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

3.2.1 · D4 · Physics › Orbital Mechanics & Astrodynamics › Two-body problem — equations of motion, reduction to one-bod

Constants jo poore note mein use honge (inhe ek baar state kar dete hain taaki baad mein koi surprise na ho):

Kisi bhi problem se pehle, teen geometric objects jo baar baar aate hain — inhe sirf naam nahi, picture mein bhi dekho:

Definition Teen vectors, is page par dobara bataye gaye

Relative position $r = r_{2} - r_{1}$ : woh arrow jo body 1 se body 2 ki taraf khicha gaya hai (dono ke beech ki "rassi").
Uski length $r = ∣ r ∣$ : ek single positive number, dono bodies ke beech ki seedhi separation. Jab bhi plain $r$ dikhe (bold nahi, arrow nahi) toh iska matlab yahi distance hai.
Unit vector $\hat{r} = \frac{r}{r}$ : wohi rassi wala arrow, length $1$ tak shrink kiya gaya. Yeh sirf body 1 se body 2 ki direction carry karta hai, koi length nahi. Hum isse use karte hain yeh kehne ke liye ki "gravity dono bodies ko join karne wali line ke along point karti hai" — bina distance ko saath liye.
Center of mass (COM) $R = \frac{m _{1} r _{1} + m _{2} r _{2}}{m _{1} + m _{2}}$ : mass-weighted average position — dono bodies ka balance point, jo bhaari body ke zyada paas hota hai. Equivalently $M R = m_{1} r_{1} + m_{2} r_{2}$ jahan $M = m_{1} + m_{2}$ .

Level 1 — Recognition

L1.1 — Vector ko naam do

Do stars positions $r_{1}$ aur $r_{2}$ par hain. Relative position vector $r$ likho aur bolo ki yeh "kis body se" aur "kis body ki taraf" point karta hai.

Recall Solution

$r = r_{2} - r_{1} .$ $r_{2}$ se $r_{1}$ subtract karne par woh arrow milta hai jo body 1 se start hota hai aur body 2 par khatam hota hai — socho jaise kid 1 se kid 2 tak khicha hua ek stretchy rope. Uski length $r = ∣ r ∣$ (ek single positive number) bodies ke beech ka separation hai.

L1.2 — Sahi $μ$ chuno

Ek $1200 kg$ satellite Earth ke orbit mein hai. $μ$ exactly likho, phir uska practical value do.

Recall Solution

Exact: $μ = G (M_{\oplus} + m_{sat}) = G (5.97 \times 1 0^{24} + 1200)$ . $1200$ Earth ki mass se lagbhag $1 0^{21}$ times chhota hai — practically invisible. Isliye $μ \approx G M_{\oplus} = 6.674 \times 1 0^{- 11} \times 5.97 \times 1 0^{24} \approx 3.985 \times 1 0^{14} m^{3} s^{- 2} .$ Yahi famous $G M_{\oplus}$ hai. Satellite ki mass drop out hoti hai sirf isliye kyunki woh tiny hai, formula usse ignore karta hai isliye nahi.

L1.3 — Equal masses ka reduced mass

Do identical stars, har ek ki mass $m$ . $μ_{red}$ ko $m$ ke terms mein compute karo.

Recall Solution

$μ_{red} = \frac{m \cdot m}{m + m} = \frac{m ^{2}}{2 m} = \frac{m}{2} .$ Product over sum: do equal masses ka reduced mass exactly aadha ek mass ke barabar hota hai.

Level 2 — Application

L2.1 — Sun–Jupiter system ka gravitational parameter

Jupiter ki mass $m_{J} = 1.898 \times 1 0^{27} kg$ hai. $μ = G (M_{⊙} + m_{J})$ compute karo, aur yeh $G M_{⊙}$ alone se kitne percent zyada hai?

