Visual walkthrough — Orbit shape from eccentricity — circle (e=0), ellipse (0 - e - 1), parabola (e=1), hyperbola (e - 1)

This page rebuilds the result $r(\theta )=\frac{p}{1+e\cos \theta }$ from nothing. No prior orbit knowledge assumed. We earn every symbol with a picture before it ever appears in an equation. By the end you will see why one number — the eccentricity $e$ — decides whether a body loops home or leaves forever.

Parent: Orbit shape from eccentricity ($0<e<1$ ellipse, $e>1$ hyperbola)

Step 1 — What are we even measuring? Set up the two rulers

WHY these two and not $x,y$? Gravity pulls straight along the line joining the two bodies — a purely radial pull. Coordinates built on that same radial line make the pull look simple. Cartesian $x,y$ would smear the pull across two equations; polar keeps it clean.

PICTURE. The big mass $M$ sits at the origin (a yellow dot). The small body is out at distance $r$, at angle $\theta$. As it moves, $r$ breathes in and out while $\theta$ winds around.

Step 2 — The two things that never change: energy and angular momentum

WHY $h$ constant helps. $h=r^2\dot{\theta }$ lets us trade time-derivatives for angle-derivatives. Any "$\frac{d}{dt}$" (rate per second) can be swapped for "$\frac{d}{d\theta }$" (rate per angle) using $\dot{\theta }=h/r^2$. This is the trick that removes time entirely, leaving a pure shape $r(\theta )$.

PICTURE. In equal time slices the body sweeps equal areas (that is what constant $h$ looks like). Near the mass it moves fast on a short arc; far away it crawls on a long arc — the shaded slivers all have the same area.

Step 3 — A clever change of variable: let $u=1/r$

WHY do this — why the reciprocal and not $r$ itself? Gravity is an inverse-square force: its strength goes like $1/r^2=u^2$. When we write the equation of motion in terms of $u$ instead of $r$, the ugly $1/r^2$ becomes a tidy $u^2$, and — as we will see next — the whole equation collapses into the single simplest oscillation in physics. Choosing $u$ is choosing the variable in which gravity's fingerprint is cleanest.

PICTURE. The same orbit plotted two ways: distance $r$ against angle (curvy, hard to read) versus $u=1/r$ against angle (a clean cosine wave). Same physics, but $u$ tells a straight story.

Step 4 — Newton's law in polar form, then the $u$-substitution — deriving Binet

Line 1 — Newton's second law, radial part. In polar coordinates the acceleration along $r$ is not just $\ddot{r}$; the spinning motion adds a $-r{\dot{\theta }}^2$ (the outward "centrifugal" bookkeeping term). Gravity supplies the pull $-\mu /r^2$ (the same $\mu =GM$ we named in Step 2). So: $\ddot{r}-r{\dot{\theta }}^2=-\frac{\mu }{r^2}$

• $\ddot{r}$ = how fast the distance's rate is changing (radial acceleration).
• $-r{\dot{\theta }}^2$ = the term the rotating frame contributes.
• $-\mu /r^2$ = inverse-square gravity, minus because it pulls inward.

Line 2 — kill time using constant $h$. From $h=r^2\dot{\theta }$ we get $\dot{\theta }=h/r^2$, so $r{\dot{\theta }}^2=r{\left(h/r^2\right)}^2=h^2/r^3$. This trades the angular speed for a pure function of $r$: $\ddot{r}-\frac{h^2}{r^3}=-\frac{\mu }{r^2}$

Line 3 — switch $r\to u=1/r$. Using the chain rule with $\dot{\theta }=h{u}^2$, two standard steps give $\dot{r}=-h\frac{du}{d\theta },\ddot{r}=-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}.$

• The $-h\frac{du}{d\theta }$ appears because $\frac{d}{dt}=\dot{\theta }\frac{d}{d\theta }=h{u}^2\frac{d}{d\theta }$ and $\frac{d}{d\theta }(1/u)=-\frac{1}{u^2}\frac{du}{d\theta }$; the $u^2$ cancels, leaving just $-hdu/d\theta$. Differentiate once more the same way to get $\ddot{r}$.

Line 4 — substitute and simplify. Put Line 3's $\ddot{r}$ into Line 2, and replace $1/r^3=u^3$, $1/r^2=u^2$: $-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}-h^2{u}^3=-\mu u^2.$ Divide every term by $-h^2{u}^2$ (allowed since $u>0$): $\frac{d^{2}u}{d{\theta }^2}+u=\frac{\mu }{h^2}.$

WHY this is exciting. Read the left side alone: "the curvature of $u$ plus $u$ equals something constant." That is exactly the equation of a mass on a spring — simple harmonic motion — with a constant offset. We already know its solutions: a steady value plus a cosine wiggle.

WHY a derivative tool at all? A derivative answers "how fast is this changing right now?" Newton's law is a statement about acceleration = second derivative. To connect gravity (a force) to shape (geometry), we must speak the language of second derivatives — nothing weaker captures "curving."

PICTURE. A block on a spring: pull it out, it accelerates back toward a rest point. Overlay the identical shape of $\frac{d^{2}u}{d{\theta }^2}+u=\text{const}$ — the "rest point" is shifted up to $\mu /h^2$.

