Kepler's Third Law: Period-Semimajor Relation

On 8 March 1618 — by his own excited account — Kepler finally saw the pattern he had hunted for a decade: the square of a planet's period is proportional to the cube of its mean distance. Mercury to Saturn, six planets spanning a factor of a hundred and twenty in period, all sitting on one line. He called it the harmonic law and buried it in a book about celestial music; today we call it Kepler's Third Law, and it is the single most-used equation in astronomy — because it is a weighing machine. Measure how long something takes to orbit and how big its orbit is, and $T^2=4\pi^2a^3/GM$ hands you the mass of whatever sits at the focus. The Sun, Jupiter, the Milky Way's central black hole, and a few thousand exoplanet host stars have all been put on this scale. Within the two-body machinery that the Kepler problem article develops, the Third Law stops being an empirical fit and becomes a two-line theorem: the Second Law says orbits sweep area at the constant rate $L/2\mu$ , the First Law says the area to be swept is $\pi ab$ , and dividing one by the other — with the ellipse geometry to eliminate $b$ and $L$ — leaves $T^2\propto a^3$ with every trace of eccentricity gone. This article runs that derivation in full, checks it against a circular orbit, and then confronts its fine print: the constant contains the total mass $M+m$ , a correction invisible for planets but essential for binary stars. The surprise worth savoring: the orbital clock reads only the orbit's size. Shape, orientation, the masses' ratio — the period ignores them all.

Key takeaways

Kepler's Third Law: $T^2\propto a^3$ — the square of the period scales as the cube of the semi-major axis, with one constant for everything orbiting the same body.
The exact two-body form is $T^2=\dfrac{4\pi^2}{G(M+m)}a^3$ : the constant contains the total mass of the pair.
It follows from dividing the ellipse's area $\pi ab$ by the Second Law's constant sweep rate $L/2\mu$ — and in that division the angular momentum, hence the eccentricity, cancels exactly.
Orbits of the same $a$ have the same period, however eccentric: a circle and a near-radial cigar keep the same time.
Read backwards, the law weighs the central body: $M=4\pi^2a^3/(GT^2)$ — how we know the masses of the Sun, planets, black holes, and exoplanet hosts.
Key idea: Equating the area of the ellipse $\pi a b$ to the constant areal sweep $(L/2\mu)\,T$ and eliminating $L$ and $b$ using the orbit's energy and angular momentum yields $T^2 = 4\pi^2 a^3 / [G(M+m)]$ — a relation in which eccentricity cancels and only total mass and $a$ survive.

Where this lives today

The exoplanet census — nearing 6,000 confirmed worlds in 2026 — is essentially a Third-Law production line. A transit survey measures one thing well: the period. The star's mass comes from its spectrum, and equation (1) then converts $T$ into $a$ — which is how we know whether a planet sits in its star's habitable zone despite never resolving the orbit. Every "Earth-like planet in the temperate zone" headline is equation (2) applied to a photometric time series; the TRAPPIST-1 system of the worked example, with seven transiting planets around one tiny star, remains the benchmark target for JWST atmosphere work precisely because its Third-Law geometry packs temperate orbits into telescope-friendly two-week periods. At the other end of the mass scale, the law weighs monsters. The star S2 completes its $e=0.88$ ellipse around Sagittarius A* every 16 years, and equation (3) prices the unseen focus at about four million solar masses — the measurement (2020 Nobel Prize) that made our galaxy's central black hole a settled fact. Gaia's DR4 release, due in late 2026, will apply the same arithmetic to tens of thousands of astrometric binaries at once.

