Conic-Section Orbits

Around 200 BC, Apollonius of Perga sliced a cone with a plane at every possible angle and catalogued the curves that appeared: the ellipse, the parabola, the hyperbola. It was geometry for geometry's sake — beautiful, useless, and destined to sit in the library for eighteen centuries. Then Newton solved the gravitational two-body problem and found that nature had been reading Apollonius all along. Every object that has ever fallen freely around another — every planet, comet, spacecraft, and interstellar visitor — moves on one of his three curves, with the attracting body parked at a focus. Better still, the whole family collapses into a single formula. The polar equation $r = p/(1 + e\cos\theta)$ describes all three conics at once, and the two constants in it are not free decorations: the size parameter $p$ is set by the orbit's angular momentum, and the shape parameter $e$ — the eccentricity — is set by energy and angular momentum together. Dial $e$ from 0 upward and the orbit morphs from circle to ellipse to parabola to hyperbola, with the transitions happening at exactly the energies you would guess: bound, marginal, unbound. This article is the geometric heart of the central-force story. You will learn to read a conic like an orbital engineer — perihelion, aphelion, semi-major axis, asymptotes — and to translate fluently between the geometry ( $e$ , $p$ , $a$ ) and the dynamics ( $E$ , $\ell$ ). By the end, two numbers scribbled on a napkin will tell you whether a newly spotted comet is a permanent resident of the solar system or a one-time visitor from another star.

Key takeaways

Motion under an attractive inverse-square force traces exactly the conic sections — ellipse, parabola, or hyperbola — with the force center at one focus.
One formula covers them all: $r(\theta) = \dfrac{p}{1 + e\cos\theta}$ , with semi-latus rectum $p = \ell^2/(\mu k)$ and eccentricity $e$ .
The dynamics fixes the geometry: $e = \sqrt{1 + 2E\ell^2/(\mu k^2)}$ — energy and angular momentum are the only inputs the shape needs.
The energy sign picks the conic: $E < 0 \Rightarrow e < 1$ (ellipse), $E = 0 \Rightarrow e = 1$ (parabola), $E > 0 \Rightarrow e > 1$ (hyperbola).
Geometry and dynamics divide the labor: the semi-major axis depends only on energy, $a = -k/(2E)$ , while angular momentum sets the width $p = a(1 - e^2)$ .
Kepler ellipses close without precessing because of an extra conserved quantity — the Laplace–Runge–Lenz vector, which points permanently at perihelion with magnitude $\mu k e$ .
Key idea: integrating the orbit equation for $V = -k/r$ gives a conic $r = p/(1+e\cos\theta)$ ; the eccentricity $e = \sqrt{1 + 2E\ell^2/(\mu k^2)}$ makes the orbit's geometric shape a direct readout of the two conserved scalars $E$ and $\ell$ .

Where this lives today

Every spacecraft trajectory begins life as a patchwork of conics. Mission designers still rough out journeys with "patched conics": an ellipse around Earth, a hyperbolic escape leg, a solar ellipse to the target, a hyperbolic capture — each segment pure Apollonius, stitched at the boundaries of each body's gravitational sphere of influence. A Hohmann transfer to Mars is nothing but an ellipse tangent to two circles, sized by equations (4)–(6) and the vis-viva relation; interplanetary gravity assists are hyperbolic flybys whose deflection angle $\sin(\delta/2) = 1/e$ is engineered by choosing the approach distance. The catalogued "orbital elements" that every satellite operator exchanges daily — $a$ , $e$ , inclination, and friends — are just this article's parameters plus orientation angles. The eccentricity formula has also become an astronomical discovery tool. Surveys now flag any object whose fitted $e$ exceeds 1 as an interstellar visitor: 'Oumuamua ( $e \approx 1.2$ ) in 2017, comet 2I/Borisov ( $e \approx 3.36$ ) in 2019, and 3I/ATLAS in 2025. In exoplanet science, the measured eccentricity distribution of thousands of systems encodes their dynamical history — hot Jupiters circularized by tides, violently scattered planets on $e > 0.9$ ellipses. And at the galaxy's center, the star S2's tight ellipse around Sagittarius A* ( $e = 0.88$ , period 16 yr) is the cleanest Kepler laboratory known: its tiny departures from a perfect closed conic — the relativistic precession of its perihelion — were measured by the GRAVITY interferometer in 2020, confirming that where the conic picture finally fails, it fails exactly as Einstein predicts.

The physical picture

Fix the attracting body — the Sun, say — and launch a projectile sideways from the same point with ever more speed. At low speed it falls around the Sun on a small ellipse. Give it more speed and the ellipse swells and elongates. At one critical speed the far end of the ellipse stretches to infinity and the curve opens into a parabola: the projectile barely escapes, coasting to a halt at infinite distance. Any faster and it escapes with speed to spare, on a hyperbola that straightens into a beeline once the Sun's pull fades. One knob — launch energy — sweeps through the entire Apollonius catalogue.

