Kepler's first law — orbits are conic sections: Worked examples

This page is the "no surprises" workbook for Kepler's first law. The parent proved the orbit equation $r (θ) = \frac{p}{1 + e c o s θ}, p = \frac{h ^{2}}{μ}, e = 1 + \frac{2 E h ^{2}}{μ ^{2}}$ Here we drive it through every kind of input it can meet: each eccentricity band, the degenerate circle, the zero-energy edge, the negative-cosine back half of the orbit, a real-world word problem, and one exam twist. If a scenario exists, it is a row below and an example further down.

The scenario matrix

Every cell below is hit by at least one worked example (its number is in brackets).

Cell class	Specific case	Which example
$e = 0$ degenerate	circle, constant $r$	Ex 1
$0 < e < 1$ front half	perihelion, $θ = 0$ , $cos θ = + 1$	Ex 2
$0 < e < 1$ back half	aphelion, $θ = π$ , $cos θ = - 1$	Ex 2
$0 < e < 1$ side / intermediate $θ$	$θ = 9 0^{\circ}$ , $cos θ = 0$ (negative-cos region probed too)	Ex 3
$e = 1$ edge	parabola — no aphelion, $r \to \infty$	Ex 4
$e > 1$ unbound	hyperbola — asymptote angle, forbidden angles	Ex 5
Inverse problem	given two $r$ values, find $e$ and $p$	Ex 6
Real-world word problem	satellite altitudes → orbit shape	Ex 7
Limiting behaviour	$e \to 1^{-}$ ellipse stretches to parabola	Ex 8
Exam twist	classify from raw energy + $h$ , sign traps	Ex 9

Definition The allowed

$θ$ -domain — where the formula is even valid This distinction is the theme of the whole page, so pin it down before any arithmetic (see figure below). Recall from the sign convention that negative $θ$ just means "swept the other way from perihelion":

Ellipse / circle ( $0 \leq e < 1$ ): the denominator $1 + e cos θ$ is always positive (since $e cos θ \geq - e > - 1$ ), so $r$ is finite and positive for every angle. Domain: $0 \leq θ < 2 π$ rad ( $0^{\circ} \leq θ < 36 0^{\circ}$ ) — the orbit closes on itself.
Parabola ( $e = 1$ ): denominator hits $0$ only at $θ = π$ rad ( $18 0^{\circ}$ ), where $r \to \infty$ . Domain: $- π < θ < π$ rad ( $- 18 0^{\circ} < θ < 18 0^{\circ}$ ) — the single missing direction points to infinity.
Hyperbola ( $e > 1$ ): denominator hits $0$ at $θ_{\infty} = arccos (- 1/ e)$ . Domain: $- θ_{\infty} < θ < θ_{\infty}$ ; beyond it $r$ would be negative — forbidden (Ex 5).

Figure — Kepler's first law — orbits are conic sections

The figure above draws all four cases from the same focus (orange dot). Notice the orange circle and teal ellipse sweep a full loop, while the plum parabola and dark hyperbola open and run off to infinity along the dashed directions — that is the $θ$ -domain rule made visible. We return to it quantitatively in Ex 5.

Ex 1 — The degenerate case: a perfect circle ( $e = 0$ )

Worked example A moon orbits at fixed

$r = 4.0 \times 1 0^{8}$ m with $e = 0$ . Find $r$ at $θ = 0$ , $9 0^{\circ}$ , $18 0^{\circ}$ , and its $p$ .

Forecast: guess before reading — does $r$ change as you go around? What is $p$ compared to that fixed radius?

Write the orbit equation with $e = 0$ . $r (θ) = \frac{p}{1 + 0 \cdot cos θ} = p$ . Why this step? Setting the dial to zero kills the $cos θ$ term entirely, so the angle no longer matters — that is the algebraic fingerprint of a circle.
Evaluate at every angle. $r (0) = r (9 0^{\circ}) = r (18 0^{\circ}) = p$ . Why this step? We must check "all quadrants." Here they are all equal — the whole point of a circle is that no direction is special.
Identify $p$ . Since $r$ is constant and equals $4.0 \times 1 0^{8}$ m, we read $p = 4.0 \times 1 0^{8}$ m. Why this step? $p$ is defined as $r$ at $θ = 9 0^{\circ}$ ; for a circle that equals the radius everywhere.

