Two-body problem — equations of motion, reduction to one-body: Visual walkthrough

3.2.1 · D2 · Physics › Orbital Mechanics & Astrodynamics › Two-body problem — equations of motion, reduction to one-bod

Hum is ek boxed result tak pahunchenge: $\ddot{r} = - \frac{μ}{r ^{2}} \hat{r}, μ = G (m_{1} + m_{2}) .$ Lekin pehle — $r$ aakhir hai kya? Uske upar ek dot ka matlab kya hai? Chalte hain bilkul zero se.

Step 0 — Alphabet (pehle har symbol earn karo)

Definition Char picture-objects jo hum draw karenge

Ek point mass ek aisi body hoti hai jo itni choti ho ki hum use ek dot ke roop mein draw karte hain. Hamare paas do hain: mass $m_{1}$ aur mass $m_{2}$ (masses sirf numbers hain kilograms mein).
Ek position vector ek arrow hota hai jo ek chosen origin $O$ se ek dot tak draw kiya jaata hai. $r_{1}$ dot 1 tak ka arrow hai; $r_{2}$ dot 2 tak ka arrow. Bold letters arrows hain (unki direction hoti hai); plain letters lengths hain.
Ek arrow $r$ ki length $r = ∣ r ∣$ likhi jaati hai — bas "yeh kitna lamba hai", ek positive number.
Upar ek dot, $\dot{r}$ , matlab hai "yeh arrow per second kitni tezi se change ho raha hai" — velocity. Do dots, $\ddot{r}$ , matlab hai "velocity kitni tezi se change ho rahi hai" — acceleration. (Pehle ek derivative, phir doosra; derivative bas rate of change hota hai, position-vs-time graph ka slope.)

Hum vectors kyun chahiye aur plain numbers kyun nahi? Kyunki gravity ki ek direction hoti hai — yeh ek body se doosri body ki taraf point karti hai. Ek plain number "which way" nahi bata sakta; ek arrow bata sakta hai. Yahi poori wajah hai ki hum yahan ordinary algebra ki jagah vectors use karte hain.

Figure — Two-body problem — equations of motion, reduction to one-body

Figure dekho: do dots, ek origin, do arrows. Is page par baaki sab kuch exactly inhi ingredients se bana hai.

Step 1 — Relative arrow $r$

KYA kiya humne: humne ek single arrow define kiya jo do bodies ko connect karta hai.

KYU: gravity sirf bodies ke beech ke gap par depend karti hai, pair space mein kahan baitha hai uس par nahi. Agar dono dots 5 metre right slide kar jayein, toh force unchanged rahegi. Toh physics poori tarah $r$ ke andar rehti hai. $r$ ko track karna koi bhi important cheez nahi chodhta.

PICTURE: figure mein, $r$ pale-yellow arrow hai. Uski length $r$ separation hai; uski direction "1 se 2 ki taraf" hai. Hum unit arrow $\hat{r} = r / r$ bhi draw karte hain — same direction, lekin length exactly 1. Ek arrow ko uski apni length se divide karna use length 1 tak shrink/stretch kar deta hai jabki direction same rehti hai; yahi "unit vector" ka matlab hai, aur hum ise sirf direction, koi length nahi bolne ke liye use karte hain.

Step 2 — Newton's law har dot par (sahi sign ke saath)

Left equation ko piece by piece padho:

$m_{1} \ddot{r}_{1}$ = (body 1 ki mass) $\times$ (uska acceleration) = body 1 par net force (yahi toh $F = ma$ hai).
$\frac{G m _{1} m _{2}}{r ^{2}}$ = gravity ki strength: $G$ Newton's constant hai, dono masses multiply hote hain, aur $1/ r^{2}$ kehta hai ki pull distance ke square ke saath kamzor hoti hai.
$\hat{r}$ = direction: body 1 ko body 2 ki taraf khicha jaata hai, jo $+ \hat{r}$ direction hai (yaad raho $\hat{r}$ 1 se 2 ki taraf point karta hai). Isliye $+$ .
Right equation mein sign flip hokar $- \hat{r}$ ho jaata hai kyunki body 2 ko wapas body 1 ki taraf khicha jaata hai — ulti direction.

