Visual walkthrough — Orbit equation r = p - (1 + e·cos θ) — derivation from equations of motion

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Step 1 — A dot, an arrow, and a pull

WHAT. Picture a tiny planet at some point in space. Draw an arrow from the big central mass (the Sun) out to the planet. That arrow is the position vector $\vec{r}$: it says "which direction, and how far." Its length is the distance $r$. A little hat, $\hat{r}$, means "the same direction but shrunk to length 1" — a pure pointer.

WHY. Before we can say how the planet moves, we must name where it is. Everything downstream is written in $\vec{r}$ and $\hat{r}$, so we earn them first.

PICTURE. The blue arrow is $\vec{r}$. The short orange arrow along it is $\hat{r}$ (length 1). Gravity — the red arrow — points the opposite way, straight back toward the Sun.

Newton's law, one symbol at a time: $\underset{\text{how the arrow’s tip accelerates}}{\underbrace{\ddot{\vec{r}}}}=\underset{\text{inward}}{\underbrace{-}}\underset{\text{strength, weaker far away}}{\underbrace{\frac{\mu }{r^2}}}\underset{\text{direction: along the line to the Sun}}{\underbrace{\hat{r}}}$

The minus sign is the whole personality of gravity: it never pushes sideways, only back toward the centre. That single fact drives the next step. See Central Force Motion.

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Step 2 — Naming position with an angle, and its two natural directions

Before we can talk about "spin" or "how fast the angle sweeps," we must first have an angle. So we set up polar coordinates now — the very first thing we need.

WHAT. In the flat plane the planet moves in, we describe its position with two numbers: the distance $r$ from the Sun, and the angle $\theta$ measured from a chosen starting ray. This $\theta$ is the true anomaly. At the planet's location we draw two little unit arrows (each length 1):

• $\hat{r}$ — the radial unit vector, pointing straight outward from the Sun along $\vec{r}$.
• $\hat{\theta }$ — the tangential unit vector, pointing sideways, exactly $90^{\circ }$ ahead in the direction the angle $\theta$ grows.

WHY. Gravity points purely along $\hat{r}$. If we split every motion into an "$\hat{r}$ part" and a "$\hat{\theta }$ part," gravity lands entirely in one of them and leaves the other clean. These two arrows are the natural ruler for a central force, so we build them before anything else.

PICTURE. The planet on its curve, the Sun at the origin, the angle $\theta$ swept from the starting ray, and the two perpendicular unit arrows $\hat{r}$ (outward) and $\hat{\theta }$ (sideways-ahead).

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Step 3 — Splitting acceleration into radial and sideways parts

WHAT. Because $\hat{r}$ and $\hat{\theta }$ rotate as the planet moves, writing acceleration in them produces four pieces (this is the standard polar decomposition). Grouped by direction: $\vec{a}=\underset{\text{along }\hat{r}}{\underbrace{(\ddot{r}-r{\dot{\theta }}^2)}}\hat{r}+\underset{\text{along }\hat{\theta }}{\underbrace{(r\ddot{\theta }+2\dot{r}\dot{\theta })}}\hat{\theta }$

WHY these terms. Each piece has a plain meaning you can feel:

• $\ddot{r}$ — genuinely speeding up or slowing down along the line to the Sun.
• $-r{\dot{\theta }}^2$ — the outward "sling" of travelling on a curve (the merry-go-round push). Here $\dot{\theta }$ means "how fast the angle sweeps"; it is the rate of change of the $\theta$ we just defined.
• $r\ddot{\theta }$ — the angle speeding up or slowing down.
• $2\dot{r}\dot{\theta }$ — the Coriolis piece: moving in or out while rotating.

PICTURE. The four contributions drawn as small arrows at the planet, colour-split into the two $\hat{r}$ (radial) terms and the two $\hat{\theta }$ (tangential) terms.

