Orbit shape from eccentricity — circle (e=0), ellipse (0 - e - 1), parabola (e=1), hyperbola (e - 1): Worked examples

This page drills the one idea from the parent note: the eccentricity $e$ decides the shape. Here we hit every case the topic can throw at you — each sign of energy, each shape, the degenerate limits, a word problem, and an exam twist. Guess before you compute.

Before we start, three symbols we will lean on, each earned in the parent note:

Recall The three formulas we reuse (all from the parent)

Orbit shape: $r (θ) = \frac{p}{1 + e cos θ}$ , with $p = \frac{h ^{2}}{μ}$ the semi-latus rectum (the distance $r$ at $θ = 9 0^{\circ}$ ).
Eccentricity from energy: $e = 1 + \frac{2 ε h ^{2}}{μ ^{2}}$ , where $ε = E / m$ is specific energy (energy per kilogram) and $h = L / m$ is specific angular momentum (per kilogram). See Specific Orbital Energy and Angular Momentum in Orbits.
$μ = GM$ is the gravitational parameter of the central mass — one number bundling $G$ and $M$ so we never carry both.

Every symbol above was built in the parent; here we only use them. If a new tool appears (e.g. why we take a square root, why $cos$ ), it gets explained on the spot.

The scenario matrix

Think of the topic as a machine: you feed it $(ε, h)$ or $(e)$ and it outputs a shape. Every "cell" below is a genuinely different behaviour of that machine. The worked examples are chosen so that together they light up every cell.

Cell	Input character	Shape	Sign of $ε$	Covered by
A	$e = 0$ exactly (degenerate: foci merge)	circle	$ε < 0$	Ex 1
B	$0 < e < 1$ (generic bound)	ellipse	$ε < 0$	Ex 2
C	$e = 1$ exactly (knife-edge)	parabola	$ε = 0$	Ex 3
D	$e > 1$ (generic unbound)	hyperbola	$ε > 0$	Ex 4
E	Sign check: given $(ε, h)$ , classify	any	any	Ex 5
F	Angle where $r \to \infty$ (all $e$ )	para/hyper	$ε \geq 0$	Ex 6
G	Word problem: real body ( $e$ near 0)	near-circle	$ε < 0$	Ex 7
H	Limit: $e \to 1^{-}$ (ellipse becoming parabola)	ellipse→parabola	$ε \to 0^{-}$	Ex 8
I	Exam twist: same $e$ , different sizes	ellipse	$ε < 0$	Ex 9
J	Degenerate: $h = 0$ (radial fall, straight line)	$e = 1$ line	$ε < 0$ possible	Ex 10

We keep numbers simple by working in units where $μ = 1$ unless a real body is named.

The figure below is our map for the whole page. It draws all four shapes from a single focus at the origin (the black dot = the central mass), using the same $p = 1$ . Watch how turning the one dial $e$ from $0$ upward inflates a circle (solid black) into an ellipse, then — at the red dashed curve $e = 1$ — the loop breaks open into a parabola, and finally into the open hyperbola ( $e = 1.6$ , dotted). The red curve is the knife-edge every example returns to.

Figure — Orbit shape from eccentricity — circle (e=0), ellipse (0 - e - 1), parabola (e=1), hyperbola (e - 1)

The worked examples

Cell A — the circle ( $e = 0$ )

Worked example Ex 1 · A circular orbit

Take $μ = 1$ and specific angular momentum $h = 2$ . The orbit has $e = 0$ . Find $p$ , and show $r$ is the same at every angle. What is that constant radius?

Forecast: Guess — will $r$ depend on $θ$ at all? What number does it settle to?

Compute $p = h^{2} / μ$ . With $h = 2, μ = 1$ : $p = 4/1 = 4$ .
- Why this step? $p$ is the only length in the shape formula; nail it first.
Plug $e = 0$ into $r (θ) = \frac{p}{1 + e cos θ}$ . The $cos θ$ term is multiplied by $e = 0$ , so it vanishes: $r = \frac{p}{1 + 0} = p$ .
- Why this step? Setting $e = 0$ is exactly the "degenerate" cell — the two foci merge and the angle-dependence disappears.
Read off the radius. $r = p = 4$ for every $θ$ .
- Why this step? A constant $r$ is the definition of a circle — this confirms $e = 0$ gives cell A.

