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Unit 10 · Lesson 10-2
Departure: From Parking Orbit to Escape Asymptote
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Departure: From Parking Orbit to Escape Asymptote

The injection burn, the Oberth effect, hyperbola geometry, DLA and RLA, parking-orbit planes and launch azimuth

New Horizons, 19 January 2006
An Atlas V with five solid boosters and an extra solid-rocket third stage sends a 478 kg spacecraft away from Earth faster than anything before it: about 16 km/s relative to Earth after burnout, a C₃ of roughly 158 km²/s². It crosses the Moon's orbit in nine hours. To reach that speed from a standstill would be hopeless. The launch vehicle does not do that. It first places the upper stages in a low parking orbit, already moving at 7.8 km/s, then burns there — deep in Earth's gravity well, where every metre per second of Δv buys the most energy. Where in that orbit to burn, and in which direction, is fixed entirely by one vector: the hyperbolic excess velocity the heliocentric trajectory requires.
Watch first · presented by Shoaib Aziz
The numbers in this film are the labs' own: from a 200 km parking orbit the single periapsis burn costs 3.611 km/s for Mars against 6.169 for escape-then-boost (a saving of 2.56), and 6.305 against 12.017 for Jupiter (5.71); the Mars hyperbola has e = 1.1431, a 122° turn and its asymptote 151° from periapsis, with 11.396 km/s after the burn against a circular 7.784. For Mars 2020 (arrival fixed on 18 February 2021), 17 July gives C₃ 14.39 km²/s² and DLA +34.7°, two planes at i = 40° and a launch azimuth of 60.7° from the Cape; 30 July DLA +23.3°; 15 August C₃ 20.53. The Jupiter injection falls from 6.305 km/s at 200 km to 6.221 at 5,000 km, then rises to 6.735 from GEO; no 30° orbit can contain a 34.7° asymptote; the 3-D departure is within 0.091° of its asymptote after a day.
The injection burn. From a circular parking orbit of radius rpr_p, a single tangential burn at what becomes the periapsis of the departure hyperbola needs
Δv=v∞2+2μrp−μrp\Delta v = \sqrt{v_\infty^2 + \frac{2\mu}{r_p}} - \sqrt{\frac{\mu}{r_p}}

Energy on the hyperbola: v_p²/2 − μ/r_p = v∞²/2. The first term is the periapsis speed, the second the circular speed already in hand.

💡The Oberth effect, priced
Compare the single periapsis burn with a naive two-step plan: first escape exactly (to a parabola), then add v∞v_\infty far away. From 200 km:
  • Mars (v∞=2.945v_\infty = 2.945 km/s): 3.611 km/s versus 6.169 — saving 2.56 km/s
  • Jupiter (v∞=8.793v_\infty = 8.793 km/s): 6.305 versus 12.017 — saving 5.71 km/s
The saving grows with v∞v_\infty: squaring vesc+δvv_{esc} + \delta v produces a cross term 2vescδv2v_{esc}\delta v that is only available where the spacecraft is already moving fast. This is why departures happen from low orbit, and why the Oberth effect, met in the lesson on bi-elliptic transfers, is not a curiosity but the central fact of interplanetary launch.
Which parking orbit? For fixed v∞v_\infty, Δv(rp)\Delta v(r_p) has a minimum at rp=2μ/v∞2r_p = 2\mu/v_\infty^2: about 92,000 km for a Mars departure but only 10,300 km (radius) for Jupiter. Higher is not always better: for Jupiter the cost falls from 6.305 km/s at 200 km to 6.221 at 5,000 km, then rises again (6.735 from GEO). In practice the parking orbit is set by the launch vehicle — low, because getting into a high orbit costs more than the injection saves.
Hyperbola geometry. The eccentricity, turn angle and asymptote angle of the departure hyperbola are
e=1+rpv∞2μ,sin⁡δ2=1e,θ∞=arccos⁡(−1e),β=arccos⁡1e=π−θ∞e = 1 + \frac{r_p v_\infty^2}{\mu},\qquad \sin\frac{\delta}{2} = \frac{1}{e},\qquad \theta_\infty = \arccos\left(-\frac{1}{e}\right),\qquad \beta = \arccos\frac{1}{e} = \pi - \theta_\infty

θ∞ is the true anomaly of the asymptote, measured from periapsis. β locates the burn: periapsis lies β from the −v∞ direction, on the far side of the planet from where the spacecraft is going.