Verify: perihelion $= a (1 - e) = a (1 - 0) = a$ and aphelion $= a (1 + 0) = a$ collapse to the same value, confirming there is no "close" or "far" point. Units: metres throughout. ✓

Ex 2 — Bound ellipse, front and back halves ( $0 < e < 1$ )

Worked example Mars:

$e = 0.0934$ , semi-major axis $a = 2.279 \times 1 0^{11}$ m. Find perihelion ( $θ = 0$ ) and aphelion ( $θ = π$ ) distances.

Forecast: which is bigger, and by roughly what fraction of $a$ ?

Relate $a$ and $p$ . Use the recapped identity $p = a (1 - e^{2})$ (derived in the [!formula] callout above). Why this step? We are given $a$ , but the orbit equation uses $p$ ; this identity converts between them, and we already proved it, so no hand-waving. $p = 2.279 \times 1 0^{11} (1 - 0.093 4^{2}) = 2.259 \times 1 0^{11}$ m.
Perihelion, $θ = 0$ (the front half, $cos θ = + 1$ ). $r_{m i n} = \frac{p}{1 + e} = a (1 - e)$ . Why this step? Largest $cos$ ⇒ largest denominator ⇒ smallest $r$ . This is the $cos θ = + 1$ corner of the matrix. $r_{m i n} = 2.279 \times 1 0^{11} (1 - 0.0934) = 2.066 \times 1 0^{11}$ m.
Aphelion, $θ = π$ (the back half, $cos θ = - 1$ ). $r_{m a x} = \frac{p}{1 - e} = a (1 + e)$ . Why this step? $cos π = - 1$ makes the denominator smallest, so $r$ is largest — the negative-cosine region of the orbit. $r_{m a x} = 2.279 \times 1 0^{11} (1 + 0.0934) = 2.492 \times 1 0^{11}$ m.

Verify: their average $\frac{1}{2} (r_{m i n} + r_{m a x}) = \frac{1}{2} \cdot a [(1 - e) + (1 + e)] = a$ . Compute: $\frac{1}{2} (2.066 + 2.492) \times 1 0^{11} = 2.279 \times 1 0^{11}$ m $= a$ . ✓ The semi-major axis really is the mean of the two extremes.

Ex 3 — Intermediate angle & the negative-cosine region ( $0 < e < 1$ )

Worked example For an orbit with

$p = 1.20 \times 1 0^{7}$ m and $e = 0.50$ , find $r$ at $θ = 9 0^{\circ}$ and at $θ = 12 0^{\circ}$ .

Forecast: at $12 0^{\circ}$ we are past sideways — is $r$ bigger or smaller than at $9 0^{\circ}$ ?

Sideways point, $θ = 9 0^{\circ}$ ( $cos = 0$ ). $r = \frac{p}{1 + e \cdot 0} = p = 1.20 \times 1 0^{7}$ m. Why this step? This is the defining point of $p$ ; a clean anchor before we push into the negative-cosine zone.
Past sideways, $θ = 12 0^{\circ}$ ( $cos 12 0^{\circ} = - 0.5$ ). $r = \frac{p}{1 + 0.50 ( - 0.5 )} = \frac{1.20 \times 1 0 ^{7}}{0.75} = 1.60 \times 1 0^{7}$ m. Why this step? Once $cos θ$ goes negative the denominator drops below 1, so $r$ grows — the body is heading toward aphelion. This covers the "intermediate $θ$ with negative cosine" cell.

Verify: the two extremes must bracket these numbers: $r_{m i n} = p /1.5 = 8.0 \times 1 0^{6}$ m and $r_{m a x} = p /0.5 = 2.4 \times 1 0^{7}$ m. Both $1.20 \times 1 0^{7}$ and $1.60 \times 1 0^{7}$ lie inside $[8.0 \times 1 0^{6}, 2.4 \times 1 0^{7}]$ . ✓ And $r (12 0^{\circ}) > r (9 0^{\circ})$ as forecast, because we moved toward aphelion.

Ex 4 — The parabolic edge ( $e = 1$ )

Worked example A comet has

$e = 1.000$ and $p = 2.0 \times 1 0^{11}$ m. Find its perihelion distance, and describe $r$ as $θ \to 18 0^{\circ}$ .

Forecast: does this comet ever reach an aphelion?