KYA: $F = ma$ ke do applications, ek per body.

SIGNS KYU: har body doosri ki taraf girti hai. Strengths identical hain ( $G m_{1} m_{2} / r^{2}$ dono mein aata hai) — yeh Newton's third law hai, action equals reaction, pehle se hi included hai. Agar signs ulte hote toh bodies alag fly karti, jo gravity ke liye galat hai.

PICTURE: do force arrows ek-doosre ki taraf point karte hue, equal length, opposite direction.

Step 3 — Equations add karo: center freely drift karta hai

KYA: humne literally Step 2 ke do boxed equations add kiye.

KYU: right-hand sides equal aur opposite hain, isliye woh annihilate ho jaate hain. Left side $m_{1} r_{1} + m_{2} r_{2}$ ka second derivative hai. Center of mass define karo: $R = \frac{m _{1} r _{1} + m _{2} r _{2}}{m _{1} + m _{2}},$ mass-weighted average position — pair ka balance point. Tab $\ddot{R} = 0$ .

$\ddot{R} = 0$ matlab zero acceleration, toh $R$ constant speed par ek straight line mein move karta hai (ya ruka rehta hai). Isolated pair par koi external force nahi lagti, isliye unka balance point accelerate nahi ho sakta.

PICTURE: do dots ghoomte hain, lekin marked balance point ek perfectly straight track par glide karta hai. Isme koi orbital information nahi hoti — isliye hum ek frame mein hop karte hain jo iske saath ride karta hai aur ise bhool jaate hain. Yeh hamare 6 unknowns mein se 3 delete kar deta hai.

Step 4 — Equations subtract karo: relative motion

Pehle har Step-2 law ko uski apni mass se divide karo (taaki hum acceleration isolate kar sakein): $\ddot{r}_{1} = + \frac{G m _{2}}{r ^{2}} \hat{r}, \ddot{r}_{2} = - \frac{G m _{1}}{r ^{2}} \hat{r} .$ Ab pehle ko doosre se subtract karo. Kyunki $r = r_{2} - r_{1}$ , do baar differentiate karne par $\ddot{r} = \ddot{r}_{2} - \ddot{r}_{1}$ milta hai:

Final result mein term by term:

$\ddot{r}$ = relative arrow ka acceleration — gap-vector kitni tezi se speed up ya turn hota hai.
$G (m_{1} + m_{2})$ = dono masses appear hote hain, kyunki dono dots accelerate karte hain. Unke do accelerations relative motion mein add up ho jaate hain.
$1/ r^{2}$ aur $\hat{r}$ = same inverse-square strength aur same "center ki taraf" direction.
Minus sign = relative arrow ko shrink hone ki taraf pull kiya jaata hai (bodies attract hoti hain), hamesha inward point karta hai.

KYA: humne do motions ko ek arrow ke ek equation mein combine kiya.

YEH SAARI BAAT KYU HAI: is equation ki form bilkul wahi hai jaise ek particle ek fixed center ki taraf gir raha ho. Humne 2-body chase ko 1-body Kepler problem mein convert kar diya.

PICTURE: do dots bhool jao; ek single "ghost" particle $r$ ki tip par draw karo, hamesha ek fixed origin ki taraf yank kiya jaata hua. Woh ghost orbit trace karta hai.

Step 5 — Reduced mass (taaki energy sahi nikle)

Hum relative law ko ek particle ke honest $F = ma$ ke roop mein bhi likh sakte hain. Real force magnitude hai $F = G m_{1} m_{2} / r^{2}$ . Kaunsi mass $μ_{red}$ se $μ_{red} \ddot{r} = F$ true hoga?

