Visual walkthrough — Kepler's second law — equal areas in equal times, from angular momentum conservation

3.2.6 · D2 · Physics › Orbital Mechanics & Astrodynamics › Kepler's second law — equal areas in equal times, from angul

Yeh parent note ka picture-companion hai. Same result, lekin yahan figures proof carry karte hain.

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Step 1 — Sun se arrow ka matlab kya hota hai

KYA HAI. Sun ko ek dot ki tarah draw karo. Planet ek aur dot hai. Seedha arrow Sun se planet ki taraf ko radius vector kehte hain, likha jaata hai $\vec{r}$. Upar chhoti arrow-hat ka matlab bas itna hai ki "yeh ek aisi cheez hai jisme length AUR direction dono hain," na ki sirf ek number.

KYUN. Is law mein sab kuch us area ke baare mein hai jo yeh ek arrow paint karta hai jab planet move karta hai. Toh area se pehle, speed se pehle, hum bilkul clear hona chahte hain ki hum kaunsa arrow dekh rahe hain. Uski length $r$ Sun–planet ki distance hai; uski direction hai "planet kidhar hai."

PICTURE. Figure mein, burnt-orange arrow $\vec{r}$ hai. Dotted circle orbit nahi hai — woh sirf yeh dikhane ke liye hai ki arrow kisi bhi direction mein point kar sakta hai. Planet arrowhead pe baitha hai.

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Step 2 — Ek tiny instant mein, arrow ek patla triangle sweep karta hai

KYA HAI. Bahut chhota time $dt$ guzarne do ($d$ = "thoda sa"). Planet aage thoda sa khisak jaata hai, toh radius vector ban jaata hai $\vec{r}+d\vec{r}$. Yahan $d\vec{r}$ chhota displacement arrow hai — woh chhoti si step jo planet ne li. Jo region arrow ne "pehle" aur "baad" ke beech mein sweep ki woh ek patla triangle hai.

KYUN. Kepler's law swept area ke baare mein hai. Sabse chhota possible swept region yahi ek patla triangle hai. Agar hum uski area nikaal sakein, toh triangles ko jod ke koi bhi swept area paa sakte hain. Isliye hum bilkul zoom in karte hain.

PICTURE. Do orange arrows Sun se bahar jaate hain: purana $\vec{r}$ aur naya $\vec{r}+d\vec{r}$. Unke tips ke beech ka teal arrow $d\vec{r}$ hai. Shaded plum sliver swept area $dA$ hai. Notice karo yeh sach mein triangle hai: Sun se do seedhe edges, ek chhota edge $d\vec{r}$ use close karta hai.

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Step 3 — Cross product: ek machine jo triangle area output karti hai

KYA HAI. Kisi bhi do arrows $\vec{a}$ aur $\vec{b}$ ke liye jo same point se shuru hote hain, ek operation hota hai jise $\vec{a}\times \vec{b}$ (kaho "a cross b") likhte hain, jiski size us parallelogram ki area ke barabar hoti hai jo unse banta hai. Us parallelogram ka aadha triangle hai. Toh:

$dA=\frac{1}{2}|\underset{\text{one edge}}{\underbrace{\vec{r}}}\times \underset{\text{other edge}}{\underbrace{d\vec{r}}}|$

Vertical bars $||$ ka matlab hai "ki length/size," direction hata ke positive area number rakh do. $\frac{1}{2}$ parallelogram ko hamara triangle banata hai.

KYA HAI — direction bhi. Cross product sirf ek size nahi hai; yeh khud ek arrow hai, aur hume batana padega woh kidhar point karta hai. Rule yeh hai: $\vec{a}\times \vec{b}$ us flat sheet ke perpendicular point karta hai jisme $\vec{a}$ aur $\vec{b}$ hain — seedha page se bahar ya seedha page mein andar. Dono mein se kaun sa? Right-hand rule: apne right hand ki ungliyan pehle arrow $\vec{a}$ ke saath point karo, phir unhe doosre arrow $\vec{b}$ ki taraf curl karo; tumhara thumb tab $\vec{a}\times \vec{b}$ ki direction mein point karega. Counter-clockwise sweep (jaise $\vec{r}$, $d\vec{r}$ ki taraf swing karta hai normal orbit mein) ek thumb deta hai jo page se bahar point karta hai — isliye Step 5 mein angular-momentum arrow orbital plane se upar nikalta hai. Sirf area ke liye hum size $|\vec{a}\times \vec{b}|$ use karte hain; direction tab important ho jaati hai jab hum ise $\vec{L}$ bulate hain.