Perihelion, $θ = 0$ . $r_{m i n} = \frac{p}{1 + 1} = \frac{p}{2} = 1.0 \times 1 0^{11}$ m. Why this step? Perihelion still behaves normally — the difference only shows up on the far side.
Approach $θ \to 18 0^{\circ}$ ( $π$ rad). Denominator $1 + cos θ \to 1 + (- 1) = 0$ , so $r \to \infty$ . Why this step? For $e = 1$ the denominator can hit zero. That is a new phenomenon absent for ellipses: the curve never closes, it escapes to infinity as it lines up opposite perihelion.
Conclusion: no aphelion exists; the comet visits once and leaves. Why this step? This is the exact boundary between "comes back" ( $e < 1$ ) and "gone forever."

Verify: for the ellipse formula $r_{m a x} = p / (1 - e)$ ; plugging $e \to 1$ gives $p /0 = \infty$ , matching the limit — the parabola is the continuous limit of a stretched ellipse. (Note $p = a (1 - e^{2})$ also blows the $a$ -picture up: $a = p / (1 - e^{2}) \to \infty$ , so a parabola is an ellipse with infinite semi-major axis.) Units: $r_{m i n}$ in metres. ✓

Ex 5 — The hyperbolic flyby ( $e > 1$ ) and forbidden angles

Worked example An interstellar object has

$e = 1.5$ , $p = 3.0 \times 1 0^{11}$ m. Find perihelion, and the asymptote angle beyond which no orbit exists.

Forecast: for a hyperbola, are all angles $θ$ physically reachable?

Perihelion, $θ = 0$ . $r_{m i n} = \frac{p}{1 + e} = \frac{3.0 \times 1 0 ^{11}}{2.5} = 1.2 \times 1 0^{11}$ m. Why this step? Same closest-approach logic; unbound orbits still have a well-defined nearest point.
Where does $r \to \infty$ ? Set denominator to zero. $1 + e cos θ_{\infty} = 0 \Rightarrow cos θ_{\infty} = - \frac{1}{e} = - \frac{1}{1.5} = - 0.6667$ . Why this step? Because $e > 1$ , the value $- 1/ e$ lies inside $[- 1, 1]$ , so a real angle exists where $r$ blows up. This is the asymptote direction. $θ_{\infty} = arccos (- 0.6667) = 131. 8^{\circ}$ .
Forbidden region. For $θ > θ_{\infty}$ (up to $18 0^{\circ}$ ), $1 + e cos θ < 0$ , giving negative $r$ — unphysical. The body physically only occupies $- 131. 8^{\circ} < θ < 131. 8^{\circ}$ (recall from the sign convention: that range covers $131. 8^{\circ}$ swept either way from perihelion). Why this step? This is the case-coverage the parent skipped: hyperbolas have angular no-go zones, unlike ellipses which sweep a full $36 0^{\circ}$ .

Verify: contrast with the ellipse ( $e < 1$ ): there $- 1/ e < - 1$ lies outside $[- 1, 1]$ , so the denominator never reaches zero, $r$ stays finite, and every angle is allowed — exactly why ellipses close. For $e = 1.5$ , $θ_{\infty}$ should exceed $9 0^{\circ}$ and be under $18 0^{\circ}$ : $131. 8^{\circ}$ ✓.

The figure shows the plum hyperbola tracing only the allowed wedge $- 131. 8^{\circ} < θ < 131. 8^{\circ}$ . The dashed teal lines are the asymptotes at $θ_{\infty}$ ; the shaded grey wedges behind them are the forbidden directions where the formula would return a negative $r$ . Compare this truncated fan to the full loops of Ex 1–3: that visual contrast is the $θ$ -domain rule from the matrix section.

Ex 6 — Inverse problem: two radii give you the shape

Worked example A probe is tracked: at

$θ = 0$ its distance is $r_{1} = 7.0 \times 1 0^{6}$ m; at $θ = 18 0^{\circ}$ it is $r_{2} = 2.1 \times 1 0^{7}$ m. Find $e$ , $p$ , and $a$ .

Forecast: with $r_{2} = 3 r_{1}$ , is this nearly circular or quite eccentric?