Step 5 — Write down the solution: a constant plus a cosine

WHY $\cos$ and not $\sin$? We are free to choose where $\theta =0$ points. We aim it at the point of closest approach, where $u$ is biggest. $\cos \theta$ peaks at $\theta =0$; $\sin$ would peak sideways. So $\cos$ is just the convenient orientation — it earns its place by putting the peak where we defined "start."

PICTURE. The cosine wave for $u$: a flat baseline at height $\mu /h^2$, with a cosine riding on top. Bigger $e$ = taller wave. Watch where the wave dips lowest — that dip is where the orbit reaches farthest.

Step 6 — Flip back to distance: the conic equation appears

WHY this is the answer. This is precisely the polar equation of a conic section (circle/ellipse/parabola/hyperbola) with a focus at the origin. Gravity did not "choose" ellipses by decree — the inverse-square law's algebra forces a conic. See Conic Sections and Kepler's First Law.

PICTURE. The plotted curve: mass at the focus (yellow), the semi-latus rectum $p$ drawn straight up at $\theta =90^{\circ }$, closest point (perihelion) on the right, farthest point on the left.

Step 7 — The decisive question: when can $r$ reach infinity?

Every case, no gaps:

• $e=0$ (circle): no wiggle at all, $r=p$ constant. Perfectly round, bound.
• $0<e<1$ (ellipse): $-1/e<-1$, so $\cos \theta =-1/e$ is impossible. Denominator never zero → $r$ stays finite for all $\theta$ → closed loop, bound.
• $e=1$ (parabola): $\cos \theta =-1$ solved only at $\theta =180^{\circ }$. $r\to \infty$ in exactly one direction — the knife-edge escape.
• $e>1$ (hyperbola): $-1/e$ lies between $-1$ and $0$, reachable at a finite angle ${\theta }_{\infty }=\arccos (-1/e)$. $r\to \infty$ there → open curve, unbound. Beyond ${\theta }_{\infty }$ the formula gives negative $r$ (unphysical) — those angles are the forbidden gap between the incoming and outgoing asymptotes.

WHY $\arccos$ here? We know the cosine of the escape angle ($=-1/e$) and want the angle itself. $\arccos$ is the tool that undoes cosine — it answers "which angle has this cosine?" That is exactly our question.

PICTURE. Four denominators plotted against $\theta$: for $e<1$ the curve floats safely above zero; at $e=1$ it just kisses zero at $180^{\circ }$; for $e>1$ it crosses zero early — and the crossing angle is the hyperbola's asymptote.

Step 8 — Tie the shape to energy — deriving $e=\sqrt{1+2\epsilon h^2/{\mu }^2}$

Line 1 — split the speed into two parts. Speed $v$ has a radial piece $\dot{r}$ and a sideways piece $r\dot{\theta }$, at right angles, so $v^2={\dot{r}}^2+(r\dot{\theta }{)}^2$. Using $r\dot{\theta }=h/r=hu$ and (from Step 4) $\dot{r}=-hdu/d\theta$: $v^2=h^2{\left(\frac{du}{d\theta }\right)}^2+h^2{u}^2.$

Line 2 — put this into $\epsilon$. Recall $\epsilon =\frac{1}{2}{v}^2-\mu /r=\frac{1}{2}{v}^2-\mu u$. So $\epsilon =\frac{h^2}{2}{\left(\frac{du}{d\theta }\right)}^2+\frac{h^2}{2}{u}^2-\mu u.$

Line 3 — insert our solution. From Step 5, $u=\frac{\mu }{h^2}(1+e\cos \theta )$, so $\frac{du}{d\theta }=-\frac{\mu }{h^2}e\sin \theta$. Substitute both: $\epsilon =\frac{h^2}{2}\frac{{\mu }^2}{h^4}{e}^2{\sin }^2\theta +\frac{h^2}{2}\frac{{\mu }^2}{h^4}(1+e\cos \theta {)}^2-\mu \frac{\mu }{h^2}(1+e\cos \theta ).$

Line 4 — clean up (the $\theta$'s cancel). Factor $\frac{{\mu }^2}{2h^2}$ and use ${\sin }^2\theta +{\cos }^2\theta =1$; every $\theta$-term cancels — as it must, since $\epsilon$ is constant: $\epsilon =\frac{{\mu }^2}{2h^2}(e^2-1).$

Line 5 — solve for $e$. Rearranging $\epsilon =\frac{{\mu }^2}{2h^2}(e^2-1)$ gives $e^2=1+\frac{2\epsilon h^2}{{\mu }^2}$, hence:

WHY the sign, physically. Negative energy means the body sits below the zero line of the potential well — no way out. Zero means it can just reach the rim with nothing to spare. Positive means it clears the rim with energy left over. The maths of $e$ merely photographs this.

PICTURE. A potential-energy well: a ball at negative energy rattles inside (ellipse), at zero energy tops out exactly at the rim (parabola), at positive energy sails over (hyperbola).

The one-picture summary

Everything above in a single frame: the same focus, three launch energies, three destinies — round loop, escape edge, flyby — all born from the one denominator $1+e\cos \theta$.

Active recall

Connections

• Conic Sections — the curves this derivation produces
• Vis-viva Equation — the speed law behind Step 8
• Specific Orbital Energy — the sign of $\epsilon$ drives the shape
• Angular Momentum in Orbits — sets $h$ and hence $p$
• Kepler's First Law — the ellipse special case
• Escape Velocity — the $e=1$ boundary