The physical picture

Kepler's discovery is best seen the way modern astronomers plot it — logarithmically. Take every planet's period and semi-major axis, plot $\log T$ against $\log a$ , and the Solar System collapses onto a single straight line of slope exactly $\tfrac32$ :

Figure

Eight planets, one straight line — the log-log plot of period against semi-major axis has slope exactly three-halves

Why should there be any clean rule linking different orbits? Dimensional analysis makes the law feel inevitable: out of the only available ingredients — the coupling $GM$ (units $\mathrm{m^3/s^2}$ ) and a length $a$ — exactly one timescale can be built, $\sqrt{a^3/GM}$ . The Third Law's real content is that the dimensionless prefactor is the same $2\pi$ for every orbit, round or ragged. A bigger orbit is slower twice over: there is more path to cover, and the planet moves more slowly out there; the two effects compound into $T\propto a^{3/2}$ rather than the naive $T\propto a$ . The part that surprises even good students is what the clock ignores. Take a circular orbit and a violently eccentric one with the same semi-major axis — same $a$ , wildly different shapes, different speeds everywhere along them:

Figure

Same semi-major axis, same period — the eccentric orbit dives fast and crawls far, and the clock cannot tell the difference

They keep identical time. The eccentric orbit encloses less area ( $\pi ab$ , and $b$ shrinks with eccentricity) — but it also carries proportionally less angular momentum, so its sweep rate $L/2\mu$ is slower in exactly the compensating ratio. That conspiracy — area and sweep rate both scaling with the same factor of $b$ — is the mechanism behind the eccentricity's disappearance, and the derivation below shows the cancellation happening line by line.

The law and its fine print

The headline result first, in its exact two-body form. Both bodies orbit their common centre of mass, and the ellipse whose semi-major axis appears here is that of the relative orbit — the separation vector's ellipse, the one the Kepler problem article reduces everything to:

T^2=\frac{4\pi^2}{G(M+m)}\,a^3 \qquad (1)

where: $T$ — the orbital period; $a$ — the semi-major axis of the relative orbit; $G$ — Newton's constant; $M,m$ — the two masses. Note what is absent: the eccentricity, the orientation, and the mass ratio. Two facts hide in the constant: it is set by the mass of the pair (the fine print, derived below), and it is the same for every body orbiting the same primary — which is why the law shows up as a single line in the figure above. For a planet around a star or a satellite around a planet, $m\ll M$ and the total mass collapses to the primary's:

\frac{T^2}{a^3}=\frac{4\pi^2}{GM}=\text{const},\qquad n\equiv\frac{2\pi}{T}=\sqrt{\frac{GM}{a^3}} \qquad (2)

where: $M$ — the dominant central mass; $n$ — the mean motion, the average angular rate $2\pi/T$ that orbit designers quote instead of the period. This is Kepler's original statement: one constant per central body. Around the Sun, with $T$ in years and $a$ in AU, the constant is exactly $1$ — the friendliest formula in astronomy, $T^2=a^3$ . Read equation (2) backwards and it becomes the weighing machine. Solve for the central mass:

M=\frac{4\pi^2a^3}{G\,T^2} \qquad (3)

where: everything on the right is observable — a size and a time. This one line, fed with Earth's $a$ and $T$ , weighs the Sun at $1.99\times10^{30}\ \mathrm{kg}$ ; fed with a moon of Jupiter, it weighs Jupiter; fed with the star S2's 16-year orbit, it weighs the Milky Way's central black hole. When the two masses are comparable, equation (1) delivers the total $M+m$ instead — the starting point of binary-star astronomy.

Derivation AThe third law from the swept area

The goal is equation (1) in full, $T^2=4\pi^2a^3/[G(M+m)]$ , derived — as the scope of this article demands — from the area sweep, with every cancellation shown. The strategy is the "lazy" route that needs no integration of the motion: the Second Law says the radius vector sweeps area at the constant rate $L/2\mu$ , so one period is simply the total area of the ellipse divided by that rate. The dynamics enters through a single hook, the semi-latus rectum $\ell=L^2/(\mu k)$ from the First Law's geometry, which converts the semi-minor axis into dynamical variables. Watch the angular momentum: it enters through both the area and the sweep rate, and its exact cancellation is the entire content of "eccentricity does not matter".

