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

Before any symbol, here is the whole cast we will meet, in plain words:

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Step 1 — One satellite, one arrow, one pull

WHAT. Put Earth's centre at the origin. Draw the satellite somewhere, and draw the arrow $\vec{r}$ from origin to satellite. Its length $r$ is how far away the satellite is.

WHY. Every orbital idea is a statement about this one arrow — how long it is and which way it points as time passes. If we can find a rule for $r$ as a function of an angle, we have the orbit.

PICTURE. The blue arrow is $\vec{r}$. Gravity (red) always points back toward the origin — that is what "central force" means: the pull lies exactly along $-\vec{r}$.

The gravitational acceleration is $\ddot{\vec{r}}=-\frac{\mu }{r^3}\vec{r}.$

Reading it term by term, right where each symbol sits:

• $\ddot{\vec{r}}$ — how the velocity arrow is changing (acceleration).
• the minus sign — points the pull inward, opposite to $\vec{r}$.
• $\mu$ — the strength of the pull.
• $1/r^3$ times $\vec{r}$ — this equals $\frac{1}{r^2}\cdot \frac{\vec{r}}{r}$: a $1/r^2$ strength along the unit direction $\vec{r}/r$. So the force weakens as the inverse square of distance.

That is the whole of two-body gravity. Everything else is consequence.

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Step 2 — The plane never tilts: angular momentum $\vec{h}$

WHAT. Define the arrow $\vec{h}=\vec{r}\times \vec{v}.$ This "$\times$" is the cross product: it produces an arrow perpendicular to both $\vec{r}$ and $\vec{v}$, whose length equals the area of the parallelogram they span.

WHY THIS TOOL. We want to prove the motion stays in one flat plane. The cross product is the exact tool that answers "what single direction is perpendicular to the plane of these two arrows?" If that perpendicular arrow never moves, the plane never tilts.

PICTURE. $\vec{r}$ (blue) and $\vec{v}$ (green) lie in a plane; $\vec{h}$ (yellow) stands straight out of it like a flagpole.

Watch it stay constant. Differentiate: $\dot{\vec{h}}=\dot{\vec{r}}\times \vec{v}+\vec{r}\times \dot{\vec{v}}=\underset{=0}{\underbrace{\vec{v}\times \vec{v}}}+\vec{r}\times (-\frac{\mu }{r^3}\vec{r})=0.$

• $\vec{v}\times \vec{v}=0$ — any arrow crossed with itself is zero.
• $\vec{r}\times \vec{r}=0$ — same reason, so the gravity term vanishes too.

So $\vec{h}$ is frozen. Its constant length $h$ and constant direction mean:

• the orbit lives in one fixed plane (the plane $\vec{h}$ is perpendicular to),
• and $h=r{v}_{\perp }$ is conserved (this is $\vec h$ and it is Kepler's second law).

From now on we work inside that single plane. The 3D tilt angles $i,\Omega$ just say how this plane sits in space — a separate story handled by the perifocal frame.

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Step 3 — A second frozen arrow points at the closest approach: $\vec{e}$

WHAT. Build the eccentricity vector $\vec{e}=\frac{\vec{v}\times \vec{h}}{\mu }-\frac{\vec{r}}{r}.$

WHY. We already have one constant arrow, $\vec{h}$, telling us the plane. To pin down the ellipse inside that plane we need a constant arrow lying in the plane, pointing at a landmark. It turns out $\vec{e}$ is exactly that: it is frozen in time and it aims from the focus toward perigee (the closest point). Its length will become the shape number $e$.

PICTURE. The yellow arrow $\vec{e}$ sits in the orbital plane and points at the near end of the ellipse. The angle from $\vec{e}$ around to $\vec{r}$ is our true anomaly $\nu$.

• $\frac{\vec{v}\times \vec{h}}{\mu }$ — a vector combining velocity and the frozen $\vec{h}$.
• $\frac{\vec{r}}{r}$ — the unit arrow pointing straight out at the satellite.
• Their difference happens to be a constant, and its length is the number we call $e$.

That $\vec{e}$ is constant is a short (grindy) derivative check, shown in the Eccentricity vector note; here we use it.

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Step 4 — Dot $\vec{e}$ into $\vec{r}$: the angle $\nu$ enters

WHAT. Take the dot product of $\vec{e}$ with $\vec{r}$.

WHY THIS TOOL. The dot product $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos (\text{angle between them})$ is precisely the tool that pulls an angle out of two arrows. We have a constant landmark $\vec{e}$ and the moving arrow $\vec{r}$; dotting them converts "where is the satellite" into "what angle $\nu$ has it swept from perigee."

PICTURE. The shaded angle between the fixed yellow $\vec{e}$ and the moving blue $\vec{r}$ is $\nu$.

Compute the dot product two ways.

Geometric way (by definition of dot product): $\vec{e}\cdot \vec{r}=er\cos \nu .$

• $e$ — length of $\vec{e}$.
• $r$ — length of $\vec{r}$.
• $\cos \nu$ — because $\nu$ is the angle between them.

Algebraic way (plug in the definition of $\vec{e}$ from Step 3 and use the vector identity $\vec{r}\cdot (\vec{v}\times \vec{h})=(\vec{r}\times \vec{v})\cdot \vec{h}=\vec{h}\cdot \vec{h}=h^2$): $\vec{e}\cdot \vec{r}=\frac{h^2}{\mu }-r.$

• $h^2/\mu$ — a pure constant built from the frozen $\vec{h}$.
• $-r$ — because $\frac{\vec{r}}{r}\cdot \vec{r}=r$.

