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Unit 5 · Lesson 5-2
Hyperbolic Orbits and Universal Variables
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Hyperbolic Orbits and Universal Variables

The hyperbolic and parabolic Kepler equations, the universal anomaly, Stumpff functions and the f and g functions

Haleakalā, Hawaii, 2017
On 19 October 2017 the Pan-STARRS survey telescope on Haleakalā, Maui, recorded a faint object moving unusually fast against the stars. Robert Weryk spotted it the next morning in the night's detections. It was catalogued as a comet, then as an asteroid. Neither label was the remarkable part. The remarkable part arrived when the observations were fitted with an orbit: the eccentricity came out at about 1.2. Not 0.99, the long thin ellipse of a comet falling in from the edge of the solar system. Greater than one. A hyperbola. The object was moving faster than the Sun's escape speed. It had come from another star, and it was leaving for good. It was named ʻOumuamua, roughly a messenger from afar, arriving first in Hawaiian, and it was the first interstellar object ever seen. The orbit software that week was handed positions and times, not an orbit. It could not assume the answer would be an ellipse. Lesson 5-1's Kepler equation could not have handled it: M=E−esin⁡EM = E - e\sin E is an equation for ellipses only. This lesson builds the version that never needs to ask.
Watch first · presented by Shoaib Aziz
The numbers in this film are the labs' own: a Mars departure from a 300 km parking orbit with 3 km/s to spare (e = 1.1508) is 25,695 km from Earth's centre after 1 hour and 338,000 km after 24 hours, turning towards its 150.3° asymptote; the universal propagator reproduces the same 25,695 km, its check f·ġ − ḟ·g comes out at 1 to 12 decimal places, and near e = 1 the classical method drifts by 31 m, then 2 km, then 22 km before failing outright.
Lesson 5-1 solved the time problem for one conic. Energy decides which conic you are on:
ε=v22−μr=−μ2a\varepsilon = \frac{v^2}{2} - \frac{\mu}{r} = -\frac{\mu}{2a}
Negative energy is an ellipse with a>0a > 0. Zero is a parabola with a=∞a = \infty. Positive energy is a hyperbola, and the same formula then gives a negative semi-major axis. Every spacecraft that leaves Earth for another planet is on the third kind, at least for a while. So is every interstellar visitor, and every gravity-assist flyby seen from the planet being passed.
The hyperbola. The eccentric anomaly came from a circle drawn around the ellipse. A hyperbola has no circle around it, but it has the same construction with hyperbolic functions in place of circular ones. The hyperbolic anomaly FF plays the role of EE:
r=a(1−ecosh⁡F),tanh⁡F2=e−1e+1 tan⁡ν2,  Mh=esinh⁡F−F=μ(−a)3 (t−tp)  r = a(1 - e\cosh F), \qquad \tanh\frac{F}{2} = \sqrt{\frac{e-1}{e+1}}\,\tan\frac{\nu}{2}, \qquad \boxed{\;M_h = e\sinh F - F = \sqrt{\frac{\mu}{(-a)^3}}\,(t - t_p)\;}

The hyperbolic Kepler equation. Same structure as the elliptic one, same Newton iteration, and just as unsolvable in closed form. With a < 0, r comes out positive. M_h grows without limit: there is no period, because the object never comes back.

Newton's method carries straight over: Fk+1=Fk−(esinh⁡Fk−Fk−Mh)/(ecosh⁡Fk−1)F_{k+1} = F_k - (e\sinh F_k - F_k - M_h)/(e\cosh F_k - 1). One difference matters in practice. sinh⁡\sinh grows exponentially, so the starting guess F0=MhF_0 = M_h that worked for ellipses can overshoot wildly for large MhM_h. A safe start is F0=arcsinh⁡(Mh/e)F_0 = \operatorname{arcsinh}(M_h/e).
The parabola. Exactly at e=1e = 1 the time equation turns into something unexpected: a cubic. With D=tan⁡(ν/2)D = \tan(\nu/2) and p=2rpp = 2r_p:
t−tp=12p3μ(D+D33)t - t_p = \frac{1}{2}\sqrt{\frac{p^3}{\mu}}\left(D + \frac{D^3}{3}\right)

Barker's equation. Cubics have a closed-form root, so the parabola is the one conic where Kepler's problem has an exact formula and needs no iteration at all.