Step 1. Period = area over sweep rate. The Second Law article derives $\dfrac{\mathrm{d}A}{\mathrm{d}t}=\dfrac{L}{2\mu}$ , constant. Over one period the swept area is the whole ellipse, $\pi ab$ , so $T=\dfrac{\pi ab}{L/2\mu}=\dfrac{2\pi\mu\,ab}{L}$ . Why: dividing a total by a constant rate is legitimate only because the rate is conserved — the one physical input so far.

Step 2. Express the semi-minor axis through the semi-latus rectum. The ellipse geometry of the First Law article gives $b=a\sqrt{1-e^2}$ and $\ell=a(1-e^2)$ , hence $b=\sqrt{a\ell}$ . Why: $b$ is a purely geometric quantity; writing it as $\sqrt{a\ell}$ funnels all shape information into $\ell$ , which is the one length dynamics knows how to set.

Step 3. Insert the dynamical hook. With $\ell=\dfrac{L^2}{\mu k}$ : $b=L\sqrt{\dfrac{a}{\mu k}}$ , so $T=\dfrac{2\pi\mu a}{L}\cdot L\sqrt{\dfrac{a}{\mu k}}$ . Why: this is where the force law enters — $\ell=L^2/\mu k$ is special to the inverse-square attraction, quoted from the Kepler problem article's orbit solution.

Step 4. Watch the cancellation. The angular momentum divides out identically: $T=2\pi\sqrt{\dfrac{\mu}{k}}\;a^{3/2}$ . Why: $L$ carried all the eccentricity dependence (at fixed $a$ , rounder orbits have larger $L$ ), so its cancellation is precisely the statement that the period ignores the shape. This is the line to remember.

Step 5. Square and restore the masses. $T^2=\dfrac{4\pi^2\mu}{k}a^3$ , and with $k=GMm$ , $\mu=\dfrac{Mm}{M+m}$ : $\dfrac{\mu}{k}=\dfrac{1}{G(M+m)}$ , giving $T^2=\dfrac{4\pi^2}{G(M+m)}a^3$ — equation (1). Why: the individual masses cancel from the ratio $\mu/k$ , leaving only their sum — the fine print that the next derivation explains physically. *For $m\ll M_\odot$ , Earth's $a=1.496\times10^{11}$ m gives $T=2\pi\sqrt{a^3/GM_\odot}=3.156\times10^7$ s $=365.3$ d ✓*

Go deeper · Derivation B · 1 minThe circular orbit as an independent check

Derivation BThe circular orbit as an independent check

Any formula that claims to hold for all eccentricities had better be easy to verify at $e=0$ , where no ellipse geometry is needed at all — and this check is worth doing because it uses completely different physics: force balance instead of area bookkeeping. The goal is to recover $T^2=4\pi^2r^3/GM$ for a circular orbit of radius $r$ from nothing but "gravity supplies the centripetal force". If the general derivation is right, setting $a=r$ (a circle's semi-major axis is its radius) and $m\ll M$ must reproduce it. The subtlety worth noticing: this route also delivers the circular speed $v=\sqrt{GM/r}$ , the anchor for the $v\propto a^{-1/2}$ scaling that makes outer orbits slow.

Step 1. Balance gravity against the centripetal requirement. Uniform circular motion at speed $v$ and radius $r$ requires the net inward force $mv^2/r$ ; gravity supplies $GMm/r^2$ . Equate: $\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$ . Why: for a circle the pull must exactly match the turning requirement at every instant — no radial motion ever develops.

Step 2. Solve for the speed. $v^2=\dfrac{GM}{r}$ — the orbiting mass $m$ cancels. Why: heavier satellites need proportionally more force and get proportionally more; all bodies orbit alike, gravity's equivalence-principle signature.

Step 3. Convert speed to period. $T=\dfrac{2\pi r}{v}=2\pi r\sqrt{\dfrac{r}{GM}}$ , so $T^2=\dfrac{4\pi^2 r^3}{GM}$ . Why: circumference over speed — the only step where the circle's geometry is used.

