Kepler's First Law: Elliptical Orbits

Plot Mars's orbit to scale and you will be underwhelmed: it looks like a circle. Stretch it by less than half a percent and it is one — no eye could tell the difference. So what, exactly, did Kepler discover in 1605 that was worth demolishing two thousand years of astronomy? Not the shape. The placement. The Sun does not sit at the middle of that near-perfect circle; it sits 21 million kilometres off centre, at a special point called a focus. Because of that offset, Mars's distance from the Sun swings by a full 21% every orbit, and its speed swings with it — which is precisely the misbehaviour in Tycho Brahe's data that circles, however cleverly arranged, could never reproduce. Kepler's First Law packages both statements: each planet moves on an ellipse, with the Sun at one focus. The dynamical machinery that forces this — the two-body reduction, the effective potential, the Binet equation and its conic solution — is developed in full in the Kepler problem article. Here we take the result, $r(\theta)=\ell/(1+e\cos\theta)$ , and learn to read it: what the eccentricity $e$ really measures (offset, not flattening), what the semi-latus rectum $\ell$ is, where perihelion and aphelion sit, and why "Sun at a focus" is something an observer can actually check. Master this geometry and you own the vocabulary of every orbit in the sky: exoplanet catalogues, comet ephemerides, and spacecraft trajectories are all written in the language of $a$ , $e$ , and $\ell$ — the three letters this article is about.

Key takeaways

Kepler's First Law: each planet moves on an ellipse with the Sun at one focus — offset from the centre by $c=ae$ , not at the middle.
The orbit in polar form is $r(\theta)=\ell/(1+e\cos\theta)$ , with $\theta$ measured from perihelion and the Sun at the origin; the full derivation lives in the Kepler problem article.
Eccentricity measures offset, not flattening: the shape deviates from a circle only at order $e^2$ , but the Sun's displacement and the distance swing are first order in $e$ .
The geometry dictionary: perihelion $a(1-e)$ , aphelion $a(1+e)$ , semi-minor axis $b=a\sqrt{1-e^2}=\sqrt{a\ell}$ , and $\ell=L^2/(\mu k)$ is the radius at $\theta=90^\circ$ .
Bound orbits ( $E<0$ ) have $0\le e<1$ ; the same conic formula continues to parabolas ( $e=1$ ) and hyperbolas ( $e>1$ ).
Key idea: For an attractive inverse-square force, integrating the orbit equation gives $r(\theta)=\ell/(1+e\cos\theta)$ — a conic section with the force centre at a focus. When $0\le e<1$ (equivalently $E<0$ ) this conic is an ellipse, which is Kepler's First Law.

Where this lives today

Every exoplanet orbit ever published is a fitted Kepler ellipse. Radial-velocity and transit surveys record time series, fit the conic $r=\ell/(1+e\cos\theta)$ dressed in orbital elements, and report $a$ and $e$ in the catalogue. The result is a standing puzzle: while the Solar System's planets keep $e\lesssim0.09$ , the exoplanet census — closing in on 6,000 confirmed worlds in 2026 — is full of giant planets on $e>0.3$ ellipses, evidence of violent scattering histories our own system avoided. ESA's Gaia mission, whose full DR4 release lands in late 2026, adds the astrometric channel: thousands of stars whose sky-paths wobble on milli-arcsecond ellipses, each a scaled mirror of an unseen companion's Kepler orbit. The law's small print earns money too: every transfer orbit is specified by its perigee and apogee — the $a(1\mp e)$ of equation (3) — and collision screening for tens of thousands of tracked objects in Earth orbit runs on focal-conic geometry evaluated millions of times per day.

