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

3.2.1 · Physics › Orbital Mechanics & Astrodynamics

KYA solve kar rahe hain hum

Do key objects:

Center of mass $R = \frac{m _{1} r _{1} + m _{2} r _{2}}{m _{1} + m _{2}}$
Relative position $r = r_{2} - r_{1}$ , jahan $r = ∣ r ∣$ .

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

KAISE: scratch se equations of motion derive karo

Step 1 — Newton's law har body par. $m_{1}$ par gravity $m_{2}$ ki taraf point karti hai, yani $+ r = r_{2} - r_{1}$ direction mein:

\qquad m_2\ddot{\mathbf r}_2 = -\frac{G m_1 m_2}{r^2}\,\hat{\mathbf r}$$ *Signs kyun?* Har body doosri ki taraf kheenchi jaati hai. $m_1$ ko $m_2$ ki direction mein kheeencha jaata hai (yeh $+\hat{\mathbf r}$ hai); $m_2$ ko $m_1$ ki taraf wapas kheencha jaata hai ($-\hat{\mathbf r}$). Newton's third law built-in hai: forces equal aur opposite hain. **Step 2 — Center of mass freely move karta hai.** Dono equations add karo: $$m_1\ddot{\mathbf r}_1 + m_2\ddot{\mathbf r}_2 = 0 \;\Rightarrow\; \frac{d^2}{dt^2}\big(m_1\mathbf r_1+m_2\mathbf r_2\big)=0 \;\Rightarrow\; \ddot{\mathbf R}=0.$$ *Yeh kyun matter karta hai:* internal forces cancel ho jaate hain, isliye system par **koi external force** nahi lagta → COM constant velocity se move karta hai. Isme koi useful orbital info nahi hai, isliye hum coordinates ko *usके saath chalte hue* set karte hain aur use bhool jaate hain. Isse 6 unknowns mein se 3 hat jaate hain. **Step 3 — Relative motion (asli cheez).** Har Newton equation ko uski mass se divide karo: $$\ddot{\mathbf r}_1 = +\frac{G m_2}{r^2}\hat{\mathbf r}, \qquad \ddot{\mathbf r}_2 = -\frac{G m_1}{r^2}\hat{\mathbf r}.$$ Subtract karo (pehla doosre se) taaki $\ddot{\mathbf r}=\ddot{\mathbf r}_2-\ddot{\mathbf r}_1$ mile: $$\ddot{\mathbf r} = -\frac{G m_1}{r^2}\hat{\mathbf r} - \frac{G m_2}{r^2}\hat{\mathbf r} = -\frac{G(m_1+m_2)}{r^2}\hat{\mathbf r}.$$ > [!formula] Reduced equation of motion > $$\boxed{\;\ddot{\mathbf r} = -\frac{\mu}{r^2}\,\hat{\mathbf r} = -\frac{\mu}{r^3}\,\mathbf r,\qquad \mu \equiv G(m_1+m_2)\;}$$ > Yeh **bilkul same** hai ek single particle ki tarah jo ek *fixed* mass ke around orbit kar rahi ho jo gravitational parameter $\mu$ produce karti hai. Two-body problem ab one-body problem ban gaya. Yahan $\hat{\mathbf r}=\mathbf r/r$. > [!mistake] "$\mu$ should be $Gm_1$ for a satellite around Earth." > *Kyun sahi lagta hai:* satellite chhota hai, Earth fixed lagti hai, isliye hum expect karte hain sirf Earth ki mass. **Exact answer hai $\mu=G(m_1+m_2)$** — dono masses aate hain, kyunki dono accelerate karte hain. *Fix/limit:* jab $m_2\ll m_1$ (satellite mass $\ll$ Earth), toh $m_1+m_2\approx m_1$, isliye $\mu\approx Gm_1$ ek achha *approximation* hai, exact statement nahi. Binary stars mein jahan comparable mass ho wahan **dono** rakhna zaroori hai. --- ## Reduced mass — energy/momentum sahi tarike se nikalna Hum relative dynamics ko ek *real* one-body problem ki tarah bhi likh sakte hain ek effective mass ke saath. > [!definition] Reduced mass > $$\mu_{\text{red}} = \frac{m_1 m_2}{m_1+m_2}.