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

Before we start, meet the five characters in the formula — each one earned in plain words:

$r$ — distance from the focus (the Sun/planet sits there), always $r > 0$ .
$θ$ — the true anomaly, the angle measured at the focus, starting from perihelion ( $θ = 0$ ).
$e$ — the eccentricity, a pure number $\geq 0$ that decides the shape.
$μ$ — the gravitational parameter, $μ = GM$ : it bundles the gravitational constant $G$ with the central mass $M$ , and is the single number that measures "how strong is the pull". Bigger $μ$ means stronger gravity.
$h$ — the specific angular momentum, $h = r^{2} \dot{θ}$ : the "spin amount per unit mass" of the orbiting body. Because gravity pulls straight inward it can never change $h$ , so $h$ stays constant along the whole orbit (see Specific Angular Momentum h).

With $μ$ and $h$ now named, the scale of the orbit is:

$p = \frac{h ^{2}}{μ}$ — the semi-latus rectum, the value of $r$ when $θ = 9 0^{\circ}$ (spin-squared divided by pull-strength).

If any of these still feel shaky, the derivation note for $r = p/(1+e\cos\theta)$ earns each one from Newton's law.

The scenario matrix

Every question about this equation lands in one of these cells. The examples below are tagged with the cell they cover.

Cell	What changes	Covered by
A. Circle	$e = 0$ (degenerate: shape term vanishes)	Ex 1
B. Ellipse, given extremes	$0 < e < 1$ , find $e, p$ from $r_{m i n}, r_{m a x}$	Ex 2
C. Quadrant sweep	$cos θ$ sign in each quadrant ( $θ = 0, 9 0^{\circ}, 18 0^{\circ}, 27 0^{\circ}$ )	Ex 3
D. Parabola limit	$e = 1$ (aphelion $\to \infty$ )	Ex 4
E. Hyperbola / escape	$e > 1$ (a forbidden angle exists)	Ex 5
F. Recover $h$ or $μ$	invert $p = h^{2} / μ$	Ex 6
G. Real-world word problem	messy units, physical sanity check	Ex 7
H. Exam twist	given $r$ and $θ$ , solve for $θ$ (two answers!)	Ex 8

Worked examples

Ex 1 — Cell A: the circle ( $e = 0$ )

Worked example

A weather satellite has $p = 8000$ km and $e = 0$ . Find $r$ at $θ = 0^{\circ}$ , $9 0^{\circ}$ , and $18 0^{\circ}$ .

Forecast: guess before reading — does $r$ change as $θ$ sweeps around?

Write the equation with $e = 0$ : $r = \frac{p}{1 + 0 \cdot cos θ} = \frac{p}{1} = p$ . Why this step? When $e = 0$ the whole $e cos θ$ term dies, so $θ$ drops out entirely. This is the degenerate case where the shape term vanishes.
Therefore $r = 8000$ km at every angle. Why this step? A constant $r$ for all $θ$ is the definition of a circle — no perihelion, no aphelion, no favoured direction.

Verify: The two extremes of $r$ occur where $cos θ$ is most extreme: since $cos θ$ ranges over $[- 1, + 1]$ , the biggest denominator (smallest $r$ ) is at $cos θ = + 1$ , i.e. $θ = 0$ , giving $r_{m i n} = p / (1 + e)$ ; the smallest denominator (largest $r$ ) is at $cos θ = - 1$ , i.e. $θ = 18 0^{\circ}$ , giving $r_{m a x} = p / (1 - e)$ . Here $r_{m i n} = p / (1 + e) = 8000/1 = 8000$ and $r_{m a x} = p / (1 - e) = 8000/1 = 8000$ . They agree — the two "extremes" collapse into one radius, exactly as a circle demands. Units: km throughout. ✓

Ex 2 — Cell B: ellipse from its two extremes

Worked example

A comet has perihelion $r_{p} = 0.6$ AU and aphelion $r_{a} = 5.4$ AU. Find $e$ and $p$ .

Forecast: will $e$ be closer to $0$ (round) or to $1$ (very stretched)?

