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Unit 10 · Lesson 10-4
Gravity Assists in the Patched-Conic Model
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Gravity Assists in the Patched-Conic Model

The flyby as a rotation of v∞, energy exchange, leading and trailing passes, powered flybys and the Tisserand parameter

Mariner 10, 5 February 1974
Mariner 10 passes about 5,800 km above the clouds of Venus, and its path around the Sun bends inward toward Mercury. No engine fires. It is the first time a spacecraft has used one planet to reach another. At the Jet Propulsion Laboratory, the Italian mathematician Giuseppe Colombo notices something about the new orbit: its period is almost exactly twice Mercury's. After each pass, Mariner 10 will meet Mercury again two Mercury-years later. The mission is extended, and the spacecraft flies past Mercury three times — a second gravity-assist idea, resonance, discovered on the job. The lesson on gravity assists and Lagrange points explained the physics of a flyby in two frames. This lesson makes it an algorithm.
The flyby as a vector operation. Inside the planet's sphere of influence the flyby is a hyperbola: the spacecraft leaves with the same ∣v⃗∞∣|\vec v_\infty| it arrived with, rotated by the turn angle δ\delta within the flyby plane. In the patched-conic model the whole encounter happens at one point and one instant, so
v⃗∞−=v⃗−−v⃗p,v⃗∞+=R(h^,δ) v⃗∞−,v⃗+=v⃗p+v⃗∞+\vec v_{\infty}^{-} = \vec v^{-} - \vec v_p, \qquad \vec v_{\infty}^{+} = \mathcal{R}(\hat h, \delta)\,\vec v_{\infty}^{-}, \qquad \vec v^{+} = \vec v_p + \vec v_{\infty}^{+}

Subtract the planet's velocity, rotate about the flyby-plane normal ĥ by δ, add the planet's velocity back. sin(δ/2) = 1/e, with e = 1 + r_p v∞²/μ.

💡What the flyby changes, and what it does not
Unchanged: ∣v⃗∞∣|\vec v_\infty|, relative to the planet. The encounter is elastic. Changed: the direction of v⃗∞\vec v_\infty, and therefore the heliocentric velocity. The heliocentric change is Δv⃗=v⃗∞+−v⃗∞−,∣Δv⃗∣=2v∞sin⁡δ2\Delta\vec v = \vec v_\infty^+ - \vec v_\infty^-, \qquad |\Delta\vec v| = 2v_\infty\sin\frac{\delta}{2} and because ∣v⃗∞∣|\vec v_\infty| is conserved, the heliocentric energy change is exactly Δε=12(∣v⃗+∣2−∣v⃗−∣2)=v⃗p⋅Δv⃗\Delta\varepsilon = \tfrac{1}{2}\left(|\vec v^+|^2 - |\vec v^-|^2\right) = \vec v_p\cdot\Delta\vec v The gain is largest when Δv⃗\Delta\vec v points along the planet's motion — when the spacecraft passes behind the planet (the trailing side). A pass in front (leading side) slows it down: Mariner 10 at Venus, MESSENGER, Parker Solar Probe.
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📐Worked Example: three ways past Jupiter
A spacecraft arrives at Jupiter (moving at 13.06 km/s) at 7.52 km/s with a small sunward and out-of-plane velocity; v∞=5.78v_\infty = 5.78 km/s. A flyby at 10 Jupiter radii turns v⃗∞\vec v_\infty by 114.6°.
  • Turn one way in the ecliptic (periapsis behind Jupiter): heliocentric speed 7.52 → 15.38 km/s, energy +90.1 km²/s², new aphelion 12.9 AU.
  • Turn the other way: 7.52 → 17.14 km/s, +118.6 km²/s², aphelion 32.8 AU.
  • Over the pole: 16.50 km/s, perihelion 5.20 AU — and an orbit tilted out of the ecliptic.
In every case ∣v⃗∞∣|\vec v_\infty| leaves at 5.777 km/s, and the energy change equals v⃗J⋅Δv⃗\vec v_J\cdot\Delta\vec v to the last digit. Arrive with v⃗∞\vec v_\infty pointing at the Sun instead, and one side of Jupiter slows the spacecraft (−40.6 km²/s²) while the other ejects it from the solar system (e=1.064e = 1.064).
Limits. The turn angle is largest for the closest safe pass and the slowest approach. At Earth with v∞=9v_\infty = 9 km/s and 300 km altitude, δmax⁡≈50°\delta_{\max} \approx 50°; at v∞=5v_\infty = 5 km/s, 89.6°. Massive planets turn much harder: Jupiter can turn a 6 km/s v∞v_\infty through well over 100° from a safe distance.
💡Powered flybys
When the incoming and outgoing legs of a tour need different ∣v⃗∞∣|\vec v_\infty|, an engine burn is needed. The cheapest place is periapsis (Oberth again). The two legs become two half-hyperbolas sharing a periapsis rpr_p: δ=arcsin⁡11+rpv∞−2/μ+arcsin⁡11+rpv∞+2/μ\delta = \arcsin\frac{1}{1 + r_p v_\infty^{-2}/\mu} + \arcsin\frac{1}{1 + r_p v_\infty^{+2}/\mu} Solve for rpr_p, then Δv=∣v∞+2+2μ/rp−v∞−2+2μ/rp∣\Delta v = \left|\sqrt{v_\infty^{+2} + 2\mu/r_p} - \sqrt{v_\infty^{-2} + 2\mu/r_p}\right|. Raising v∞v_\infty from 9.0 to 9.4 km/s during a 40° turn at Earth costs 280 m/s at a 2,689 km periapsis, against 400 m/s far from the planet. Some turns are simply impossible above a safe altitude — the solver must say so.
The Tisserand parameter. In 1889 Félix Tisserand gave astronomers a way to recognise a comet after a close encounter with Jupiter had scrambled its orbit. With respect to a planet on a circular orbit of radius apa_p,
Tp=apa+2cos⁡i aap(1−e2)  ≈  const,v∞=vp3−TpT_p = \frac{a_p}{a} + 2\cos i\,\sqrt{\frac{a}{a_p}(1 - e^2)} \;\approx\; \text{const}, \qquad v_\infty = v_p\sqrt{3 - T_p}

