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Unit 2 · Lesson 2-3
Conic Sections and Orbit Geometry
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Conic Sections and Orbit Geometry

Circle, ellipse, parabola and hyperbola: geometry, energy and angular momentum

Alexandria, roughly 200 BC
Apollonius of Perga wrote eight books about what happens when you slice a cone with a flat plane. He worked out the whole family — circle, ellipse, parabola, hyperbola — and he gave them the names we still use. He did it for its own sake. There was no application, and he expected none. Eighteen centuries later Kepler took Tycho Brahe's observations of Mars, tried circle after circle, failed by eight arcminutes — a gap smaller than the width of a fingernail at arm's length — and refused to ignore it. What finally fitted was one of Apollonius's curves. And four centuries after that, we can say why. Not "the data fits an ellipse." The equation of motion admits no other solution. The orbit equation you derived last lesson is the polar form of a conic section, and ee is the tilt of Apollonius's plane. Every trajectory ever flown, by anything, anywhere, is a slice of that cone.
Watch first · presented by Shoaib Aziz
The worked example in this film is the lesson's own: the Molniya orbit, 500 km by 39,850 km, gives a = 26,546 km, e = 0.7412, ε = −7.51 km²/s² and h = 69,060 km²/s, and the master formula hands back the same 0.7412. The speeds, 10.05 km/s at perigee and 1.49 km/s at apogee, come from vis-viva.
Now, because they could not be disregarded, these eight minutes alone will lead us along a path to the reform of the whole of Astronomy.
– Johannes Kepler, Astronomia nova, 1609
Figure 2.6
what the slice producescirclee = 0ellipse0 < e < 1parabolae = 1hyperbolae > 1tilt the plane — that tilt is e
Apollonius's construction. A plane perpendicular to the axis cuts a circle. Tilt it and the circle stretches into an ellipse. Tilt until the plane is parallel to the cone's side and the curve tears open into a parabola. Tilt past that and you get a hyperbola. The tilt angle is e.
Read the four cases straight off r=p/(1+ecos⁡ν)r = p/(1 + e\cos\nu) by asking one question: can the denominator reach zero? If it can, rr runs to infinity and the orbit is open.
eShapeDenominatorεaPhysical meaning
0CircleAlways exactly 1−μ/2a= rBound, constant radius, constant speed
0 < e < 1EllipseNever reaches 0< 0> 0Bound; r cycles between r_p and r_a forever
e = 1ParabolaReaches 0 at ν = 180°0∞The exact escape threshold; arrives at infinity with zero speed
e > 1HyperbolaReaches 0 before ν = 180°> 0< 0Escapes with speed left over; a flyby, one pass, never returns
Figure 2.7
p — shared by all fourOne parameter, four fatescirclebound · ε < 0e = 0ellipsebound · ε < 0e = 0.55parabolathreshold · ε = 0e = 1hyperbolaescape · ε > 0e = 1.5same focus, same p — only e differs
The four families drawn from a common focus with the same p. As e grows the curve stretches, opens, and finally tears apart. Nothing discontinuous happens at e = 1 — the shape passes through it smoothly, which is why "parabolic" is a knife-edge case no real trajectory ever exactly occupies.
⚠The negative semi-major axis of a hyperbola
For e>1e > 1, a=p/(1−e2)a = p/(1-e^2) is negative. Students assume they have made an arithmetic error. They have not. The sign is doing real work. Energy is ε=−μ/2a\varepsilon = -\mu/2a, so a negative aa gives a positive energy — which is exactly the definition of an escape trajectory. The bookkeeping is consistent all the way through; you simply have to accept that aa stops being a length you can measure with a ruler and becomes a parameter that carries a sign. Some texts dodge this by writing ∣a∣|a| everywhere for hyperbolas. Do not follow them. Keep the sign, and every formula in the course — vis-viva included — continues to work unchanged.
For the bound case, the ellipse's geometry is worth having at your fingertips, because every manoeuvre calculation in Unit 6 is built from these five quantities.
Figure 2.8
abc = aepcentrefocusempty focusr_p = a(1−e)r_a = a(1+e)b = a√(1−e²) · c = ae · p = a(1−e²)
Semi-major axis a, semi-minor axis b, centre-to-focus distance c = ae, and the semi-latus rectum p standing vertically over the focus. Earth sits at the focus — never at the centre, a mistake that survives in a great many textbook diagrams.
rp=a(1−e),ra=a(1+e),b=a1−e2,c=ae,p=a(1−e2)r_{p} = a(1-e), \quad r_{a} = a(1+e), \quad b = a\sqrt{1-e^2}, \quad c = ae, \quad p = a(1-e^2)

The complete geometry of an ellipse in five expressions. Note a = (rpr_{p} + rar_{a})/2 and e = (rar_{a} − rpr_{p})/(rar_{a} + rpr_{p}) run the relationships backwards.

💡Two vocabularies for the same thing
You will meet both, often on the same page, and they mean exactly the same points: Periapsis / apoapsis — the generic terms, for orbits around anything. Perigee / apogee — Earth. Perihelion / aphelion — the Sun. Perilune / apolune — the Moon. Periareion / apoareion — Mars, and yes, people really do write it. The line joining them passes through the focus and is called the line of apsides. It is the long axis of the ellipse, it has length 2a2a, and it points along e⃗\vec{e} — which is precisely why the eccentricity vector was worth deriving.
Now the payoff Unit 1 promised, and the reason this unit exists.
At the end of 1-4 the claim was: give me hh and ε\varepsilon and the orbit is completely determined. We can now write that down explicitly. Size comes from energy alone:
a=−μ2εa = -\frac{\mu}{2\varepsilon}

Energy fixes the size. Nothing else does.