The extremes are $r_{p} = \frac{p}{1 + e}$ (at $θ = 0$ ) and $r_{a} = \frac{p}{1 - e}$ (at $θ = π$ ). Why this step? $cos θ$ is largest ( $+ 1$ ) at $θ = 0$ giving the smallest $r$ , and smallest ( $- 1$ ) at $θ = π$ giving the largest $r$ . This is only valid for $e < 1$ ; we will confirm that at the end.
Divide the two: $\frac{r _{a}}{r _{p}} = \frac{1 + e}{1 - e}$ , so $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{5.4 - 0.6}{5.4 + 0.6} = \frac{4.8}{6.0} = 0.8$ . Why this step? Dividing cancels the unknown $p$ , leaving a single equation in $e$ only.
Now $p = r_{p} (1 + e) = 0.6 (1.8) = 1.08$ AU. Why this step? With $e$ known, either extreme equation gives $p$ directly.

Verify: back-substitute into aphelion: $p / (1 - e) = 1.08/ (1 - 0.8) = 1.08/0.2 = 5.4$ AU $= r_{a}$ . ✓ And $0 < e = 0.8 < 1$ confirms this really is a (very stretched) ellipse, so Step 1 was legitimate.

Ex 3 — Cell C: sweeping through all four quadrants

Worked example

An orbit has $p = 10000$ km and $e = 0.5$ . Tabulate $r$ at $θ = 0^{\circ}, 9 0^{\circ}, 18 0^{\circ}, 27 0^{\circ}$ and describe the motion.

Forecast: at which of these four angles is the body closest to the focus? Farthest?

The figure below plots exactly this orbit: the orange star is the focus, the violet curve is the path, and the four dashed radii are the four angles we compute. Watch how the radius arrow is shortest pointing to the right ( $θ = 0$ ) and longest pointing left ( $θ = 18 0^{\circ}$ ).

$θ = 0^{\circ}$ (perihelion): $cos 0 = + 1$ , so $r = \frac{10000}{1 + 0.5 ( 1 )} = \frac{10000}{1.5} = 6666.7$ km — smallest. This is the short magenta arrow pointing right in the figure. Why this step? Adding the full $+ e$ makes the denominator biggest, so $r$ is smallest. This is the closest point.
$θ = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0$ , so $r = \frac{10000}{1 + 0} = 10000$ km $= p$ . This is the upward orange arrow of length $p$ . Why this step? At $9 0^{\circ}$ the cosine term vanishes and $r$ equals the semi-latus rectum — the "sideways" reference distance. Same at $27 0^{\circ}$ by symmetry (Step 4).
$θ = 18 0^{\circ}$ (aphelion): $cos 18 0^{\circ} = - 1$ , so $r = \frac{10000}{1 - 0.5} = \frac{10000}{0.5} = 20000$ km — largest. This is the long navy arrow pointing left. Why this step? Subtracting the full $e$ makes the denominator smallest, so $r$ is largest — the farthest point.
$θ = 27 0^{\circ}$ : $cos 27 0^{\circ} = 0$ , so $r = 10000$ km again — the downward orange arrow, a mirror of Step 2. Why this step? Cosine is even about the perihelion axis, so $θ$ and $- θ$ (i.e. $27 0^{\circ} = - 9 0^{\circ}$ ) give identical $r$ . The orbit is mirror-symmetric top-to-bottom, which you can see directly in the figure.

Verify: $r$ rises monotonically $6666.7 \to 10000 \to 20000$ as $θ$ goes $0 \to 90 \to 180$ , then falls back symmetrically — exactly the smooth in-and-out of a closed ellipse. Also $r_{m i n}$ check: $p / (1 + e) = 10000/1.5 = 6666.7$ ✓.

Ex 4 — Cell D: the parabola limit ( $e = 1$ )

Worked example

A probe is launched on a parabolic escape trajectory with $p = 12000$ km ( $e = 1$ ). Find $r$ at $θ = 0^{\circ}$ , $θ = 9 0^{\circ}$ , and describe what happens as $θ \to 18 0^{\circ}$ .

Forecast: does the probe ever come back? What does $r_{m a x}$ equal?

Perihelion, $θ = 0$ : $r = \frac{12000}{1 + 1 ( 1 )} = \frac{12000}{2} = 6000$ km. Why this step? Even at $e = 1$ the perihelion formula $p / (1 + e)$ is finite and well behaved — the parabola still has a closest point.
$θ = 9 0^{\circ}$ : $r = \frac{12000}{1 + 0} = 12000$ km $= p$ . Why this step? The " $r = p$ at $9 0^{\circ}$ " rule holds for every conic, because it comes from $cos 9 0^{\circ} = 0$ , independent of $e$ .
As $θ \to 18 0^{\circ}$ : denominator $1 + cos θ \to 1 + (- 1) = 0$ , so $r \to \frac{12000}{0 ^{+}} \to + \infty$ . Why this step? With $e = 1$ exactly, the aphelion " $p / (1 - e)$ " would divide by zero — the orbit never closes. The probe recedes forever, so there is no aphelion. This is the borderline between bound ( $e < 1$ ) and unbound ( $e > 1$ ).