The physical picture

Hold Mars's real orbit up against a perfect circle of the same size. The two curves lie on top of each other — the ellipse is narrower by $0.4\%$ , thinner than the ink line you would draw it with. Every popular diagram of a "squashed" planetary orbit is a lie. What is not a lie is the position of the Sun:

Figure

Mars's orbit is a near-perfect circle — with the Sun conspicuously off centre

This figure is the whole observational content of the First Law. The shape deviates from a circle at second order in the eccentricity — for Mars, $1-b/a\approx e^2/2\approx0.4\%$ . But the Sun's offset from the centre is first order: $c=ae$ , a full $9.3\%$ of the orbit's size. The consequences are first order too: Mars's solar distance swings between $1.38$ and $1.67\,\mathrm{AU}$ — a $21\%$ variation — and the sunlight falling on it changes by $45\%$ over each orbit. It was this off-centre signature, not any visible squashing, that killed the circle in Tycho's data. The second picture to install is the family the ellipse belongs to. Fix the angular momentum — equivalently, fix $\ell$ — and dial the energy up from the circular minimum: the circle stretches into ever-longer ellipses and finally snaps open into a parabola, but every member of the family threads the same two points, a distance $\ell$ directly "above" and "below" the Sun at $\theta=\pm90^\circ$ . The perihelion side barely moves while the aphelion side does all the stretching. That anchoring is why $\ell$ — the semi-latus rectum — is the natural size label on a conic: it is the one length angular momentum fixes by itself, before energy has any say.

Reading the orbit equation

The dynamics behind the First Law is settled elsewhere: the Kepler problem article develops the full two-body reduction, builds the effective potential, and integrates the Binet orbit equation $u''+u=\mu k/L^2$ (with $u=1/r$ ) to obtain the trajectory. What that machinery delivers is one line of geometry, and this article's job is to read it fluently. Here is the line:

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

where: $r,\theta$ — polar coordinates with the force centre (the Sun) at the origin and $\theta$ measured from perihelion; $\ell$ — the semi-latus rectum, the value of $r$ at $\theta=90^\circ$ ; $e$ — the eccentricity, $0\le e<1$ for a bound orbit. Read the denominator as a "closeness dial": it is largest at $\theta=0$ (perihelion, $r=\ell/(1+e)$ ), passes through $1$ at $\theta=90^\circ$ (where $r=\ell$ exactly), and is smallest at $\theta=180^\circ$ (aphelion, $r=\ell/(1-e)$ ). For $e<1$ the denominator never reaches zero, so $r$ stays finite: the orbit is closed. The two constants in equation (1) are not free decoration — each is owned by a conserved quantity. Angular momentum alone sets the size scale, and energy (with $L$ ) sets the shape; both identifications are derived in the Kepler problem article and simply quoted here:

\ell=\frac{L^2}{\mu k},\qquad e=\sqrt{1+\frac{2EL^2}{\mu k^2}} \qquad (2)

where: $L$ — orbital angular momentum of the relative motion; $E$ — its total energy; $\mu=m_1m_2/(m_1+m_2)$ — the reduced mass; $k=Gm_1m_2$ — the strength of the attraction. The reading matters more than the algebra: $\ell$ knows only about $L$ , which is why every orbit in a fixed- $L$ family threads the same points at $\theta=\pm90^\circ$ ; and $e$ is an energy dial that hits exactly $1$ when $E=0$ — eccentricity is where the bound/unbound classification lives. Equation (1) says "conic with focus at the origin", but for a bound orbit we want the familiar ellipse vocabulary — semi-major axis, semi-minor axis, centre, two foci. Here is the dictionary, laid out on an exaggerated $e=0.6$ ellipse:

Figure

Anatomy of the Kepler ellipse — reading every length in the orbit equation

The apsidal distances come straight from evaluating equation (1) at its extremes, and averaging them defines the semi-major axis. Two lines of algebra (done properly in the second derivation below) give the working set every orbit calculation uses:

r_\mathrm{min}=a(1-e),\qquad r_\mathrm{max}=a(1+e),\qquad a=\frac{\ell}{1-e^2} \qquad (3)

where: $r_\mathrm{min}$ — the perihelion distance, reached at $\theta=0$ ; $r_\mathrm{max}$ — the aphelion distance, at $\theta=180^\circ$ ; $a$ — the semi-major axis, half the long diameter. Notice what averaging hides: $a$ is the mean of the two apsidal distances, so "orbit radius $\approx a$ " is a fine mental shortcut — but the instantaneous distance roams across the full $(1\pm e)$ range, a swing of $\pm e$ about the mean and a factor $(1+e)/(1-e)$ from end to end; the worked examples put honest numbers on this. The remaining lengths — the semi-minor axis and the centre-to-focus offset — complete the dictionary and carry the article's central moral:

