Kepler's first law — orbits are conic sections

The law (in one line): The orbit of a planet is an ellipse with the Sun at one focus (not the center!). More generally, any two-body gravitational orbit is a conic section — ellipse, parabola, or hyperbola.

Why does this law even exist?

WHAT we want: prove that the shape $r(\theta)$ of an orbit is a conic. WHY it matters: it tells us where a planet/satellite is, lets us classify trajectories (bound vs escape), and underpins all of astrodynamics. HOW we get there: combine conservation of angular momentum + Newton's gravity, and solve the radial equation with a clever substitution.

Building blocks

Derivation from scratch (the heart of the note)

We have a small body of mass $m$ orbiting a large mass $M$ at the origin. Let $\mu = GM$.

Step 1 — Equation of motion

Newton's second law with gravity: $\ddot{\vec r} = -\frac{\mu}{r^2}\,\hat r$ Why this step? Gravity points from $m$ toward $M$ (toward the origin), magnitude $GM/r^2$ per unit mass.

Step 2 — Use angular momentum conservation

Because the force is central, $h = r^2\dot\theta$ is constant. We'll use this to change variables from time $t$ to angle $\theta$. Why? Time is awkward; we want the shape $r(\theta)$, not the timetable. Eliminating $t$ gives geometry directly.

Step 3 — The magic substitution $u = 1/r$

Let $u = 1/r$. Then: $\dot\theta = \frac{h}{r^2} = h u^2$ $\dot r = \frac{dr}{d\theta}\dot\theta = \frac{d}{d\theta}\!\left(\frac1u\right) h u^2 = -\frac{1}{u^2}\frac{du}{d\theta}\,h u^2 = -h\frac{du}{d\theta}$ $\ddot r = \frac{d}{dt}\!\left(-h\frac{du}{d\theta}\right) = -h\frac{d^2u}{d\theta^2}\dot\theta = -h^2 u^2 \frac{d^2u}{d\theta^2}$ Why this step? $u=1/r$ turns the ugly nonlinear $1/r^2$ equation into a clean linear one — this is the famous trick that makes the whole thing solvable by hand.

Step 4 — Radial equation of motion

The radial acceleration (in polar coordinates) is $\ddot r - r\dot\theta^2$. Setting it equal to the radial force per mass: $\ddot r - r\dot\theta^2 = -\frac{\mu}{r^2}$ Substitute $\ddot r = -h^2u^2\,u''$, $r = 1/u$, $\dot\theta^2 = h^2u^4$: $-h^2u^2 u'' - \frac{1}{u}\,h^2u^4 = -\mu u^2$ Divide through by $-h^2u^2$: $u'' + u = \frac{\mu}{h^2}$ Why this step? Look at this — it's just simple harmonic motion with a constant drive! We already know how to solve it.

Step 5 — Solve the linear ODE

$u(\theta) = \frac{\mu}{h^2} + A\cos(\theta - \theta_0)$ Choose the angle origin so $\theta_0 = 0$ (measure $\theta$ from perihelion). Then back to $r = 1/u$: $\boxed{ r(\theta) = \frac{h^2/\mu}{1 + (Ah^2/\mu)\cos\theta} }$

Step 6 — Read off the conic

Compare with the conic $r = \dfrac{p}{1 + e\cos\theta}$: $p = \frac{h^2}{\mu}, \qquad e = \frac{A h^2}{\mu}$ This IS a conic section. Kepler's first law proven. ∎

Connecting eccentricity to energy (so we know which conic)

The total energy per unit mass is $\mathcal E = \tfrac12 v^2 - \mu/r$. Working it through (using $v^2 = \dot r^2 + r^2\dot\theta^2$ and the solution above) gives the beautiful relation: $e = \sqrt{1 + \frac{2\mathcal E h^2}{\mu^2}}$

Worked examples

Common mistakes (Steel-manned)

Flashcards

Connections

• Kepler's second law — equal areas in equal times (conservation of $h$, same ingredient)
• Kepler's third law — period vs semi-major axis (follows from the ellipse geometry)
• Conservation of angular momentum in central forces
• Vis-viva equation ($v^2 = \mu(2/r - 1/a)$ — energy version of this orbit)
• Conic sections — geometry of ellipse, parabola, hyperbola
• Two-body problem and reduced mass
• Escape velocity and orbital energy
• Inverse-square law and Bertrand's theorem (why only $1/r^2$ closes orbits)

Concept Map

Hinglish (regional understanding)