Write both extremes. $r_{1} = \frac{p}{1 + e}$ , $r_{2} = \frac{p}{1 - e}$ . Why this step? Two data points, two unknowns ( $p, e$ ) — perihelion and aphelion are the cleanest equations because the cosines are $\pm 1$ .
Divide to kill $p$ . $\frac{r _{2}}{r _{1}} = \frac{1 + e}{1 - e} = \frac{2.1 \times 1 0 ^{7}}{7.0 \times 1 0 ^{6}} = 3$ . Why this step? Ratios eliminate the size, isolating the shape parameter $e$ .
Solve the linear equation for $e$ . Cross-multiply $\frac{1 + e}{1 - e} = 3$ to get $3 (1 - e) = 1 + e$ , then expand and collect: $3 - 3 e = 1 + e \Rightarrow 2 = 4 e \Rightarrow e = 0.5$ . Why this step? Cross-multiplying clears the fraction so $e$ appears linearly; gathering the $e$ -terms on one side is the only way to isolate the single unknown left after step 2.
Back-substitute for $p$ . $p = r_{1} (1 + e) = 7.0 \times 1 0^{6} (1.5) = 1.05 \times 1 0^{7}$ m. Why this step? With $e$ known, either extreme equation now yields $p$ directly; perihelion is the simplest.
Semi-major axis. $a = \frac{1}{2} (r_{1} + r_{2}) = \frac{1}{2} (7.0 \times 1 0^{6} + 2.1 \times 1 0^{7}) = 1.4 \times 1 0^{7}$ m. Why this step? $a$ is the mean of the extremes (proven in Ex 2's verify).

Verify: check $p = a (1 - e^{2}) = 1.4 \times 1 0^{7} (1 - 0.25) = 1.05 \times 1 0^{7}$ m ✓, and $r_{2} = p / (1 - e) = 1.05 \times 1 0^{7} /0.5 = 2.1 \times 1 0^{7}$ m matches the input. ✓

Ex 7 — Real-world word problem: satellite from altitudes

Worked example A satellite around Earth (radius

$R_{E} = 6.371 \times 1 0^{6}$ m) has perigee altitude $500$ km and apogee altitude $5000$ km. Find $e$ and $p$ . (Altitude = height above the surface.)

Forecast: the classic trap — do we use altitudes or distances-from-centre in the orbit equation?

Convert altitudes to focus-distances. The focus is Earth's centre, so add $R_{E}$ : $r_{m i n} = 6.371 \times 1 0^{6} + 5.00 \times 1 0^{5} = 6.871 \times 1 0^{6}$ m, $r_{m a x} = 6.371 \times 1 0^{6} + 5.00 \times 1 0^{6} = 1.1371 \times 1 0^{7}$ m. Why this step? $r$ in $r = p / (1 + e cos θ)$ is measured from the focus, never from the surface — forgetting $R_{E}$ is the #1 satellite-exam error.
Eccentricity from the ratio. $\frac{r _{m a x}}{r _{m i n}} = \frac{1 + e}{1 - e} = \frac{1.1371 \times 1 0 ^{7}}{6.871 \times 1 0 ^{6}} = 1.6549$ . Solve: $e = \frac{1.6549 - 1}{1.6549 + 1} = 0.2467$ . Why this step? Same divide-out trick as Ex 6; the ratio strips away the size and Earth's radius offset is already baked into the true $r$ values.
Semi-latus rectum. $p = r_{m i n} (1 + e) = 6.871 \times 1 0^{6} (1.2467) = 8.568 \times 1 0^{6}$ m. Why this step? With $e$ known, perihelion is the cleanest equation to invert for $p$ .

Verify: $r_{m a x} = p / (1 - e) = 8.568 \times 1 0^{6} /0.7533 = 1.137 \times 1 0^{7}$ m, matching step 1 ✓. Sanity: $e = 0.25$ is a mild ellipse, reasonable for a real transfer-style orbit. Units: metres throughout. ✓

Ex 8 — Limiting behaviour: stretching an ellipse toward a parabola

Worked example Keep perihelion fixed at

$r_{m i n} = 1.0 \times 1 0^{11}$ m while raising $e$ through $0.9, 0.99, 0.999$ . Track the aphelion and the ratio $r_{m a x} / r_{m i n}$ .

Forecast: as $e \to 1$ , does $r_{m a x}$ approach a finite value or run away?

Express aphelion via perihelion. $r_{m a x} = r_{m i n} \cdot \frac{1 + e}{1 - e}$ . Why this step? Holding $r_{m i n}$ fixed isolates the effect of $e$ on the far side alone.
Tabulate.
- $e = 0.9$ : ratio $= 1.9/0.1 = 19$ , so $r_{m a x} = 1.9 \times 1 0^{12}$ m.
- $e = 0.99$ : ratio $= 1.99/0.01 = 199$ , so $r_{m a x} = 1.99 \times 1 0^{13}$ m.
- $e = 0.999$ : ratio $= 1.999/0.001 = 1999$ , so $r_{m a x} = 1.999 \times 1 0^{14}$ m. Why this step? Each factor-of-10 closer to $e = 1$ multiplies the aphelion by ~10 — the ellipse elongates without bound.
Take the limit. As $e \to 1^{-}$ , $r_{m a x} \to \infty$ : the closed ellipse opens into the parabola of Ex 4. Why this step? This is the continuity of the whole conic family — the matrix rows are one dial, not separate universes.