And shape comes from combining the two. Substituting p=h2/μp = h^2/\mu and p=a(1−e2)p = a(1-e^2) into a=−μ/2εa = -\mu/2\varepsilon and solving for ee:
  e=1+2εh2μ2  \boxed{\;e = \sqrt{1 + \frac{2\varepsilon h^2}{\mu^2}}\;}

The claim from Unit 1, proved. Two conserved scalars in, the complete shape of the orbit out — and the sign of ε alone decides which of the four families you are in.

💡Interpreting the result
Every one of the four families is visible in a single square root. ε<0\varepsilon < 0 makes the term under the root less than 1, so e<1e < 1: bound. Push ε\varepsilon toward −μ2/2h2-\mu^2/2h^2 and the root goes to zero: a circle, the minimum-eccentricity orbit for a given angular momentum. Raise ε\varepsilon to exactly zero and e=1e = 1 exactly: parabolic, the knife-edge. Above zero, e>1e > 1: hyperbolic, gone. One square root, four destinies. And notice what is not in it — no mass, no orientation, no starting point, no time. Two numbers, and the shape of the path is settled.
📐Worked Example: Molniya, from first principles
The Soviet Molniya orbit was designed so a communications satellite could hang over the high Arctic for hours at a time — geostationary orbits are useless above ~70° latitude, because from there GEO sits on the horizon. Given: perigee altitude 500 km, apogee altitude 39,850 km. rp=6371+500=6871r_p = 6371 + 500 = 6871 km, ra=6371+39850=46,221r_a = 6371 + 39850 = 46{,}221 km. Size. a=(6871+46221)/2=26,546a = (6871 + 46221)/2 = \mathbf{26{,}546} km. Shape. e=(46221−6871)/(46221+6871)=39350/53092=0.7412e = (46221 - 6871)/(46221 + 6871) = 39350/53092 = \mathbf{0.7412}. Semi-minor. b=265461−0.74122=26546×0.6713=17,820b = 26546\sqrt{1 - 0.7412^2} = 26546 \times 0.6713 = 17{,}820 km. Scale. p=26546(1−0.5494)=11,963p = 26546(1 - 0.5494) = 11{,}963 km. Energy. ε=−398600/(2×26546)=−7.508\varepsilon = -398600/(2 \times 26546) = \mathbf{-7.508} km²/s². Angular momentum. h=μp=398600×11963=69,060h = \sqrt{\mu p} = \sqrt{398600 \times 11963} = \mathbf{69{,}060} km²/s. Check the master formula: e=1+2(−7.508)(69060)2/3986002=1−0.4506=0.5494=0.7412e = \sqrt{1 + 2(-7.508)(69060)^2/398600^2} = \sqrt{1 - 0.4506} = \sqrt{0.5494} = 0.7412. ✓ Speeds. From vis-viva: vp=10.05v_p = 10.05 km/s, va=1.49v_a = 1.49 km/s. The satellite crosses perigee nearly seven times faster than it crosses apogee, which is the whole design: it rushes through the southern half of the orbit and loiters over the north for roughly eight of its twelve hours.
Figure 2.9
a fixed — so T and ε never changee = 0r_p = 1.00 ae = 0.3r_p = 0.70 ae = 0.6r_p = 0.40 ae = 0.8r_p = 0.20 ae = 0.92r_p = 0.08 aperigee plunges toward the surfacewhile apogee climbs away — and theperiod is identical throughout.
Hold a fixed and sweep e from 0 to 0.95. The period never changes — that is Kepler's third law, T = 2π√(a³/μ), and Unit 5 will derive it. What changes is everything else: perigee plunges toward the atmosphere while apogee climbs, and the speed ratio between them blows up.
🛰ʻOumuamua, October 2017
A survey telescope on Haleakalā caught a fast-moving object and the orbit fit came back with e=1.20e = 1.20. That single number was the entire story. e>1e > 1 means ε>0\varepsilon > 0 means it is not bound to the Sun — it did not come from our solar system, it is not staying, and it will never be back. It was moving at 26 km/s relative to the Sun while still far outside the planets, when a bound object at that distance should have been crawling. Every comet ever catalogued before it had e<1e < 1, some agonisingly close to 1 but always below. ʻOumuamua was the first interstellar object ever observed, and it was identified by a number computed from the equation you derived last lesson.
↓But there is a problem
So the two-body problem is finished. We can say the shape (ee), the size (aa), the scale (pp), the speed anywhere (vis-viva), and prove that no other trajectory is possible. Now try to actually use it. "The satellite is on an ellipse with a=26,546a = 26{,}546 km and e=0.741e = 0.741." Where do I point the antenna? You can't answer. Everything derived so far lives inside the orbital plane, and the orbital plane has been floating in a featureless void with no up, no north, no reference direction of any kind. The ellipse is a shape on a page. The sky is three-dimensional and turning. We need to nail that page to the sky — and it takes exactly three more numbers to do it. Not two, not four. Three, and then one more to say where along the curve the satellite actually is. Six numbers. Unit 3.