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

What we are deriving

HOW: derivation from first principles

Step 1 — Newton's law for the relative motion

Newton's gravity gives the acceleration of the orbiting body toward the central mass: $\ddot{\vec r} = -\frac{\mu}{r^2}\,\hat{r}, \qquad \mu \equiv GM.$

Why this step? This is just $\vec F = m\vec a$ with $F = GMm/r^2$ pointing inward ($-\hat r$); dividing by $m$ removes the orbiting mass, so all bodies follow the same path for given $\mu$.

Step 2 — Angular momentum is conserved (central force)

The specific angular momentum is $\vec h = \vec r \times \dot{\vec r}$. Differentiate: $\frac{d\vec h}{dt} = \dot{\vec r}\times\dot{\vec r} + \vec r\times\ddot{\vec r} = 0 + \vec r\times\left(-\frac{\mu}{r^2}\hat r\right) = 0.$

Why this step? $\vec r$ and $\hat r$ are parallel, so their cross product is zero. Hence $\vec h$ is constant ⇒ motion is planar, and its magnitude is $h = r^2\dot\theta = \text{const.}$

Step 3 — Substitution $u = 1/r$ (the trick that linearizes it)

Working in the plane with polar $(r,\theta)$, the radial acceleration is $\ddot r - r\dot\theta^2 = -\frac{\mu}{r^2}.$

We want $r$ as a function of $\theta$, not $t$. Use $h = r^2\dot\theta$, so $\dot\theta = h u^2$ where $u=1/r$.

Convert time derivatives to $\theta$ derivatives. Why? Because $\theta$ is monotonic (it always increases), so it's a clean independent variable. $\dot r = \frac{dr}{d\theta}\dot\theta = \frac{d}{d\theta}\!\left(\frac1u\right) h u^2 = -\frac1{u^2}\frac{du}{d\theta}\cdot h u^2 = -h\frac{du}{d\theta}.$ Differentiate again: $\ddot r = \frac{d}{dt}\!\left(-h\frac{du}{d\theta}\right) = -h\frac{d^2u}{d\theta^2}\dot\theta = -h\frac{d^2u}{d\theta^2}\,h u^2 = -h^2 u^2 \frac{d^2u}{d\theta^2}.$

Why this step? The $u$-substitution turns the nasty $1/r^2$ force into a constant, converting a hard nonlinear ODE into a simple linear one.

Step 4 — The linear oscillator equation

Plug $\ddot r$ and $r\dot\theta^2 = \frac1u (hu^2)^2 = h^2 u^3$ into the radial equation: $-h^2u^2\frac{d^2u}{d\theta^2} - h^2 u^3 = -\mu u^2.$ Divide by $-h^2 u^2$: $\boxed{\frac{d^2u}{d\theta^2} + u = \frac{\mu}{h^2}}$

Why this step? This is the equation of a simple harmonic oscillator with a constant driving term $\mu/h^2$. We know its solution by heart.

Step 5 — Solve and interpret

General solution = particular + homogeneous: $u(\theta) = \frac{\mu}{h^2} + A\cos(\theta - \theta_0).$ Choose $\theta_0 = 0$ (measure angle from where $r$ is smallest, i.e. perihelion). Then $\frac1r = \frac{\mu}{h^2}\bigl(1 + e\cos\theta\bigr), \qquad e \equiv \frac{A h^2}{\mu}.$ Invert: $\boxed{\,r = \frac{h^2/\mu}{1 + e\cos\theta} = \frac{p}{1+e\cos\theta}\,}$

Why this step? $A$ is an integration constant set by initial conditions; bundling it into the dimensionless $e$ gives the standard conic form. $p = h^2/\mu$ falls out automatically.

Worked examples

Common mistakes (steel-manned)

Active recall

Flashcards

Connections

• Kepler's Laws — orbit equation is the first law; $h=$const is the second.
• Conic Sections — $r=p/(1+e\cos\theta)$ is the polar conic about a focus.
• Specific Angular Momentum h — sets $p$ and the area-sweep rate.
• Vis-viva Equation — energy version that pairs with this geometry version.
• Eccentricity and Orbital Energy — $e=\sqrt{1+2Eh^2/\mu^2}$ links shape to energy.
• Central Force Motion — the general framework this is a special case of.

Concept Map

Hinglish (regional understanding)