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Unit 8 · Lesson 8-1
Hyperbolic Trajectories and Escape
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Hyperbolic Trajectories and Escape

Hyperbolic excess speed, C₃, the sphere of influence and an introduction to patched conics

What escape velocity means
"Escape velocity is 11.2 kilometres per second." Everything about that sentence is misleading, and unpicking it is the first job of this unit. It is not a velocity. It is a speed — direction is irrelevant. Fire straight up, fire sideways, fire at 37° to the horizon: if you have escape speed and you do not hit anything, you leave. The direction changes where you go, not whether you go. It is not a limit. Nothing enforces it. A rocket climbing at 200 m/s can reach the Moon, given enough propellant and enough patience. Escape speed is what you need to leave without further thrust, coasting the rest of the way — it is an energy threshold, not a barrier. It depends entirely on where you are. 11.2 km/s is escape speed from Earth's surface. From the ISS it is 10.85. From geostationary altitude it is 4.35. From the Moon's distance it is 1.44. And escaping to infinity is not a thing that happens. There is no infinity in a solar system with a Sun in it. Leave Earth and you do not become free — you become an object in solar orbit, at 1 astronomical unit, moving at very nearly Earth's own 29.8 km/s. You did not escape. You changed owner. This lesson is about what leaving actually means, what it costs, and about the approximation that made the whole enterprise computable in an era when a computer was a room full of women with mechanical calculators.
Watch first · presented by Shoaib Aziz
The numbers in this film are the labs' own: escape speed 11.19 km/s from the surface, 10.85 km/s from the space station's 400 km and 4.35 km/s from geostationary altitude; from a 400 km parking orbit (7.67 km/s), bare escape costs 3.18 km/s against 3.86 km/s to reach GEO, and a Mars departure with v∞ = 2.95 km/s (C₃ = 8.7 km²/s²) costs 3.57 km/s — only 0.39 more; 1 km/s of excess costs 0.046 km/s extra and 3 km/s costs 0.41; Earth's sphere of influence reaches 924,648 km (145 Earth radii) and Neptune's 87 million km, yet at its edge the Sun still pulls about 13 times harder than the Earth; and a flyby 300 km above the Earth turns a 1 km/s hyperbola through 159° but a 12 km/s one through only 34°.
Unit 2 derived the four conic families, Unit 1 gave the energy criterion, and Lesson 5-2 gave the clock along a hyperbola. Escape is simply ε≥0\varepsilon \geq 0:
ε=v22−μr≥0⟺v≥vesc=2μr=2 vcirc\varepsilon = \frac{v^2}{2} - \frac{\mu}{r} \geq 0 \quad\Longleftrightarrow\quad v \geq v_{esc} = \sqrt{\frac{2\mu}{r}} = \sqrt{2}\,v_{circ}

Escape speed is exactly √2 times circular speed at the same radius — a 41.4% increase, which through the rocket equation is a substantial but not impossible price.

If you have exactly escape speed, ε=0\varepsilon = 0 and the trajectory is parabolic: you coast to infinity and arrive with nothing left. Useless, because you arrive nowhere in infinite time.
Anything more and ε>0\varepsilon > 0, the trajectory is hyperbolic, and you arrive with speed to spare. That leftover speed is the number interplanetary mission design is actually built around:
v∞=2ε=v2−2μr,C3=v∞2=−μav_\infty = \sqrt{2\varepsilon} = \sqrt{v^2 - \frac{2\mu}{r}}, \qquad C_{3} = v_\infty^2 = -\frac{\mu}{a}

Hyperbolic excess speed, and the characteristic energy C₃. Note a is negative for a hyperbola, so C₃ comes out positive — the sign convention from Unit 2 earning its keep.