Figure

One focus, one formula, four orbits — the conic family r = p/(1 + e cos θ) as eccentricity dials up

The single most useful habit is to read $r = p/(1 + e\cos\theta)$ dynamically. The denominator oscillates between $1 + e$ (at $\theta = 0$ , perihelion — closest, fastest) and $1 - e$ (at $\theta = 180°$ , aphelion — farthest, slowest). If $e < 1$ the denominator never vanishes, $r$ stays finite everywhere, and the orbit is a closed loop. At $e = 1$ the denominator hits zero exactly at $\theta = 180°$ : the aphelion has escaped to infinity — that is the parabola. For $e > 1$ the denominator vanishes before $\theta$ reaches $180°$ , at $\cos\theta_\infty = -1/e$ ; the orbit only exists inside the wedge $|\theta| < \theta_\infty$ and runs off to infinity along two asymptotic directions — the hyperbola of a flyby. Where does the shape come from physically? Picture the effective potential $V_{\mathrm{eff}}(r) = -k/r + \ell^2/(2\mu r^2)$ from The effective potential: a centrifugal wall close in, a gravity well farther out. A horizontal line at height $E$ cuts that curve at the turning radii. Cut it twice (possible only for $E < 0$ ) and the radius oscillates between two bounds — a bound ellipse. Cut it once and the particle comes in, turns at perihelion, and leaves forever. The eccentricity simply measures how high your energy line sits relative to the bottom of the well: $e = 0$ at the very bottom (circle, no radial oscillation at all), $e \to 1$ as the line rises to zero, $e > 1$ above. The systematic energy taxonomy lives in Classification of orbits by energy; here we take the geometry road.

The conic equation and its two dials

The starting point is the orbit formula derived by solving Binet's equation in The orbit equation — and re-derived below by a completely independent energy route:

r(\theta) = \frac{p}{1 + e\cos\theta} \qquad (1)

where: $r, \theta$ — polar coordinates of the relative position measured from the focus, with $\theta = 0$ at perihelion; $p$ — the semi-latus rectum, the radius at $\theta = \pm 90°$ ; $e$ — the eccentricity. This is precisely the focus-centered polar equation of a conic section that Apollonius classified: an ellipse for $0 \le e < 1$ , a parabola for $e = 1$ , a hyperbola for $e > 1$ . One equation, the entire zoo. Each of the two constants answers to exactly one dynamical master. The size dial belongs to angular momentum:

p = \frac{\ell^2}{\mu k} \qquad (2)

where: $\ell$ — the conserved angular momentum, $\mu$ — the reduced mass of the two-body system (see The two-body problem), and $k$ — the strength of the attraction, $V(r) = -k/r$ , with $k = Gm_1m_2$ for gravity. More angular momentum means the orbit is fatter sideways: $p$ is literally how far out you fly at right angles to the perihelion direction. Notice $E$ appears nowhere — two orbits with wildly different energies but equal $\ell$ cross the $\theta = 90°$ line at the same radius, the shared point in the hero figure above. The shape dial is set jointly by energy and angular momentum:

e = \sqrt{1 + \frac{2E\ell^2}{\mu k^2}} \qquad (3)

where: $E$ — the conserved total energy of the equivalent one-body problem. Read the formula's anatomy: the "1" under the root is the parabola threshold; the correction term carries the sign of $E$ . Negative energy drags $e$ below 1 (ellipse), zero energy lands exactly on $e = 1$ (parabola), positive energy pushes $e$ above 1 (hyperbola). The most bound possible state at fixed $\ell$ , $E = -\mu k^2/(2\ell^2)$ , makes the root vanish: $e = 0$ , the circle. The formula even polices itself — plug in any $E$ below the circular minimum and the argument goes negative, the mathematics refusing to build an orbit that the effective potential forbids.

Derivation AThe conic orbit from energy conservation

We are deriving the orbit equation (1), together with the eccentricity formula (3), using only the two conservation laws — no Binet equation, no oscillator analogy. The strategy: use $E$ and $\ell$ to express both $\dot r$ and $\dot\theta$ , divide one by the other to eliminate time, and integrate the resulting relation between $r$ and $\theta$ . The technical heart is a completing-the-square step that turns the integral into a pure $\arccos$ — and the two constants of the conic will emerge from the center and amplitude of that square, which is exactly how the dynamics imprints itself on the geometry. Having two independent derivations of the same result (this one and Binet's) is not redundancy; it is how you know the result is load-bearing.