YEH TOOL KYUN, na ki sirf base×height? Kyunki base×height ke liye hume dhundhni padti hai perpendicular height har instant — jab arrows tilt karte hain to annoying lagta hai. Cross product yeh automatically karta hai: $|\vec{a}\times \vec{b}|=ab\sin \varphi$, jahan $\varphi$ arrows ke beech ka angle hai aur $\sin \varphi$ pehle se hi perpendicular part extract kar leta hai. Ek formula, har angle, har quadrant. Isliye hum ise choose karte hain. (Zyada detail ke liye Cross Product and Area.)

PICTURE. Poora parallelogram faintly outlined hai; plum triangle woh aadha hai jo hume chahiye. Right-angle mark height $b\sin \varphi$ dikhata hai jo cross product secretly compute karta hai. Chhota circled dot mark karta hai $\vec{a}\times \vec{b}$ ko jo right-hand rule se page se bahar aa raha hai.

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Step 4 — Time se divide karo area per second paane ke liye

KYA HAI. Kepler ko area per second ki parwah hai, isliye tiny triangle ko tiny time se divide karo:

$\frac{dA}{dt}=\frac{1}{2}\left|\vec{r}\times \frac{d\vec{r}}{dt}\right|=\frac{1}{2}|\vec{r}\times \vec{v}|$

Woh piece $\frac{d\vec{r}}{dt}$ hai "tiny step divided by tiny time" — yeh exactly velocity $\vec{v}$ hai: planet kitni fast aur kis direction mein move karta hai. Humne $1/dt$ ko cross product ke andar slide kiya kyunki cross product har arrow ko linearly treat karta hai (ek arrow ko scale karo toh answer scale ho jaata hai).

KYUN. Yeh ek static triangle ko rate mein convert karta hai — areal velocity, woh quantity jo Kepler kehta hai constant hai. Ab fixed pieces $\vec{r}$ aur $\vec{v}$ hain, dono real physical arrows jo hum track kar sakte hain.

PICTURE. Teal velocity arrow $\vec{v}$ planet pe orbit ka tangent draw kiya gaya hai. Swept-per-second triangle $\frac{1}{2}|\vec{r}\times \vec{v}|$ hai — same shape jaise pehle, ab clock ke against measure kiya gaya.

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Step 5 — Cross product ko rename karo: angular momentum enter karta hai

KYA HAI. Physicists ke paas pehle se ek naam hai $m$ (planet ki mass) times $\vec{r}\times \vec{v}$ ke liye. Yeh angular momentum hai, ek vector jise $\vec{L}$ likhte hain:

$\vec{L}=\vec{r}\times m\vec{v}⟹\vec{r}\times \vec{v}=\frac{\vec{L}}{m}.$

Step 4 ka hamaara area formula size $|\vec{r}\times \vec{v}|$ use karta hai, toh hum dono sides ki size lete hain. $L\equiv |\vec{L}|$ likhte hain us length ke liye (ek plain positive number — vector $\vec{L}$ ki magnitude). Tab $|\vec{r}\times \vec{v}|=L/m$, aur Step 4 mein substitute karte hue:

$\boxed{\frac{dA}{dt}=\frac{1}{2}\cdot \frac{L}{m}=\frac{L}{2m}}$

Term by term: $L=|\vec{L}|$ = spin-vector ki length (ek scalar), $m$ = planet mass (ek fixed number), toh $L/2m$ hai "half the spin per mass." Yahan sirf lengths hain, isliye right side par sab kuch ek ordinary number hai.

YEH RENAME KYUN? Kyunki $\vec{L}$ ek reason se famous hai jo hum aage use karenge: sahi condition mein yeh change nahi kar sakta. Agar $L=|\vec{L}|$ frozen hai aur $m$ fixed hai, toh $L/2m$ frozen hai — aur yahi Kepler's second law hai. Humne ek geometry question ko ek physics conservation question mein convert kar diya. (Dekho Angular Momentum Conservation.)

PICTURE. $\vec{L}$ ek plum arrow ki tarah draw kiya gaya hai jo orbital plane se bahar point karta hai — woh "out of page" direction exactly woh right-hand-rule output hai Step 3 se, $\vec{r}$ ko $\vec{v}$ ki taraf curling apply karne se. Uski length $L=mrv\sin \varphi$ hai.

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Step 6 — Central force kuch twist nahi karta (torque = 0)

KYA HAI. Poochho ki $\vec{L}$ time mein kaise change karta hai. Hum product $\vec{r}\times m\vec{v}$ ko differentiate karte hain (product rule cross products ke liye bhi kaam karta hai, order maintain karte hue):

$\frac{d\vec{L}}{dt}=\frac{d}{dt}(\vec{r}\times m\vec{v})=\underset{=\vec{v}\times m\vec{v}=0}{\underbrace{\frac{d\vec{r}}{dt}\times m\vec{v}}}+\vec{r}\times m\frac{d\vec{v}}{dt}=\vec{r}\times m\vec{a}=\vec{r}\times \vec{F}$

Symbol by symbol: $\frac{d\vec{r}}{dt}=\vec{v}$ (Step 4 ki definition), toh pehla piece $\vec{v}\times m\vec{v}$ hai; aur $\frac{d\vec{v}}{dt}=\vec{a}$ woh acceleration hai jo abhi define ki, toh doosra piece $\vec{r}\times m\vec{a}$ hai. Pehla term mar jaata hai kyunki $\vec{v}\times \vec{v}=0$ (ek arrow khud se cross kiya jaaye toh zero area span karta hai — angle $\varphi =0$, $\sin 0=0$).