💡C₃ is the currency of interplanetary launch
Launch vehicles are not sold by payload mass alone. They are sold by payload mass as a function of C3C_3, and every launch provider publishes that curve. C3=0C_3 = 0 is exactly parabolic escape. C3≈9C_3 \approx 9 km²/s² reaches Mars on a minimum-energy transfer. C3≈80C_3 \approx 80 km²/s² reaches Jupiter directly. C3≈160C_3 \approx 160 km²/s² was New Horizons' launch energy — the fastest departure ever achieved, leaving Earth at 16.26 km/s and passing the Moon's orbit in nine hours instead of three days. And the curve is steep. An Atlas V 551 can put roughly 6,000 kg at C3=0C_3 = 0, about 2,000 kg at C3=30C_3 = 30, and a few hundred kilograms at C3=100C_3 = 100. Every unit of C3C_3 you can avoid needing is payload you get to keep, which is why the entire discipline of gravity assists exists. Worth noting: C3C_3 has units of energy per unit mass, but is quoted as a squared speed. When a mission designer says "C3 of twelve", they mean the spacecraft arrives at the edge of Earth's influence doing 12=3.46\sqrt{12} = 3.46 km/s relative to Earth.
Figure 8.1
v∞ inv∞ outperiapsis r_pe = 1.7 — an escape trajectoryasymptote at ν∞ = ±126.0°turn δ = 72.1°sin(δ/2) = 1/ea is negative for a hyperbola,which makes ε = −μ/2a positive.That sign is doing real work —do not take its absolute value.A slower, closer pass turnsmore sharply. VIII-3 exploits it.
A hyperbolic trajectory and its asymptotes. The spacecraft comes in along one asymptote, whips around perigee, and leaves along the other. The angle between them is the turn angle δ, and the whole of the next two lessons depends on it. Note that a is negative and measured to the *other* branch of the hyperbola — a piece of bookkeeping, not geometry you can point at.
e=1+rpv∞2μ,sin⁡δ2=1ee = 1 + \frac{r_p v_\infty^2}{\mu}, \qquad \sin\frac{\delta}{2} = \frac{1}{e}

Eccentricity and turn angle of a hyperbolic pass. A slow, close pass turns sharply; a fast, distant one barely bends. Lesson 8-3 turns this single relation into a propulsion system.

📐Worked Example: what leaving Earth costs
From a 400 km circular parking orbit, r=6771r = 6771 km, vc=7.673v_c = 7.673 km/s. Escape speed here. vesc=2×398600.44/6771=117.74=10.851v_{esc} = \sqrt{2 \times 398600.44/6771} = \sqrt{117.74} = 10.851 km/s. Cost of bare escape. Δv=10.851−7.673=3.178\Delta v = 10.851 - 7.673 = \mathbf{3.178} km/s. Compare with 3.86 km/s to reach geostationary orbit — escaping Earth entirely is cheaper than moving to GEO. That is not a trick; it is the shape of the 1/r1/r potential well, which is steep near the bottom and very flat further out. Now a real departure. A Mars transfer needs C3=8.7C_3 = 8.7 km²/s², so v∞=2.95v_\infty = 2.95 km/s. Required perigee speed: vp=v∞2+2μ/rp=8.7+117.74=126.44=11.245v_p = \sqrt{v_\infty^2 + 2\mu/r_p} = \sqrt{8.7 + 117.74} = \sqrt{126.44} = 11.245 km/s. Δv=11.245−7.673=3.572\Delta v = 11.245 - 7.673 = \mathbf{3.572} km/s. Read those two numbers side by side. Bare escape costs 3.178 km/s. Escape plus 2.95 km/s of leftover speed costs 3.572 — only 0.39 km/s more, for nearly 3 km/s of hyperbolic excess. That is the Oberth effect from Unit 6, and at escape energies it is enormous. Burning deep in the well converts propellant into useful energy with a leverage of nearly eight to one. It is also exactly why every interplanetary mission departs from the lowest practical parking orbit rather than climbing first.
⚠Leaving Earth's sphere of influence
So you have escaped Earth with v∞=2.95v_\infty = 2.95 km/s. What are you now doing? Earth orbits the Sun at 29.78 km/s. Add your excess along Earth's direction of travel and you are in a heliocentric orbit at 32.73 km/s, at a radius of 1 AU. Run that through vis-viva around the Sun and it is an ellipse with aphelion at Mars. You never became free. You spent 3.57 km/s to change which body you orbit, and the new one is a great deal bigger. This is why "escape velocity" is such a misleading phrase in a solar-system context. There is a speed that escapes the Sun too — about 42.1 km/s at Earth's distance, so 16.6 km/s on top of Earth's orbital motion — and only five spacecraft have ever had it. Voyager 1 and 2, Pioneer 10 and 11, and New Horizons. Every one of them got it by stealing from Jupiter, which is Lesson 8-3.
Now the hard part, and the approximation that made interplanetary flight possible before computers were up to the real problem.
A spacecraft between Earth and Mars is being pulled by the Sun, by Earth, by Mars, and by everything else. That is a four-body problem at minimum, and there is no analytic solution to anything past two.
So do something outrageous: pretend only one body exists at a time.
rSOI=aplanet(mplanetm⊙)2/5r_{SOI} = a_{\text{planet}}\left(\frac{m_{\text{planet}}}{m_{\odot}}\right)^{2/5}

The sphere of influence, from Laplace. Inside it, treat the planet as the only gravitating body; outside it, the Sun. Earth's is about 924,000 km — around 145 Earth radii, and 2.4 times the distance to the Moon.

