Exercises — Patched conic method — interplanetary trajectory design

L1.1

• Patch 1 = departure hyperbola, central body = Earth.
• Patch 2 = transfer ellipse (Hohmann), central body = the Sun.
• Patch 3 = arrival hyperbola, central body = Mars. A hyperbola is an open curve (energy $>0$, escapes); an ellipse is closed (energy $<0$, bound). See Hohmann transfer orbit.

L1.2

• $v$ = the spacecraft's speed at that instant.
• $\mu$ = $G\times M_{\text{central}}$, the pull-strength of the body you orbit.
• $r$ = current distance from the central body.
• $a$ = semi-major axis (half the longest diameter) of the orbit — it fixes the total energy. The heliocentric transfer ellipse (patch 2) uses $\mu ={\mu }_{\odot }$, because there the Sun is central.

L2.1 Mars's circular speed: $v_{c,2}=\sqrt{\frac{1.327\times 10^{20}}{2.279\times 10^{11}}}=\sqrt{5.823\times 10^8}\approx 2.413\times 10^4\frac{\text{m}}{\text{s}}=24.13\frac{\text{km}}{\text{s}}.$ Why: Mars moves slower than Earth (farther out ⇒ weaker pull ⇒ gentler speed), so this is below Earth's 29.78 km/s.

L2.2 First $a_t=\frac{1.496+2.279}{2}\times 10^{11}=1.8875\times 10^{11}$ m. Then at aphelion $r=r_2$: $v_{t,2}=\sqrt{1.327\times 10^{20}\left(\frac{2}{2.279\times 10^{11}}-\frac{1}{1.8875\times 10^{11}}\right)}.$ Inside the bracket: $8.776\times 10^{-12}-5.298\times 10^{-12}=3.478\times 10^{-12}$. Multiply: $1.327\times 10^{20}\times 3.478\times 10^{-12}=4.615\times 10^8$. So $v_{t,2}=\sqrt{4.615\times 10^8}\approx 2.148\times 10^4\frac{\text{m}}{\text{s}}=21.48\frac{\text{km}}{\text{s}}.$ Why slower than $v_{c,2}$? At aphelion the transfer ship is at the top of its climb — it has traded kinetic for potential energy, so it is dawdling, moving slower than a body on a circle there.

L2.3 Since the ship arrives slower than Mars, Mars catches up from behind: $v_{\infty }^{arr}=|24.13-21.48|=2.65\frac{\text{km}}{\text{s}}.$ This is the speed with which the ship enters Mars's SOI relative to Mars, i.e. the hyperbolic excess of the arrival hyperbola.

L3.1 $\frac{{\mu }_M}{{\mu }_{\odot }}=\frac{4.283\times 10^{13}}{1.327\times 10^{20}}=3.228\times 10^{-7}.$ Raise to the $2/5$ power. Take logs: $\ln (3.228\times 10^{-7})=-14.948$; times $0.4$ gives $-5.979$; exponentiate: $e^{-5.979}=2.532\times 10^{-3}$. Then $r_{SOI}=2.279\times 10^{11}\times 2.532\times 10^{-3}\approx 5.77\times 10^8\text{m}\approx 577,000\text{km}.$ Compare to $R_M=3.39\times 10^6$ m: the SOI is about $\frac{5.77\times 10^8}{3.39\times 10^6}\approx 170$ times Mars's radius. Comment: the SOI is huge compared to the planet, yet tiny compared to $r_2$ ($577,000$ km vs $228$ million km ≈ $0.25\%$). So the ship enters the SOI essentially at Mars's position in the Sun frame — that is exactly why we can "patch" position continuously and just add Mars's orbital velocity.

L3.2 With $r_p=2.0\times 10^7$ m and $v_{\infty }=2.94\times 10^3$ m/s: $v_{circ}=\sqrt{\frac{3.986\times 10^{14}}{2.0\times 10^7}}=\sqrt{1.993\times 10^7}\approx 4.464\times 10^3\frac{\text{m}}{\text{s}}.$ $v_p=\sqrt{v_{\infty }^2+\frac{2{\mu }_E}{r_p}}=\sqrt{(2940{)}^2+\frac{2(3.986\times 10^{14})}{2.0\times 10^7}}=\sqrt{8.644\times 10^6+3.986\times 10^7}.$ $=\sqrt{4.850\times 10^7}\approx 6.964\times 10^3\frac{\text{m}}{\text{s}}.$ $\Delta v=6.964-4.464=2.50\frac{\text{km}}{\text{s}}.$ Wait — that's less than 3.62 km/s? Look again: the burn got cheaper in km/s? No — it is the opposite lesson. The Oberth effect says burning deeper (smaller $r_p$) makes $v_{\infty }$ cheaper. Here we went higher ($r_p$ larger). At the higher orbit both $v_{circ}$ and $v_p$ shrink, but the gap $\Delta v$ also shrinks numerically here because the well is shallower and $v_{\infty }$ dominates less... be careful: compare the useful energy delivered per m/s of burn. Deep burns convert burn-$\Delta v$ into far more $v_{\infty }$. The correct summary: a burn made deep in the gravity well (small $r_p$, high $v_p$) buys more hyperbolic excess per m/s than the same burn made high up. The raw $\Delta v$ number being smaller at $20,000$ km does not contradict Oberth — Oberth is about efficiency (energy per burn), and to fairly compare you must also account for the $\Delta v$ it cost to raise the parking orbit in the first place. Once you add that raising cost, the low orbit wins overall.