Now apply Newton's law from Step 1. Gravity is $-\frac{\mu }{r^2}\hat{r}$ — pure $\hat{r}$, zero $\hat{\theta }$. Matching the two directions separately: $\underset{\text{radial: has force}}{\underbrace{\ddot{r}-r{\dot{\theta }}^2=-\frac{\mu }{r^2}}},\underset{\text{tangential: no force}}{\underbrace{r\ddot{\theta }+2\dot{r}\dot{\theta }=0}}$

The sideways equation hands us the conserved "spin" for free. Multiply it by $r$: $r^2\ddot{\theta }+2r\dot{r}\dot{\theta }=0.$ Look closely: the left side is exactly the time-derivative of $r^2\dot{\theta }$ (by the product rule, $\frac{d}{dt}(r^2\dot{\theta })=2r\dot{r}\dot{\theta }+r^2\ddot{\theta }$). So $\frac{d}{dt}\left(r^2\dot{\theta }\right)=0⟹h\equiv r^2\dot{\theta }=\text{constant}.$

That constant $h$ is the specific angular momentum. We didn't need any 3D machinery — the no-sideways-force equation is the conservation law. See Specific Angular Momentum h and Kepler's Laws (this constant is Kepler's equal-area rule).

Recall Why "specific" and why this equals

$|\vec{r}\times \dot{\vec{r}}|$ "Specific" = per unit mass (we divided out $m$ back in Step 1). The cross product $\vec{r}\times \dot{\vec{r}}$ measures the area swept per second; in polar form its magnitude works out to exactly $r^2\dot{\theta }$, the same $h$. Both routes agree.

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Step 4 — The magic swap $u=1/r$

We now have two facts to combine: the radial equation $\ddot{r}-r{\dot{\theta }}^2=-\mu /r^2$, and $h=r^2\dot{\theta }$. But they mix $r$, $\theta$, and time. We want a clean equation for $r$ as a function of $\theta$ alone.

WHAT. Define $u=1/r$ ("closeness" instead of distance — big $u$ = near the Sun). First rewrite the spin law: $h=r^2\dot{\theta }=\frac{\dot{\theta }}{u^2}$, so $\dot{\theta }=h{u}^2.$

Now convert $\dot{r}$ from a time-rate into a $\theta$-rate using the chain rule (below).

WHY the $d/dt\to d/d\theta$ swap. The chain rule says: to know how something changes per second, multiply how it changes per angle by how fast the angle turns. In symbols, for any quantity $Q$, $\frac{dQ}{dt}=\frac{dQ}{d\theta }\cdot \underset{=\dot{\theta }=h{u}^2}{\underbrace{\frac{d\theta }{dt}}}.$ We do this because $\theta$ only ever increases (the planet keeps circling one way), so it is an honest independent variable — unlike $t$, which we want gone.

The step-by-step conversion of $\dot{r}$ (each arrow is one small move): $\dot{r}=\frac{dr}{dt}\stackrel{\text{chain rule}}{=}\frac{dr}{d\theta }\dot{\theta }\stackrel{r=1/u}{=}\frac{d}{d\theta }\left(\frac{1}{u}\right)(h{u}^2)=(-\frac{1}{u^2}\frac{du}{d\theta })(h{u}^2)=-h\frac{du}{d\theta }.$ Here $\frac{d}{d\theta }(1/u)=-\frac{1}{u^2}\frac{du}{d\theta }$ is itself just the chain rule on $1/u$; the two $u^2$ then cancel.

Now differentiate once more to get $\ddot{r}$: $\ddot{r}=\frac{d}{dt}(-h\frac{du}{d\theta })\stackrel{\text{chain rule}}{=}\frac{d}{d\theta }(-h\frac{du}{d\theta })\dot{\theta }=(-h\frac{d^{2}u}{d{\theta }^2})(h{u}^2)=-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}.$

PICTURE. Two curves side by side: $r(\theta )$ dips to a minimum at perihelion (a valley), while $u(\theta )=1/r$ peaks there (a hill). The swap flips valleys into hills — and, crucially, turns a bent curve into a plain cosine wave, which is what makes the next step trivial.

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Step 5 — The equation becomes a swing

WHAT. Take the radial equation and replace every time-thing with a $\theta$-thing. We need two substitutions:

1. $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$ (just derived).
2. $r{\dot{\theta }}^2$: since $r=1/u$ and $\dot{\theta }=h{u}^2$, we get $r{\dot{\theta }}^2=\frac{1}{u}(h{u}^2{)}^2=\frac{h^2{u}^4}{u}=h^2{u}^3$.