b=a\sqrt{1-e^2}=\sqrt{a\ell},\qquad c=ae=\sqrt{a^2-b^2} \qquad (4)

where: $b$ — the semi-minor axis, the half-width of the ellipse; $c$ — the distance from the ellipse's centre to either focus. Expand for small $e$ : $b\approx a(1-e^2/2)$ but $c=ae$ . The flattening is second order in $e$ while the focal offset is first order — this asymmetry is why planetary orbits look circular yet the Sun sits visibly off centre, and why eccentricity should always be read as "offset", never "squashedness".

Derivation AWhy the force centre sits at a focus

The goal is the geometric heart of the First Law: to prove that the polar curve $r=\ell/(1+e\cos\theta)$ , with $0\le e<1$ , really is an ellipse, and that its focus — not its centre — sits at the origin where the Sun lives. The strategy is direct translation: convert the polar equation to Cartesian coordinates centred on the Sun, complete the square, and compare with the textbook ellipse. The payoff of brute force is that the focal offset $c=ae$ falls out as the shift needed to complete the square — the Sun ends up displaced from the ellipse's centre by exactly $ae$ , the whole observational content of the law. Watch the conventions: perihelion is placed on the $+x$ axis, so the centre lands at $x=-c$ , on the aphelion side.

Step 1. Clear the denominator. Multiply equation (1) out: $r+er\cos\theta=\ell$ . With Cartesian coordinates $x=r\cos\theta$ , $y=r\sin\theta$ centred on the Sun, this reads $r=\ell-ex$ . Why: the polar conic is linear in $r\cos\theta$ , so one multiplication converts it to a statement about $x$ — this is the step that makes the focus-centred origin tractable.

Step 2. Square away the radical. Since $r^2=x^2+y^2$ , squaring gives $x^2+y^2=\ell^2-2e\ell x+e^2x^2$ , i.e. $(1-e^2)x^2+2e\ell x+y^2=\ell^2$ . Why: squaring is safe here because both sides of $r=\ell-ex$ are non-negative on a bound orbit — $r\le r_\mathrm{max}=\ell/(1-e)$ keeps $ex$ below $\ell$ .

Step 3. Complete the square in x. Divide by $(1-e^2)$ (positive for $e<1$ ): $x^2+\dfrac{2e\ell}{1-e^2}x+\dfrac{y^2}{1-e^2}=\dfrac{\ell^2}{1-e^2}$ . Completing the square in $x$ adds $e^2\ell^2/(1-e^2)^2$ to both sides, and the right side collects into $\ell^2/(1-e^2)^2$ exactly. The result is $\dfrac{(x+c)^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a=\dfrac{\ell}{1-e^2}$ , $b=\dfrac{\ell}{\sqrt{1-e^2}}$ , $c=\dfrac{e\ell}{1-e^2}=ae$ . Why: completing the square is what exposes the centre of the curve — and it does not sit at the origin, but at $x=-c$ , shifted toward aphelion.

Step 4. Identify the geometry. The result is the standard ellipse with semi-axes $a$ and $b=a\sqrt{1-e^2}$ , centred at $(-c,0)$ . The origin — where the force centre sits — is at distance $c=ae$ from the centre, and $c^2=a^2-b^2$ : that is precisely the defining property of a focus of the ellipse. Why: an ellipse's foci lie on the major axis at $\pm\sqrt{a^2-b^2}$ from the centre; our origin lands exactly on one of them, with the empty focus at $-2c$ .