Step 4. Compare with the general law. This matches equation (1) with $a=r$ and $M+m\to M$ — an independent confirmation through force balance rather than areas. Why: two disjoint derivations agreeing on the same exponent and prefactor is strong protection against algebra slips. *For the ISS at $a=6{,}771$ km: $T=2\pi\sqrt{(6.771\times10^6)^3/3.986\times10^{14}}=5.54\times10^3$ s $=92$ min ✓.*

Go deeper · Derivation C · 2 minWhy the total mass sets the clock

Derivation CWhy the total mass sets the clock

The goal is the law's fine print: the gravitational parameter in equation (1) is $G(M+m)$ — the total mass — and this derivation shows where the companion's mass sneaks in. The strategy is to go back one step before the reduced-mass formalism: write Newton's second law for each body separately and subtract, obtaining the equation of motion for the separation vector directly. The physical picture to hold: both bodies accelerate toward each other, so the separation closes faster than either body's individual motion would suggest — the relative orbit runs on the sum of the two pulls. Watch for what this means observationally: using the naive $GM$ form for a system where $m$ is not negligible makes the predicted period too long.

Step 1. Write both equations of motion. With $\mathbf r=\mathbf r_1-\mathbf r_2$ the separation (body 1 of mass $m$ , body 2 of mass $M$ ): $m\ddot{\mathbf r}_1=-\dfrac{GMm}{r^2}\hat{\mathbf r}$ and $M\ddot{\mathbf r}_2=+\dfrac{GMm}{r^2}\hat{\mathbf r}$ . Why: Newton's third law fixes the equal-and-opposite pair; each body feels the same magnitude of force.

Step 2. Divide and subtract. $\ddot{\mathbf r}_1=-\dfrac{GM}{r^2}\hat{\mathbf r}$ , $\ddot{\mathbf r}_2=+\dfrac{Gm}{r^2}\hat{\mathbf r}$ , so the separation obeys $\ddot{\mathbf r}=\ddot{\mathbf r}_1-\ddot{\mathbf r}_2=-\dfrac{G(M+m)}{r^2}\hat{\mathbf r}$ . Why: each body's acceleration is set by the OTHER body's mass; subtracting adds the two accelerations, and the masses combine into the total.

Step 3. Read off the effective one-body problem. The separation vector moves exactly like a test particle around a fixed mass $M+m$ : every Kepler-problem result applies with the substitution $GM\to G(M+m)$ — in particular $T^2=4\pi^2a^3/[G(M+m)]$ . Why: the equation of Step 2 is formally identical to the test-mass equation; only the constant changed. This is the same conclusion the reduced-mass route reaches via $\mu/k=1/[G(M+m)]$ , by less algebra.

Step 4. Size the correction. The period at fixed $a$ scales as $(M+m)^{-1/2}$ , so the fractional shortening relative to the naive formula is $\approx m/2M$ for small $m/M$ . Why: this converts the fine print into a number you can budget for — negligible at $10^{-6}$ for Jupiter–Sun, but a measurable $0.6\%$ for the Moon. *For the Moon ( $a=3.844\times10^8$ m): the naive formula gives $T=2\pi\sqrt{a^3/GM_\oplus}=27.45$ d; including the total-mass correction with $m/M_\oplus=0.0123$ gives $T=27.45/\sqrt{1.0123}=27.29$ d — against the observed sidereal month $27.32$ d, the total-mass correction closes most of the gap (the residue is solar perturbation).*

Limiting cases

Case	Result	Meaning
Dominant primary, $m\ll M$	$T^2=\dfrac{4\pi^2}{GM}a^3$	One constant per central body — Kepler's original law; exact to $m/2M$ .
Sun, year–AU units	$T^2=a^3$	The constant is 1 by construction: Earth calibrates it.
Circular orbit	$T=2\pi\sqrt{r^3/GM}$ , $v=\sqrt{GM/r}$	Force balance reproduces the law at $e=0$ ; speeds fall as $a^{-1/2}$ .
Comparable masses	$T^2=\dfrac{4\pi^2}{G(M+m)}a^3$	Binary stars: the law weighs the pair; $a$ is the relative-orbit axis.
General power law $F\propto r^{\,n}$	$T\propto a^{(1-n)/2}$	Only $n=-2$ gives $3/2$ ; Hooke's $n=1$ gives $T\propto a^0$ — period independent of size.