Verify: Test $θ = 0, 9 0^{\circ}, 18 0^{\circ}$ : denominator is $1$ each time, so $r = 4$ always. Units: $h^{2} / μ$ has units of length (as it must, since $r$ is a length). ✓

Cell B — the generic ellipse ( $0 < e < 1$ )

Intuition Deriving

$r_{m i n}$ , $r_{m a x}$ and $p = a (1 - e^{2})$ from scratch Everything about an ellipse's endpoints falls straight out of the shape formula $r (θ) = \frac{p}{1 + e cos θ}$ .

Step 1 — the near and far points. $r$ is smallest where the denominator is biggest. $cos θ$ tops out at $+ 1$ (at $θ = 0$ ), so $r_{m i n} = \frac{p}{1 + e} .$ $r$ is largest where the denominator is smallest: $cos θ = - 1$ (at $θ = 18 0^{\circ}$ ), so $r_{m a x} = \frac{p}{1 - e} .$

Step 2 — connect to size $a$ . By our definition $a = \frac{r _{m i n} + r _{m a x}}{2}$ . Add the two fractions: $r_{m i n} + r_{m a x} = \frac{p}{1 + e} + \frac{p}{1 - e} = \frac{p ( 1 - e ) + p ( 1 + e )}{( 1 + e ) ( 1 - e )} = \frac{2 p}{1 - e ^{2}} .$ Halving, $a = \frac{p}{1 - e ^{2}}$ , i.e. $p = a (1 - e^{2})$ .

Step 3 — endpoints in terms of $a, e$ . Substitute $p = a (1 - e^{2}) = a (1 + e) (1 - e)$ back: $r_{m i n} = \frac{a ( 1 + e ) ( 1 - e )}{1 + e} = a (1 - e), r_{m a x} = \frac{a ( 1 + e ) ( 1 - e )}{1 - e} = a (1 + e) .$ So the famous formulas $r_{m i n} = a (1 - e)$ and $r_{m a x} = a (1 + e)$ are derived, not assumed. The red long-axis in the figure below marks these two endpoints; the red vertical segment is $p$ , the sideways width at $θ = 9 0^{\circ}$ .

Worked example Ex 2 · Perihelion and aphelion of an ellipse

An orbit has $e = 0.6$ and semi-major axis $a = 10$ (units where $μ = 1$ ). Find the closest distance $r_{m i n}$ and farthest distance $r_{m a x}$ , and the semi-latus rectum $p$ .

Forecast: Guess — is the far point more or less than twice the near point?

Closest approach $r_{m i n} = a (1 - e)$ . $= 10 (1 - 0.6) = 10 (0.4) = 4$ .
- Why this step? $θ = 0$ makes $cos θ = + 1$ , the biggest denominator $1 + e$ , hence smallest $r$ — the formula we just derived.
Farthest point $r_{m a x} = a (1 + e)$ . $= 10 (1.6) = 16$ .
- Why this step? $θ = 18 0^{\circ}$ gives $cos θ = - 1$ , smallest denominator $1 - e$ , largest $r$ .
Semi-latus rectum $p = a (1 - e^{2})$ . $= 10 (1 - 0.36) = 10 (0.64) = 6.4$ .
- Why this step? Using the identity we just derived above ( $p = a (1 + e) (1 - e)$ ); it links the two views (shape formula vs $a, e$ ), and we'll need it in Ex 6.

Verify: $r_{m i n} + r_{m a x} = 4 + 16 = 20 = 2 a$ ✓ (major axis = near + far). Also $p$ should equal $r$ at $θ = 9 0^{\circ}$ : $p / (1 + 0.6 cos 9 0^{\circ}) = 6.4/1 = 6.4 = p$ ✓. See Kepler's First Law.

Cell C — the parabola ( $e = 1$ )

Worked example Ex 3 · A parabolic comet

A comet is on a parabola, $e = 1$ , with $p = 2$ (so $μ = 1, h = 2$ ). Find $r$ at perihelion ( $θ = 0$ ) and describe what happens as $θ \to 18 0^{\circ}$ .

Forecast: Guess — does $r$ reach a finite maximum, or run off to infinity?