Recall Solution

$μ = 6.674 \times 1 0^{- 11} (1.989 \times 1 0^{30} + 1.898 \times 1 0^{27}) .$ Masses ka sum $= 1.990898 \times 1 0^{30} kg$ , toh $μ = 1.3287 \times 1 0^{20} m^{3} s^{- 2} .$ $G M_{⊙}$ se fractional excess: $\frac{m _{J}}{M _{⊙}} = \frac{1.898 \times 1 0 ^{27}}{1.989 \times 1 0 ^{30}} = 9.54 \times 1 0^{- 4}$ , matlab lagbhag $0.095%$ . Chhota hai — lekin Jupiter ke liye measurable hai, satellite wale case ke unlike.

L2.2 — Center of mass kahan hai?

Ek Earth ( $M_{\oplus}$ ) aur ek Moon ( $m_{M} = 7.35 \times 1 0^{22} kg$ ) ke beech separation $r = 3.84 \times 1 0^{8} m$ hai. Earth ke centre se COM kitna dur hai? (Earth ki radius $6.37 \times 1 0^{6} m$ hai — kya COM Earth ke andar hai ya bahar?)

Recall Solution

Pehle, distance khud derive karte hain. Earth (mass $M_{\oplus}$ ) ko $r_{1}$ par aur Moon (mass $m_{M}$ ) ko $r_{2}$ par rakh do. COM mass-weighted average hai: $R = \frac{M _{\oplus} r _{1} + m _{M} r _{2}}{M _{\oplus} + m _{M}} .$ COM se Earth ki distance $∣ r_{1} - R ∣$ hai. Subtract karo: $r_{1} - R = r_{1} - \frac{M _{\oplus} r _{1} + m _{M} r _{2}}{M _{\oplus} + m _{M}} = \frac{( M _{\oplus} + m _{M} ) r _{1} - M _{\oplus} r _{1} - m _{M} r _{2}}{M _{\oplus} + m _{M}} = \frac{m _{M} ( r _{1} - r _{2} )}{M _{\oplus} + m _{M}} = - \frac{m _{M}}{M _{\oplus} + m _{M}} r .$ Lengths lete hain (aur $r = ∣ r ∣$ separation hai): $d_{\oplus} = ∣ r_{1} - R ∣ = \frac{m _{M}}{M _{\oplus} + m _{M}} r .$ Intuitive samajh: Earth ka "lever" mein share doosre body ke mass fraction se set hota hai. Kyunki Moon halka hai, woh fraction tiny hai, toh Earth COM se bahut kam move karti hai. $d_{\oplus} = \frac{7.35 \times 1 0 ^{22}}{5.97 \times 1 0 ^{24} + 7.35 \times 1 0 ^{22}} \times 3.84 \times 1 0^{8} .$ Fraction $0.01216$ hai, toh $d_{\oplus} = 4.67 \times 1 0^{6} m$ . Kyunki $4.67 \times 1 0^{6} < 6.37 \times 1 0^{6}$ (Earth ki radius), COM Earth ke andar hai — surface tak ka lagbhag $73%$ tak. Earth is internal point ke around wobble karti hai.

L2.3 — Binary ke individual orbit radii

Do stars jinki masses $m_{1} = 2 \times 1 0^{30}$ kg aur $m_{2} = 6 \times 1 0^{30}$ kg hain, aur separation $r = 1.0 \times 1 0^{11}$ m hai. Har star ki COM se distance nikalo.

Recall Solution

$M = m_{1} + m_{2} = 8 \times 1 0^{30}$ kg. $d_{1} = \frac{m _{2}}{M} r = \frac{6}{8} \times 1 0^{11} = 7.5 \times 1 0^{10} m,$ $d_{2} = \frac{m _{1}}{M} r = \frac{2}{8} \times 1 0^{11} = 2.5 \times 1 0^{10} m .$ Check: $d_{1} + d_{2} = 1 0^{11} = r$ ✓. Bhaari star ( $m_{2}$ ) COM ke zyada paas ( $2.5 \times 1 0^{10}$ ) baitha hai — uska coefficient $m_{1} / M$ chhoti mass fraction hai.