T is (nearly) conserved through a flyby of that planet, because v∞ is. In the patched-conic model with a circular planetary orbit the relation to v∞ is exact.

💡What Tisserand tells a mission designer
A flyby moves the spacecraft along a curve of constant TT — it cannot jump to a different curve. So the question "can a flyby of Earth get me to Jupiter?" has an immediate answer: only if v∞v_\infty at Earth is at least 8.79 km/s (raising aphelion to 5.2 AU along Earth's velocity). From Venus it would take 11.39 km/s. A single Earth flyby with v∞=5v_\infty = 5 km/s can reach no further than 2.14 AU. Tisserand graphs (Strange & Longuski, 2002) plot these curves for several planets in periapsis–apoapsis space. Where curves of two planets cross, a flyby of one can lead to an encounter with the other — the whole space of possible tours, on one chart.
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Figure 10.7
in Jupiter's framespeed unchanged, direction turned|v∞| = 10|v∞| = 10 — unchangedδ = 79.2°in the Sun's framethe same turn, now an energy changev_Jupiter 13.07before 6.33 km/safter 18.66 km/sΔv_helio = +12.34 km/s, unpoweredrotating a fixed-length vector changes the length of the sum — the energy comes from Jupiter's orbit
The same flyby seen from the planet (an elastic bounce) and from the Sun (a change of speed) — the picture from the lesson on gravity assists and Lagrange points, now with numbers attached.
↓But there is a problem
A flyby rotates v⃗∞\vec v_\infty; Tisserand says which orbits it can reach; a powered flyby pays for any mismatch. Now chain them. Galileo left Earth in 1989 with far too little energy to reach Jupiter. It went inward to Venus, came back to Earth, went round the Sun once, came back to Earth again — and only then had the energy for Jupiter. Each leg is a Lambert arc; each encounter is a flyby that must join two of them. The lesson on multiple gravity-assist trajectory design puts this unit, and the unit on Lambert's problem, to work.