Perihelion, $θ = 0$ . $r = \frac{p}{1 + 1 \cdot cos 0} = \frac{2}{1 + 1} = \frac{2}{2} = 1$ .
- Why this step? The closest point is always at $θ = 0$ ; it's the one distance we can pin down exactly.
Approach $θ = 18 0^{\circ}$ . Denominator $1 + cos 18 0^{\circ} = 1 - 1 = 0$ , so $r \to \infty$ .
- Why this step? This zero-denominator is the signature of $e = 1$ : the orbit opens. Unlike an ellipse it never closes.
Interpret the energy. For $e = 1$ , the relation $e = 1 + 2 ε h^{2} / μ^{2}$ forces $2 ε h^{2} / μ^{2} = 0$ , i.e. $ε = 0$ .
- Why this step? Zero specific energy = barely escapes, arriving at infinity with zero speed. This is Escape Velocity's boundary.

Verify: With $μ = 1$ , $p = h^{2} / μ = h^{2} \Rightarrow h = 2 \Rightarrow h^{2} = 2 = p$ ✓. And $ε = 0$ gives $e = 1 + 0 = 1$ ✓, matching the shape we assumed.

Cell D — the generic hyperbola ( $e > 1$ )

Worked example Ex 4 · A hyperbolic flyby

An interstellar object has $e = 2$ and $p = 3$ . Find perihelion $r_{m i n}$ and the angle at which $r \to \infty$ (its asymptote).

Forecast: Guess — will the "escape angle" be more or less than the parabola's $18 0^{\circ}$ ?

Perihelion, $θ = 0$ . $r_{m i n} = \frac{p}{1 + e} = \frac{3}{1 + 2} = \frac{3}{3} = 1$ .
- Why this step? Same closest-point logic; $cos 0 = 1$ maximises the denominator.
Where does $r \to \infty$ ? Need denominator $= 0$ : $1 + 2 cos θ = 0 \Rightarrow cos θ = - \frac{1}{2}$ .
- Why this step? This is the "forbidden angle" — beyond it the formula gives negative $r$ , which is unphysical. It marks where the body flies off to infinity.
Solve $cos θ = - 0.5$ . $θ = 12 0^{\circ}$ (and by symmetry $- 12 0^{\circ}$ on the way in).
- Why this step? $cos$ is the tool because the shape formula packs the angle inside a cosine; we ask "which angle has this cosine?" — that inverse question is $arccos$ .

Verify: $cos 12 0^{\circ} = - 0.5$ so $1 + 2 (- 0.5) = 0$ ✓. The object only exists for $- 12 0^{\circ} < θ < 12 0^{\circ}$ , a $24 0^{\circ}$ arc — an open curve, as a hyperbola must be. Energy: $e > 1 \Rightarrow ε > 0$ , unbound. ✓

Cell E — classify from $(ε, h)$

Worked example Ex 5 · Which shape? Three inputs

With $μ = 1$ , classify the orbit for each: (a) $ε = - 0.5, h = 1$ ; (b) $ε = 0, h = 1$ ; (c) $ε = + 0.5, h = 1$ .

Forecast: Guess — before computing, which one is unbound?

Case (a). $e = 1 + 2 (- 0.5) (1)^{2} /1 = 1 - 1 = 0 = 0$ .
- Why this step? The sign of $ε$ lives under a square root as $1 + (that)$ . Here it drives the whole thing to $0$ → circle (cell A).
Case (b). $e = 1 + 2 (0) (1) /1 = 1 = 1$ → parabola.
- Why this step? $ε = 0$ is the knife-edge; always $e = 1$ regardless of $h$ .
Case (c). $e = 1 + 2 (0.5) (1) /1 = 2 \approx 1.414$ → hyperbola (unbound).
- Why this step? Positive energy pushes the term under the root above $1$ , so $e > 1$ .

Verify: Signs march in step with shape: $ε < 0 \Rightarrow e < 1$ , $ε = 0 \Rightarrow e = 1$ , $ε > 0 \Rightarrow e > 1$ — exactly the parent's table. Numerically $2 \approx 1.41421$ ✓. See Vis-viva Equation for how $ε$ is measured from speed.

Cell F — the angle where distance blows up

Worked example Ex 6 · No blow-up for the ellipse

For the ellipse of Ex 2 ( $e = 0.6, p = 6.4$ ), show there is no angle where $r \to \infty$ . Contrast with $e = 1$ and $e = 1.5$ .

Forecast: Guess — is there any $θ$ that makes the denominator zero when $e = 0.6$ ?