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⚠Three equations means two cliffs
A program built from the three classical equations has to decide which one to use, and it decides by computing ee or aa from noisy data and comparing against 1. Near e=1e = 1 both outer forms fall apart. As e→1e \to 1 from below, a→∞a \to \infty, the mean motion n→0n \to 0, and M=E−esin⁡EM = E - e\sin E becomes the tiny difference of two nearly equal numbers, so it loses digits. The hyperbolic form does the same from above. Exactly at e=1e = 1 both divide by zero. This is not a theoretical corner. Long-period comets have e=0.999e = 0.999 and beyond. A spacecraft leaving Earth at almost exactly escape speed sits on it. The third lab shows the classical route drifting by kilometres while the universal route stays exact.
The universal anomaly. The repair is to stop asking. Define one new variable χ\chi, the universal anomaly, through how it advances in time:
dχdt=μr\frac{d\chi}{dt} = \frac{\sqrt{\mu}}{r}
It is a clock that runs fast near periapsis and slowly far away. For an ellipse it is just a ΔE\sqrt{a}\,\Delta E. For a hyperbola it is −a ΔF\sqrt{-a}\,\Delta F. For a parabola it is p ΔD\sqrt{p}\,\Delta D. One variable, three disguises. The conic type now appears only through the reciprocal of the semi-major axis, which is perfectly finite everywhere:
α=1a=2r0−v02μ,z=αχ2\alpha = \frac{1}{a} = \frac{2}{r_0} - \frac{v_0^2}{\mu}, \qquad z = \alpha\chi^2

α > 0 for an ellipse, α = 0 for a parabola, α < 0 for a hyperbola, and it passes smoothly through zero. Nothing becomes infinite. The sign of z tells you the conic after the fact.

Every cos⁡\cos, sin⁡\sin, cosh⁡\cosh and sinh⁡\sinh in the three equations can be written with just two functions of zz, named after the German astronomer Karl Stumpff:
C(z)=1−cos⁡zz=12!−z4!+z26!−⋯ ,S(z)=z−sin⁡zz 3=13!−z5!+z27!−⋯C(z) = \frac{1 - \cos\sqrt{z}}{z} = \frac{1}{2!} - \frac{z}{4!} + \frac{z^2}{6!} - \cdots, \qquad S(z) = \frac{\sqrt{z} - \sin\sqrt{z}}{\sqrt{z}^{\,3}} = \frac{1}{3!} - \frac{z}{5!} + \frac{z^2}{7!} - \cdots

The Stumpff functions. For z < 0 the same formulas hold with cosh and sinh; at z = 0 they are simply 1/2 and 1/6. The series is one formula for all three conics.

With them, the time problem for any conic, starting from any position and velocity (not necessarily periapsis), is a single equation in a single unknown:
  μ Δt=r0vr0μ χ2C(z)+(1−αr0) χ3S(z)+r0χ  \boxed{\;\sqrt{\mu}\,\Delta t = \frac{r_0 v_{r0}}{\sqrt{\mu}}\,\chi^2 C(z) + (1 - \alpha r_0)\,\chi^3 S(z) + r_0\chi\;}

The universal Kepler equation. r₀ and v_r0 are the starting radius and radial velocity. Solve it for χ with Newton's method, exactly as before; one branch-free loop covers every orbit in the solar system.

