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Unit 8 · Lesson 8-3
Gravity Assists and Lagrange Points
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Gravity Assists and Lagrange Points

Flyby geometry, energy exchange in the Sun's frame, the Voyager Grand Tour and libration points

The Voyager 2 energy budget
20 August 1977. A Titan IIIE lifts 722 kilograms off Cape Canaveral with a departure energy barely sufficient to reach Jupiter. Twelve years later, that spacecraft flies past Neptune — 30 astronomical units out, on a trajectory that no rocket ever built could have provided directly. Along the way it has visited Jupiter, Saturn and Uranus, returning the first close images humanity has ever had of four planets and forty-eight moons. The energy for that journey did not come from the Titan. It did not come from Voyager's own thrusters, which carried barely 100 m/s of Δv for course corrections. It was taken from the planets. Each flyby transferred a small amount of orbital energy from a planet to the spacecraft. Jupiter, in slowing Voyager 2's departure, was itself slowed — by roughly 5×10−215\times10^{-21} metres per second, an amount so small it will never be measured. But Jupiter has 1.9×10271.9\times10^{27} kg and Voyager has 722. The exchange is not fair, and it was never meant to be. Here is the part that makes this genuinely hard to accept: in Jupiter's own reference frame, Voyager's speed was completely unchanged. It arrived at some speed relative to Jupiter and left at exactly the same speed. Not approximately. Exactly — by conservation of energy in the two-body problem, which you proved in Unit 1. So where did the energy come from?
Watch first · presented by Shoaib Aziz
The numbers in this film are the labs' own: Voyager 2 meets Jupiter at v∞ ≈ 10 km/s with closest approach 720,000 km, so e = 1.568 and the turn is 79.3°, worth at most 12.76 km/s around the Sun — against a propellant budget of about 100 m/s — while Jupiter slows by about 5 × 10⁻²¹ m/s; a pass at 3 Jupiter radii turns the spacecraft 132° and lifts its heliocentric speed from 9.71 to 20.53 km/s with v∞ unchanged to machine precision, and at 30 radii passing behind gains 7.3 km/s where passing in front loses 3.6; Sun–Earth L2 lies 1.5 million km out and Earth–Moon L1 58,000 km from the Moon, and L4/L5 are stable once the primary is about 25 times the secondary (Earth/Moon: 81); the simplified Grand Tour's aphelion grows from 5.24 AU to 28 AU at Jupiter and the orbit opens at Saturn.
The answer is that there are two frames, and the whole effect lives in the difference between them.
In the planet's frame, a flyby is a hyperbolic pass, exactly as in 8-1. The spacecraft comes in with v∞v_\infty, whips around periapsis, and leaves with v∞v_\infty. The energy is conserved because the two-body problem conserves it. All that changes is direction — the velocity vector is rotated by the turn angle δ\delta.
∣v⃗∞,in∣=∣v⃗∞,out∣,sin⁡δ2=1e=11+rpv∞2/μp|\vec{v}_{\infty,\text{in}}| = |\vec{v}_{\infty,\text{out}}|, \qquad \sin\frac{\delta}{2} = \frac{1}{e} = \frac{1}{1 + r_{p} v_\infty^2/\mu_p}

In the planet's frame: same speed in, same speed out, direction rotated by δ. Nothing has been gained. This is the whole of what happens locally.

But the planet's frame is itself moving, at 13.07 km/s in Jupiter's case, around the Sun. And the Sun's frame is the one that matters, because that is where the spacecraft's actual trajectory through the solar system is defined.
Convert between them by vector addition:
v⃗helio=v⃗planet+v⃗∞\vec{v}_{\text{helio}} = \vec{v}_{\text{planet}} + \vec{v}_\infty

The bridge between frames. Rotating v∞ leaves its magnitude alone — but the magnitude of the sum changes, and that sum is the heliocentric speed.

