398600\times 5916.6^2}{4{\pi }^2}\right)}^{1/3}=7077\text{km}$. Why subtract $R_E$: altitude is height above the surface, $h=7077-6371=706\text{km}$. (c) Why $\times 111.2$: to turn a degree of equatorial arc into km we multiply by the km-per-degree factor $2\pi R_E/360=111.2\text{km/deg}$. $\delta =360°/233=1.5451°\Rightarrow \delta =1.5451\times 111.2=171.8\text{km}$. (d) $W=185\text{km}\ge \delta =171.8\text{km}$ → gap-free, with overlap margin $185-171.8=13.2\text{km}$ (each strip overlaps its neighbour by $\approx 13$ km). Revisit $\le 16$ days everywhere at the equator.

(a) Which branch: since $i=98.2°>90°$ the orbit is retrograde, so use ${\varphi }_{\max }=180°-i$ (equivalently $\min (98.2°,81.8°)$): ${\varphi }_{\max }=180°-98.2°=81.8°$. The track sweeps to $\pm 81.8°$ latitude, leaving polar caps unimaged — a signature of near-polar Sun-synchronous orbits. Note the branch is essential: the naive "${\varphi }_{\max }=i=98.2°$" would claim a latitude beyond the North Pole, which is impossible.

(b) — Derive the $\cos \varphi$ factor first. Look at Fig 2. Two neighbouring tracks are separated by a fixed longitude difference $\Delta \lambda$ (an angle around Earth's spin axis). We want the ground distance between them, and here is the key geometric fact: a circle of constant latitude $\varphi$ is smaller than the equator. Its radius is not $R_E$ but $r(\varphi )=R_E\cos \varphi$ — because the latitude circle is a slice of the sphere taken a height $R_E\sin \varphi$ up the axis, and by the right triangle (axis, radius-to-surface, horizontal radius) the horizontal radius is $R_E\cos \varphi$. Why $\cos$ and not $\sin$: the horizontal radius is the side adjacent to the latitude angle $\varphi$ measured at Earth's centre, and adjacent/hypotenuse $=\cos \varphi$. A fixed longitude gap $\Delta \lambda$ therefore subtends a shorter arc on this smaller circle: $\text{ground gap}(\varphi )=r(\varphi )\Delta \lambda =R_E\cos \varphi \Delta \lambda =\underset{=\delta \text{at equator}}{\underbrace{R_E\Delta \lambda }}\times \cos \varphi .$ So the effective spacing is ${\delta }_{\varphi }=\delta \cos \varphi$ — the same longitude spacing simply lands on a smaller circle at higher latitude. In Fig 2 the double-headed arrow between the two tracks visibly shrinks as you climb, and the dotted guide lines show the meridians crowding toward the pole.

Compute at $\varphi =60°$: ${\delta }_{\varphi }=\delta \cos \varphi =171.8\times \cos 60°=171.8\times 0.5=85.9\text{km}.$ With $W=185\text{km}\gg 85.9\text{km}$, tracks overlap by more than double at $60°$ — high-latitude coverage is far denser (and revisit faster) than at the equator. This is why polar-region imaging is a strength of Sun-synchronous missions: the very convergence that limits ${\varphi }_{\max }$ also tightens coverage right up to it.

Spacing per orbit ::: $S=360°\times T/T_{\oplus }$ Track spacing after full repeat ::: $\delta =360°/N$ Ground spacing at latitude $\varphi$ ::: ${\delta }_{\varphi }=\delta \cos \varphi$ Period from altitude ::: $T=2\pi \sqrt{a^3/\mu }$ Repeat condition ::: $NT=D{T}_{\oplus }$, $N,D$ coprime Swath width ::: $W=2R_E{\lambda }_s$, ${\lambda }_s=\arcsin \left(\frac{(R_E+h)\sin \eta }{R_E}\right)-\eta$ Max latitude ::: ${\varphi }_{\max }=\min (i,180°-i)$