Step 5. Check the circle limit. Set $e=0$ : then $a=b=\ell$ , $c=0$ , and both foci collapse onto the centre — the orbit $r=\ell$ is a circle with the Sun at its middle, the only case where "centre" is the right word. Why: a limiting case that reduces to something obviously correct is the cheapest full-strength test of the algebra. *Verified: The identifications $a=\ell/(1-e^2)$ , $b=a\sqrt{1-e^2}$ , $c=ae$ match the standard focal-conic geometry in Goldstein Ch. 3 and Taylor Ch. 8, and reproduce equation (4).*

Go deeper · Derivation B · 2 minThe geometry dictionary from the apsides

Derivation BThe geometry dictionary from the apsides

Here we derive the working formulas (3) and (4) — the dictionary between the conic constants $(\ell,e)$ and the observer's quantities $(a,b,r_\mathrm{min},r_\mathrm{max})$ — using nothing but equation (1) evaluated at special angles. This is deliberately the practitioner's route: apsidal distances are what astronomers actually measure, so the dictionary is most useful built from the apsides inward. The subtlety to watch is which combinations are energy-like and which are angular-momentum-like: the sum of the apsidal distances gives $a$ (pure energy, as the Kepler problem article shows via $a=-k/2E$ ), while their geometric mean gives $b$ (which carries the angular momentum through $\ell$ ).

Step 1. Evaluate the orbit equation at its extremes. At $\theta=0$ the denominator of (1) is maximal: $r_\mathrm{min}=\dfrac{\ell}{1+e}$ . At $\theta=180^\circ$ it is minimal: $r_\mathrm{max}=\dfrac{\ell}{1-e}$ . Why: $\cos\theta$ ranges over $[-1,1]$ , so these two angles bracket every distance on the orbit.

Step 2. Sum for the semi-major axis. The apsides sit at opposite ends of the long diameter, so $2a=r_\mathrm{min}+r_\mathrm{max}=\dfrac{\ell(1-e)+\ell(1+e)}{1-e^2}=\dfrac{2\ell}{1-e^2}$ , giving $a=\dfrac{\ell}{1-e^2}$ and, inverting, $r_\mathrm{min}=a(1-e)$ , $r_\mathrm{max}=a(1+e)$ . Why: perihelion and aphelion are collinear with both foci, so their sum is exactly the major axis — no trigonometry needed.

Step 3. Difference for the focal offset. Similarly $r_\mathrm{max}-r_\mathrm{min}=2ae=2c$ : the offset of the Sun from the centre is half the spread between the apsidal distances, and dividing the two results gives the observer's eccentricity formula $e=\dfrac{r_\mathrm{max}-r_\mathrm{min}}{r_\mathrm{max}+r_\mathrm{min}}$ . Why: this quotient needs no absolute distance scale — ratios of angular sizes or brightnesses can pin $e$ even when the absolute scale is uncertain.

Step 4. Geometric mean for the semi-minor axis. From Derivation A, $b^2=a^2-c^2=a^2(1-e^2)$ ; substituting $a(1-e)\,a(1+e)=r_\mathrm{min}r_\mathrm{max}$ gives $b=\sqrt{r_\mathrm{min}r_\mathrm{max}}$ , and with $a=\ell/(1-e^2)$ also $b=\sqrt{a\ell}$ . Why: the arithmetic mean of the apsides is $a$ , their geometric mean is $b$ — a mnemonic that also flags $b$ as the carrier of angular momentum, since $\ell=L^2/\mu k$ enters only here.

Step 5. Close the loop on the semi-latus rectum. From $b^2=a\ell$ : $\ell=b^2/a$ , and evaluating (1) at $\theta=90^\circ$ confirms $r=\ell$ there — the "latus rectum" is literally the chord through the Sun perpendicular to the major axis, of half-length $\ell$ . Why: this gives $\ell$ an instrument-ready meaning: it is the orbit's width at the Sun, not at the centre. *Verified: For Mars, $r_\mathrm{min}=1.3817$ , $r_\mathrm{max}=1.6663\,\mathrm{AU}$ give $a=1.524\,\mathrm{AU}$ and $e=(1.6663-1.3817)/(1.6663+1.3817)=0.0934$ — the catalogue values; and $b=\sqrt{1.3817\times1.6663}=1.517\,\mathrm{AU}=a\sqrt{1-e^2}$ checks.*

Limiting cases

One formula, equation (1), covers every regime as $e$ moves through its range:

Case	Result	Meaning
$e=0$	$r=\ell$ everywhere; $a=b=\ell$ , $c=0$	The circle: both foci merge at the centre — the only orbit where "Sun at the centre" is true.
$0<e\ll1$	shape error $1-b/a\approx e^2/2$ ; offset $c=ae$	Planetary regime: the orbit looks circular but the Sun is measurably off centre — offset beats flattening by a factor $2/e$ .
$e\to1^-$ , fixed $\ell$	$r_\mathrm{max}=\ell/(1-e)\to\infty$ , $r_\mathrm{min}\to\ell/2$	The ellipse stretches without bound on the aphelion side only; at $e=1$ it opens into a parabola ( $E=0$ ).
$e>1$	$r\to\infty$ at $\cos\theta=-1/e$	Hyperbolic flyby: the conic formula survives, but the orbit is unbound — the First Law's "ellipse" clause no longer applies.
$e\to1$ , $\ell\to0$	degenerate segment through the focus	Radial fall: zero angular momentum collapses the conic to a straight line — a reminder that $e=1$ alone does not specify an orbit.

Common trapThe Sun sits at the centre, and eccentricity means squashedness

Both halves are wrong, and they reinforce each other. The Sun sits at a focus, displaced from the centre by $c=ae$ — 21 million km for Mars, nearly the entire semi-major axis for Halley's comet. And eccentricity does not primarily measure squashedness: the flattening is second order ( $1-b/a\approx e^2/2$ , an invisible $0.4\%$ for Mars) while the focal offset and the distance swing are first order in $e$ ( $9.3\%$ and $21\%$ ). The textbook drawing of a visibly squashed oval with the Sun at the middle commits both errors at once; the honest picture is a near-circle with the Sun off centre. One more trap rides along: nothing sits at the empty focus — it is a geometric shadow, not a physical point.

Worked examples

1 · Mars by the numbers (confidence builder)

Mars has semi-major axis $a=1.524\,\mathrm{AU}$ and eccentricity $e=0.0934$ . Find its perihelion and aphelion distances, the Sun's offset from the orbit centre, the shape deviation from a circle, and the variation in sunlight it receives.

Step 1. Apsides from equation (3): $r_\mathrm{min}=a(1-e)=1.524\times0.9066=\mathbf{1.382\ \mathrm{AU}}$ and $r_\mathrm{max}=a(1+e)=1.524\times1.0934=\mathbf{1.666\ \mathrm{AU}}$ . The solar distance swings by $r_\mathrm{max}/r_\mathrm{min}=1.206$ — a hefty $21\%$ .

Step 2. Focal offset: $c=ae=1.524\times0.0934=0.1423\,\mathrm{AU}=\mathbf{2.13\times10^{7}\ \mathrm{km}}$ — the Sun sits 21 million kilometres from the centre of Mars's orbit, $9.3\%$ of $a$ .

Step 3. Shape deviation: $b=a\sqrt{1-e^2}=1.524\times0.99563=1.517\,\mathrm{AU}$ , so $1-b/a=\mathbf{0.44\%}$ . Drawn to scale, the ellipse is indistinguishable from a circle — yet the Sun's off-centre position (Step 2) is twenty times larger as a fraction of $a$ .

Step 4. Sunlight variation: solar flux scales as $1/r^2$ , so $\dfrac{F_\mathrm{peri}}{F_\mathrm{aph}}=\left(\dfrac{1.666}{1.382}\right)^2=\mathbf{1.45}$ — Mars at perihelion receives $45\%$ more sunlight than at aphelion, which is why its southern-hemisphere summers (near perihelion) drive the planet's global dust-storm season.

2 · Rosetta's comet, geometry from the apsides (exam level)

Comet 67P/Churyumov–Gerasimenko — the comet ESA's Rosetta orbited — has perihelion $r_\mathrm{min}=1.243\,\mathrm{AU}$ and aphelion $r_\mathrm{max}=5.683\,\mathrm{AU}$ . Reconstruct the full orbital geometry: $a$ , $e$ , $c$ , $b$ , $\ell$ , and the true anomaly at which the comet's distance equals $a$ .