💡Why a flyby changes heliocentric energy
Here is the whole thing in one picture. Draw v⃗planet\vec{v}_{\text{planet}} as a fixed arrow. From its tip, draw v⃗∞\vec{v}_\infty — a second arrow of fixed length but free direction. The heliocentric velocity is the arrow from the origin to the tip of the second one. Now rotate v⃗∞\vec{v}_\infty. Its tip sweeps a circle of radius ∣v⃗∞∣|\vec{v}_\infty| centred on the tip of v⃗planet\vec{v}_{\text{planet}}. The length of the resultant changes. Rotate v⃗∞\vec{v}_\infty to point along the planet's motion and the heliocentric speed is vp+v∞v_p + v_\infty. Rotate it to oppose the planet and you get vp−v∞v_p - v_\infty. So a manoeuvre that changes nothing in one frame changes everything in the other. The maximum possible gain is: Δvmax⁡=2v∞sin⁡δ2\Delta v_{\max} = 2v_\infty\sin\frac{\delta}{2} which for a full 180° reversal would be 2v∞2v_\infty. And the bookkeeping closes. The energy comes from the planet's orbital motion. Jupiter loses exactly as much as the spacecraft gains — momentum is conserved between the two of them — but with a mass ratio of 2.6×10242.6\times10^{24} to one, Jupiter's share is unmeasurable and the spacecraft's is transformative. A gravity assist is not free energy. It is a loan from a planet, on terms so lopsided that the planet will never notice.
Figure 8.7
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 two-panel picture that explains everything. Left: the planet's frame — a symmetric hyperbola, equal speeds in and out, direction rotated by δ. Right: the Sun's frame — the same rotation, now added to the planet's velocity, producing a heliocentric speed that is visibly longer than it was. Same event, two frames, and the energy appears only in the second one.
⚠Trailing edge speeds you up, leading edge slows you down
Which way you gain depends on which side of the planet you pass. Pass behind the planet — through the space it is moving away from — and the planet's gravity pulls you forward along its direction of travel. You gain heliocentric energy. Pass in front and you are pulled backwards relative to the planet's motion. You lose energy. Both are used constantly, and losing energy is often the harder problem. Reaching Mercury requires shedding an enormous amount of the orbital energy Earth gave you for free, and the only affordable way to do that is repeated retrograde flybys. MESSENGER, to Mercury: one Earth flyby, two Venus flybys, three Mercury flybys. Six and a half years, 4.9 billion miles, to reach a planet 0.4 AU away. BepiColombo, currently en route: one Earth, two Venus, six Mercury flybys. Seven years. Getting to the inner solar system is harder than getting to the outer one, because you arrive with far too much energy and must give it away.
📐Worked Example: Voyager 2 at Jupiter
Voyager 2 arrives at Jupiter with v∞≈10v_\infty \approx 10 km/s relative to the planet. Jupiter orbits the Sun at 13.07 km/s. Flyby periapsis is around 720,000 km from Jupiter's centre. μJ=1.267×108\mu_J = 1.267\times10^8 km³/s². Eccentricity of the flyby hyperbola. e=1+rpv∞2/μJ=1+(720000)(100)/(1.267×108)=1+0.568=1.568e = 1 + r_p v_\infty^2/\mu_J = 1 + (720000)(100)/(1.267\times10^8) = 1 + 0.568 = 1.568. Turn angle. δ=2arcsin⁡(1/1.568)=2arcsin⁡(0.6378)=2(39.6°)=79.3°\delta = 2\arcsin(1/1.568) = 2\arcsin(0.6378) = 2(39.6°) = \mathbf{79.3°}. Jupiter turned the spacecraft's velocity by nearly eighty degrees without a single gram of propellant. Maximum heliocentric gain. Δv=2v∞sin⁡(δ/2)=2(10)(0.6378)=12.76\Delta v = 2v_\infty\sin(\delta/2) = 2(10)(0.6378) = \mathbf{12.76} km/s. Put that in context. Voyager's entire onboard propellant budget was about 100 m/s. Jupiter provided over a hundred times the spacecraft's total propulsive capability, in a single unpowered pass lasting a few hours. And what Jupiter paid. Momentum conservation: ΔvJ=−mscmJΔvsc=−7221.898×1027×12760≈−5×10−21\Delta v_J = -\frac{m_{sc}}{m_J}\Delta v_{sc} = -\frac{722}{1.898\times10^{27}} \times 12760 \approx -5\times10^{-21} m/s. To slow Jupiter by one millimetre per second would take about 2×10172\times10^{17} flybys like Voyager's. The loan will not be called in.
Figure 8.8
Jupiter1979Saturn1981Uranus1986Neptune1989Earth 1977Voyager 2, 1977–1989Launched with barely enoughenergy to reach Jupiter.Each flyby bent the path andstretched the orbit furtherout. The arcs between areplain conics.The 176-year alignment wasspotted by a summer studentin 1965. Next one: 2153.radial scale compressed
The Voyager 2 trajectory, 1977 to 1989. Each planetary encounter bends the path and stretches the orbit outward. Note that the arcs between planets are simple conics — the entire Grand Tour is patched conics from 8-1, strung together, with the flybys doing all the work.
🕰Minovitch, Flandro and the 1977 alignment
In 1965, Gary Flandro was a graduate student on a summer assignment at JPL, asked to look into missions to the outer planets. Plotting the positions of Jupiter, Saturn, Uranus and Neptune forward in time, he noticed that in the late 1970s all four would be arranged so that a single spacecraft could chain flybys from one to the next. A direct mission to Neptune would take thirty years. Using the alignment: twelve. The alignment recurs every 176 years. The previous one was around 1801, when Jefferson was president and Neptune had not been discovered. The next is around 2153. NASA proposed a four-spacecraft "Grand Tour" and Congress cancelled it as too expensive. JPL rebuilt the proposal as a modest two-spacecraft mission to Jupiter and Saturn only — and quietly designed Voyager 2's hardware and trajectory so that if it survived Saturn, it could continue. It survived. The Uranus and Neptune encounters were approved after launch, on a spacecraft already in flight, using an alignment discovered by a summer student a decade earlier. Both Voyagers are still transmitting from interstellar space, forty-eight years on, powered by plutonium and running on 70 kilobytes of memory.