Force laana. Newton's second law kehti hai ki net force mass times acceleration ke barabar hai, $\vec{F}=m\vec{a}$ — toh jahan bhi $m\vec{a}$ dikhta hai hum $\vec{F}$ likh sakte hain. Yahi woh single substitution hai jo $\vec{r}\times m\vec{a}$ ko $\vec{r}\times \vec{F}$ mein turn karti hai, hamaara torque, ek vector jise $\vec{\tau }$ ("tau") likhte hain:

$\vec{\tau }\equiv \vec{r}\times \vec{F}(\text{woh "twisting" jo ek force Sun ke baare mein apply karta hai}).$

Toh upar ka box simply kehta hai $\frac{d\vec{L}}{dt}=\vec{\tau }$: angular momentum tabhi change hota hai jab koi cheez ise twist kare.

Ab gravity ki force $\vec{F}=-\frac{GMm}{r^2}\hat{r}$ hai, jo seedha $\vec{r}$ ke saath back Sun ki taraf point karti hai ($\hat{r}$ = $\vec{r}$ ke saath unit arrow; minus = "inward"). Do parallel arrows koi area span nahi karte, toh

$\vec{\tau }=\vec{r}\times \vec{F}=0⟹\frac{d\vec{L}}{dt}=0⟹\vec{L}=\text{const}\Rightarrow L=|\vec{L}|=\text{const}.$

KYUN. Yahi poora secret hai. Ek central force (jo center ki taraf aimed ho) planet ko us center ke baare mein kabhi twist nahi kar sakti, toh woh kabhi angular momentum drain nahi kar sakti. Koi leak nahi $\Rightarrow$ $\vec{L}$ frozen $\Rightarrow$ area rate frozen. (Dekho Central Forces and Torque.)

PICTURE. Orange $\vec{r}$ aur red gravity arrow $\vec{F}$ same line par hain (opposite directions). Unke beech ka angle $180^{\circ }$ hai; $\sin 180^{\circ }=0$; toh torque $\vec{\tau }$ — unka cross product — ek zero-length arrow hai, draw kiya gaya ek tiny crossed-out dot ki tarah.

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Step 7 — Swept angle ka naam, phir har case

KYA HAI — pehle, angle define karo. $\vec{r}$ aur $\vec{v}$ ko alag-alag track karne ki bajaye, $\vec{r}$ ki direction ko ek fixed reference line se plain angle ki tarah mark karo. Use $\theta$ (the polar angle) kaho: yeh planet ki angular position hai, jaise Sun ko center maan ke ek clock ki hour-hand reading. Jab planet orbit karta hai, $\theta$ badhta hai.

Ek tiny time mein arrow ek tiny angle $d\theta$ se swing karta hai, aur Step 2 ka patla triangle radius $r$ ke circle ka ek slice ban jaata hai: base $\approx rd\theta$ (arc), height $r$, toh

$dA=\frac{1}{2}\underset{\text{height}}{\underbrace{r}}\cdot \underset{\text{arc base}}{\underbrace{rd\theta }}=\frac{1}{2}{r}^{2}d\theta .$

Base $rd\theta$ kyun hai? Asli swept region slightly curved arc se bounded hai, lekin infinitesimally small angle $d\theta$ pe arc aur straight chord mein farq sirf second order par hai $d\theta$ mein (gap $d{\theta }^2$ ki tarah shrink karta hai, $d\theta$ se kahin zyada fast). Leading order par woh identical hain, toh hum clean arc length $rd\theta$ use kar sakte hain — exact result integrate karne ke baad milta hai. Yeh wahi "itna zoom in karo ki curves straight lagne lagein" move hai jo poori calculus ko kaam karti hai.