Step 1. Write the conserved energy with the angular part expressed through $\ell$ :

E = \tfrac12\mu\dot r^2 + \frac{\ell^2}{2\mu r^2} - \frac{k}{r}

Why the middle term: the tangential kinetic energy $\tfrac12\mu r^2\dot\theta^2$ becomes $\ell^2/(2\mu r^2)$ after substituting $\dot\theta = \ell/(\mu r^2)$ — the centrifugal piece of the effective potential.

Step 2. Solve for the radial velocity:

\dot r = \sqrt{\frac{2}{\mu}\left(E + \frac{k}{r} - \frac{\ell^2}{2\mu r^2}\right)}

Why the radicand matters: its roots are the turning points $r_{\min}, r_{\max}$ — where the radial motion reverses and the integrand below will blow up integrably, as it should at the edges of an oscillation.

Step 3. Eliminate time by dividing the two rates:

\frac{\mathrm{d}\theta}{\mathrm{d}r} = \frac{\dot\theta}{\dot r} = \frac{\ell/(\mu r^2)}{\dot r}

Step 4. Substitute $u = 1/r$ (so $\mathrm{d}u = -\mathrm{d}r/r^2$ ), which collapses the integral to a standard form:

\theta = \int\frac{\mathrm{d}u}{\sqrt{\dfrac{2\mu E}{\ell^2} + \dfrac{2\mu k}{\ell^2}u - u^2}}

Why this is progress: the radicand is now an upside-down parabola in $u$ — a quadratic under a square root is the signature of an inverse-cosine integral.

Step 5. Complete the square. The quadratic $-u^2 + (2\mu k/\ell^2)u + 2\mu E/\ell^2$ has center $u_c = \mu k/\ell^2$ and, after collecting constants, amplitude $u_c\,e$ with $e$ exactly as in (3). The integral then evaluates to

\cos\theta = \frac{u - \mu k/\ell^2}{(\mu k/\ell^2)\,e}

Why the constants were forced: the center of the square is the average of the turning values of $u$ , and the amplitude is their half-spread — geometry-to-be, dictated entirely by $E$ and $\ell$ .

Step 6. Invert for the orbit:

u = \frac{\mu k}{\ell^2}\left(1 + e\cos\theta\right) \quad\Longrightarrow\quad r = \frac{p}{1 + e\cos\theta},\qquad p = \frac{\ell^2}{\mu k}

Verified: cross-checked against Goldstein eqs. (3.55)–(3.57) and Landau–Lifshitz §15; identical to the Binet-equation result, and the $e = 0$ case reduces to the circular radius $r_c = \ell^2/(\mu k)$ at $E = -\mu k^2/(2\ell^2)$ .

Anatomy of the bound orbit

For $e < 1$ the conic closes into an ellipse, and a small dictionary of lengths unlocks every practical question. The extremes of (1) sit on the major axis:

r_{\min} = \frac{p}{1 + e} = a(1 - e), \qquad r_{\max} = \frac{p}{1 - e} = a(1 + e) \qquad (4)

where: $r_{\min}$ — perihelion, at $\theta = 0$ ; $r_{\max}$ — aphelion, at $\theta = \pi$ ; and $a$ — the semi-major axis, defined geometrically as half the long diameter, $a = (r_{\min} + r_{\max})/2$ . Memorize the pair $a(1 \mp e)$ : they convert between the astronomer's catalogue numbers ( $a$ , $e$ ) and the physicist's turning points instantly.

Figure

Anatomy of an ellipse with e = 0.6 — every named length in one figure

The axes themselves split the dynamical labor with surgical cleanness:

a = -\frac{k}{2E}, \qquad b = \frac{\ell}{\sqrt{2\mu|E|}} \qquad (5)

where: $a$ — the semi-major axis, and $b = a\sqrt{1 - e^2}$ — the semi-minor axis. Stare at the first relation: the semi-major axis knows only the energy. Two satellites with equal $E$ but different $\ell$ trace differently squashed ellipses of exactly the same length — and, by Kepler's third law $T^2 \propto a^3$ , exactly the same period. This is why "energy of the orbit" and "size of the orbit" are interchangeable currencies in astrodynamics, and it is derived in three lines below. The geometric consistency relation tying the family together is

p = a(1 - e^2) = \frac{b^2}{a} \qquad (6)

where: the first equality converts between the polar equation's $p$ and the astronomer's $(a, e)$ ; the second is Apollonius's own chord relation. With (4), (5), (6) you can translate any pair of orbital facts into any other — the worked examples drill exactly this fluency.

Go deeper · Derivation B · 1 minThe semi-major axis depends only on energy

Derivation BThe semi-major axis depends only on energy

The goal is the cornerstone relation $a = -k/(2E)$ : the size of a bound orbit is a pure function of its energy, with angular momentum nowhere in sight. The strategy is almost embarrassingly simple — the turning points of the radial motion are the roots of a quadratic equation, and the semi-major axis is half the sum of those roots, which Vieta's formula reads off without ever solving anything. Watch how $\ell$ enters the quadratic only through its constant term: it moves the two roots individually (squeezing perihelion and aphelion together or apart) but leaves their sum untouched — that is the entire theorem.