$μ_{red} \ddot{r} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} (- \frac{G ( m _{1} + m _{2} )}{r ^{2}} \hat{r}) = - \frac{G m _{1} m _{2}}{r ^{2}} \hat{r} = F . ✓$

KYA / KYU: same motion ki do alag repackagings — ek $μ$ (strength) use karta hai bare acceleration ke saath, doosra $μ_{red}$ (ek real effective mass) use karta hai real force ke saath. Doosra kinetic energy ko cleanly split karta hai: $T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}, M = m_{1} + m_{2},$ "drifting center ki energy" $+$ "internal orbit ki energy," bina kisi cross-term ke.

PICTURE: mass $μ_{red}$ ka ek effective particle force $F$ ke end par swing kar raha hai.

Step 6 — Degenerate & edge cases (koi gap mat chhodho)

Figure teeno dikhata hai: heavy-light (tiny wobble), equal (twin circles), aur collapse cone jahan $r \to 0$ .

Recall Woh coefficients check karo jo dots recover karte hain

$r_{1} = R - \frac{m _{2}}{M} r$ aur $r_{2} = R + \frac{m _{1}}{M} r$ . Heavier body ka smaller coefficient hota hai, isliye woh COM ke paas rehti hai. Yeh kaam kyun karte hain? ::: Inhe subtract karo: $r_{2} - r_{1} = \frac{m _{1} + m _{2}}{M} r = r$ ✓. Mass-weight karke add karo: $r$ -terms cancel ho jaate hain, sirf $M R$ bachta hai ✓.

Ek-picture summary

Arrow ke left: do dots, do arrows, do coupled equations, 6 unknowns. Arrow ke right: ek drifting balance point $R$ (straight, boring) $+$ ek ghost particle on $r$ jo $\ddot{r} = - μ \hat{r} / r^{2}$ obey karta hai. Two-body mess → ek Kepler problem.

Recall Feynman: plain words mein poora walkthrough

Do bacche ice par rope pakad ke spin kar rahe hain. Dono bacchon ko dekhne ki jagah (mushkil — sab kuch move karta hai), do simpler cheezein dekho. Pehle, rope ka middle: woh sirf ek straight line par slide karta hai, bilkul boring, kyunki bahar se pair ko koi push nahi kar raha — yeh center of mass hai, aur hum ise throw away kar dete hain. Doosra, rope khud, ek bacche se doosre tak ka arrow: yeh stretch aur swing karta hai, aur isme saari spinning information hoti hai. Jab hum har bacche ke liye Newton's push likhte hain aur unhe add karte hain, to do pushes cancel ho jaate hain (isliye middle freely drift karta hai). Jab hum unhe subtract karte hain, to do pushes pile up ho jaate hain, isliye rope ko dono bacchon ke weights mila ke pull feel hoti hai — yahi $m_{1} + m_{2}$ hai. Result: do chasing bacchon ki jagah, hamare paas ek kalpnik bead hai jo rope par slide karta hai, ek invisible center ki taraf pull hota hua. Woh single bead hi poora orbit hai. Heavy baccha barely hilta hai, light baccha wide swing karta hai — exactly yahi reason hai ki ek planet fixed lagti hai jabki uska moon tezi se around race karta hai.

Connections

Parent topic — poora statement aur worked examples.
Kepler's laws — $\ddot{r} = - μ \hat{r} / r^{2}$ solve karne par conic orbits milte hain.
Conservation of angular momentum (central force) — kyun ghost particle ek plane mein rehta hai.
Vis-viva equation — reduced one-body system ki energy.
Center of mass frame — woh "ride-along" frame jo humne Step 3 mein use kiya.
Reduced mass in molecular vibrations — same $μ_{red}$ trick do atoms ke liye spring par.
Three-body problem — kya break hota hai jab teesra dot add hota hai.