Common trapT squared over a cubed is the same constant for everything

It is one constant per central body, not one constant for the universe. $T^2/a^3=4\pi^2/GM$ is identical for all the Sun's planets because they share the same $M$ — but Earth's satellites obey a different constant (set by $M_\oplus$ ), Jupiter's moons yet another. Comparing a lunar orbit to a heliocentric one with the year–AU rule is a classic exam disaster. Two more pieces of fine print: for comparable masses the constant contains the total $M+m$ , so a massive companion measurably shortens the period (the Moon by $0.6\%$ , a binary star by a lot); and $a$ always means the semi-major axis of the relative orbit — not the radius of either body's path about the barycentre, and not the orbit's radius "on average over time" (the time-averaged separation is actually $a(1+e^2/2)$ ).

Worked examples

1 · The geostationary altitude (confidence builder)

At what altitude does a satellite hover over a fixed point on Earth's equator? Use $GM_\oplus=3.986\times10^{14}\ \mathrm{m^3/s^2}$ and the sidereal day $T=86{,}164$ s.

Step 1. "Hovers over a fixed point" means the orbital period equals Earth's rotation period — the sidereal day of $86{,}164$ s, not the $86{,}400$ -s solar day (Earth must not be allowed the extra $\sim4$ minutes it gains from its motion around the Sun).

Step 2. Invert equation (2) for the semi-major axis: $a=\left(\dfrac{GM_\oplus T^2}{4\pi^2}\right)^{1/3}=\left(\dfrac{3.986\times10^{14}\times(86164)^2}{4\pi^2}\right)^{1/3}$ .

Step 3. Evaluate: the argument is $7.496\times10^{22}\ \mathrm{m^3}$ , and its cube root gives $a=\mathbf{4.216\times10^{7}\ \mathrm{m}}=42{,}164$ km from Earth's centre.

Step 4. Subtract Earth's equatorial radius $6{,}378$ km: altitude $=42{,}164-6{,}378=\mathbf{35{,}786\ \mathrm{km}}$ — the crowded ring where every geostationary communications and weather satellite lives, all of them pinned to one number by the Third Law.

2 · Weighing a star with one planet (exam level)

The planet TRAPPIST-1b transits its star every $T=1.5109$ days on an orbit of semi-major axis $a=0.01154\ \mathrm{AU}$ . Weigh the star. ( $G=6.674\times10^{-11}$ SI, $1\ \mathrm{AU}=1.496\times10^{11}$ m, $M_\odot=1.989\times10^{30}$ kg.)

Step 1. Convert to SI: $a=0.01154\times1.496\times10^{11}=1.726\times10^{9}\ \mathrm{m}$ ; $T=1.5109\times86400=1.305\times10^{5}\ \mathrm{s}$ .

Step 2. The planet's mass is utterly negligible ( $\sim10^{-5}$ of the star's), so the weighing formula (3) applies: $M=\dfrac{4\pi^2a^3}{GT^2}$ .

Step 3. Numerator: $4\pi^2a^3=4\pi^2\times(1.726\times10^9)^3=2.03\times10^{29}\ \mathrm{m^3}$ . Denominator: $GT^2=6.674\times10^{-11}\times(1.305\times10^5)^2=1.137\times10^{0}\ \mathrm{m^3/kg}$ .

Step 4. $M=\dfrac{2.03\times10^{29}}{1.137}=\mathbf{1.79\times10^{29}\ \mathrm{kg}}=0.090\ M_\odot$ — a star one-eleventh the Sun's mass, barely above the hydrogen-burning limit. Cross-check in astronomer's units: $M/M_\odot=a_\mathrm{AU}^3/T_\mathrm{yr}^2=(0.01154)^3/(0.004137)^2=0.0898$ ✓.