Step 1. Semi-major axis is the arithmetic mean of the apsides: $a=\tfrac12(1.243+5.683)=\mathbf{3.463\ \mathrm{AU}}$ .

Step 2. Eccentricity from the apsidal quotient (Derivation B, Step 3): $e=\dfrac{5.683-1.243}{5.683+1.243}=\dfrac{4.440}{6.926}=\mathbf{0.641}$ .

Step 3. Focal offset: $c=ae=3.463\times0.641=\mathbf{2.220\ \mathrm{AU}}$ — here the Sun's displacement is $64\%$ of the orbit's size, and the ellipse is visibly eccentric: $b=\sqrt{r_\mathrm{min}r_\mathrm{max}}=\sqrt{1.243\times5.683}=\mathbf{2.658\ \mathrm{AU}}$ , a flattening $1-b/a=23\%$ .

Step 4. Semi-latus rectum: $\ell=b^2/a=2.658^2/3.463=\mathbf{2.040\ \mathrm{AU}}$ . Check against $\ell=a(1-e^2)=3.463\times(1-0.411)=2.040\,\mathrm{AU}$ ✓. This is the comet's distance from the Sun when it stands at $\theta=90^\circ$ , a quarter-turn past perihelion.

Step 5. Where does $r=a$ ? Set $\ell/(1+e\cos\theta)=a=\ell/(1-e^2)$ , so $1+e\cos\theta=1-e^2$ , giving $\cos\theta=-e=-0.641$ and $\theta=\mathbf{129.9^\circ}$ . Note it is not at $90^\circ$ : the comet is already well past the quarter-turn point when its distance passes through the "average" value — a direct consequence of the Sun's off-centre seat.

Origins · 1 minJohannes Kepler and how this was discovered

Johannes Kepler

1571–1630 · Prague

Kepler came to the ellipse kicking and screaming. Handed Tycho Brahe's Mars observations — accurate to two arcminutes, the finest naked-eye data ever taken — he spent five years forcing circles, offset circles, and equant constructions onto the planet's path. His best circular model failed by just eight arcminutes, a quarter of the full Moon's width; every astronomer before him would have shrugged it into the noise. Kepler instead declared that "upon these eight minutes I will build a universe", discarded the circle, fought through an oval hypothesis that tormented him for a year, and recognized the ellipse — announcing in Astronomia Nova (1609) that Mars orbits the Sun on an ellipse with the Sun at one focus. It was the first precise law of nature extracted from data over theoretical prejudice, and it made Newton's dynamics possible.

History

1600

— Kepler joins Tycho Brahe in Prague and is assigned the orbit of Mars — by luck, the planet with the largest eccentricity among the well-observed ones.

1605

— After failed circles, equants, and a year lost to an egg-shaped oval, Kepler recognizes that Mars's orbit is an ellipse with the Sun at a focus.

1609

— Astronomia Nova publishes the First (and Second) Law, built on the famous eight-arcminute discrepancy he refused to ignore.

1687

— Newton's Principia derives the ellipse-at-a-focus from the inverse-square law, turning Kepler's description into a consequence of dynamics.

2017

— The interstellar visitor 1I/'Oumuamua is tracked on an

e\approx1.2

hyperbola — the unbound end of the same conic family, fitted with the same equation.

Going further

References. 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 are standard Kepler-problem results. A verifier should reproduce Binet's equation $u''+u=\mu k/L^2$ and its conic solution (Goldstein §3.5; Morin Ch. 7), confirm $e=\sqrt{1+2EL^2/\mu k^2}$ and $a=-k/2E$ (Landau–Lifshitz §15; Taylor Ch. 8), and verify $\mathrm{d}\vec A/\mathrm{d}t=0$ for $F=-k/r^2$ (Goldstein §3.9). Numbers: Earth $e=0.0167$ , $a=1\,\mathrm{AU}=1.496\times10^8$ km; Halley's comet $e\approx0.967$ , period $\approx76$ yr; Mercury anomalous precession $43''$ /century — all standard astronomical/textbook values.