$$ > (Symbol $\mu$ se alag! Ek mass hai, doosra $G\cdot M_\text{tot}$ hai. Log $\mu$ reuse karte hain — context dekho.) *Kyun aata hai, yeh derive karte hain.* Bodies ke beech force ka magnitude hai $F=Gm_1m_2/r^2$. Relative coordinate $\mu_{\text{red}}\,\ddot{\mathbf r}=\mathbf F$ follow karta hai: $$\mu_{\text{red}}\ddot{\mathbf r}=\frac{m_1m_2}{m_1+m_2}\Big(-\frac{G(m_1+m_2)}{r^2}\hat{\mathbf r}\Big)=-\frac{Gm_1m_2}{r^2}\hat{\mathbf r}=\mathbf F.\;\checkmark$$ Toh relative motion *exactly* mass $\mu_{\text{red}}$ ke ek particle jaisi hai gravitational force $\mathbf F$ mein. Total kinetic energy cleanly split hoti hai: $$T=\tfrac12 m_1\dot{\mathbf r}_1^2+\tfrac12 m_2\dot{\mathbf r}_2^2 = \underbrace{\tfrac12 M\dot{\mathbf R}^2}_{\text{COM}} + \underbrace{\tfrac12\mu_{\text{red}}\dot{\mathbf r}^2}_{\text{internal}},\quad M=m_1+m_2.$$ *Iss tarah kyun split karte hain?* Kyunki $\mathbf R$ aur $\mathbf r$ independent coordinates hain jinke cross-terms zero hote hain — energy decouple ho jaati hai, isliye orbit problem (internal part) akele solve ho sakta hai. --- ## WORKED EXAMPLES > [!example] 1 — Earth ka $\mu$ ek low satellite ke liye > $m_\oplus=5.97\times10^{24}\,$kg, satellite $m=1000\,$kg. > $\mu=G(m_\oplus+m)=6.674\times10^{-11}(5.97\times10^{24}+10^3)$. > *$10^3$ kyun drop karte hain?* Yeh $10^{21}$ times chhota hai — negligible. Isliye $\mu\approx 3.986\times10^{14}\,\text{m}^3/\text{s}^2$ (standard $GM_\oplus$). > **Lesson:** satellite ki mass invisible hai; planet–Sun *comparable* case mein dono matter karte hain. > [!example] 2 — Equal-mass binary stars > $m_1=m_2=m$. Toh $\mu=G(2m)$ aur $\mu_{\text{red}}=m^2/2m=m/2$. > COM kahan hai? Bilkul **beech mein** dono ke, kyunki masses equal hain. Agar $\mathbf r$ ki length $r$ hai toh har star COM ke around $r/2$ radius ke circle par orbit karta hai. > *Radius $r/2$ kyun?* $\mathbf r_1-\mathbf R = -\frac{m_2}{M}\mathbf r=-\tfrac12\mathbf r$, isliye star 1 COM se $r/2$ distance par baitha hai. Har star relative orbit ka ek chhota scaled copy trace karta hai. > [!example] 3 — $\mathbf R,\mathbf r$ se individual positions recover karna > $\mathbf R$ aur $\mathbf r$ diye hain, 2×2 linear system solve karo: > $$\mathbf r_1=\mathbf R-\frac{m_2}{M}\mathbf r,\qquad \mathbf r_2=\mathbf R+\frac{m_1}{M}\mathbf r.$$ > *Yeh coefficients kyun?* $\mathbf r=\mathbf r_2-\mathbf r_1$ mein plug karo: $\frac{m_1+m_2}{M}\mathbf r=\mathbf r$ milta hai ✓. $m_1\mathbf r_1+m_2\mathbf r_2$ mein plug karo: $\mathbf r$ terms cancel ho jaate hain aur $M\mathbf R$ milta hai ✓. Bhaari body COM ke *paas* rehti hai (chhota coefficient). --- > [!recall]- Feynman: ek 12-saal ke bache ko explain karo > Socho do bache barf par ek rope pakad ke ghoom rahe hain. Dono bacchon ko track karne ki bajaye, tum *beech ka point* dekho (woh sirf seedha smoothly slide karta hai — boring) aur *unke beech ki rope* dekho (yeh tumhe spin ke baare mein sab kuch batati