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

3.2.7 · D2 · Physics › Orbital Mechanics & Astrodynamics › Kepler's third law — T² ∝ a³ — derivation

Step 1 — Ek planet ek dot hai jo circle par move kar raha hai

KYA. Ek choti ball (planet) ko imagine karo jo ek badi central ball (Sun ya Earth) ke around ghoom rahi hai. Dono ke centres ke beech ki distance ko hum $r$ kehte hain — ise ek string ki length ki tarah socho jo planet ko centre se baandh rahi hai. Ek poore loop mein lagni wali time ko hum period $T$ kehte hain.

KYUN. Kisi bhi physics se pehle, hum characters ko identify karna zaroori hai. Neeche har symbol in picture-things mein se ek hai — ek distance, ek time, ek speed, ya ek pull. Agar tum use drawing mein point kar sako, toh woh allowed hai.

PICTURE. Red dot planet hai. Black circle uska track hai. $r$ length wali straight red line centre se planet tak ki string hai.

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

Step 2 — Planet straight line mein kyun nahi uda jaata

KYA. Ek moving cheez, agar akela chhod do, straight line mein jaati hai (yeh sirf inertia hai — ek rolling marble seedha rollta rehta hai). Us straight line ko circle mein modne ke liye, kuch cheez planet ko har pal inward, centre ki taraf, kheenchti rehni chahiye.

KYUN. Hume jaanna hai ki kaunsi force kaam kar rahi hai, pehle hum uska equation likh sakein. Picture dikhati hai ki planet "seedha jaana chahta" hai (dashed grey arrow) lekin continuously circle par vapas kheencha jaata hai. "Jo inward pull ek circle maangta hai" uska naam centripetal hai (Latin: centre-seeking).

PICTURE. Grey dashed arrow = planet bina force ke kahan jaata (tangent par seedha). Red arrow = actual inward pull jo use black circle par wapas curve karta hai.

Step 3 — Do forces ko naam dena jo equal honi chahiye

KYA. Inward pull ke baare mein do alag facts hain, aur ek steady circle ke liye woh same number hone chahiye:

$gravity kitni zor se kheenchti hai \frac{GM m}{r ^{2}} = circle ko kitni pulling chahiye \frac{m v ^{2}}{r}$

Term by term:

$G$ — nature ki ek fixed number (gravitational constant), har jagah same.
$M$ — central mass (jise orbit kiya ja raha hai).
$m$ — planet ki mass (jo orbit kar raha hai).
Left par $r^{2}$ — gravity distance ke square ke saath kamzor hoti jaati hai.
Right par $m v^{2} / r$ — "circle tax": ek bhaari ( $m$ ) ya tez ( $v$ ) planet tighter ( $r$ ) circle par zyada inward pull maangta hai.

KYUN. Agar left side badi hoti, planet andar spiral karta; agar choti hoti, woh bahar drift karta. Stable orbit precisely woh case hai jahan dono balance karte hain. Yahi balance woh ek equation hai jis par hum sab kuch build karte hain. (Gravity ka form Newton's Law of Universal Gravitation se aata hai; right side Centripetal Force & Circular Motion se.)

PICTURE. Left red arrow = gravity ka pull $\frac{GM m}{r ^{2}}$ . Right red arrow = circle ki demand $\frac{m v ^{2}}{r}$ . Dono same length ke draw hain kyunki woh equal hain.

Step 4 — Planet ki mass cancel ho jaati hai (magic)

KYA. Dono sides par planet ki mass $m$ hai. Use divide kar do:

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

$m$ dono sides se gayab ho jaata hai — woh kabhi wapas nahi aata.
Jo bacha, $v^{2} = \frac{GM}{r}$ , kehta hai: speed-squared sirf central mass $M$ aur distance $r$ par depend karta hai.

KYUN yeh matter karta hai. Kyunki $m$ cancel ho jaata hai, ek bhaari planet aur ek halka planet same distance par exactly same speed se move karte hain. Yahi reason hai ki vacuum mein ek feather aur hammer saath girate hain. Yeh hume pehle se bata deta hai ki final law mein $m$ nahi hoga.

PICTURE. $v^{2} = GM / r$ ko ek curve ki tarah padho: jaise $r$ badhta hai, $v$ ghatta hai. Bahut door ka planet ek slow planet hota hai.

Step 5 — Speed ko time mein badalna (ek lap)

KYA. Ek period $T$ mein planet circle ke around ek baar travel karta hai. Us trip ki length circumference $2 π r$ hai. Speed = distance ÷ time, toh:

$v = \frac{2 π r}{T}$

$2 π r$ — poore loop ki length (yahi hai jahan se famous $4 π^{2}$ baad mein aayega, jab hum ise square karenge).
$T$ — us ek loop ke liye time.

KYUN. Hum ek law chahte hain time ( $T$ ) ke baare mein, lekin Step 4 ne speed ( $v$ ) diya. Yeh chhota formula bridge hai: yeh speed ko time se exchange karta hai. Yeh "longer track" penalty bhi inject karta hai — bada $r$ matlab cover karne ke liye lamba circumference.

PICTURE. Circle ko $2 π r$ length ki ek straight strip mein unroll karo; planet woh strip ek baar time $T$ mein chalti hai.