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Step 5 — Set the two expressions equal: the orbit equation appears

WHAT. The left sides are identical, so the right sides must match: $er\cos \nu =\frac{h^2}{\mu }-r.$

WHY. This single line already is the orbit — it links distance $r$ to angle $\nu$. We just tidy it.

Solve for $r$: bring both $r$ terms together, $r+er\cos \nu =\frac{h^2}{\mu }⟹r(1+e\cos \nu )=\frac{h^2}{\mu },$ $\boxed{r=\frac{h^2/\mu }{1+e\cos \nu }}.$

Name the constant on top the semi-latus rectum: $p\equiv \frac{h^2}{\mu },\text{so}r=\frac{p}{1+e\cos \nu }.$

PICTURE. This is a conic section. Watch $r$ sweep as $\nu$ goes around: it is smallest when $\cos \nu =+1$ (perigee) and largest when $\cos \nu =-1$ (apogee).

Term by term:

• $p$ — the "size scaffold": it is literally $r$ when $\nu =90^{\circ }$ (then $\cos \nu =0$).
• $e\cos \nu$ — the "squash": it grows the denominator near apogee (making $r$ big) and shrinks it near perigee (making $r$ small).
• $r$ — the distance we wanted, now a clean function of one angle.

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Step 6 — Reading off perigee, apogee, and connecting to $a$

WHAT. Plug in the two special angles.

At $\nu =0$ ($\cos \nu =+1$): $r_p=\frac{p}{1+e}$ — perigee, the closest point. At $\nu =180^{\circ }$ ($\cos \nu =-1$): $r_a=\frac{p}{1-e}$ — apogee, the farthest point.

WHY. These give us the physical meaning of $e$ and let us swap $p$ for the more familiar semi-major axis $a$.

Add and halve the two apside distances (this defines $a$ as their average): $a=\frac{r_p+r_a}{2}=\frac{p}{1-e^2}⟹p=a(1-e^2).$

Substituting back gives the parent note's headline form: $\boxed{r=\frac{a(1-e^2)}{1+e\cos \nu }}.$

PICTURE. The ellipse with $r_p$ (perigee, red) and $r_a$ (apogee, green) marked, and $a$ shown as the half-length of the long axis.

• $e$ now reads as a squash ratio: $e=\frac{r_a-r_p}{r_a+r_p}$.
• $a$ is the average distance / half the long axis. (See Vis-viva equation for how $a$ fixes the energy.)

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Step 7 — Every case of $e$: circle, ellipse, parabola, hyperbola

WHAT. The single formula $r=\frac{p}{1+e\cos \nu }$ silently contains all conic shapes, just by changing $e$. We must check every one so no reader hits a surprise.

WHY. The denominator $1+e\cos \nu$ can be always positive, or reach zero, or go negative — and that is what switches the shape.

PICTURE. Four orbits sharing the same focus, one per value of $e$.

Walk the cases:

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The one-picture summary

Everything above, compressed: two frozen arrows ($\vec{h}$ out of plane, $\vec{e}$ in plane toward perigee), the moving arrow $\vec{r}$ at angle $\nu$ from $\vec{e}$, and the resulting conic whose radius obeys $r=p/(1+e\cos \nu )$.

Recall Feynman: the whole walkthrough in plain words

Start with nothing but a satellite being pulled straight toward Earth's centre. First trick: cross the position arrow with the velocity arrow — that makes a flagpole arrow $\vec{h}$ that never moves, which proves the whole orbit lives in one flat, non-tilting plane. Second trick: mix velocity and $\vec{h}$ in just the right recipe and subtract the outward unit arrow — you get another frozen arrow, $\vec{e}$, that always points at the closest approach. Now the clever bit: "dot" that landmark arrow into the position arrow. Doing the dot two different ways — once geometrically (which coughs up $\cos \nu$) and once algebraically (which coughs up a constant) — and setting them equal is the orbit. Clean it up and you get $r=p/(1+e\cos \nu )$: distance as a simple function of one angle. The top constant $p$ is the size; the $e\cos \nu$ term is the squash. Dial $e$ from $0$ upward and the same formula becomes a circle, an ellipse, a parabola, then a hyperbola — you never needed a new equation, just a new number.

Recall Quick self-test

What tool converts two arrows into an angle, and why did we need it in Step 4? ::: The dot product, because $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta$ isolates the angle $\nu$ between the fixed $\vec{e}$ and moving $\vec{r}$. Why does the orbit lie in a fixed plane? ::: Because $\vec{h}=\vec{r}\times \vec{v}$ has zero time-derivative, so its direction is frozen and the motion stays perpendicular to it. What is $p$ physically? ::: The radius $r$ when $\nu =90^{\circ }$ (the semi-latus rectum), equal to $h^2/\mu$. Which form of the orbit equation survives $e=1$? ::: $r=p/(1+e\cos \nu )$ with $p=h^2/\mu$; the $a(1-e^2)$ form fails because $a\to \infty$.

Related: Kepler's equation and mean anomaly (turning $\nu$ into a clock), State vectors to orbital elements (going backwards from $\vec{r},\vec{v}$), Orbital perturbations (J2) (what slowly un-freezes $\vec{e}$ and $\vec{h}$).