L4.1(a) Hyperbolic periapsis speed uses Mars's ${\mu }_M$: $v_p=\sqrt{(v_{\infty }^{arr}{)}^2+\frac{2{\mu }_M}{r_p}}=\sqrt{(2650{)}^2+\frac{2(4.283\times 10^{13})}{3.890\times 10^6}}.$ $=\sqrt{7.0225\times 10^6+2.2021\times 10^7}=\sqrt{2.9044\times 10^7}\approx 5.389\times 10^3\frac{\text{m}}{\text{s}}=5.39\frac{\text{km}}{\text{s}}.$ Why: even a slow SOI entry (2.65 km/s) speeds up enormously as it falls into Mars's well — the $2{\mu }_M/r_p$ term dominates.

L4.1(b) Circular speed there: $v_{circ}=\sqrt{\frac{{\mu }_M}{r_p}}=\sqrt{\frac{4.283\times 10^{13}}{3.890\times 10^6}}=\sqrt{1.1011\times 10^7}\approx 3.318\times 10^3\frac{\text{m}}{\text{s}}=3.32\frac{\text{km}}{\text{s}}.$

L4.1(c) Capture burn (retro): $\Delta v_{cap}=v_p-v_{circ}=5.389-3.318=2.07\frac{\text{km}}{\text{s}}.$ This burn removes 2.07 km/s at periapsis to drop from the fly-by hyperbola into a bound circle.

L4.2 Rough total: $\Delta v_{total}\approx 3.62+2.07=5.69\frac{\text{km}}{\text{s}}.$ This is the classic "≈ 5.7 km/s in-space budget" for an impulsive Hohmann Mars orbiter (before atmospheric-drag aerobraking, which can slash the capture cost dramatically).

L5.1(a) For an inward Hohmann the ellipse touches Earth's orbit (its aphelion, the far point) and Venus's orbit (its perihelion, the near point): $a_t=\frac{r_1+r_V}{2}=\frac{1.496\times 10^{11}+1.082\times 10^{11}}{2}=1.289\times 10^{11}\text{m}.$

L5.1(b) At Earth we are at aphelion ($r=r_1$): $v_{t,1}=\sqrt{1.327\times 10^{20}\left(\frac{2}{1.496\times 10^{11}}-\frac{1}{1.289\times 10^{11}}\right)}.$ Bracket: $1.3369\times 10^{-11}-7.7580\times 10^{-12}=5.611\times 10^{-12}$. Product: $1.327\times 10^{20}\times 5.611\times 10^{-12}=7.446\times 10^8$. $v_{t,1}=\sqrt{7.446\times 10^8}\approx 2.729\times 10^4\frac{\text{m}}{\text{s}}=27.29\frac{\text{km}}{\text{s}}.$

L5.1(c) Earth's circular speed (same as parent): $v_{c,1}=\sqrt{\frac{1.327\times 10^{20}}{1.496\times 10^{11}}}\approx 2.978\times 10^4\frac{\text{m}}{\text{s}}=29.78\frac{\text{km}}{\text{s}}.$

L5.1(d) Departure excess: $v_{\infty }^{dep}=|27.29-29.78|=2.49\frac{\text{km}}{\text{s}}.$ Direction: the transfer needs the ship to be slower than Earth in the Sun frame ($v_{t,1}<v_{c,1}$), so $v_{\infty }^{dep}$ points opposite to Earth's orbital velocity (a "retrograde" departure). You brake against Earth's motion so you fall Sun-ward toward Venus — the mirror image of the outward Mars case where you sped up.

L5.2 Transfer time: $t=\pi \sqrt{\frac{(1.289\times 10^{11}{)}^3}{1.327\times 10^{20}}}=\pi \sqrt{\frac{2.142\times 10^{33}}{1.327\times 10^{20}}}=\pi \sqrt{1.614\times 10^{13}}.$ $=\pi \times 4.018\times 10^6\approx 1.262\times 10^7\text{s}\approx 146\text{days}.$ Shorter than the 259-day Mars trip because the inward ellipse is smaller. See Launch windows & synodic period for when such a window opens.