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

Intuition

This page is a ladder. Each rung asks a little more of you than the last. Start at L1 (can you recognise the pieces?) and climb to L5 (can you build something new from them?). Every problem has a full solution hidden inside a collapsible box — try it yourself first, then reveal. All the physics lives in the parent note; here you use it.

One symbol reminder before we start, so nothing is unearned:

$r$ = distance from the focus (where the Sun/planet sits) to the orbiting body.
$θ$ = the true anomaly, the angle measured at the focus, starting from the closest point (perihelion, $θ = 0$ ).
$p$ = the semi-latus rectum, $p = h^{2} / μ$ — the value of $r$ when you are "sideways" ( $θ = 9 0^{\circ}$ ).
$e$ = eccentricity, a pure number ( $e \geq 0$ ) that sets the shape.
$μ$ = gravitational parameter, $μ = GM$ .
$h$ = specific angular momentum, $h = r^{2} \dot{θ}$ .

The master equation for every problem below: $r = \frac{p}{1 + e c o s θ}, p = \frac{h ^{2}}{μ} .$

Prerequisites you may want open: Conic Sections, Specific Angular Momentum h, Eccentricity and Orbital Energy, Vis-viva Equation, Kepler's Laws, Central Force Motion.

Level 1 — Recognition

Can you read the equation and identify its parts?

L1.1

In the orbit equation $r = \frac{p}{1 + e cos θ}$ , what physical distance does $r$ represent, and from which point is it measured?

Recall Solution

$r$ is the distance from the focus (the location of the central mass, e.g. the Sun) to the orbiting body — not from the geometric centre of the ellipse. This is the single most important reading of the equation. See the focus vs. centre picture:

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

The red dot is the focus; every $r$ arrow starts there.

L1.2

For an orbit with $e = 0$ , what shape is the trajectory, and why does $r$ no longer depend on $θ$ ?

Recall Solution

With $e = 0$ the equation becomes $r = \frac{p}{1 + 0} = p$ : a constant. A constant distance from the focus is a circle. Because the $cos θ$ term is multiplied by $e = 0$ , the angle drops out entirely — every direction is the same distance away.

L1.3

Match each eccentricity to its conic: $e = 0$ , $0 < e < 1$ , $e = 1$ , $e > 1$ .

Recall Solution

$e$	shape
$0$	circle
$0 < e < 1$	ellipse
$1$	parabola
$> 1$	hyperbola

Mnemonic: Cats Eat Plump Herring as $e$ climbs $0 \to> 1$ . See Conic Sections.

Level 2 — Application

Plug in numbers correctly, watch the signs.

L2.1

A satellite has $p = 8000$ km and $e = 0.25$ . Find $r$ at $θ = 0^{\circ}$ (perihelion) and $θ = 18 0^{\circ}$ (aphelion).

Recall Solution

Perihelion ( $θ = 0$ , $cos 0 = 1$ ): $r_{m i n} = \frac{p}{1 + e} = \frac{8000}{1.25} = 6400 km .$ Aphelion ( $θ = 18 0^{\circ}$ , $cos 18 0^{\circ} = - 1$ ): $r_{m a x} = \frac{p}{1 - e} = \frac{8000}{0.75} \approx 10667 km .$ Why the signs? $cos$ swings from $+ 1$ at $θ = 0$ down to $- 1$ at $θ = 18 0^{\circ}$ , so the denominator is largest (→ smallest $r$ ) at perihelion and smallest (→ largest $r$ ) at aphelion.

L2.2

Same satellite ( $p = 8000$ km, $e = 0.25$ ). Find $r$ at $θ = 9 0^{\circ}$ .

Recall Solution

$cos 9 0^{\circ} = 0$ , so $r = \frac{8000}{1 + 0.25 ( 0 )} = 8000 km = p .$ At $9 0^{\circ}$ you sit exactly one semi-latus rectum from the focus — this is what $p$ means.

L2.3

An orbit has $p = 7000$ km, $e = 0.4$ . Find $r$ at $θ = 12 0^{\circ}$ .

Recall Solution

$cos 12 0^{\circ} = - \frac{1}{2}$ , so $r = \frac{7000}{1 + 0.4 ( - 0.5 )} = \frac{7000}{1 - 0.2} = \frac{7000}{0.8} = 8750 km .$ Note $θ = 12 0^{\circ}$ is past $9 0^{\circ}$ , so $cos$ is negative and $r > p$ — you are on the "far" half of the orbit heading toward aphelion.