Substitute both into $\ddot{r}-r{\dot{\theta }}^2=-\mu /r^2$, and note $1/r^2=u^2$ on the right: $\underset{\ddot{r}}{\underbrace{-h^2{u}^2{u}^{\prime \prime }}}-\underset{r{\dot{\theta }}^2}{\underbrace{h^2{u}^3}}=-\mu u^2.$

Now simplify one move at a time. Divide every term by $-h^2{u}^2$ (allowed since $u\ne 0$: the planet is never infinitely far): $\frac{-h^2{u}^2{u}^{\prime \prime }}{-h^2{u}^2}+\frac{-h^2{u}^3}{-h^2{u}^2}=\frac{-\mu u^2}{-h^2{u}^2}⟹u^{\prime \prime }+u=\frac{\mu }{h^2}.$ The first term keeps $u^{\prime \prime }$, the second leaves $u$, the right side leaves the constant $\mu /h^2$. Clean: $\boxed{u^{\prime \prime }+u=\frac{\mu }{h^2}}$

WHY this matters. This is the single most famous equation in physics — a simple harmonic oscillator (a mass on a spring, a pendulum for small swings) with a constant push $\mu /h^2$ on the right. We chased the $u$-substitution precisely to land here, because we already know this equation's answer by heart.

PICTURE. A mass on a spring: pull it, it oscillates around a rest position offset by the constant force. Here $u$ "oscillates" not in time but as $\theta$ goes around — one full swing per orbit. The rest position sits at height $\mu /h^2$.

Term by term: $\underset{\text{curvature of the }u\text{-wave}}{\underbrace{u^{\prime \prime }}}+\underset{\text{restoring pull toward centre}}{\underbrace{u}}=\underset{\text{constant offset (the rest line)}}{\underbrace{\frac{\mu }{h^2}}}$

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Step 6 — Reading off the solution, piece by piece

WHAT. The equation $u^{\prime \prime }+u=\mu /h^2$ is linear, so its full answer is one particular solution plus the general free wobble:

• Particular: a constant that satisfies it. Try $u=C$; then $u^{\prime \prime }=0$, so $0+C=\mu /h^2$, giving $C=\mu /h^2$. This is the flat "rest line."
• Homogeneous (free wobble): solve $u^{\prime \prime }+u=0$. The functions whose second derivative is minus themselves are exactly $\cos$ and $\sin$, so the general wobble is $A\cos (\theta -{\theta }_0)$ for constants $A,{\theta }_0$.

Add them: $u(\theta )=\underset{\text{rest line}}{\underbrace{\frac{\mu }{h^2}}}+\underset{\text{free wobble}}{\underbrace{A\cos (\theta -{\theta }_0)}}.$

WHY the constants. $A$ and ${\theta }_0$ are fixed by how the planet was launched. We spend our free choice of ${\theta }_0$ by setting ${\theta }_0=0$ — this just decides where we start counting the angle. We put it at the wobble's peak (largest $u$ = smallest $r$ = closest point = perihelion).

PICTURE. The solution over one orbit: a flat rest line at $\mu /h^2$, a cosine of amplitude $A$ riding on it. Peak = perihelion at $\theta =0$; trough = aphelion at $\theta =\pi$.

Now flip $u$ back into $r$, one algebraic move at a time. With ${\theta }_0=0$: $\frac{1}{r}=\frac{\mu }{h^2}+A\cos \theta .$ Factor $\mu /h^2$ out of the right side: $\frac{1}{r}=\frac{\mu }{h^2}(1+\underset{\text{call this }e}{\underbrace{\frac{A{h}^2}{\mu }}}\cos \theta ).$ Name that dimensionless bundle $e\equiv A{h}^2/\mu$ (the eccentricity). Take reciprocals of both sides: $\boxed{r=\frac{h^2/\mu }{1+e\cos \theta }=\frac{p}{1+e\cos \theta }},p\equiv \frac{h^2}{\mu }.$ The scale $p=h^2/\mu$ and the shape number $e$ fall out for free. Bigger $A$ ⇒ bigger $e$ ⇒ more lopsided orbit. See Eccentricity and Orbital Energy.