Step 1. At a turning point $\dot r = 0$ , so the energy equation of the previous derivation becomes $E = -k/r + \ell^2/(2\mu r^2)$ . Multiply through by $r^2$ :

E\,r^2 + k\,r - \frac{\ell^2}{2\mu} = 0

Why a quadratic: the effective potential has exactly two competing terms; equating it to a constant energy can balance at most at two radii — perihelion and aphelion.

Step 2. Apply Vieta's formula for the sum of the roots:

r_{\min} + r_{\max} = -\frac{k}{E}

Why the sign is healthy: for a bound orbit $E < 0$ , so $-k/E > 0$ — a positive total length, as geometry demands. And note: $\ell$ sits only in the constant term of the quadratic, which controls the roots' product, not their sum.

Step 3. Identify the semi-major axis as half the long diameter:

a = \frac{r_{\min} + r_{\max}}{2} = -\frac{k}{2E}

Step 4. Cross-check against the eccentricity formula. Vieta's product gives $r_{\min}r_{\max} = -\ell^2/(2\mu E) = b^2$ , and combining with (3):

p = a(1 - e^2) = -\frac{k}{2E}\left(-\frac{2E\ell^2}{\mu k^2}\right) = \frac{\ell^2}{\mu k}

Why this closes the loop: the geometric relation (6) reproduces the dynamical $p$ of equation (2) exactly — energy owns $a$ , angular momentum owns $p$ , and the two bookkeeping systems agree. *Verified: matches Morin Ch. 7 and Taylor Ch. 8; the energy-only dependence of $a$ is the foundation of Kepler's third law $T^2 \propto a^3$ and of the vis-viva equation $v^2 = (k/\mu)(2/r - 1/a)$ .*

The hidden compass: why the ellipse never drifts

A generic central force lets the orbit's turning points wander: each radial oscillation completes at a slightly different angle, and the "ellipse" slowly pivots into a rosette. The inverse-square force does not — perihelion stays nailed to one direction forever. Behind every exact degeneracy hides an extra conservation law, and here it is a full conserved vector:

\mathbf{A} = \mathbf{p}\times\mathbf{L} - \mu k\,\hat{\mathbf{r}}, \qquad |\mathbf{A}| = \mu k\,e \qquad (7)

where: $\mathbf{A}$ — the Laplace–Runge–Lenz vector, built from the momentum $\mathbf{p}$ and angular momentum $\mathbf{L}$ of the relative motion; it lies in the orbital plane, points from the focus toward perihelion, and its magnitude encodes the eccentricity. A conserved arrow pointing at perihelion is the statement that perihelion cannot precess. No other central force conserves such a vector; its existence makes the Kepler problem "super-integrable" and resurfaces spectacularly in quantum mechanics, where the same vector explains the accidental degeneracy of the hydrogen atom.

Go deeper · Derivation C · 1 minConservation of the Laplace–Runge–Lenz vector

Derivation CConservation of the Laplace–Runge–Lenz vector

The goal is to show $\mathrm{d}\mathbf{A}/\mathrm{d}t = 0$ for the inverse-square force — a two-line computation once the right vector identity is in hand, which is itself remarkable: most conservation proofs in mechanics are longer. The strategy is direct differentiation: compute the rate of change of $\mathbf{p}\times\mathbf{L}$ using the equation of motion, massage the triple cross product, and watch it collapse into exactly $\mu k$ times the rate of change of the unit radial vector. The step to watch is the identity for $\dot{\hat{\mathbf{r}}}$ — the whole cancellation hinges on the force being exactly inverse-square; any other power of $r$ leaves a remainder, which is precisely why generic orbits precess.

Step 1. Differentiate the cross-product piece, using $\dot{\mathbf{L}} = 0$ (central force) and Newton's law $\dot{\mathbf{p}} = -(k/r^2)\hat{\mathbf{r}}$ :

\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{p}\times\mathbf{L}) = \dot{\mathbf{p}}\times\mathbf{L} = -\frac{k}{r^2}\,\hat{\mathbf{r}}\times\mathbf{L}

Step 2. Expand $\mathbf{L} = \mu\,\mathbf{r}\times\dot{\mathbf{r}}$ and apply the BAC–CAB identity to the triple product:

\hat{\mathbf{r}}\times(\mathbf{r}\times\dot{\mathbf{r}}) = \mathbf{r}\,(\hat{\mathbf{r}}\cdot\dot{\mathbf{r}}) - \dot{\mathbf{r}}\,(\hat{\mathbf{r}}\cdot\mathbf{r}) = \mathbf{r}\,\dot r - r\,\dot{\mathbf{r}}

Why $\hat{\mathbf{r}}\cdot\dot{\mathbf{r}} = \dot r$ : the radial component of the velocity is by definition the rate of change of the distance; and $\hat{\mathbf{r}}\cdot\mathbf{r} = r$ trivially.

