Kepler's third law — T² ∝ a³ — derivation: Visual walkthrough

Step 1 — A planet is a dot moving on a circle

WHAT. Picture a small ball (the planet) going round and round a big central ball (the Sun or the Earth). We call the distance between their centres $r$ — think of it as the length of a string tying the planet to the centre. We call one full loop's worth of time the period $T$ .

WHY. Before any physics, we must pin down the cast of characters. Every symbol below is one of these picture-things — a distance, a time, a speed, or a pull. If you can point to it in the drawing, it's allowed.

PICTURE. The red dot is the planet. The black circle is its track. The straight red line of length $r$ is the string from centre to planet.

Figure — Kepler's third law — T² ∝ a³ — derivation

Step 2 — Why the planet doesn't fly off in a straight line

WHAT. A moving thing, left alone, goes in a straight line (that's just inertia — a rolling marble keeps rolling straight). To bend that straight line into a circle, something must keep tugging the planet inward, toward the centre, at every instant.

WHY. We need to know what force is at work before we can write an equation for it. The picture shows the planet "wants" to go straight (the dashed grey arrow) but is continuously yanked back onto the circle. The name for "the inward pull a circle demands" is centripetal (Latin: centre-seeking).

PICTURE. Grey dashed arrow = where the planet would go with no force (straight off on a tangent). Red arrow = the actual inward pull that curves it back onto the black circle.

Step 3 — Naming the two forces that must be equal

WHAT. Two separate facts describe the inward pull, and for a steady circle they must be the same number:

$how hard gravity pulls \frac{GM m}{r ^{2}} = how hard a circle needs pulling \frac{m v ^{2}}{r}$

Term by term:

$G$ — a fixed number of nature (the gravitational constant), the same everywhere.
$M$ — the central mass (the thing being orbited).
$m$ — the planet's mass (the thing doing the orbiting).
$r^{2}$ on the left — gravity gets weaker with the square of distance.
$m v^{2} / r$ on the right — the "circle tax": a heavier ( $m$ ) or faster ( $v$ ) planet on a tighter ( $r$ ) circle needs a bigger inward pull.

WHY. If the left side were bigger, the planet would spiral in; if smaller, it would drift out. A stable orbit is precisely the case where they balance. That balance is the one equation we build everything on. (Gravity's form comes from Newton's Law of Universal Gravitation; the right side from Centripetal Force & Circular Motion.)

PICTURE. Left red arrow = gravity's pull $\frac{GM m}{r ^{2}}$ . Right red arrow = the circle's demand $\frac{m v ^{2}}{r}$ . They're drawn the same length because they are equal.

Step 4 — The planet's mass cancels (the magic)

WHAT. Both sides carry the planet's mass $m$ . Divide it away:

$\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} ⟹ \frac{GM}{r ^{2}} = \frac{v ^{2}}{r} ⟹ v^{2} = \frac{GM}{r}$

The $m$ vanishes from both sides — it never comes back.
What's left, $v^{2} = \frac{GM}{r}$ , says: speed-squared depends only on the central mass $M$ and the distance $r$ .

WHY this matters. Because $m$ cancels, a heavy planet and a light planet at the same distance move at exactly the same speed. This is the same reason a feather and a hammer fall together in a vacuum. It also tells us in advance that the final law will not contain $m$ .

PICTURE. Read $v^{2} = GM / r$ as a curve: as $r$ grows, $v$ shrinks. A planet far out is a slow planet.

Step 5 — Turning speed into time (one lap)

WHAT. In one period $T$ the planet travels once around the circle. The length of that trip is the circumference $2 π r$ . Speed is distance ÷ time, so:

$v = \frac{2 π r}{T}$

$2 π r$ — the full loop's length (this is where the famous $4 π^{2}$ will later come from, once we square it).
$T$ — the time for that one loop.

WHY. We want a law about time ( $T$ ), but Step 4 gave us speed ( $v$ ). This little formula is the bridge: it trades speed for time. It also injects the "longer track" penalty — a bigger $r$ means a longer circumference to cover.

PICTURE. Unroll the circle into a straight strip of length $2 π r$ ; the planet walks that strip once in time $T$ .