Level 3 — Analysis

Combine relations; go backwards from data.

L3.1

Mars has perihelion $r_{p} = 2.067 \times 1 0^{8}$ km and aphelion $r_{a} = 2.492 \times 1 0^{8}$ km. Find $e$ and $p$ .

Recall Solution

Use $r_{p} = \frac{p}{1 + e}$ and $r_{a} = \frac{p}{1 - e}$ . Divide: $\frac{r _{a}}{r _{p}} = \frac{1 + e}{1 - e} \Rightarrow e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{0.425 \times 1 0 ^{8}}{4.559 \times 1 0 ^{8}} \approx 0.0932.$ Then $p = r_{p} (1 + e) = 2.067 \times 1 0^{8} (1.0932) \approx 2.259 \times 1 0^{8}$ km. Why divide? Dividing cancels the unknown $p$ and isolates $e$ in one clean step.

L3.2

For an orbit around Earth ( $μ = 3.986 \times 1 0^{5} km^{3} / s^{2}$ ) with $p = 7000$ km, find the specific angular momentum $h$ .

Recall Solution

Invert $p = h^{2} / μ$ : $h = μ p = 3.986 \times 1 0^{5} \times 7000 \approx 5.28 \times 1 0^{4} km^{2} / s .$ This $h$ fixes the area-sweep rate $\frac{1}{2} h$ (Kepler's 2nd law — see Kepler's Laws and Specific Angular Momentum h).

L3.3

An ellipse has $e = 0.6$ and semi-major axis $a = 10000$ km. Find $p$ , then $r$ at $θ = 9 0^{\circ}$ . (Use $p = a (1 - e^{2})$ .)

Recall Solution

$p = a (1 - e^{2}) = 10000 (1 - 0.36) = 10000 (0.64) = 6400 km .$ At $θ = 9 0^{\circ}$ : $r = p = 6400$ km. Why $p = a (1 - e^{2})$ ? $p$ is the half-chord through the focus, $a$ is half the long axis; they only coincide when $e = 0$ .

Level 4 — Synthesis

Build several relations together; degenerate and limiting cases.

L4.1

A comet is on a parabolic orbit ( $e = 1$ ) with $p = 1.2 \times 1 0^{8}$ km. Find $r$ at $θ = 0^{\circ}$ , $θ = 9 0^{\circ}$ , and describe what happens as $θ \to 18 0^{\circ}$ .

Recall Solution

$θ = 0$ : $r = \frac{p}{1 + 1} = \frac{p}{2} = 6.0 \times 1 0^{7}$ km (perihelion).
$θ = 9 0^{\circ}$ : $r = p = 1.2 \times 1 0^{8}$ km.
As $θ \to 18 0^{\circ}$ : $cos θ \to - 1$ , denominator $1 + e cos θ \to 1 - 1 = 0$ , so $r \to \infty$ .

This is the signature of a parabola: it has no aphelion — the comet escapes to infinity along an open arc. The direction $θ = 18 0^{\circ}$ is unreachable (it would require infinite $r$ ). See the limiting-case figure:

L4.2

A hyperbolic flyby has $e = 1.5$ , $p = 9000$ km. There are angles where $r$ becomes infinite (the asymptotes). Find the true anomaly $θ_{\infty}$ where $r \to \infty$ .

Recall Solution

$r \to \infty$ when the denominator vanishes: $1 + e cos θ_{\infty} = 0 \Rightarrow cos θ_{\infty} = - \frac{1}{e} = - \frac{1}{1.5} = - 0.6667.$ $θ_{\infty} = arccos (- 0.6667) \approx 131. 8^{\circ} (and by symmetry - 131. 8^{\circ}) .$ For angles beyond $\pm 131. 8^{\circ}$ the denominator would go negative → unphysical negative $r$ . So the body only ever occupies $∣ θ ∣ < 131. 8^{\circ}$ : the two branches of the "swoosh." This is why hyperbolas are open.

Why does $cos θ = - 1/ e$ only have a solution when $e > 1$ ? Because $- 1/ e$ must lie in $[- 1, 1]$ ; that needs $e \geq 1$ . For an ellipse ( $e < 1$ ) the denominator never reaches zero, so $r$ stays finite everywhere — a closed orbit.

L4.3

Show that for a circular orbit ( $e = 0$ ) the equation is consistent with $r = a$ , and find $r$ if $μ = 3.986 \times 1 0^{5} km^{3} / s^{2}$ and $h = 6.0 \times 1 0^{4} km^{2} / s$ .