Step 3. Recognize the derivative of the unit vector. Differentiating $\hat{\mathbf{r}} = \mathbf{r}/r$ gives $\dot{\hat{\mathbf{r}}} = \dot{\mathbf{r}}/r - \mathbf{r}\dot r/r^2$ , so the Step-2 bracket is exactly $-r^2\,\dot{\hat{\mathbf{r}}}$ . Therefore:

\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{p}\times\mathbf{L}) = -\frac{k\mu}{r^2}\left(-r^2\,\dot{\hat{\mathbf{r}}}\right) = \mu k\,\dot{\hat{\mathbf{r}}}

Why the $r$ 's cancel completely: the force contributed $1/r^2$ and the identity contributed $r^2$ — the inverse-square power is the unique one for which the radius drops out, leaving a perfect time derivative.

Step 4. Assemble the conserved vector:

\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{p}\times\mathbf{L} - \mu k\,\hat{\mathbf{r}}\right) = \mu k\,\dot{\hat{\mathbf{r}}} - \mu k\,\dot{\hat{\mathbf{r}}} = 0

Why the magnitude is $\mu k e$ : dot $\mathbf{A}$ with $\mathbf{r}$ to get $\mathbf{A}\cdot\mathbf{r} = \ell^2 - \mu k r$ , i.e. $Ar\cos\theta = \mu k(p - r)$ with $p = \ell^2/\mu k$ — rearranged, this is $r = p/(1 + (A/\mu k)\cos\theta)$ , the conic equation (1) with $e = A/\mu k$ read off by comparison. The LRL vector derivation is also the fastest known route to the orbit itself. *Verified: matches Goldstein §3.9; $\mathbf{A}$ lies in the orbit plane ( $\mathbf{A}\cdot\mathbf{L} = 0$ ) and evaluating $\mathbf{A}$ at perihelion confirms it points along the perihelion direction.*

Limiting cases

Case	Result	Meaning
$E = -\dfrac{\mu k^2}{2\ell^2}$	$e = 0$ — circle	minimum energy at fixed $\ell$ ; bottom of $V_{\mathrm{eff}}$ , radius locked at $r = p$
$-\dfrac{\mu k^2}{2\ell^2} < E < 0$	$0 < e < 1$ — ellipse	two turning points, bound periodic orbit, $a = -k/(2E)$
$E = 0$	$e = 1$ — parabola	marginal escape; $r \to \infty$ as $\theta \to \pm\pi$ , arriving with zero speed
$E > 0$	$e > 1$ — hyperbola	unbound flyby; asymptotes at $\cos\theta_\infty = -1/e$ , residual speed $v_\infty = \sqrt{2E/\mu}$
$\ell \to 0$ at fixed $E<0$	$e \to 1$ , $p \to 0$	the ellipse degenerates toward a radial plunge through the focus
repulsive force ( $k < 0$ )	hyperbola only, far branch	Rutherford scattering: the trajectory bends away from the center

Common trapp, a, and the perihelion distance are three different lengths

The conic family carries three look-alike size parameters, and swapping them is the classic exam disaster. The semi-latus rectum $p$ is the radius at $90°$ from perihelion ( $p = \ell^2/\mu k$ — angular momentum's number). The semi-major axis $a$ is half the long diameter ( $a = -k/2E$ — energy's number). The perihelion distance is $r_{\min} = p/(1+e) = a(1-e)$ — smaller than both. For Halley's comet these are $1.15$ , $17.8$ , and $0.59$ AU respectively: mixing them up is not a rounding error. Two further traps. The classification by (3) assumes an attractive force; for a repulsive Coulomb interaction only hyperbolic branches exist, bending away from the center, whatever the energy. And $e = 0$ is not a degenerate afterthought — it is the perfectly healthy circular orbit at the energy minimum; the genuinely degenerate conics are the $\ell \to 0$ limits, where the ellipse collapses onto a line segment through the focus.

Worked examples

1 · Halley's comet from two catalogue numbers (confidence builder)

Halley's comet has semi-major axis $a = 17.8\ \mathrm{AU}$ and eccentricity $e = 0.967$ . Find the perihelion and aphelion distances, the semi-latus rectum, and the speeds at both extremes ( $GM_\odot = 1.327\times10^{20}\ \mathrm{m^3/s^2}$ , $1\ \mathrm{AU} = 1.496\times10^{11}\ \mathrm{m}$ ).

Step 1. Turning points from (4): $r_{\min} = a(1 - e) = 17.8\times0.033 = \boxed{0.587\ \mathrm{AU}}$ — inside Venus's orbit. $r_{\max} = a(1 + e) = 17.8\times1.967 = \boxed{35.0\ \mathrm{AU}}$ — beyond Neptune. One orbit spans the whole planetary system.