Step 6 — Combine karo, aur law nikalta hai

KYA. Humare paas $v$ ke baare mein do facts hain. Bridge ko square karke Step 4 se match karke unhe equal set karo:

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

Toh:

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

Cross-multiply karo ( $T^{2}$ upar, $r$ across):

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

$T^{2} = \frac{4 π ^{2}}{GM} r^{3}$

$4 π^{2}$ — Step 5 ka squared circumference factor.
$GM$ — akela "main kise orbit kar raha hoon" wali information; koi $m$ nahi.
$r^{3}$ — distance, cubed. Time-squared, space-cubed se match karta hai.

KYUN. Yahi poora point hai. Dono penalties milke — lamba path ( $\propto r$ ) aur slow speed ( $\propto r^{- 1/2}$ ) — ek lap time $\propto r \cdot r^{1/2} = r^{3/2}$ deta hai. Use square karo aur tum $r^{3}$ paate ho. Exponent $3$ literally $1 + \frac{1}{2}$ hai, double kiya gaya.

PICTURE. Ek log-log plot: $lo g T$ ko $lo g r$ ke against plot karo aur tumhe slope $\frac{3}{2}$ ki ek perfectly straight red line milegi — $T \propto r^{3/2}$ ki signature.

Step 7 — Edge case: ek squashed circle (ellipse)

KYA. Real orbits perfect circles nahi hote; woh ellipses hote hain — squashed circles. Ek ellipse ki ek longest half-width hoti hai jise semi-major axis $a$ kehte hain, aur planet ki actual distance ek nearest point (perihelion, $r_{m i n}$ ) aur ek farthest point (aphelion, $r_{m a x}$ ) ke beech swing karti hai, jahan

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

Hairani ki baat: law identical hai, bas $r$ ki jagah $a$ replace karo:

$T^{2} = \frac{4 π ^{2}}{GM} a^{3} .$

KYUN. Sun ke paas planet speed up karta hai; door slow down karta hai (yeh Kepler's Second Law (equal areas) hai). Ek poore lap mein yeh exactly trade off karte hain, toh total time sirf average size $a$ par depend karta hai, orbit kitna squashed hai us par nahi. Squashiness (eccentricity) completely cancel ho jaati hai.

PICTURE. Ek circle aur ek ellipse jinka same $a$ hai: woh same time $T$ lete hain, chahe ek round ho aur ek stretched. Red segment dono mein $a$ mark karta hai.

Ek-picture summary

Upar sab kuch, ek flow mein: force balance → $m$ cancel → speed → time → law.

Recall Feynman retelling — kisi dost ko batao

Ek ball ko string par ghoomte hue imagine karo. Gravity woh string hai, hamesha seedha andar kheench rahi hai. Ball circle par rehne ke liye, string ka pull circle ki "yank" se exactly match karna chahiye — na zyada, na kam. Dono likho aur equal set karo. Ab yahan trick hai: ball ka apna weight us equality ki dono sides par hai, toh woh bas cancel ho jaata hai — isliye ek bowling ball aur ek marble same distance par lockstep mein orbit karte hain. Jo bacha woh kehta hai ball slower move karti hai jitni door ho. Phir pucho: ek lap mein kitna time? Distance (loop, $2 π r$ ) divided by speed. Loop distance ke saath badata hai, aur speed distance ke saath ghatti hai — double slowdown. Unhe multiply karo aur ek lap lagbhag $r^{3/2}$ time leta hai. Us neat " $3/2$ " ko square karo aur headline milti hai: time squared equals distance cubed. 4× door jao aur tumhara year 4× lamba nahi — 8× lamba hoga. Last twist: circle ko oval mein squash karo bina average width change kiye, aur lap time nahi badlega — star ke paas fast wala hissa aur door slow wala hissa perfectly cancel ho jaate hain. Sirf average size $a$ matter karta hai.

Recall Checkpoints (answers cover karo)

Step 3 mein kaunse do arrows equal set kiye jaate hain? ::: gravity ka pull $\frac{GM m}{r ^{2}}$ aur circle ki demand $\frac{m v ^{2}}{r}$ $m$ cancel karne ke baad, $v^{2}$ kya hai? ::: $v^{2} = GM / r$ Speed aur time ko bridge karne wala formula kya hai? ::: $v = 2 π r / T$ $4 π^{2}$ kahan se aata hai? ::: circumference relation $v = 2 π r / T$ ko square karne se Ellipse ke liye $r$ ki jagah kya aata hai? ::: semi-major axis $a$ ; eccentricity cancel ho jaati hai 4× door jao — year kitna lamba hoga? ::: $4^{3/2} = 8$ times lamba

Connections

Newton's Law of Universal Gravitation — Step 3 mein left arrow.
Centripetal Force & Circular Motion — Step 3 mein right arrow.
Kepler's Second Law (equal areas) — kyun eccentricity Step 7 mein cancel hoti hai.
Orbital Energy & Vis-viva Equation — $v = GM / r$ ko ellipse par kisi bhi point tak generalize karta hai.
Geostationary & Geosynchronous Orbits — $T$ se $r$ find karne ke liye law ko invert karna.
Reduced Mass & Two-Body Problem — kya badlta hai jab $m$ negligible nahi hota.