Recall Solution

With $e = 0$ : $r = p = h^{2} / μ$ for all $θ$ — a constant, so the orbit is a circle of radius $a = p$ . $r = \frac{h ^{2}}{μ} = \frac{( 6.0 \times 1 0 ^{4} ) ^{2}}{3.986 \times 1 0 ^{5}} = \frac{3.6 \times 1 0 ^{9}}{3.986 \times 1 0 ^{5}} \approx 9032 km .$

Level 5 — Mastery

Design and prove; connect to energy and full derivation.

L5.1

Starting from $r = \frac{p}{1 + e cos θ}$ with $p = a (1 - e^{2})$ , prove that the sum $r_{m i n} + r_{m a x} = 2 a$ for an ellipse.

Recall Solution

$r_{m i n} = \frac{p}{1 + e} = \frac{a ( 1 - e ^{2} )}{1 + e} = \frac{a ( 1 - e ) ( 1 + e )}{1 + e} = a (1 - e) .$ $r_{m a x} = \frac{p}{1 - e} = \frac{a ( 1 - e ^{2} )}{1 - e} = a (1 + e) .$ Add: $r_{m i n} + r_{m a x} = a (1 - e) + a (1 + e) = 2 a .$ This is exactly the long axis of the ellipse — confirming $a$ is the semi-major axis. Notice the factorisation $1 - e^{2} = (1 - e) (1 + e)$ is what makes it collapse so cleanly.

L5.2

A spacecraft is measured to have $r = 7500$ km at true anomaly $θ = 6 0^{\circ}$ , and $r = 9000$ km at $θ = 12 0^{\circ}$ . Determine $e$ and $p$ .

Recall Solution

Write the equation as $\frac{1}{r} = \frac{1}{p} (1 + e cos θ)$ . Two data points give two linear equations in unknowns $x = 1/ p$ and $y = e / p$ : $\frac{1}{r} = x + y cos θ .$

$θ = 6 0^{\circ}, cos = 0.5$ : $\frac{1}{7500} = x + 0.5 y$ .
$θ = 12 0^{\circ}, cos = - 0.5$ : $\frac{1}{9000} = x - 0.5 y$ .

Add the two: $2 x = \frac{1}{7500} + \frac{1}{9000} = 1.3333 \times 1 0^{- 4} + 1.1111 \times 1 0^{- 4} = 2.4444 \times 1 0^{- 4}$ , so $x = 1.2222 \times 1 0^{- 4} \Rightarrow p = \frac{1}{x} \approx 8182 km .$ Subtract: $y = \frac{1}{0.5} (\frac{1}{7500} - x) = 2 (1.3333 \times 1 0^{- 4} - 1.2222 \times 1 0^{- 4}) = 2.2222 \times 1 0^{- 5}$ , and $e = y \cdot p = y / x$ : $e = \frac{2.2222 \times 1 0 ^{- 5}}{1.2222 \times 1 0 ^{- 4}} \approx 0.1818.$ So $p \approx 8182$ km, $e \approx 0.182$ . Why linearise in $1/ r$ ? Because the equation is linear in $x$ and $y$ — turning a nasty fit into simple simultaneous equations.

L5.3

Using $p = h^{2} / μ$ and $p = a (1 - e^{2})$ , show that $h = μ a (1 - e^{2})$ , then evaluate for an Earth orbit with $a = 10000$ km, $e = 0.2$ , $μ = 3.986 \times 1 0^{5} km^{3} / s^{2}$ .

Recall Solution

Set the two expressions for $p$ equal: $\frac{h ^{2}}{μ} = a (1 - e^{2}) \Rightarrow h = μ a (1 - e^{2}) .$ Plug in: $h = 3.986 \times 1 0^{5} \times 10000 \times (1 - 0.04) = 3.986 \times 1 0^{5} \times 9600 .$ $h = 3.8266 \times 1 0^{9} \approx 6.186 \times 1 0^{4} km^{2} / s .$ This links geometry ( $a, e$ ) directly to the conserved $h$ — the bridge to the Vis-viva Equation and Eccentricity and Orbital Energy.

Active recall

Recall One-line self-tests

$r$ at $θ = 9 0^{\circ}$ ? ::: $r = p$ .
$r_{m i n} + r_{m a x}$ for an ellipse? ::: $2 a$ .
Where does $r \to \infty$ on a hyperbola? ::: $cos θ = - 1/ e$ .
Relation $h (a, e)$ ? ::: $h = μ a (1 - e^{2})$ .