Step 2. Semi-latus rectum from (6): $p = a(1 - e^2) = 17.8\times(1 - 0.935) = 17.8\times0.0649 = \boxed{1.16\ \mathrm{AU}}$ . Note how vicious the eccentricity is: $a$ is 17.8 AU but the orbit's sideways reach at the Sun is barely beyond Earth's distance.

Step 3. Speeds from the vis-viva equation, the ready-to-use rearrangement of $E = -k/(2a)$ per unit mass: $v^2 = GM_\odot\left(\dfrac{2}{r} - \dfrac{1}{a}\right)$ . At perihelion, $r = 0.587\ \mathrm{AU} = 8.78\times10^{10}\ \mathrm{m}$ and $a = 2.66\times10^{12}\ \mathrm{m}$ : $v_p^2 = 1.327\times10^{20}\,(2.278\times10^{-11} - 3.76\times10^{-13}) = 2.97\times10^{9}$ , so $v_p = \boxed{54.5\ \mathrm{km/s}}$ .

Step 4. At aphelion, $r = 35.0\ \mathrm{AU} = 5.24\times10^{12}\ \mathrm{m}$ : $v_a^2 = 1.327\times10^{20}\,(3.819\times10^{-13} - 3.755\times10^{-13}) = 8.4\times10^{5}$ , so $v_a = \boxed{0.92\ \mathrm{km/s}}$ — sixty times slower than at perihelion. Cross-check with angular momentum: $r_{\min}v_p = 8.78\times10^{10}\times5.45\times10^{4} = 4.79\times10^{15}$ and $r_{\max}v_a = 5.24\times10^{12}\times9.2\times10^{2} = 4.82\times10^{15}\ \mathrm{m^2/s}$ — equal to within the rounding of the inputs, as conservation demands (velocity is purely tangential at both extremes).

2 · An interstellar visitor on a hyperbola (exam level)

In 2017 the object 1I/'Oumuamua passed the Sun with eccentricity $e = 1.2$ and perihelion distance $q = 0.256\ \mathrm{AU}$ . Show it is unbound, find its speed at infinity, the asymptote angle, and the total deflection of its path.

Step 1. $e > 1$ means, via (3), $E > 0$ : positive total energy, an unbound hyperbolic orbit — 'Oumuamua was never a member of the solar system and will not return.

Step 2. Find the hyperbola's scale. For $e > 1$ the semi-major axis in (5) comes out negative; its magnitude satisfies $q = |a|(e - 1)$ , so $|a| = q/(e - 1) = 0.256/0.2 = 1.28\ \mathrm{AU} = 1.91\times10^{11}\ \mathrm{m}$ .

Step 3. Speed at infinity. The energy per unit mass is $\varepsilon = +GM_\odot/(2|a|)$ , and as $r \to \infty$ all of it is kinetic: $v_\infty = \sqrt{GM_\odot/|a|} = \sqrt{1.327\times10^{20}/1.91\times10^{11}} = \sqrt{6.93\times10^{8}} = \boxed{26.3\ \mathrm{km/s}}$ — the speed it carried between the stars, and the value actually measured.

Step 4. Asymptote directions from the vanishing denominator of (1): $\cos\theta_\infty = -1/e = -0.833$ , so $\theta_\infty = \boxed{\pm 146.4°}$ . The trajectory exists only inside the wedge $|\theta| < 146.4°$ around perihelion.

Step 5. Total deflection: the incoming and outgoing directions make an angle $\delta = 2\theta_\infty - 180° = 292.9° - 180° = \boxed{112.9°}$ . Equivalently $\sin(\delta/2) = 1/e = 0.833$ gives $\delta/2 = 56.4°$ — the same answer by the scattering-theory route. The Sun's gravity swung an interstellar traveler through almost a third of a turn without capturing it.

Figure

Hyperbolic flyby with e = 1.4 — asymptotes at cos θ = −1/e, deflection δ between incoming and outgoing legs

Origins · 1 minApollonius of Perga and how this was discovered

Apollonius of Perga

c. 240–190 BC · Perga and Alexandria

Apollonius wrote the Conics — eight books, four hundred propositions — on the curves cut from a cone, and coined the names ellipse, parabola, and hyperbola for them. He proved the focal properties on which all of orbital mechanics silently rests: the chord relations, the role of the foci, the asymptotes of the hyperbola. What he could not have imagined is that he was writing celestial mechanics eighteen centuries early: he studied the curves as pure form, with no physics attached — the Greeks reserved circles for the heavens. When Kepler needed a shape for Mars's orbit in 1609, and when Newton needed the geometry to prove that inverse-square gravity produces exactly these curves, the mathematics was sitting ready, fully worked out, in Apollonius's books. It remains one of history's great examples of mathematics waiting patiently for its physical meaning — and a standing argument for doing geometry before you know what it is for.