Step 5. Reality check the inputs: a $1.5$ -day period at $0.0115$ AU means an orbital speed $v\approx2\pi a/T=83$ km/s — nearly three times Earth's — around a star of $9\%$ solar mass. Tight, fast, and entirely consistent: small $M$ and small $a$ conspire to keep $T$ short.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · Cambridge

Newton is the reason the Third Law has a right-hand side. Kepler's $T^2\propto a^3$ was a pattern; Newton's 1666 "moon test" turned it into physics: if gravity falls off as the inverse square, the Moon — sixty Earth radii away — should fall toward Earth $3600$ times more feebly than an apple, and it does. In the Principia (1687) he closed the loop in both directions, deriving $T^2\propto a^3$ from the inverse-square law and, conversely, showing that Kepler's harmonic law forces the exponent $-2$ . He also read the fine print no one had suspected: the constant contains the total mass, so the law is a balance scale — he used it to compare the Sun's mass with Jupiter's and Saturn's from their moons' orbits, the first weighing of worlds. Characteristically, he did it while inventing the calculus the problem required and telling almost no one for twenty years.

History

1618

— Kepler, after a decade of failed harmonies, finds that the squares of the periods match the cubes of the mean distances for all six planets.

1619

— Harmonices Mundi publishes the harmonic law, wedged between speculations on the music of the spheres.

1666

— Newton's moon test: the Moon's centripetal acceleration is the apple's divided by 60² — the inverse square, checked against the Third Law.

1687

— The Principia derives

T^2\propto a^3

with the total-mass fine print, and weighs Jupiter against the Sun using its moons.

1995

— 51 Pegasi b: a 4.23-day Doppler wobble becomes, through the Third Law, a

0.052

-AU orbit — the first exoplanet found around a Sun-like star.

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 · Taylor, Classical Mechanics, Ch. 8 The intermediate results worth noting: the period formula takes the form $T = 2\pi\sqrt{\mu/k}\,a^{3/2}$ ; the reduced-mass ratio satisfies $\mu/k = 1/[G(M+m)]$ ; Earth's orbit yields the solar mass $\approx 1.989\times10^{30}$ kg; and the geostationary radius works out to $\approx 42{,}164$ km from Earth's center. Constants used throughout: $G = 6.674\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ , $1\,\mathrm{AU} = 1.496\times10^{11}$ m, sidereal day $= 86{,}164$ s.

Builds on: Kepler's First Law: Elliptical Orbits · Kepler's Second Law: Equal Areas · Angular Momentum Conservation in Central Fields · The Two-Body Problem and Reduced Mass · Conic-Section Orbits Leads to: Orbital Elements and Parameterization · The Kepler Problem · Classification of Orbits by Energy

Why does the eccentricity vanish from the period — is it a coincidence?

It is an exact cancellation with a clean mechanism, not a numerical accident. The period is total area over sweep rate: $T=\pi ab\,/\,(L/2\mu)$ . At fixed $a$ , making the orbit more eccentric shrinks the enclosed area (through $b=a\sqrt{1-e^2}$ ) — but it also shrinks the angular momentum in exactly the same proportion, because $L=\sqrt{\mu k\,a(1-e^2)}=\sqrt{\mu k\,\ell}$ carries the same $\sqrt{1-e^2}$ . Numerator and denominator lose identical factors, and the quotient forgets the shape. Derivation A shows the moment it happens (Step 4). The deeper reason: the period, like the energy $E=-k/2a$ , depends only on $a$ — the Kepler problem article traces both degeneracies to the same hidden symmetry of the inverse-square force.

Why does the total mass appear instead of just the big central mass?

Because both bodies fall toward each other. The relative separation — whose ellipse the law describes — closes under the sum of the two accelerations: body 1 accelerates at $GM/r^2$ toward body 2, body 2 at $Gm/r^2$ toward body 1, and subtracting the vector equations (Derivation C) gives $\ddot{\mathbf r}=-G(M+m)\hat{\mathbf r}/r^2$ . A more massive companion makes the separation close faster, shortening the period at fixed $a$ . For planets the correction is invisible ( $m/2M\sim10^{-6}$ even for Jupiter), for the Moon it is a measurable $0.6\%$ , and for a binary star it is the whole point: the observed $T$ and $a$ weigh the pair, and only additional information (like the ratio of the two stars' barycentric orbits) splits the total into individual masses.