Verify: at $e = 0.999$ , $r_{m a x} / r_{m i n} = 1999$ and indeed $(1 + e) / (1 - e) = 1.999/0.001 = 1999$ ✓. The trend is monotonic and diverges, confirming the parabolic limit.

Ex 9 — Exam twist: classify straight from energy and angular momentum

Worked example A body around the Sun (

$μ = 1.327 \times 1 0^{20} m^{3} / s^{2}$ ) has specific energy $E = + 4.0 \times 1 0^{8} J/kg$ and $h = 4.5 \times 1 0^{15} m^{2} / s$ . Classify the orbit and find $e$ and $p$ .

Forecast: positive energy — bound or unbound? Guess the conic before computing.

Sign check on energy. $E > 0$ ⇒ unbound. Predict $e > 1$ , a hyperbola. Why this step? The sign of $E$ decides the conic before any arithmetic — the fastest exam move and a common trap (people compute needlessly).
Compute $e$ . $e = 1 + \frac{2 E h ^{2}}{μ ^{2}}$ . Inside: $\frac{2 ( 4.0 \times 1 0 ^{8} ) ( 4.5 \times 1 0 ^{15} ) ^{2}}{( 1.327 \times 1 0 ^{20} ) ^{2}} = \frac{2 ( 4.0 \times 1 0 ^{8} ) ( 2.025 \times 1 0 ^{31} )}{1.7609 \times 1 0 ^{40}} = \frac{1.620 \times 1 0 ^{40}}{1.7609 \times 1 0 ^{40}} = 0.9200$ . So $e = 1.9200 = 1.386$ . Why this step? Confirms the sign-prediction quantitatively; $e > 1$ ⇒ hyperbola. ✓
Semi-latus rectum. $p = \frac{h ^{2}}{μ} = \frac{( 4.5 \times 1 0 ^{15} ) ^{2}}{1.327 \times 1 0 ^{20}} = \frac{2.025 \times 1 0 ^{31}}{1.327 \times 1 0 ^{20}} = 1.526 \times 1 0^{11}$ m. Why this step? $p$ depends only on $h$ and $μ$ (parent's Step 6), independent of energy.
Closest approach as a reality check. $r_{m i n} = \frac{p}{1 + e} = \frac{1.526 \times 1 0 ^{11}}{2.386} = 6.40 \times 1 0^{10}$ m. Why this step? Even an escaping flyby has a well-defined perihelion; computing it confirms the hyperbola is a physical trajectory that grazes the Sun and leaves.

Verify: $E > 0 \Leftrightarrow e > 1$ ; we got $e = 1.386 > 1$ — consistent. Units of $p$ : $(m^{2} / s)^{2} / (m^{3} / s^{2}) = m$ ✓. The asymptote angle $arccos (- 1/1.386) = arccos (- 0.7215) = 136. 2^{\circ}$ lies in $(9 0^{\circ}, 18 0^{\circ})$ , as every hyperbola must.

Recall check

Recall Which cosine value gives perihelion, which gives aphelion?

$cos θ = + 1$ (at $θ = 0$ ) → perihelion (smallest $r$ ); $cos θ = - 1$ (at $θ = π$ rad $= 18 0^{\circ}$ ) → aphelion (largest $r$ ). Aphelion only exists when $e < 1$ .

Recall For a hyperbola (

$e > 1$ ), what angle marks the asymptote? $cos θ_{\infty} = - 1/ e$ ; beyond it the denominator goes negative and no orbit exists.

Recall A satellite problem gives you altitudes. First move?

Add the planet's radius: $r = altitude + R_{E}$ , because $r$ is measured from the focus (centre).

Recall For which conics is every angle allowed?

Only the circle and ellipse ( $0 \leq e < 1$ ) allow all of $0 \leq θ < 2 π$ rad. Parabola loses $θ = π$ rad; hyperbola is confined to $- arccos (- 1/ e) < θ < arccos (- 1/ e)$ .

Related: Vis-viva equation · Escape velocity and orbital energy · Conic sections — geometry of ellipse, parabola, hyperbola · Conservation of angular momentum in central forces