Builds on: The Orbit Equation and Binet's Equation · The Kepler Problem · Angular Momentum Conservation in Central Fields · Conic-Section Orbits Leads to: Kepler's Second Law: Equal Areas · Kepler's Third Law: Period-Semimajor Relation · Orbital Elements and Parameterization · Classification of Orbits by Energy

If planetary orbits look like circles anyway, why does the First Law matter?

Because the law's content is the focus, not the flattening. A circle centred on the Sun predicts constant distance, speed, and brightness; the real orbit — a near-circle with the Sun offset by $c=ae$ — predicts a distance swinging by the factor $(1+e)/(1-e)$ , with the speed swinging in step. For Mars that is a $21\%$ distance variation and a $45\%$ sunlight variation, flatly incompatible with a centred circle; Kepler's eight-arcminute discrepancy was that signature's footprint in Tycho's data. The "looks circular" intuition confuses the second-order shape correction ( $e^2/2$ ) with the first-order offset ( $e$ ) — the article's central distinction.

What is at the empty focus of the ellipse?

Nothing at all — no mass, no force, no special dynamics. The ellipse, as a geometric shape, comes with two foci; the physics selects one for the force centre and ignores the other. The empty focus is still worth knowing about: the string construction pins both foci through the focal-sum property $r+r'=2a$ , and historically Ptolemy's equant point sat suspiciously close to where the empty focus belongs — his machinery was approximating a focal ellipse without knowing it. But dynamically it is a shadow: a planet crossing that region of space feels nothing special there.

Why does the Sun sit at a focus rather than at the centre — is there a physical reason?

The mathematics never offers the centre as an option. The orbit equation is solved in coordinates centred on the force centre — the Sun is the origin of the $r$ in $r=\ell/(1+e\cos\theta)$ — and translating that polar curve to Cartesian form (the first derivation) puts the centre of the resulting ellipse a distance $ae$ away from the origin. Physically, the asymmetry is forced by dynamics: angular momentum conservation makes the planet fastest where it is closest to the attractor, and an ellipse traversed about its own centre would have a speed pattern symmetric between the two ends of the major axis — which a $1/r^2$ force cannot produce. Only the focal placement makes geometry and dynamics consistent.

Is Kepler's First Law exactly true for real planets?

No — it is exact only for the idealized two-body inverse-square problem. Real orbits feel the pulls of the other planets, the Sun's slight oblateness, and general-relativistic corrections, so the actual trajectory is an ellipse whose elements drift slowly: astronomers describe it as an osculating ellipse, the instantaneous Kepler orbit that best matches the motion at each moment. The drifts are tiny — Mercury's perihelion direction, the most famous case, advances by only 574 arcseconds per century, of which 43 are the relativistic share — and their very smallness is what makes them precious: deviations from a static First-Law ellipse are the standard probe for unseen masses and post-Newtonian physics alike.

What exactly does the semi-latus rectum tell me that a and b do not?

It is the one length angular momentum fixes by itself: $\ell=L^2/(\mu k)$ , with no reference to energy. That gives it two clean roles. Geometrically, $\ell$ is the orbit's half-width measured at the Sun (the distance at $\theta=90^\circ$ ), whereas $b$ is the half-width at the centre — different places because the Sun is off centre. Dynamically, $\ell$ is what all orbits of the same angular momentum share: a fixed- $L$ family threads the same two points at distance $\ell$ , from the circle through the parabola. When you know how hard something was "thrown sideways" but not how fast it is overall, $\ell$ is the piece of geometry you already know.

Do comets on hyperbolic paths violate Kepler's First Law?

No — they fall outside its "bound orbit" clause while obeying the same underlying equation. The conic formula covers all cases: $e<1$ gives the closed ellipses of the First Law, $e=1$ a parabola, and $e>1$ a hyperbola on which the denominator hits zero at $\cos\theta=-1/e$ — the two asymptotic directions along which the object escapes. Interstellar visitors like 1I/'Oumuamua ( $e\approx1.2$ ) and 2I/Borisov ( $e\approx3.4$ ) are textbook hyperbolic conics with the Sun still at the focus. The classification is set by the orbit's energy, as the Kepler problem article derives: $e>1$ simply means the object arrived with speed to spare.