Is the constant in T² = a³ universal — can I use year-and-AU units for anything?

Only for objects orbiting the Sun. The tidy $T^2=a^3$ works because years and AU are defined by Earth's orbit, so the constant $4\pi^2/GM_\odot$ becomes exactly 1 in those units. Around any other primary the constant is different: for Earth satellites, $4\pi^2/GM_\oplus$ — which is why the geostationary example had to work in SI. The transferable statement is "one constant per central body": all of Jupiter's moons share a $T^2/a^3$ , all of Earth's satellites share a different one, and the ratio of such constants gives the ratio of central masses — precisely how Newton weighed Jupiter against the Sun in the Principia.

Does the third law tell me the mass of the orbiting planet itself?

Essentially no — and this surprises everyone at first. In the test-mass limit the orbiting body's mass cancels (Derivation B, Step 2: heavier bodies feel proportionally more force), so Mars's orbit weighs the Sun, not Mars. The companion's mass enters only through the $M+m$ fine print, at the level $m/2M$ — usually far below measurement error. To weigh a planet you need something orbiting it: Jupiter's mass comes from Io through Callisto, Earth's from the Moon and from artificial satellites. Exoplanets are weighed differently again — by the reflex wobble they induce in their star (radial velocity), which measures $m$ directly rather than through the period.

Where does the 3/2 exponent actually come from — is there a way to see it without the full derivation?

Dimensional analysis gets you there in three lines. The only quantities available are $GM$ (units $\mathrm{m^3/s^2}$ — the $\mathrm{m^3}$ is the fingerprint of the inverse-square law) and the size $a$ . The unique time you can build is $T\sim\sqrt{a^3/GM}$ : the $3$ in the exponent is inherited directly from the $\mathrm{m^3}$ in $G$ 's units. A more physical split: the path length grows as $a$ , and the orbital speed falls as $a^{-1/2}$ (from $v^2=GM/r$ ), so the time grows as $a/a^{-1/2}=a^{3/2}$ — big orbits are slow twice over. Any other force law changes the speed scaling and hence the exponent: $T\propto a^{(1-n)/2}$ for $F\propto r^n$ , as the limiting-cases table records.

If a is the semi-major axis of the relative orbit, what do the two bodies actually trace?

Each body orbits the common centre of mass on its own ellipse — scaled copies of the relative orbit, with sizes in inverse proportion to the masses: $a_1=\frac{M}{M+m}a$ and $a_2=\frac{m}{M+m}a$ , so $a=a_1+a_2$ . All three ellipses share the same period and eccentricity. For the Earth–Sun pair the Sun's ellipse is a 450-km wobble about a point inside its own body; for a comparable-mass binary both ellipses are large and neither star sits still. The Third Law's $a$ is always the relative one — the sum — which is what a visual binary's angular separation actually measures. Confusing $a$ with one star's barycentric semi-axis underestimates the total mass by the cube of the ratio you dropped.

Do galaxies obey the third law — and what do flat rotation curves mean for it?

The law as written assumes essentially all the mass sits at the focus, well inside the orbit. In a galaxy the mass is spread out, so a star's orbit responds to the mass enclosed within its radius, and the clean $T^2\propto a^3$ becomes $v^2 = GM_{\mathrm{enc}}(r)/r$ for near-circular orbits. That turns the law into a mass profiler: if visible matter were everything, orbital speeds in a galaxy's outskirts should fall off as $v\propto r^{-1/2}$ , the Keplerian decline of Derivation B. Observed rotation curves instead stay flat — $v\approx\text{const}$ — implying $M_{\mathrm{enc}}\propto r$ : enormous halos of unseen mass. The rotation-curve case for dark matter is, at bottom, a report that galaxies refuse to be Keplerian.