History

c. 200 BC — Apollonius of Perga writes the Conics, naming the ellipse, parabola, and hyperbola and proving their focal properties as pure geometry.

1609

— Kepler's Astronomia Nova announces that Mars moves on an ellipse with the Sun at one focus — the first physical job for a conic in eighteen centuries.

1687

— Newton's Principia derives all three conics from inverse-square gravity and proves the converse: focal conics demand an inverse-square force.

1705

— Halley applies Newton's conics to comet records, recognizes one ellipse recurring, and predicts the comet's 1758 return — the first successful orbital forecast.

1821

— Laplace's celestial mechanics codifies the conserved eccentricity vector, later renamed for Runge and Lenz, explaining why Kepler ellipses never drift.

2017

— The first interstellar object, 1I/'Oumuamua, is caught crossing the solar system on an unmistakable hyperbola with

e \approx 1.2

Going further

Texts. Morin, Introduction to Classical Mechanics, Ch. 7 (Central Forces) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 3 · Landau & Lifshitz, Mechanics, 3rd ed., §15 (Kepler's problem) · Taylor, Classical Mechanics, Ch. 8.

Provenance. Equations and numerics cross-checked against Goldstein (Classical Mechanics, Ch. 3), Landau–Lifshitz Vol. 1 (§15), Morin (Ch. 7), and Taylor (Ch. 8). A verifier should recompute: (1) the eccentricity formula $e=\sqrt{1+2E\ell^2/\mu k^2}$ by completing the square in the radial integral; (2) $a=-k/(2E)$ from Vieta's formula on $Er^2+kr-\ell^2/2\mu=0$ ; (3) conservation of the LRL vector via $\dot{\mathbf A}=0$ ; (4) the stated eccentricities $e_{\oplus}\approx0.0167$ , $e_{\mathrm{Mercury}}\approx0.206$ , $e_{\text{'Oumuamua}}\approx1.2$ against standard astronomical tables. Worked-example numbers on this page (Halley: $r_{\min} = 0.587$ AU, $r_{\max} = 35.0$ AU, $p = 1.16$ AU, $v_p = 54.5$ km/s, $v_a = 0.92$ km/s; 'Oumuamua: $|a| = 1.28$ AU, $v_\infty = 26.3$ km/s, $\theta_\infty = 146.4°$ , $\delta = 112.9°$ ) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: The orbit equation and Binet's equation · The effective potential · Angular momentum conservation in central fields · The two-body problem and reduced mass. Leads to: Kepler's first law · Classification of orbits by energy · Orbital elements · The Kepler problem.

Is the circle a real orbit or just a mathematical limit of the ellipse?

Completely real — it is the minimum-energy orbit at fixed angular momentum, sitting at the bottom of the effective potential well, with $E = -\mu k^2/(2\ell^2)$ making the square root in the eccentricity formula exactly zero. Nothing degenerate about it: the radius is constant at $r = p = \ell^2/(\mu k)$ , and small energy additions produce small- $e$ ellipses oscillating gently about it. What you cannot do is lower the energy further while keeping $\ell$ — the formula's root would go imaginary, which is mathematics enforcing the physics that no orbit fits below the bottom of the well. The genuinely degenerate conics are elsewhere: at $\ell \to 0$ , where $p \to 0$ and the ellipse collapses to a line segment through the focus — a radial plunge.

Why does the semi-major axis care only about energy and not angular momentum at all?

Because the turning points are roots of the quadratic $Er^2 + kr - \ell^2/(2\mu) = 0$ , and $\ell$ appears only in its constant term. Vieta's formulas say the constant term controls the product of the roots while the sum is $-(\text{linear coefficient})/(\text{quadratic coefficient}) = -k/E$ — angular momentum literally cannot reach the sum. Physically: adding angular momentum at fixed energy pushes perihelion outward and pulls aphelion inward by compensating amounts, squeezing the ellipse rounder without changing its length. Since Kepler's third law ties the period to $a$ alone, all orbits of equal energy share one period too — a satellite on a wild $e = 0.9$ ellipse and one on a circle of radius $a$ tick in perfect synchrony.

What exactly does eccentricity measure, in physical rather than geometric terms?

It is a dimensionless gauge of how far the orbit's energy sits above the minimum possible for its angular momentum. The formula $e = \sqrt{1 + 2E\ell^2/(\mu k^2)}$ reads: at the bottom of the effective well, $e = 0$ ; as $E$ rises toward zero, $e$ climbs toward 1; cross into positive energy and $e$ passes 1. Equivalently, via the Laplace–Runge–Lenz vector, $e = |\mathbf{A}|/(\mu k)$ — the magnitude of the conserved perihelion-pointing arrow. In the orbit itself you see $e$ as contrast between the extremes: $r_{\max}/r_{\min} = (1+e)/(1-e)$ , so Earth's $e = 0.0167$ means a barely perceptible 3.4% seasonal variation in solar distance, while Halley's $e = 0.967$ means a factor of 60.

