Kepler's third law — T² ∝ a³ — derivation

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WHAT the law claims

WHY does the proportionality constant only depend on $M$? Because the planet's own mass $m$ cancels — heavy and light planets at the same distance orbit in the same time. (Same reason all things fall at the same rate: gravitational mass = inertial mass.)

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HOW: derivation from scratch (circular orbit)

We derive the clean case (circle, $a = r$); the ellipse result is identical with $r \to a$.

Step 1 — What keeps the planet in orbit? Gravity supplies the centripetal force that bends the straight-line motion into a circle.

$\underbrace{\frac{GMm}{r^2}}_{\text{gravity pulls in}} = \underbrace{\frac{mv^2}{r}}_{\text{needed to curve}}$

Why this step? A circular orbit is just "falling without getting closer." The inward pull (gravity) must exactly equal the inward force a circle demands ($mv^2/r$), no more, no less — otherwise the radius would change.

Step 2 — Cancel $m$, solve for speed. $\frac{GM}{r^2} = \frac{v^2}{r} \;\Rightarrow\; v^2 = \frac{GM}{r}$

Why this step? The orbiting mass $m$ appears on both sides, so it cancels. This is why the constant in the final law won't contain $m$. Notice $v^2 \propto 1/r$: farther out = slower. (Strike #2 from the intuition.)

Step 3 — Relate speed to period. The planet covers one circumference $2\pi r$ in one period $T$: $v = \frac{2\pi r}{T}$

Why this step? Period is just distance-per-orbit divided by speed. This injects the "longer track" effect (strike #1).

Step 4 — Combine and isolate $T^2$. Substitute $v = 2\pi r/T$ into $v^2 = GM/r$: $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}$ Cross-multiply: $4\pi^2 r^3 = GM\,T^2$ $\boxed{T^2 = \frac{4\pi^2}{GM}\,r^3}$

Why this step? Both penalties (longer path $\propto r$, slower speed $\propto 1/\sqrt r$) combine: $T = \dfrac{2\pi r}{v} = \dfrac{2\pi r}{\sqrt{GM/r}} \propto r^{3/2}$. Square it → $r^3$. The 3/2 is literally (path power 1) − (speed power −1/2) = 3/2.

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From circle to ellipse (the honest version)

Real orbits are ellipses. The full result, $T^2 = \dfrac{4\pi^2}{GM}a^3$, with $a$ the semi-major axis, comes from integrating the area-sweep (Kepler's 2nd law: equal areas in equal times). In one period the radius vector sweeps the whole ellipse area $\pi a b$ at constant areal rate $\frac{dA}{dt}=\frac{L}{2m}$: $T = \frac{\pi a b}{L/2m} = \frac{2\pi m\, a b}{L}.$ Using $L^2 = GMm^2 a(1-e^2)$ and $b = a\sqrt{1-e^2}$ collapses everything to the same boxed formula with $r\to a$. The eccentricity $e$ vanishes — only $a$ survives.

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Worked examples

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Common mistakes (steel-manned)

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Active recall

Recall Forecast-then-verify (cover the answers!)

1. Before deriving: will doubling $a$ more-than-double or less-than-double $T$? → More than ($\times 2^{1.5}=2.83$).
2. Which two physical facts must you equate in Step 1? → Gravity $=$ centripetal force.
3. Why does $m$ vanish? → Cancels between $\frac{GMm}{r^2}$ and $\frac{mv^2}{r}$.
4. For an ellipse, $r \to ?$ → semi-major axis $a$; eccentricity drops out.

Recall Feynman: explain to a 12-year-old

Imagine runners on circular tracks around a campfire. The runners far from the fire have a bigger loop to run, so that alone takes longer. But there's a twist: the farther runners are also lazier — they jog more slowly (because the fire's "pull" is weaker out there). Slow runner + long track = a really long lap. If you go 4 times farther out, your lap isn't 4 times longer — it's 8 times longer. That magic "8" is the law: square the time, and it matches the cube of the distance.

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Connections

• Newton's Law of Universal Gravitation — supplies the $1/r^2$ force this whole derivation starts from.
• Centripetal Force & Circular Motion — provides $F = mv^2/r$, Step 1's right-hand side.
• Kepler's Second Law (equal areas) — needed to extend the result to ellipses.
• Orbital Energy & Vis-viva Equation — $v^2 = GM(2/r - 1/a)$ generalizes Step 2.
• Geostationary & Geosynchronous Orbits — direct application (Example 3).
• Reduced Mass & Two-Body Problem — replaces $M$ with $M_1+M_2$ for comparable masses.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Kepler ka Third Law bolta hai bas itna: planet ka year ka square, uske Sun se distance (semi-major axis $a$) ke cube ke proportional hota hai — yaani $T^2 \propto a^3$. Iska matlab agar koi planet 4 guna door hai, to uska year 4 guna nahi, balki $4^{1.5} = 8$ guna lamba ho jaata hai. Reason simple hai: door wale planet ko bada raasta bhi tay karna padta hai aur gravity weak hone se woh dheere bhi chalta hai — dono milke time ko bahut zyada bada kar dete hain.

Derivation ka core ek hi line hai: gravity hi centripetal force ka kaam karti hai. To $\frac{GMm}{r^2} = \frac{mv^2}{r}$. Yahan planet ka mass $m$ dono side cancel ho jaata hai — isliye final formula mein planet ka weight aata hi nahi (heavy ya light, same time lagega). Isse $v = \sqrt{GM/r}$ nikalta hai. Phir $v = 2\pi r / T$ daal do (ek lap mein circumference cover karta hai), thoda algebra, aur seedha $T^2 = \frac{4\pi^2}{GM} r^3$ aa jaata hai.

Practical baat: yehi formula se geostationary satellite ka radius nikalta hai (24 ghante ka period chahiye to ~42,200 km radius). Aur ellipse ke case mein $r$ ki jagah $a$ (semi-major axis) use karo — eccentricity matter hi nahi karti, sirf average size matter karta hai. Exam mein ratio form sabse fast hai: $\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}$, constant cancel ho jaata hai.