hai). Two-body problem yahi karta hai: beech ka point = center of mass (freely drift karta hai), rope = relative vector $\mathbf r$. Sirf rope dekh ke, do ghoomti cheezein ek simple cheez ban jaati hain jo ek invisible center ke around jaati hai. Bhaari baccha barely hilta hai; halka baccha zyada jhoolता hai — isliye planet "fixed" lagti hai jabki satellite zoom karta hai. > [!mnemonic] Do $\mu$'s yaad rakho > "**Force ke liye SUM, mass ke liye PRODUCT-over-sum.**" > Gravitational parameter $\mu=G(m_1+m_2)$ **sum** use karta hai; reduced mass $\mu_{\text{red}}=\dfrac{m_1m_2}{m_1+m_2}$ **product over sum** use karta hai. --- ## #flashcards/physics Two-body problem mein relative position vector define karo ::: $\mathbf r=\mathbf r_2-\mathbf r_1$ (body 1 se body 2 tak). Center of mass constant velocity se kyun move karta hai? ::: Dono Newton equations add karne par, internal forces cancel ho jaate hain ($m_1\ddot{\mathbf r}_1+m_2\ddot{\mathbf r}_2=0$), isliye $\ddot{\mathbf R}=0$ — koi external force nahi. Reduced equation of motion kya hai? ::: $\ddot{\mathbf r}=-\dfrac{\mu}{r^3}\mathbf r$ jahan $\mu=G(m_1+m_2)$. Gravitational parameter $\mu$ define karo ::: $\mu=G(m_1+m_2)$ — exact two-body value; jab ek mass dominate kare toh $GM$ ban jaata hai. Reduced mass define karo ::: $\mu_{\text{red}}=\dfrac{m_1 m_2}{m_1+m_2}$. Total kinetic energy kaise split hoti hai? ::: $T=\tfrac12 M\dot{\mathbf R}^2+\tfrac12\mu_{\text{red}}\dot{\mathbf r}^2$ (COM motion + internal motion, decoupled). $\mathbf R,\mathbf r$ se $\mathbf r_1$ recover karo ::: $\mathbf r_1=\mathbf R-\dfrac{m_2}{M}\mathbf r$. Satellites ke liye $\mu\approx GM_\oplus$ valid kyun hai? ::: Kyunki $m_\text{sat}\ll m_\oplus$, isliye $m_1+m_2\approx m_1$; comparable masses ke liye dono rakhne zaroori hain. Equal-mass binary mein COM kahan hai aur har orbit radius kya hai? ::: COM beech mein; har star $r/2$ radius par orbit karta hai. --- ## Connections - [[Kepler's laws]] — reduced equation directly conic-section orbits deta hai. - [[Conservation of angular momentum (central force)]] — orbit planar kyun hoti hai. - [[Vis-viva equation]] — reduced one-body system ki energy. - [[Center of mass frame]] — yahan jo coordinate trick use hui hai. - [[Reduced mass in molecular vibrations]] — quantum/diatomic systems mein same math. - [[Three-body problem]] — jab teesri mass add hoti hai toh kya toot jaata hai (koi clean reduction nahi). ## 🖼️ Concept Map ```mermaid flowchart TD TB[Two-body problem] N1[Newton on m1] N2[Newton on m2] R[Relative vector r = r2 - r1] COM[Center of mass R] ADD[Add equations] SUB[Subtract equations] FREE[COM moves freely, R-ddot = 0] EOM[Reduced EOM r-ddot = -mu/r^3 r] MU[mu = G m1 + m2] KEP[One-body Kepler problem] TB -->|gravity gives| N1 TB -->|gravity gives| N2 N1 -->|equal opposite| N2 N1 --> ADD N2 --> ADD ADD -->|internal forces cancel| FREE FREE -->|removes 3 unknowns| COM N1 --> SUB N2 --> SUB SUB -->|yields| EOM R -->|governs| EOM MU -->|defines| EOM EOM -->|identical to| KEP ```