Set denominator to zero. $1 + e cos θ = 0 \Rightarrow cos θ = - 1/ e$ .
- Why this step? $r = p / (denominator)$ can only explode where the denominator vanishes.
Plug $e = 0.6$ . $cos θ = - 1/0.6 \approx - 1.667$ .
- Why this step? Cosine can only take values in $[- 1, 1]$ . Here we need $- 1.667$ , which is impossible — so no such $θ$ exists. The ellipse stays finite everywhere (cell B, bounded).
Contrast. $e = 1 \Rightarrow cos θ = - 1$ (reachable at $18 0^{\circ}$ , exactly one direction). $e = 1.5 \Rightarrow cos θ = - 0.667$ (reachable at $θ \approx 131. 8^{\circ}$ ).
- Why this step? This single inequality $∣ - 1/ e ∣ \leq 1$ is the whole bound-vs-unbound dividing line: it holds only when $e \geq 1$ .

Verify: $∣ - 1/ e ∣ \leq 1$ requires $e \geq 1$ . Check $e = 1.5$ : $arccos (- 0.667) \approx 131.8 1^{\circ}$ , and $cos 131.8 1^{\circ} \approx - 0.667$ ✓.

Cell G — real-world word problem

Worked example Ex 7 · Earth's near-circular orbit

Earth's orbit has $e = 0.0167$ . By what percentage is Earth closer to the Sun at perihelion than aphelion? Use $r_{m a x} / r_{m i n} = (1 + e) / (1 - e)$ .

Forecast: Guess — a few percent, or tens of percent?

Form the ratio. $\frac{r _{m a x}}{r _{m i n}} = \frac{1 + 0.0167}{1 - 0.0167} = \frac{1.0167}{0.9833}$ .
- Why this step? The $a$ cancels in the ratio, so we don't even need Earth's actual distance — pure geometry from $e$ .
Divide. $\frac{1.0167}{0.9833} \approx 1.0340$ .
- Why this step? This says aphelion is about $3.4%$ farther than perihelion.
Interpret. A $\sim 3.4%$ distance swing — tiny. Seasons come from axial tilt, not this.
- Why this step? Ties the number to physical reality, the point of a word problem.

Verify: $(1.0167/0.9833 - 1) \times 100 \approx 3.40%$ ✓. Small $e$ → near-circle (cell A limit), consistent with $ε < 0$ deeply bound.

Cell H — the limit $e \to 1^{-}$

Worked example Ex 8 · An ellipse stretching toward a parabola

Fix $p = 2$ and let $e$ climb: $e = 0.9, 0.99, 0.999$ . Track $r_{m a x} = \frac{p}{1 - e}$ (aphelion). What happens in the limit $e \to 1^{-}$ ?

Forecast: Guess — does $r_{m a x}$ approach a finite ceiling or run away?

$e = 0.9$ : $r_{m a x} = \frac{2}{1 - 0.9} = \frac{2}{0.1} = 20$ .
- Why this step? Aphelion sits at $θ = 18 0^{\circ}$ where the denominator is $1 - e$ ; as $e \to 1$ this shrinks.
$e = 0.99$ : $r_{m a x} = \frac{2}{0.01} = 200$ . $e = 0.999$ : $r_{m a x} = \frac{2}{0.001} = 2000$ .
- Why this step? Each extra "9" multiplies the far distance by $\sim 10$ — it grows without bound.
Limit. As $e \to 1^{-}$ , $1 - e \to 0^{+}$ so $r_{m a x} \to \infty$ : the ellipse's far end runs off to infinity and the closed loop opens into the parabola of Ex 3.
- Why this step? This is why $e = 1$ is a genuine change of kind, not "a thinner ellipse" — the mistake the parent warns about.

Verify: $2/0.1 = 20$ , $2/0.01 = 200$ , $2/0.001 = 2000$ ✓ — each 10× as $e$ gains a nine.

Cell I — exam twist: same shape, different size

Worked example Ex 9 · Two ellipses, one eccentricity

Orbit P has $a = 5$ ; orbit Q has $a = 50$ ; both have $e = 0.4$ . Show they have the same shape but different sizes by comparing $r_{m i n} / a$ and $r_{m a x} / r_{m i n}$ .

Forecast: Guess — do the two ratios differ between P and Q?