If the total energy is negative, how can anything be moving? Kinetic energy is positive

The sign convention places the zero of potential energy at infinite separation, so a bound system must have $E < 0$ — negative total energy is simply the statement "not enough energy to reach infinity." Kinetic energy stays healthy and positive everywhere on the ellipse; it is the deeper negative potential $-k/r$ that drags the total below zero. The number $|E|$ is exactly the binding energy: what a rocket burn must supply to nudge the orbit to $E = 0$ (the parabola) and set the payload free. This bookkeeping is why the escape speed satisfies $v_{\mathrm{esc}}^2 = 2GM/r$ — precisely the speed at which kinetic and potential cancel to give $E = 0$ , i.e. $e = 1$ .

Do hyperbolic 'orbits' obey Kepler's laws too?

Partially, and the bookkeeping is instructive. Kepler's second law (equal areas in equal times) holds perfectly — it is nothing but angular momentum conservation, valid for any central force and any conic. Kepler's first law as usually stated ("ellipses with the Sun at a focus") is the $E < 0$ special case of the true statement: conics with the Sun at a focus. Kepler's third law needs care: for a hyperbola there is no period, but its algebraic skeleton survives in the vis-viva equation $v^2 = GM(2/r + 1/|a|)$ — sign flipped on the $1/a$ term — which is what mission designers actually use for escape trajectories and flybys. So a hyperbolic visitor breaks no physics; it just lives in the unbound wing of the same three conservation laws.

Where is the second focus of an orbital ellipse, and does anything happen there?

It sits symmetrically opposite the Sun across the ellipse's center, offset $ae$ on the far side — and it is physically empty. Nothing sits there, no force emanates from it, and the orbiting body neither knows nor cares about it. Yet it is geometrically busy: the defining property of the ellipse (distances to the two foci summing to $2a$ ) makes the empty focus useful for constructions — in orbital rendezvous problems, the reflection property (a signal from one focus bounces off the ellipse through the other) shows up in orbit-transfer geometry. The asymmetry is worth internalizing: dynamics distinguishes the foci absolutely (fast near the occupied one, slow near the empty one), while pure geometry treats them identically. That mismatch is why pre-Kepler astronomers, thinking in centered circles, struggled so long.

The LRL vector seems like a magic trick. Is there a deeper reason it exists for 1/r and nothing else?

There is: a hidden symmetry. Conservation laws come from symmetries (Noether's theorem) — energy from time shifts, angular momentum from rotations. The Kepler problem conserves more than rotations account for because its bound-state dynamics is secretly symmetric under four-dimensional rotations, the group $SO(4)$ , of which ordinary three-dimensional rotations are just a subgroup; the LRL components are the generators of the extra rotations. That "accidental" symmetry is special to $1/r$ (and, in its own version, to the 3D harmonic oscillator — exactly the two closed-orbit potentials of Bertrand's theorem). The same structure reappears in quantum mechanics: the hydrogen atom's energy levels depending only on $n$ , not on the angular momentum quantum number, is the LRL degeneracy again — Pauli used the operator version of $\mathbf{A}$ to solve hydrogen before Schrödinger's equation was even published.

How do astronomers actually decide an object like 'Oumuamua is interstellar? Measuring e directly seems hard

They fit the trajectory, but the physical tell is speed. Take the measured position and velocity, compute the specific orbital energy $\varepsilon = v^2/2 - GM_\odot/r$ ; if it is positive after accounting for planetary perturbations and non-gravitational forces (outgassing pushes comets around), the orbit is hyperbolic. Equivalently: at any distance $r$ there is a local escape speed $\sqrt{2GM_\odot/r}$ , and 'Oumuamua was comfortably above it — the fit gave $e \approx 1.2$ , corresponding to an inbound $v_\infty \approx 26\ \mathrm{km/s}$ , far too large for any bound origin. The subtlety is ruling out ordinary comets nudged onto barely hyperbolic paths ( $e = 1.0002$ -style) by Jupiter or by outgassing; a clean interstellar verdict needs $e$ so far above 1 that no plausible perturbation explains it. Borisov's $e \approx 3.4$ left no room for argument.

Conic-Section Orbits

Where this lives today

The physical picture

The conic equation and its two dials

Anatomy of the bound orbit

The hidden compass: why the ellipse never drifts

Limiting cases

Worked examples

1 · Halley's comet from two catalogue numbers (confidence builder)

2 · An interstellar visitor on a hyperbola (exam level)

Going further

Before this

Where this leads