Kepler's third law — T² ∝ a³ — derivation

3.2.7 · Physics › Orbital Mechanics & Astrodynamics

LAW kya claim karta hai

Proportionality constant sirf $M$ par kyun depend karta hai? Kyunki planet ki khud ki mass $m$ cancel ho jaati hai — same distance par heavy aur light planets same time mein orbit karte hain. (Same reason sabhi cheezein same rate se girती hain: gravitational mass = inertial mass.)

KAISE: scratch se derivation (circular orbit)

Hum clean case derive karte hain (circle, $a = r$ ); ellipse result identical hai $r \to a$ ke saath.

Step 1 — Planet ko orbit mein kya rakhta hai? Gravity centripetal force supply karta hai jo straight-line motion ko circle mein bend karta hai.

$gravity andar kheenchti hai \frac{GM m}{r ^{2}} = curve karne ke liye chahiye \frac{m v ^{2}}{r}$

Yeh step kyun? Circular orbit bas "getting closer ke bina girna" hai. Inward pull (gravity) exactly equal hona chahiye us inward force ke jo ek circle demand karta hai ( $m v^{2} / r$ ), na zyada, na kam — warna radius change ho jaata.

Step 2 — $m$ cancel karo, speed solve karo. $\frac{GM}{r ^{2}} = \frac{v ^{2}}{r} \Rightarrow v^{2} = \frac{GM}{r}$

Yeh step kyun? Orbiting mass $m$ dono sides par appear hoti hai, isliye cancel ho jaati hai. Yahi reason hai ki final law mein $m$ nahi hoga. Notice karo $v^{2} \propto 1/ r$ : farther out = slower. (Intuition se Strike #2.)

Step 3 — Speed ko period se relate karo. Planet ek period $T$ mein ek circumference $2 π r$ cover karta hai: $v = \frac{2 π r}{T}$

Yeh step kyun? Period bas distance-per-orbit divided by speed hai. Yeh "longer track" effect inject karta hai (strike #1).

Step 4 — Combine karo aur $T^{2}$ isolate karo. $v = 2 π r / T$ substitute karo $v^{2} = GM / r$ mein: $\frac{4 π ^{2} r ^{2}}{T ^{2}} = \frac{GM}{r}$ Cross-multiply: $4 π^{2} r^{3} = GM T^{2}$ $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$

Yeh step kyun? Dono penalties (longer path $\propto r$ , slower speed $\propto 1/ r$ ) combine hote hain: $T = \frac{2 π r}{v} = \frac{2 π r}{GM / r} \propto r^{3/2}$ . Square karo → $r^{3}$ . 3/2 literally hai (path power 1) − (speed power −1/2) = 3/2.

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

Circle se ellipse tak (honest version)

Real orbits ellipses hoti hain. Pura result, $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ , jahan $a$ semi-major axis hai, area-sweep integrate karne se aata hai (Kepler's 2nd law: equal areas in equal times). Ek period mein radius vector poori ellipse area $π ab$ sweep karta hai constant areal rate $\frac{d A}{d t} = \frac{L}{2 m}$ par: $T = \frac{π ab}{L /2 m} = \frac{2 π m ab}{L} .$ $L^{2} = GM m^{2} a (1 - e^{2})$ aur $b = a 1 - e^{2}$ use karne se sab kuch same boxed formula mein collapse ho jaata hai $r \to a$ ke saath. Eccentricity $e$ vanish ho jaata hai — sirf $a$ bachta hai.

Worked examples

Common mistakes (steel-manned)

Active recall

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

Derive karne se pehle: $a$ double karne se $T$ more-than-double hoga ya less-than-double? → More than ( $\times 2^{1.5} = 2.83$ ).
Step 1 mein kaunse do physical facts equate karne hain? → Gravity $=$ centripetal force.
$m$ kyun vanish hoti hai? → $\frac{GM m}{r ^{2}}$ aur $\frac{m v ^{2}}{r}$ ke beech cancel hoti hai.
Ellipse ke liye, $r \to ?$ → semi-major axis $a$ ; eccentricity drop out ho jaati hai.

Kepler's Third Law ek equation mein?

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

Law mein

a

kya represent karta hai?

semi-major axis (circle ke liye orbit radius)

Derive karne ke liye kaunsi do forces equal set ki jaati hain (circular case)?

gravitational force = centripetal force,

GM m / r^{2} = m v^{2} / r

Orbiting mass

m

final law mein kyun appear nahi karta?

force balance ke dono sides par cancel ho jaata hai

Circular orbit ki orbital speed?

v = GM / r

(farther out slower)

Agar

a

4 se multiply ho, toh

T

kis factor se change hoga?

4^{3/2} = 8

Kya orbital period eccentricity par depend karta hai?

Nahi — sirf semi-major axis

a

par

AU aur years mein Sun ke around, law simplify ho jaata hai?

T^{2} = a^{3}

T

a

se faster kyun badhta hai?

longer path (

\propto a

) AUR slower speed (

\propto a^{- 1/2}

) combine karke

a^{3/2}

banate hain

4 π^{2}

kahan se aata hai?

v = 2 π r / T

square karne se (circumference relation)

Recall Feynman: ek 12-saal ke bachche ko explain karo

Ek campfire ke around circular tracks par runners imagine karo. Fire se door ke runners ke paas bada loop run karne ka hai, toh akele isse zyada time lagega. Lekin ek twist hai: door ke runners aalsi bhi hain — woh zyada slow jog karte hain (kyunki fire ki "pull" wahan bahut kamzor hai). Slow runner + long track = bahut hi lamba lap. Agar tum 4 guna door jaao, tumhara lap 4 guna longer nahi hoga — yeh 8 guna longer hoga. Yahi magic "8" hai law: time square karo, aur yeh distance ke cube se match karta hai.

Connections

Newton's Law of Universal Gravitation — $1/ r^{2}$ force supply karta hai jisse yeh poori derivation shuru hoti hai.
Centripetal Force & Circular Motion — $F = m v^{2} / r$ provide karta hai, Step 1 ka right-hand side.
Kepler's Second Law (equal areas) — result ko ellipses tak extend karne ke liye zaroori hai.
Orbital Energy & Vis-viva Equation — $v^{2} = GM (2/ r - 1/ a)$ Step 2 ko generalize karta hai.
Geostationary & Geosynchronous Orbits — direct application (Example 3).
Reduced Mass & Two-Body Problem — comparable masses ke liye $M$ ko $M_{1} + M_{2}$ se replace karta hai.