Verify: $r_{m i n} = p / (1 + e) = 12000/2 = 6000$ km matches Step 1. And $lim_{θ \to π} p / (1 + cos θ) = \infty$ confirms the open, never-returning path. ✓

Ex 5 — Cell E: hyperbola with a forbidden angle ( $e > 1$ )

Worked example

An interstellar object flies past the Sun on a hyperbola with $p = 2$ AU and $e = 2$ . Find $r$ at $θ = 0^{\circ}$ and $θ = 6 0^{\circ}$ , then find the angle beyond which $r$ becomes negative (physically unreachable).

Forecast: for a hyperbola, is there a maximum angle the object can ever be seen at?

The figure shows the open magenta hyperbola: the orbit only exists between the two dotted navy asymptote rays. As the object races out toward either ray, its distance $r$ shoots to infinity — those rays are the forbidden angles we are about to compute.

Perihelion, $θ = 0$ : $r = \frac{2}{1 + 2 ( 1 )} = \frac{2}{3} \approx 0.667$ AU (the violet dot nearest the focus in the figure). Why this step? Closest approach still uses $p / (1 + e)$ ; larger $e$ pulls the object in tighter at perihelion.
$θ = 6 0^{\circ}$ : $cos 6 0^{\circ} = 0.5$ , so $r = \frac{2}{1 + 2 ( 0.5 )} = \frac{2}{2} = 1$ AU. Why this step? Straightforward substitution; still positive because the denominator is still positive.
Find where the denominator hits zero: $1 + e cos θ = 0 \Rightarrow cos θ = - \frac{1}{e} = - \frac{1}{2}$ , so $θ = 12 0^{\circ}$ — exactly the dotted asymptote ray in the figure. Why this step? Beyond this angle $1 + e cos θ < 0$ , making $r < 0$ — impossible. So the object exists only for $∣ θ ∣ < 12 0^{\circ}$ ; the two rays $θ = \pm 12 0^{\circ}$ are the asymptotes of the hyperbola. This "forbidden angle" only exists when $e > 1$ (since we need $1/ e < 1$ ).

Verify: $cos 12 0^{\circ} = - 0.5 = - 1/ e$ , and $1 + 2 (- 0.5) = 0$ ✓ — denominator vanishes exactly there. For $e < 1$ the equation $cos θ = - 1/ e$ has no solution (since $- 1/ e < - 1$ ), which is why ellipses have no forbidden angle. Consistent with Ex 2–3. ✓

Ex 6 — Cell F: recovering $h$ from the geometry

Worked example

A satellite around Earth has measured $p = 7000$ km. Earth's $μ = 3.986 \times 1 0^{5} km^{3} / s^{2}$ . Find the specific angular momentum $h$ and the area-sweep rate.

Forecast: will $h$ have units of km²/s or km³/s²?

Invert the definition $p = \frac{h ^{2}}{μ}$ : $h = μ p$ . Why this step? $p$ came out of the derivation as $h^{2} / μ$ ; solving for $h$ recovers the physical constant of motion from the orbit's shape alone.
Compute: $h = (3.986 \times 1 0^{5}) (7000) = 2.7902 \times 1 0^{9} \approx 5.282 \times 1 0^{4} km^{2} / s$ . Why this step? Units: $(km^{3} / s^{2}) (km) = km^{4} / s^{2} = km^{2} / s$ — an area per time, as $h$ must be.
Kepler's equal-areas rate is $\frac{d A}{d t} = \frac{h}{2} \approx 2.641 \times 1 0^{4} km^{2} / s$ . Why this step? The area swept by the radius vector per unit time is exactly $h /2$ — this is Kepler's second law falling straight out of $h = r^{2} \dot{θ}$ .

Verify: square $h$ back: $h^{2} / μ = (5.282 \times 1 0^{4})^{2} / (3.986 \times 1 0^{5}) = 2.790 \times 1 0^{9} /3.986 \times 1 0^{5} = 7000$ km $= p$ ✓. Units check in Step 2. Links to Specific Angular Momentum h.

Ex 7 — Cell G: real-world word problem (Earth's orbit)

Worked example

Earth's perihelion is $1.471 \times 1 0^{8}$ km and aphelion is $1.521 \times 1 0^{8}$ km. Find $e$ , $p$ , and check that $p$ is about 1 AU ( $1 AU = 1.496 \times 1 0^{8}$ km).