Loading interactive…
📐Worked Example: a Mars departure from 200 km
v∞=2.945v_\infty = 2.945 km/s, rp=6578r_p = 6578 km. e=1+6578×2.9452/398600=1.1431e = 1 + 6578 \times 2.945^2/398600 = 1.1431. Turn angle δ=2arcsin⁡(1/e)=122.0°\delta = 2\arcsin(1/e) = 122.0°. Asymptote at θ∞=151.0°\theta_\infty = 151.0° from periapsis, so β=29.0°\beta = 29.0°. Periapsis speed 2.9452+2μ/rp=11.396\sqrt{2.945^2 + 2\mu/r_p} = 11.396 km/s; circular speed 7.784 km/s; Δv=3.611\Delta v = 3.611 km/s. The burn happens 151° before the point on the hyperbola where the spacecraft is heading along v⃗∞\vec v_\infty — on the side of Earth facing away from the departure direction.
In three dimensions. v⃗∞\vec v_\infty fixes a direction S^\hat S in space. The departure hyperbola may lie in any plane containing S^\hat S: a one-parameter family of planes rotating about the asymptote. The parking orbit must be one of them. In Earth's equatorial frame, write the asymptote as DLA (declination) and RLA (right ascension). A plane of inclination ii and node Ω\Omega has normal h^=(sin⁡isin⁡Ω, −sin⁡icos⁡Ω, cos⁡i)\hat h = (\sin i\sin\Omega,\ -\sin i\cos\Omega,\ \cos i) and contains S^\hat S when h^⋅S^=0\hat h\cdot\hat S = 0:
sin⁡i (Sxsin⁡Ω−Sycos⁡Ω)=−cos⁡i Sz\sin i\,(S_x\sin\Omega - S_y\cos\Omega) = -\cos i\,S_z

Solvable only if |tan(DLA)| ≤ tan i, i.e. i ≥ |DLA|; then there are two nodes — the asymptote is reached on an ascending or a descending pass.

💡From the asymptote to the launch pad
  • Inclination. i≥∣DLA∣i \ge |\text{DLA}| (plane must contain the asymptote) and i≥ϕi \ge \phi (a launch site at latitude ϕ\phi cannot reach a lower inclination directly).
  • Azimuth. For a direct ascent into inclination ii, sin⁡Az=cos⁡i/cos⁡ϕ\sin A_z = \cos i/\cos\phi. From Cape Canaveral (28.5°N), i=40°i = 40° needs Az=60.7°A_z = 60.7°; i=55°i = 55° needs 40.7°40.7°. Range-safety corridors limit the usable azimuths (roughly 35°–120° on the Eastern Range).
  • Time of day. The plane is fixed in inertial space; the launch site rotates with Earth. It passes under the plane twice a day — the two daily launch opportunities. A coast in the parking orbit before the injection burn lets the rocket launch over a window of minutes rather than an instant.
📐Worked Example: Mars 2020 across its launch period
With arrival fixed on 18 February 2021 (as in the lesson on launch windows and porkchop plots):
  • 17 July: C3=14.39C_3 = 14.39, DLA =+34.7°= +34.7°. Minimum parking inclination 34.7° (DLA dominates). For i=40°i = 40° the two possible planes have nodes at 250.2° and 318.8°.
  • 30 July: C3=14.46C_3 = 14.46, DLA =+23.3°= +23.3°. Now the site latitude dominates: i≥28.5°i \ge 28.5°.
  • 15 August: C3=20.53C_3 = 20.53, DLA =+13.8°= +13.8°.
Early in the period the asymptote points high into the northern sky; late in the period C₃ climbs. Each constraint closes one end.
Placing the burn. Given S^\hat S and the chosen plane normal h^\hat h, the periapsis direction is S^\hat S rotated about h^\hat h by −θ∞-\theta_\infty, and the post-burn velocity is v∞2+2μ/rp\sqrt{v_\infty^2 + 2\mu/r_p} along h^×r^p\hat h\times\hat r_p. In the third lab you construct this state for Mars 2020 and fly it outward: a day later the velocity is within 0.09° of S^\hat S; at 1,000 hours, indistinguishable from it.
🛰Upper stages that restart
The two-burn profile — burn to a parking orbit, coast, burn again to escape — needs an upper stage that can restart in space after a coast. Centaur, with its liquid hydrogen and oxygen, made it routine for American planetary missions from the 1960s. The coast also solves the time-of-day problem: by varying its length, the injection point can be placed where the hyperbola needs it, whatever the exact minute of liftoff.
↓But there is a problem
Departure has generous freedom: any plane through the asymptote, either node, a window of minutes every day. Arrival has none. The spacecraft must come in on a hyperbola with exactly the right periapsis altitude — too high and the capture burn is wasted, too low and it meets the atmosphere — and in exactly the right plane for the science orbit or the landing site. And it must be aimed months in advance, from hundreds of millions of kilometres away. Mission designers aim at a plane that does not exist.