Worked examples — Orbital elements (Keplerian) — semi-major axis a, eccentricity e, inclination i, RAAN Ω, argument of perigee ω, true ano

This is the drill page for the parent topic. The parent taught you the six numbers and the formulas. Here we throw every kind of case at those formulas — every sign, every degenerate shape, every quadrant of an angle, plus a real-world problem and an exam trap — and grind each one out by hand.

Before any symbol appears, remember the toolkit the parent built (all defined there):

Recall The formulas we will hammer (all from the parent)

• Radii of the ends: perigee $r_p=a(1-e)$, apogee $r_a=a(1+e)$. Here $a$ is the semi-major axis (half the long diameter) and $e$ is the eccentricity (how squashed, $0$ = circle).
• Recover $a,e$: $a=\frac{r_p+r_a}{2}$, $e=\frac{r_a-r_p}{r_a+r_p}$.
• Vis-viva: $v^2=\mu \left(\frac{2}{r}-\frac{1}{a}\right)$, with $\mu$ the gravitational parameter (Earth's "gravity strength number").
• Orbit shape: $r=\frac{p}{1+e\cos \nu }$, semi-latus rectum $p=a(1-e^2)$.
• Angular momentum magnitude $h$ (area-sweeping rate $\times 2$) links to $p$ by $h=\sqrt{\mu p}$, so $p=h^2/\mu$.
• True anomaly $\nu$ = angle from perigee to the satellite, measured at Earth's centre.
• Energy: $\epsilon =-\frac{\mu }{2a}$ (negative = bound, zero = escape, positive = flyby).

Everywhere on this page ${\mu }_E=3.986\times 10^5{\text{km}}^3/{\text{s}}^2$ and Earth radius $R_E=6378\text{km}$.

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The scenario matrix

Every problem this topic can throw is one of these cells. Each worked example below is tagged with the cell it fills.

Cell	What makes it special	Example
A. Standard ellipse	$0<e<1$, plain numbers	Ex 1
B. Circle (degenerate $e=0$)	perigee = apogee, $\nu$ meaningless	Ex 2
C. Escape (limiting $e\to 1$)	$\epsilon =0$, $a\to \infty$	Ex 3
D. Hyperbolic flyby ($e>1$)	$a<0$, energy positive	Ex 4
E. Angle quadrant trap	recover $\nu$ where $\cos$ alone is ambiguous	Ex 5
F. Orientation angles: signs & quadrants	$i$, $\Omega$ from a vector, all sign cases	Ex 6
G. Real-world word problem	GEO / mission framing	Ex 7
H. Exam twist	mixes two formulas, hidden step	Ex 8

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A. Standard ellipse

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B. The circle — a degenerate case

When $e=0$ the ellipse collapses to a circle. Two things quietly break: perigee and apogee become the same point, and the argument of perigee $\omega$ and true anomaly $\nu$ lose their meaning (there's no "closest point" to measure from). The figure below shows the collapse — the two black dots that were once perigee and apogee now sit at the same distance from the focus, and the red loop is the same radius all the way round.

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C. The escape limit — $e\to 1$

As $e$ climbs toward $1$ the orbit stops being a closed loop. Watch what the formulas do at exactly $e=1$: energy hits zero and $a$ runs off to infinity.

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D. Hyperbolic flyby — $e>1$ and a negative $a$

Now push past escape. For $e>1$ the "ellipse" becomes a hyperbola, energy is positive, and the semi-major axis $a$ comes out negative. It still works — you just have to keep the sign. In the figure the red curve is the open hyperbolic path: it swings in past perigee (black dot) and flies back out to infinity, never closing — the visual signature of an unbound $e>1$ orbit.

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E. The angle-quadrant trap — recovering $\nu$

Here is the classic pitfall. Given $r$, the orbit equation gives you $\cos \nu$ — but $\cos$ can't tell "going away from perigee" ($0<\nu <180^{\circ }$) from "coming back" ($180^{\circ }<\nu <360^{\circ }$), because $\cos \nu =\cos (-\nu )$. You need a second piece of information to pick the quadrant. In the figure, one red radius points to $\nu =104^{\circ }$ (outbound, above the axis) and the other to $\nu =256^{\circ }$ (inbound, below the axis) — both hit the same radius, which is exactly the ambiguity we must resolve.

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F. Orientation angles — every sign of $i$ and $\Omega$

The tilt $i$ (0°–180°) and the swing $\Omega$ (0°–360°) come from the angular-momentum vector $\vec{h}$. Getting them right means handling all quadrants and both hemispheres. The 3D figure fixes the geometry: the black arrow is Earth's spin axis $Z$ (North), the faint plane is the equator, and the red arrow is $\vec{h}$ — the angle $i$ between them is the inclination.

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G. Real-world word problem

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H. The exam twist

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Recall Self-test: match the cell

Circle has $e=0$ and speed ::: $v=\sqrt{\mu /r}$, constant everywhere. Parabolic escape speed relates to circular speed by factor ::: $\sqrt{2}$. For a hyperbola the semi-major axis $a$ is ::: negative, and energy $\epsilon$ is positive. Given only $\cos \nu$, to pick the right $\nu$ you also need ::: the sign of $\sin \nu$ (radial-velocity direction), best done with atan2. To get $\Omega$ from the node vector you flip past $180^{\circ }$ when ::: the node's $Y$-component $n_y<0$. The point where orbital speed equals local circular speed is at ::: $r=a$, i.e. the minor-axis ends, $\cos \nu =-e$.

See also: Kepler's equation and mean anomaly for turning $\nu$ into time, State vectors to orbital elements for the full $\vec{r},\vec{v}\to$ elements pipeline, Perifocal coordinate frame and Angular momentum vector h for the geometry behind Ex 6, and Eccentricity vector for a sign-clean way to get $e$ and $\omega$.