Step 6 — Combine, and out pops the law

WHAT. We have two facts about $v$ . Set them equal by squaring the bridge and matching to Step 4:

$v^{2} = \frac{GM}{r} and v^{2} = (\frac{2 π r}{T})^{2} = \frac{4 π ^{2} r ^{2}}{T ^{2}}$

So:

$\frac{4 π ^{2} r ^{2}}{T ^{2}} = \frac{GM}{r}$

Cross-multiply (move $T^{2}$ up, move $r$ across):

$4 π^{2} r^{2} \cdot r = GM T^{2} ⟹ 4 π^{2} r^{3} = GM T^{2}$

$T^{2} = \frac{4 π ^{2}}{GM} r^{3}$

$4 π^{2}$ — the squared circumference factor from Step 5.
$GM$ — the only "who am I orbiting" information left; no $m$ in sight.
$r^{3}$ — the distance, cubed. Time-squared matches space-cubed.

WHY. This is the whole point. Both penalties multiplied together — longer path ( $\propto r$ ) and slower speed ( $\propto r^{- 1/2}$ ) — give a lap time $\propto r \cdot r^{1/2} = r^{3/2}$ . Square that and you get $r^{3}$ . The exponent $3$ is literally $1 + \frac{1}{2}$ , doubled.

PICTURE. A log-log plot: plot $lo g T$ against $lo g r$ and you get a perfectly straight red line of slope $\frac{3}{2}$ — the signature of $T \propto r^{3/2}$ .

Step 7 — Edge case: a squashed circle (the ellipse)

WHAT. Real orbits aren't perfect circles; they're ellipses — squashed circles. An ellipse has a longest half-width called the semi-major axis $a$ , and the planet's actual distance swings between a nearest point (perihelion, $r_{m i n}$ ) and a farthest point (aphelion, $r_{m a x}$ ), with

$a = \frac{r _{m i n} + r _{m a x}}{2} .$

The astonishing fact: the law is identical, you just replace $r$ with $a$ :

$T^{2} = \frac{4 π ^{2}}{GM} a^{3} .$

WHY. Near the Sun the planet speeds up; far away it slows down (this is Kepler's Second Law (equal areas)). Over one full lap these exactly trade off, so the total time depends only on the average size $a$ , not on how squashed the orbit is. The squashiness (eccentricity) cancels out completely.

PICTURE. A circle and an ellipse with the same $a$ : they take the same time $T$ , even though one is round and one is stretched. The red segment marks $a$ in both.

The one-picture summary

Everything above, in one flow: force balance → cancel $m$ → speed → time → law.

Recall Feynman retelling — say it to a friend

Picture a ball whizzing round on a string. Gravity is that string, always pulling straight in. For the ball to stay on the circle, the string's pull must exactly match the "yank" a circle demands — no more, no less. Write both down and set them equal. Now here's the trick: the ball's own weight sits on both sides of that equality, so it just cancels — which is why a bowling ball and a marble at the same distance orbit in lockstep. What's left says the ball moves slower the farther out it is. Then ask: how long for one lap? Distance (the loop, $2 π r$ ) divided by speed. The loop grows with distance, and the speed shrinks with distance — a double slowdown. Multiply them and one lap takes about $r^{3/2}$ worth of time. Square that neat little " $3/2$ " and you get the headline: time squared equals distance cubed. Go 4× farther out and your year isn't 4× longer — it's 8× longer. Last twist: squash the circle into an oval without changing its average width, and the lap time doesn't budge — the fast bit near the star and the slow bit far away perfectly cancel. Only the average size $a$ matters.

Recall Checkpoints (cover the answers)

Which two arrows are set equal in Step 3? ::: gravity's pull $\frac{GM m}{r ^{2}}$ and the circle's demand $\frac{m v ^{2}}{r}$ After cancelling $m$ , what is $v^{2}$ ? ::: $v^{2} = GM / r$ What formula bridges speed and time? ::: $v = 2 π r / T$ Where does the $4 π^{2}$ come from? ::: squaring the circumference relation $v = 2 π r / T$ For an ellipse, what replaces $r$ ? ::: the semi-major axis $a$ ; eccentricity cancels Go 4× farther out — how much longer is the year? ::: $4^{3/2} = 8$ times longer

Connections

Newton's Law of Universal Gravitation — the left arrow in Step 3.
Centripetal Force & Circular Motion — the right arrow in Step 3.
Kepler's Second Law (equal areas) — why eccentricity cancels in Step 7.
Orbital Energy & Vis-viva Equation — generalises $v = GM / r$ to any point on an ellipse.
Geostationary & Geosynchronous Orbits — inverting the law to find $r$ from $T$ .
Reduced Mass & Two-Body Problem — what changes when $m$ is not negligible.