Look again at that dot. That's here. That's home. That's us.
– Carl Sagan, on the image Voyager 1 took of Earth from 6 billion kilometres, Pale Blue Dot, 1994
One last piece, and it points at where the field goes next.
Almost everything in this course has been two bodies. (Lesson 4-5 broke the rule once, in the frame that turns with the Moon, and found two of the points below.) Add a third and the problem becomes unsolvable — but the restricted three-body problem, where the third mass is negligible, has structure worth knowing. In particular it has five points where a small object can hold station relative to two large ones:
Figure 8.9
SunEarthL1L2L3L4L5L1/L2 offset exaggeratedfive equilibriaL1, L2, L3 — saddlesunstable; need station-keepingL1: SOHO, DSCOVR, ACEL2: JWST, Gaia, EuclidL4, L5 — genuinely stableif M₁/M₂ > 24.96Jupiter holds over a millionTrojan asteroids there.Both sit 1.5 million km out —1% of the way to the Sun, sothe gap here is not to scale.
The five Lagrange points of the Earth–Sun system. L1, L2 and L3 lie on the line through both bodies and are unstable — a saddle, requiring active station-keeping. L4 and L5 lead and trail by 60° and are genuinely stable, which is why Jupiter has accumulated over a million Trojan asteroids at its own L4 and L5.
PointLocationStabilityOccupants
L11.5 million km sunward of EarthUnstable (saddle)SOHO, DSCOVR, ACE — continuous solar view
L21.5 million km anti-sunwardUnstable (saddle)JWST, Gaia, Euclid, Planck — Sun, Earth and Moon all behind you
L3Opposite side of the SunUnstableNothing; permanently hidden from Earth
L4 / L560° ahead and behind in Earth's orbitStable if M₁/M₂ > 24.96Dust; Jupiter's Trojans; a captured Earth Trojan, 2010 TK₇
🛰JWST does not orbit L2
The James Webb Space Telescope is described as sitting at L2. It does not. L2 is an unstable equilibrium — a saddle point. An object placed exactly there falls away within weeks. Worse, sitting exactly at L2 would put the telescope in Earth's shadow permanently, with no solar power at all. So JWST flies a halo orbit: a large, roughly elliptical loop around the empty point, about 800,000 km across, with a period of six months. The orbit exists only because of the combined Earth–Sun gravity field, and it is a solution that has no counterpart in two-body mechanics at all. It is unstable, so JWST performs a station-keeping burn every three weeks — and only ever outward, away from the Sun. A burn in the wrong direction could push the telescope onto the unstable manifold toward the Sun, and the sunshield would be lost. The design deliberately makes an over-burn recoverable and an under-burn merely inefficient. The propellant loaded at launch is what sets the mission lifetime. Ariane 5's insertion was so accurate that JWST saved enough correction propellant to roughly double its expected life, from ten years to more than twenty.
Figure 8.10
from the sideL2Sun, Earth, Moonall behind the shieldfrom the SunL2JWST does not sit at L2 — L2 is an unstable saddle, and sitting there would put thetelescope in permanent Earth shadow with no power. It flies a ~800 000 km loop aroundthe empty point, with a six-month period, and burns every three weeks — always outward,because a burn the wrong way could push it onto the manifold toward the Sun.
JWST's halo orbit about Earth–Sun L2, viewed from two angles. The telescope never enters Earth's shadow, keeps its solar arrays permanently illuminated, and keeps Sun, Earth and Moon all on the same side behind the sunshield — a geometry that only exists in a three-body gravity field.
And beyond halo orbits lies the current frontier. The unstable manifolds of these orbits — the surfaces along which objects naturally fall toward and away from them — connect. They form a network of low-energy pathways threading the solar system: the Interplanetary Transport Network.
Travel along it is nearly free in Δv terms, and extraordinarily slow. Japan's Hiten used a low-energy route to reach lunar orbit in 1991 with propellant that should not have been enough. ESA's SMART-1 combined it with ion propulsion. It is how comets are captured and released by Jupiter, and how material moves between the moons of the giant planets.
It is also completely outside the two-body framework, and there is no closed-form theory for it at all.
↓But there is a problem
Every result in this unit came with an asterisk. The interplanetary Hohmann transfer assumed circular, coplanar planetary orbits and a launch on exactly the right day. The porkchop plot was a picture of solutions we had not yet learned to compute. The gravity assist was an elegant argument about frames, not a calculation you could hand to a navigator. The next unit removes the first asterisk. Given any two positions and any flight time, it finds the orbit that connects them — the problem Lambert stated in 1761 and Gauss used to find a lost asteroid. With it, a porkchop plot becomes something you compute, and a launch period something you design. The unit after that removes the rest: it puts the planets back at both ends of the arc, aims at them, and chains them into tours.