💡The f and g functions
Once χ\chi is known, you never have to compute an orbital element at all. In two-body motion the satellite stays in the plane of its starting position r⃗0\vec r_0 and velocity v⃗0\vec v_0, so the new position must be a mix of those two vectors: r⃗=f r⃗0+g v⃗0,v⃗=f˙ r⃗0+g˙ v⃗0\vec r = f\,\vec r_0 + g\,\vec v_0, \qquad \vec v = \dot f\,\vec r_0 + \dot g\,\vec v_0 f=1−χ2r0C,g=Δt−χ3μS,f˙=μrr0(αχ3S−χ),g˙=1−χ2rCf = 1 - \frac{\chi^2}{r_0}C, \quad g = \Delta t - \frac{\chi^3}{\sqrt{\mu}}S, \quad \dot f = \frac{\sqrt{\mu}}{r r_0}\left(\alpha\chi^3 S - \chi\right), \quad \dot g = 1 - \frac{\chi^2}{r}C These are the Lagrange coefficients. They come with a built-in self-check: conservation of angular momentum forces fg˙−f˙g=1f\dot g - \dot f g = 1 for every conic and every time step. A propagator that breaks this identity has a bug. State vector in, state vector out, no elements, no branches. This is the form tracking data arrives in, and it is the form you will need again for Lambert's problem in Unit 9.
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📐Worked Example: leaving Earth for Mars
A spacecraft leaves a 300 km parking orbit on a hyperbola with v∞=3v_\infty = 3 km/s, roughly what a Mars transfer needs. Where is it 1 hour after perigee? The hyperbola. rp=6,678r_p = 6{,}678 km. a=−μ/v∞2=−398,600/9=−44,289a = -\mu/v_\infty^2 = -398{,}600/9 = -44{,}289 km. e=1+rpv∞2/μ=1.1508e = 1 + r_p v_\infty^2/\mu = 1.1508. Perigee speed vp=v∞2+2μ/rp=11.330v_p = \sqrt{v_\infty^2 + 2\mu/r_p} = 11.330 km/s. Mean motion. n=μ/(−a)3=6.7737×10−5n = \sqrt{\mu/(-a)^3} = 6.7737\times10^{-5} rad/s, so after 3,600 s, Mh=0.24385M_h = 0.24385. Solve. Start at F0=arcsinh⁡(Mh/e)=0.210F_0 = \operatorname{arcsinh}(M_h/e) = 0.210. Newton converges in 7 passes to F=0.83903F = 0.83903. Position. r=a(1−ecosh⁡F)=25,695r = a(1 - e\cosh F) = 25{,}695 km. tanh⁡(F/2)=0.0701 tan⁡(ν/2)\tanh(F/2) = \sqrt{0.0701}\,\tan(\nu/2) gives ν=112.5°\nu = 112.5°. Speed from energy: v=v∞2+2μ/r=6.33v = \sqrt{v_\infty^2 + 2\mu/r} = 6.33 km/s. Read the numbers. In 1 hour the spacecraft has climbed 19,000 km, 3 Earth radii, and lost almost half its speed. After 24 hours it is 338,000 km out, close to the Moon's distance, and the true anomaly has crept to 146.3°. It can never pass ν∞=arccos⁡(−1/e)=150.3°\nu_\infty = \arccos(-1/e) = 150.3°, the direction of the asymptote. The universal propagator, started from perigee with the same state, lands on the same 25,695 km without ever being told the orbit is a hyperbola.
🕰Gauss, Stumpff and the guidance computers
The f and g functions go back to Lagrange and Gauss, who used them as power series to fit short arcs of comet and asteroid observations. In the mid-twentieth century the German astronomer Karl Stumpff gave the series the form that now carries his name. The formulation became standard because of spaceflight. A guidance computer cannot afford a routine that branches on the conic type and occasionally takes the wrong branch, so the universal form went into the navigation software of the Apollo era. Unit 9 tells that part of the story. Today it is the textbook propagator in Vallado's and Curtis's books, and the one you have just written.
↓But there is a problem
One equation. Any orbit, from any state, with no elements and no branches. Give it a position, a velocity and a time, and it gives you the future. But only one kind of future. Every formula in this unit so far, the universal one included, is an exact solution of the two-body problem: a point-mass Earth, nothing else in the universe. The real Earth bulges at the equator. The real orbit brushes the top of the atmosphere. The Moon pulls. Add any of those, and there is no closed form left to solve. Which leaves the other way of seeing the future: stop solving the orbit, and simulate the physics, a few seconds at a time.