Figure 8.2
10⁰10¹10⁵10⁶10⁷distance from the Sun (AU)sphere of influence (km)MercuryVenusEarthMarsJupiterSaturnUranusNeptuner = a (m/M☉)^(2/5)Earth: 925×10³ km145 Earth radii2.4× the lunar distanceGravity does not stop at aboundary. This is a modellingradius from Laplace, 1799 —where the planet's pull*relative to the Sun's* dropsbelow a chosen threshold.
Spheres of influence across the solar system, on a logarithmic scale. Jupiter's reaches 48 million km and Neptune's, far from the Sun, 87 million; Mercury's only 112,000. The SOI is not a physical surface — gravity does not stop at a boundary — it is the radius at which the planet's pull *relative to the Sun's* drops below a chosen threshold.
💡Patched conics
The method, in three steps: 1. Departure. Inside Earth's SOI, ignore the Sun. The trajectory is a hyperbola about Earth, leaving with v∞v_\infty. 2. Cruise. Outside every SOI, ignore the planets. The trajectory is an ellipse about the Sun. At the boundary, the spacecraft's heliocentric velocity is Earth's velocity plus v⃗∞\vec{v}_\infty — a straight vector addition. 3. Arrival. Inside Mars's SOI, ignore the Sun again. The trajectory is a hyperbola about Mars, arriving with an excess speed set by the mismatch between the transfer ellipse and Mars's own orbit. Three two-body problems, glued at the seams. Every one of them has a closed-form solution from Units 2 through 5, so the entire interplanetary trajectory can be computed by hand. And it is wrong. The SOI is a fiction, the transitions are discontinuous, and both bodies pull at once near the boundary. Errors are typically thousands of kilometres at Mars. But thousands of kilometres out of 200 million is good enough to design a mission, and once you have a design you refine it numerically. Every interplanetary trajectory ever flown, from Mariner to Perseverance, started life as a patched-conic sketch.
Figure 8.3
Sundeparture hyperbolaarrival hyperbolaheliocentric ellipsethree two-body problems, glued at the seams — and every one has a closed-form solutionat each seam the velocities simply add: v_helio = v_planet + v∞
The three-piece trajectory. An Earth-centred hyperbola, a Sun-centred ellipse, and a Mars-centred hyperbola, joined at the SOI boundaries. At each seam the velocity is a vector sum: heliocentric velocity equals planet velocity plus hyperbolic excess.
🕰The human computers of JPL
Before the mid-1960s, patched-conic trajectory design was done by hand, and "computer" was a job title. At JPL, a group of women — Barbara Paulson, Sue Finley, Helen Ling and others — computed the trajectories for Explorer 1, the Ranger and Surveyor lunar probes, and the Mariner missions to Venus and Mars, using Friden mechanical calculators and graph paper. Sue Finley was hired in 1958 and was still working at JPL in 2019, making her the longest-serving woman in NASA's history. At Langley, Katherine Johnson computed the trajectory for Alan Shepard's flight and verified the orbital equations for John Glenn's — Glenn specifically asked for her to check the electronic computer's output before he would fly. The reason the patched-conic method matters historically is precisely that it is tractable by hand. A method that required numerical integration of a four-body problem would have made interplanetary flight impossible until the 1970s. Laplace's approximation from 1799 is what put spacecraft at Mars in 1965.
↓But there is a problem
So we can leave. A 3.57 km/s burn from low Earth orbit and the spacecraft crosses the sphere of influence with 2.95 km/s to spare, joins the Sun's family, and coasts outward on an ellipse whose aphelion touches the orbit of Mars. Touches the orbit of Mars. Mars itself is not a place. It is a 6,779 km rock moving at 24.1 kilometres per second, and it will be at the aphelion point for roughly four minutes out of every 687 days. So the transfer is not the problem. The transfer was solved in Unit 6. The problem is arranging for the planet to be there when you arrive — which means departing at exactly the right moment, when Earth and Mars hold precisely the right angle between them. That angle recurs every 26 months. Miss it, and there is nothing you can do: no amount of propellant fixes a phase error of that magnitude. You wait two years and two months for the geometry to come back. Every Mars mission ever flown left within a few weeks of every other one. There is a reason for that, and it is drawn on a chart that looks like a piece of meat.