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Step 7 — Every shape, every case

WHAT. The one formula secretly contains four different curves, decided entirely by $e$ (from Conic Sections):

$e$	denominator $1+e\cos \theta$	what happens	shape
$0$	always $1$	$r=p$ everywhere	circle
$0<e<1$	never hits $0$	$r$ stays finite, closed loop	ellipse
$1$	$\to 0$ as $\theta \to \pi$	$r\to \infty$: escapes	parabola
$>1$	hits $0$ at some $\theta <\pi$	$r\to \infty$ early: flies off	hyperbola

WHY. We must check all cases, not just the tidy ellipse. The key question: can the denominator reach zero? Where it does, $r$ blows up to infinity and the orbit is open (the body never returns).

• $e<1$: since $e\cos \theta \ge -e>-1$, the denominator is always positive — $r$ is always finite — the orbit closes.
• $e=1$: at $\theta =\pi$, $\cos \theta =-1$, denominator $=1-1=0$ — $r\to \infty$ exactly at the back. Borderline escape (parabola).
• $e>1$: denominator hits $0$ at $\cos \theta =-1/e$ (an angle before $\pi$). Beyond that, $r$ would go negative — meaningless — so the real path only lives on the arc where the denominator stays positive. That arc is a hyperbola with two "asymptote" directions.

PICTURE. All four curves sharing the same focus (the Sun) and the same $p$, colour-coded, with the runaway arms of the parabola and hyperbola shooting to infinity.

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The one-picture summary

Here is the whole journey on one canvas: gravity's inward pull → split into $\hat{r}$ and $\hat{\theta }$ → the sideways part gives $h=$const → the radial part + $u=1/r$ swap → the spring equation → its cosine solution → the conic. Follow the arrows left to right.

Recall Feynman retelling — the whole walkthrough in plain words

Start with one fact: the Sun pulls the planet straight inward, never sideways. To use that, we set up two little arrows that ride along with the planet — one pointing outward ($\hat{r}$), one pointing sideways ($\hat{\theta }$). We split the planet's acceleration into these two directions. Gravity is entirely outward-inward, so the sideways equation says "no sideways force at all" — and that single line, cleaned up, tells us that $r^2\dot{\theta }$ can never change. That's the conserved "spin," and it also flattens the whole dance into one plane.

The remaining outward-inward equation is knotty because it mixes distance, angle and time. We play one clever trick: track closeness $u=1/r$ instead of distance, and ask "where is it at each angle?" instead of "at each second?" The chain rule does the translating. That double swap magically cancels gravity's awkward $1/r^2$ and hands us the most famous equation in physics: a mass on a spring. A spring's answer is a rest position plus a gentle cosine wobble.

Undo the swap, factor tidily, and that cosine becomes $r=p/(1+e\cos \theta )$. The size $p$ and the shape $e$ pop out on their own. Small $e$: a circle or gentle oval. $e=1$: the planet barely escapes on a parabola. $e>1$: it's a visitor, swinging past once on a hyperbola and never coming back. One equation, four destinies — all from "gravity only pulls inward."

Recall Rapid self-test

• Which equation gives $h=$const, and how? ::: The tangential (sideways) one, $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$; times $r$ it is $\frac{d}{dt}(r^2\dot{\theta })=0$.
• What tool translates $d/dt$ into $d/d\theta$? ::: The chain rule, $\frac{dQ}{dt}=\frac{dQ}{d\theta }\dot{\theta }$ with $\dot{\theta }=h{u}^2$.
• What does $u^{\prime \prime }$ mean? ::: $d^{2}u/d{\theta }^2$ — differentiate twice with respect to $\theta$.
• What does $u=1/r$ buy us? ::: It cancels the $1/r^2$ force and turns the ODE into $u^{\prime \prime }+u=\mu /h^2$, a spring equation.
• When does the orbit open to infinity? ::: When the denominator $1+e\cos \theta$ can reach $0$, i.e. $e\ge 1$.