Ab $\dot{\theta }\equiv \frac{d\theta }{dt}$ likho — upar ki chhoti dot ka matlab hai "time mein rate of change," toh $\dot{\theta }$ hai angle kitni fast sweep hota hai (angular speed). $dA$ ko $dt$ se divide karte hue:

$\frac{dA}{dt}=\frac{1}{2}{r}^2\dot{\theta }.$

Yeh Step 5 se match kyun karta hai? Velocity ka woh part jo $\vec{r}$ ke perpendicular hai (sirf wahi part jo area sweep karta hai) exactly $v\sin \varphi$ hai, aur geometrically woh perpendicular speed arc rate $r\dot{\theta }$ ke barabar hai. Toh $v\sin \varphi =r\dot{\theta }$, aur $mr$ se multiply karte hue:

$L=mrv\sin \varphi =m{r}^2\dot{\theta }.$

$L$ ke dono forms agree karte hain — ek arrows-ke-beech-angle $\varphi$ se likha, doosra sweep-rate $\dot{\theta }$ se.

KYUN (sab cases). $\frac{1}{2}{r}^2\dot{\theta }=L/2m$ har jagah fixed hai, hume check karna hai ki yeh every orbit position pe survive karta hai:

• Perihelion (sabse paas): $r$ chhota, toh $\dot{\theta }$ bada hona chahiye — ek chhota, mota triangle. Yahan sabse fast.
• Aphelion (sabse door): $r$ bada, toh $\dot{\theta }$ chhota — ek lamba, patla triangle. Yahan sabse slow.
• Apsides par, velocity exactly $\vec{r}$ ke perpendicular hoti hai ($\varphi =90^{\circ }$, $\sin \varphi =1$), toh $L=mvr$ cleanly — sirf yahi jagah $\sin \varphi$ drop kar sakte ho. Isliye $v_p{r}_p=v_a{r}_a$.
• Kahin aur $\varphi \ne 90^{\circ }$, toh ZAROOR rakhna $L=mrv\sin \varphi =m{r}^2\dot{\theta }$. Off-apsis $L=mvr$ use karna classic error hai.
• Degenerate case — radial fall ($L=0$): agar planet seedha Sun ki taraf giraya jaaye, $\vec{v}\parallel \vec{r}$, toh $\vec{r}\times \vec{v}=0$, $\dot{\theta }=0$, aur $dA/dt=0$. Law tab bhi hold karta hai — yeh zero area per second sweep karta hai, consistently, hamesha. Koi contradiction nahi; bas boundary value.

PICTURE. Ek ellipse par teen plum slices draw hain — Sun ke paas moti, side pe medium, door se lambi-aur-patli — sab equal area labeled hain. Reference line se $\vec{r}$ ka angle $\theta$ aur $\vec{v}$ ke saath arrow ka angle $\varphi$ dono mark hain; $\varphi =90^{\circ }$ sirf dono apsides par.

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Ek-picture summary

Upar sab kuch, compressed: ek central force $\Rightarrow$ zero torque $\Rightarrow$ constant $\vec{L}$ $\Rightarrow$ constant $dA/dt=L/2m$ $\Rightarrow$ equal areas in equal times.

Recall Poore walkthrough ki Feynman retelling

Socho Sun se planet tak ek rubber band hai. Har second woh band ek chhota triangle of area paint karta hai. Step 1 mein humne band ko $\vec{r}$ naam diya. Step 2 mein humne dekha ek pal mein woh ek patla triangle paint karta hai. Step 3 mein humne ek magic ruler dhundha — cross product — jo kisi bhi triangle ki area measure karta hai chahe woh kaise bhi tilted ho, aur right-hand rule se apna answer seedha page se bahar point karta hai. Step 4 mein humne poochha "kitni area per second?" aur usne planet ki speed ko andar kheench liya. Step 5 mein humne notice kiya ki "area per second" actually sirf planet ke spin-vector $\vec{L}$ ki length hai disguise mein, uski mass ke twice se divide ki gayi. Phir punchline, Step 6: gravity hamesha planet ko seedha Sun ki taraf kheenchti hai (Newton's $\vec{F}=m\vec{a}$ inward point karta hai), aur ek seedha-andar pull kabhi ise Sun ke baare mein twist nahi kar sakta — torque zero hai — toh spin-amount locked ho jaata hai. Step 7 ne sweep ko ek naam diya, angle $\theta$: moti slices jab close ho aur whirling ho ($\dot{\theta }$ bada), patli slices jab door aur dawdling ho ($\dot{\theta }$ chhota) — lekin hamesha, hamesha utni hi pie. Yahi poora law hai, aur ise kabhi jaanne ki zaroorat nahi padi ki gravity inverse-square thi; ise sirf yeh jaanna tha ki gravity Sun ki taraf point karti hai.

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Connections

• Angular Momentum Conservation — woh conserved quantity jo saara kaam karti hai
• Central Forces and Torque — $\vec{r}\times \vec{F}=0$ kyun
• Cross Product and Area — Step 3 ki triangle-area machine
• Kepler's First Law (Ellipse) — jahan inverse-square actually matter karta hai
• Kepler's Third Law (T² ∝ a³) — $\pi ab=(L/2m)T$ use karta hai
• Vis-viva Equation — kisi bhi radius par speed