Neeche ka figure bilkul yahi configuration draw karta hai: dono dashed circles woh paths hain jo har star COM (origin par navy dot) ke around trace karta hai. Notice karo ki bhaari violet body $m_{2}$ centre ke paas hoti hai jabki halki magenta body $m_{1}$ bade circle par swing karti hai — orange arrow dono ko span karta hai jo poora separation $r = d_{1} + d_{2}$ hai.

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

Level 3 — Analysis

L3.1 — COM seedhi line mein kyun drift karta hai

Do Newton equations se shuru karo (yahan $\hat{r} = r / r$ unit direction hai body 1 se body 2 ki taraf, page ke upar define kiya gaya) $m_{1} \ddot{r}_{1} = + \frac{G m _{1} m _{2}}{r ^{2}} \hat{r}, m_{2} \ddot{r}_{2} = - \frac{G m _{1} m _{2}}{r ^{2}} \hat{r},$ prove karo ki $\ddot{R} = 0$ aur physically explain karo ki "no external force" ka matlab kya hai.

Recall Solution

Dono equations add karo. Right sides equal aur opposite hain, toh cancel ho jaate hain: $m_{1} \ddot{r}_{1} + m_{2} \ddot{r}_{2} = 0.$ Kyunki $M R = m_{1} r_{1} + m_{2} r_{2}$ (COM ki definition times total mass) aur $M$ constant hai, do baar differentiate karne par: $M \ddot{R} = m_{1} \ddot{r}_{1} + m_{2} \ddot{r}_{2} = 0 \Rightarrow \ddot{R} = 0.$ Physical meaning: sirf woh forces present hain jo dono bodies ek doosre par laga rahi hain (internal). Newton's third law se woh equal aur opposite hote hain, toh sum mein cancel ho jaate hain. System ke bahar se koi force nahi hai, toh COM accelerate nahi kar sakta — woh forever constant velocity se glide karta hai. Isliye hum uske saath ride kar sakte hain aur use ignore kar sakte hain.

L3.2 — $μ$ ka limit check

Dikhao ki $μ = G (m_{1} + m_{2})$ collapse hokar $G M_{\oplus}$ ban jaata hai jab $m_{2} / m_{1} \to 0$ , aur $1000$ kg satellite ke liye error estimate karo.

Recall Solution

$m_{1}$ factor out karo: $μ = G m_{1} (1 + \frac{m _{2}}{m _{1}})$ . Jab mass ratio $m_{2} / m_{1} \to 0$ , bracket $\to 1$ , toh $μ \to G m_{1} = G M_{\oplus}$ . Satellite ke liye, fractional error exactly $\frac{m _{2}}{m _{1}} = \frac{1000}{5.97 \times 1 0 ^{24}} = 1.68 \times 1 0^{- 22}$ hai (pure ratio, koi units nahi). Fraction ko percent mein convert karne ke liye $100$ se multiply karo: $1.68 \times 1 0^{- 22} \times 100 = 1.68 \times 1 0^{- 20} %$ . Dono tarah se yeh kisi bhi instrument se measure nahi ho sakta. Toh "fixed Earth" approximation essentially exact hai yahan, lekin equation khud $m_{2}$ kabhi drop nahi karti.

L3.3 — Reduced mass sahi force deta hai

Verify karo ki relative coordinate ko $μ_{red}$ mass wale ek particle ki tarah treat karne par true gravitational force magnitude $G m_{1} m_{2} / r^{2}$ wapas milta hai.

Recall Solution

Relative equation hai $\ddot{r} = - \frac{G ( m _{1} + m _{2} )}{r ^{2}} \hat{r}$ (jahan $\hat{r} = r / r$ ). Dono sides ko $μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ se multiply karo: $μ_{red} \ddot{r} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} \cdot (- \frac{G ( m _{1} + m _{2} )}{r ^{2}}) \hat{r} = - \frac{G m _{1} m _{2}}{r ^{2}} \hat{r} .$ $(m_{1} + m_{2})$ factors exactly cancel ho jaate hain, Newton ki gravity magnitude $G m_{1} m_{2} / r^{2}$ bach jaati hai. Toh "mass $μ_{red}$ jisko real force $F$ push kar rahi hai" relative motion reproduce karta hai — isliye energy $μ_{red}$ use karti hai jabki orbit shape $μ$ use karta hai.