Forecast: is Earth's orbit nearly circular or noticeably squashed?

$e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{1.521 - 1.471}{1.521 + 1.471} = \frac{0.050}{2.992} \approx 0.01671$ . Why this step? Same extreme-ratio trick as Ex 2; small $e$ means nearly circular.
$p = r_{p} (1 + e) = 1.471 \times 1 0^{8} (1.01671) \approx 1.4956 \times 1 0^{8}$ km. Why this step? Plug $e$ back into the perihelion equation to get the scale.
Compare to 1 AU: $1.4956 \times 1 0^{8} /1.496 \times 1 0^{8} \approx 0.9997$ , essentially 1 AU. Why this step? Sanity check against a known real number — a physicist always asks "does this match reality?"

Verify: $e \approx 0.0167$ is famously Earth's eccentricity (nearly circular, as expected). Aphelion check: $p / (1 - e) = 1.4956 \times 1 0^{8} / (0.98329) \approx 1.521 \times 1 0^{8}$ km $= r_{a}$ ✓.

Ex 8 — Cell H: exam twist — solve for the angle (two answers)

Worked example

An orbit has $p = 9000$ km, $e = 0.3$ . At what true anomaly $θ$ is $r = 10000$ km?

Forecast: how many angles in $[0^{\circ}, 36 0^{\circ})$ satisfy this — one, or two?

Start from $r = \frac{p}{1 + e cos θ}$ and solve for $cos θ$ : $1 + e cos θ = \frac{p}{r}$ , so $cos θ = \frac{1}{e} (\frac{p}{r} - 1)$ . Why this step? We are inverting the formula — instead of "angle in, distance out" we go "distance in, angle out".
Plug in: $cos θ = \frac{1}{0.3} (\frac{9000}{10000} - 1) = \frac{1}{0.3} (0.9 - 1) = \frac{- 0.1}{0.3} = - 0.3333$ . Why this step? Numbers substituted; the negative value tells us we are past $9 0^{\circ}$ (beyond the semi-latus rectum, moving outward toward aphelion).
$θ = arccos (- 0.3333) \approx 109.4 7^{\circ}$ — and also $θ = 36 0^{\circ} - 109.4 7^{\circ} = 250.5 3^{\circ}$ . Why this step? Cosine gives the same value for $θ$ and $- θ$ , so two points on the orbit are at this distance — one on the way out (approaching aphelion) and one on the way back in. Missing the second root is the classic exam slip.

Verify: at $θ = 109.4 7^{\circ}$ : $r = 9000/ (1 + 0.3 (- 0.3333)) = 9000/ (1 - 0.1) = 9000/0.9 = 10000$ km ✓. At $θ = 250.5 3^{\circ}$ : $cos 250.5 3^{\circ} = - 0.3333$ too, so $r = 9000/ (1 + 0.3 (- 0.3333)) = 9000/0.9 = 10000$ km again ✓. Both angles are valid physical positions. As a range check, $r = 10000$ km must lie between $r_{m i n} = p / (1 + e) = 9000/1.3 = 6923$ km and $r_{m a x} = p / (1 - e) = 9000/0.7 = 12857$ km — and it does. Conclusion: the satellite is at $r = 10000$ km at exactly two true anomalies, $θ \approx 109.4 7^{\circ}$ (outbound) and $θ \approx 250.5 3^{\circ}$ (inbound). ✓

Active recall

Recall Which cell does each fact come from?

$e = 0$ makes $θ$ vanish, giving a constant $r$ → Cell A (circle).
The angle where $1 + e cos θ = 0$ only exists if $e > 1$ → Cell E (hyperbola asymptote).
$e = 1$ sends $r \to \infty$ at $θ = 18 0^{\circ}$ → Cell D (parabola, no aphelion).
Solving for $θ$ from a given $r$ gives two answers → Cell H.

Recall Self-test

Given $r = r_{m i n}$ observed, which $θ$ ? ::: $θ = 0$ (perihelion). For $e = 2$ , the forbidden angle satisfies? ::: $cos θ = - 1/ e = - 0.5$ , i.e. $θ = 12 0^{\circ}$ . Why do distance-to-angle problems have two solutions? ::: Because $cos θ = cos (- θ)$ , so $θ$ and $36 0^{\circ} - θ$ share the same $r$ .

See also Conic Sections, Vis-viva Equation, Eccentricity and Orbital Energy, and Central Force Motion for where these numbers come from.