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

This page builds every symbol the parent derivation leans on, starting from nothing. If a symbol appears in the parent, it is defined here first — with a picture and a reason it exists.

0 — The picture we will keep pointing at

Before any symbol, look at the scene. A big mass (call it the Sun) sits fixed. A small body (call it a planet) moves around it. We stand at the Sun and describe where the planet is.

Everything below is a label glued onto some part of this one drawing.

1 — Position by distance and angle: $r$ and $\theta$

Why does the topic need this and not $x,y$? Gravity points along the arrow — toward the Sun. A coordinate system whose first direction is "along the arrow" makes the physics fall out cleanly. In $x,y$ the force would smear across both axes; in $(r,\theta )$ it lives entirely in the $r$ direction. See Central Force Motion.

2 — The two direction-arrows: $\hat{r}$ and $\hat{\theta }$

Why does the topic need these? Because they let us split any motion into "toward/away from Sun" (the $\hat{r}$ part) and "swinging around" (the $\hat{\theta }$ part). Gravity is pure $\hat{r}$; that is the whole reason the sideways swing is left untouched — which becomes angular momentum in §6.

3 — Rates of change: the dot, $\dot{}$ and $\ddot{}$

Picture: $\dot{r}$ is the speedometer for the length of the arrow; $\dot{\theta }$ is the speedometer for the sweeping of the arrow.

Why does the topic need this? Motion is change. To write "gravity causes acceleration" we need a symbol for acceleration, and that is $\ddot{\vec{r}}$. The dot is just shorthand so equations stay short.

4 — Vectors and the arrow-multiply: $\vec{r}$, dot, and cross product

Two arrows can be "multiplied" in two different ways, and the topic uses one of them:

Two facts we will reuse:

• If the two arrows are parallel (same or opposite direction), the parallelogram is squashed flat — zero area — so $\vec{a}\times \vec{a}=0$ and $\vec{a}\times (-k\vec{a})=0$.
• The result's length measures how much sideways swing is packed between the arrows.

Why does the topic need the cross product and not ordinary multiplication? The topic must prove that "sideways swing" never changes. Only the cross product measures sideways-ness (through that perpendicular area). Ordinary number multiplication cannot see direction at all. This is the tool built for exactly this question. It powers Specific Angular Momentum h.

5 — Where gravity comes from: $G$, $M$, $m$, and $\mu$

Why bundle into $\mu$? Because $G$ and $M$ always appear glued together as $GM$. Giving the pair one name $\mu$ shortens every later formula and reminds us the planet's own mass has cancelled out — every planet at the same spot feels the same acceleration. See Kepler's Laws.

6 — The star of the show: specific angular momentum $h$

Why does the topic need $h$? Two reasons, both crucial:

1. Since gravity is parallel to $\vec{r}$ (both along $\hat{r}$), the cross product $\vec{r}\times \ddot{\vec{r}}=0$ (parallel arrows, §4), so $\vec{h}$ never changes. A constant is gold: we can trade the messy time variable for the angle.
2. $h=r^2\dot{\theta }$ becomes the bridge that converts "rate per second" into "change per angle" in the derivation.

7 — Radial acceleration in polar form: $\ddot{r}-r{\dot{\theta }}^2$

Here is the one line that looks scary but is pure geometry of the turning arrows from §2.

• $\ddot{r}$ = the genuine "distance is accelerating" part.
• $-r{\dot{\theta }}^2$ = the centripetal term: even if $r$ held steady, swinging in a circle is itself an inward acceleration (this is the same $v^2/r$ you feel on a merry-go-round, since $v_{\theta }=r\dot{\theta }$).

Why the extra term? Because $\hat{r}$ rotates (§2 mistake box). When you write acceleration in a spinning frame of arrows, the rotation leaks a $-r{\dot{\theta }}^2$ into the radial direction. It is not new physics — it is bookkeeping for the turning coordinates. The right side is just gravity from §5.

8 — The two dials of the final curve: $p$ and $e$

These are the payoffs — the numbers the whole derivation is chasing.

Why two separate dials? Because a conic section needs exactly two pieces of information: how big and how stretched. $p$ and $e$ carry precisely those, and both drop out of the physics automatically (from $h$, $\mu$, and one integration constant). Deeper meaning lives in Eccentricity and Orbital Energy, Conic Sections, and Vis-viva Equation.

Prerequisite map

Read it top-down: the coordinate and calculus symbols build the acceleration line; the cross product builds $h$; gravity builds the inward pull; all three streams merge into the orbit equation, with $p$ and $e$ as the output dials.

Equipment checklist

Cover the right side and test yourself. If any answer is fuzzy, re-read that section before the derivation.