Ratio $r_{m i n} / a = (1 - e)$ . For both: $1 - 0.4 = 0.6$ .
- Why this step? Dividing by $a$ strips out the size, leaving only shape info (which depends on $e$ alone).
Ratio $r_{m a x} / r_{m i n} = (1 + e) / (1 - e) = 1.4/0.6 \approx 2.333$ . Same for both.
- Why this step? Again $a$ cancels; identical $e$ → identical proportions.
Absolute sizes differ. P: $r_{m i n} = 5 (0.6) = 3$ . Q: $r_{m i n} = 50 (0.6) = 30$ .
- Why this step? This exposes the parent's warning — $e$ is a ratio, not a length; Q is 10× bigger yet equally "squashed."

Verify: $1.4/0.6 \approx 2.3333$ ✓; $5 \times 0.6 = 3$ , $50 \times 0.6 = 30$ ✓. Same $e$ , same shape, sizes scale with $a$ .

Cell J — degenerate: zero angular momentum (radial fall)

Worked example Ex 10 · Straight-down fall,

$h = 0$ A body is dropped from rest (no sideways motion), so its specific angular momentum $h = 0$ . With $μ = 1$ and specific energy $ε = - 0.5$ , compute $e$ and interpret the "orbit."

Forecast: Guess — with no sideways swing, is this an ellipse, a line, or something odd?

Plug $h = 0$ into $e = 1 + 2 ε h^{2} / μ^{2}$ . The $h^{2}$ kills the whole second term: $e = 1 + 0 = 1$ .
- Why this step? No matter the energy sign, $h = 0$ forces $e = 1$ because the energy term is multiplied by $h^{2}$ .
Check the shape formula. $p = h^{2} / μ = 0$ , so the numerator of $r = p / (1 + cos θ)$ is $0$ : the "conic" has collapsed to a straight line through the focus (a degenerate parabola).
- Why this step? $p = 0$ means the curve has no sideways width; it's the limiting radial line. The body falls straight in.
Interpret the physics. Even though $ε = - 0.5 < 0$ (bound, so energetically it should be an ellipse), the path isn't a fat ellipse — it is a line segment along which the body plunges toward the centre and would swing back out. So the shortcut " $ε < 0 \Rightarrow$ ellipse" needs the caveat "and $h \neq = 0$ ."
- Why this step? This is the true degenerate corner the shape table hides: eccentricity alone ( $e = 1$ here) does not tell the whole story when $h = 0$ .

Verify: $e = 1 + 2 (- 0.5) (0) /1 = 1 = 1$ ✓, independent of $ε$ . And $p = 0^{2} /1 = 0$ ✓ → a degenerate radial line, not an area-filling conic. Consistency check: a genuine ellipse needs $h \neq = 0$ to give $p > 0$ ; here $p = 0$ correctly flags the degeneracy.

Active recall

Which cell does

e = 0.6, p = 6.4

never reach infinity, and why?

Cell F/B — needs

cos θ = - 1/0.6 \approx - 1.667

, outside

[- 1, 1]

, impossible.

For

e = 2, p = 3

, what is perihelion and the asymptote angle?

r_{m i n} = 1

;

r \to \infty

θ = 12 0^{\circ}

What does

h = 0

do to eccentricity, whatever the energy?

Forces

e = 1

(energy term

\times h^{2} = 0

) and

p = 0

: a degenerate radial line.

e \to 1^{-}

with fixed

p

, what happens to aphelion?

r_{m a x} = p / (1 - e) \to \infty

; the ellipse opens into a parabola.

Two orbits same

e

, different

a

— same or different shape?

Same shape (all ratios depend only on

e

); different absolute size (scales with

a

Why is

p = a (1 - e^{2})

Because

r_{m i n} + r_{m a x} = 2 p / (1 - e^{2}) = 2 a

, so

a = p / (1 - e^{2})

, i.e.

p = a (1 - e^{2})

What is the semi-major axis

a

Half the long axis,

a = (r_{m i n} + r_{m a x}) /2

— the orbit's size.

Connections

Parent topic
Conic Sections — the geometry behind every cell
Vis-viva Equation — where $ε$ comes from (speed at any $r$ )
Specific Orbital Energy — the sign that picks the shape
Angular Momentum in Orbits — $h$ sets $p$ and the $h = 0$ degenerate case
Kepler's First Law — the ellipse cell in nature
Escape Velocity — the $e = 1$ boundary of Ex 3