Level 4 — Synthesis

L4.1 — Full circular binary

Do stars, $m_{1} = m_{2} = m = 3 \times 1 0^{30}$ kg, apne common COM ke around circle mein orbit kar rahe hain, separation $r = 2 \times 1 0^{11}$ m. (a) $μ$ nikalo, (b) orbital period $T$ nikalo Kepler's third law ke reduced form $T^{2} = \frac{4 π ^{2}}{μ} r^{3}$ use karke.

Recall Solution

(a) $μ = G (m_{1} + m_{2}) = 6.674 \times 1 0^{- 11} \times 6 \times 1 0^{30} = 4.0044 \times 1 0^{20} m^{3} s^{- 2} .$ (b) Relative Kepler law mein plug karo (yeh separation $r$ use karta hai, har orbit radius nahi): $T^{2} = \frac{4 π ^{2}}{μ} r^{3} = \frac{4 π ^{2}}{4.0044 \times 1 0 ^{20}} (2 \times 1 0^{11})^{3} .$ Numerator: $4 π^{2} \times 8 \times 1 0^{33} = 3.1583 \times 1 0^{35}$ . Divide karo: $T^{2} = 7.887 \times 1 0^{14} s^{2}$ , toh $T = 2.808 \times 1 0^{7} s \approx 0.89 years .$ Yeh third law hai reduced one-body problem ke liye likha gaya — note karo ki $μ$ (sum-based wala) appear karta hai, bilkul waise jaise parent note ne build kiya tha.

L4.2 — Dono positions recover karo

Ek instant par, COM $R = (0, 0)$ par hai aur relative vector $r = (3 \times 1 0^{8}, 0)$ m hai, jahan $m_{1} = 1 \times 1 0^{24}$ kg, $m_{2} = 3 \times 1 0^{24}$ kg. $r_{1}$ aur $r_{2}$ nikalo, aur confirm karo ki $r_{2} - r_{1} = r$ .

Recall Solution

$M = 4 \times 1 0^{24}$ kg. $r_{1} = R - \frac{m _{2}}{M} r$ aur $r_{2} = R + \frac{m _{1}}{M} r$ use karte hain: $r_{1} = - \frac{3}{4} (3 \times 1 0^{8}, 0) = (- 2.25 \times 1 0^{8}, 0) m,$ $r_{2} = + \frac{1}{4} (3 \times 1 0^{8}, 0) = (7.5 \times 1 0^{7}, 0) m .$ Check: $r_{2} - r_{1} = (7.5 \times 1 0^{7} - (- 2.25 \times 1 0^{8}), 0) = (3 \times 1 0^{8}, 0) = r$ ✓. Bhaari body 2 COM ke zyada paas hai ( $7.5 \times 1 0^{7}$ vs $2.25 \times 1 0^{8}$ ), jo L2.3 ke rule se match karta hai.

Neeche ka figure is instant ko plot karta hai: navy dot origin par COM hai, orange arrow relative vector $r = r_{2} - r_{1}$ hai, aur dono bodies COM ke opposite sides par hain. Dekho kaise halki magenta body $m_{1}$ door left par hai jabki bhaari violet body $m_{2}$ centre ke just right par hai — COM formula separation ko mass ratio $3 : 1$ mein split karta hai.

Level 5 — Mastery

L5.1 — Kinetic-energy split scratch se derive karo

$T = \frac{1}{2} m_{1} \dot{r}_{1}^{2} + \frac{1}{2} m_{2} \dot{r}_{2}^{2}$ se shuru karo, aur coordinate transforms $r_{1} = R - \frac{m _{2}}{M} r$ , $r_{2} = R + \frac{m _{1}}{M} r$ use karte hue prove karo: $T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2} .$

Recall Solution

Transforms differentiate karo: $\dot{r}_{1} = \dot{R} - \frac{m _{2}}{M} \dot{r}$ , $\dot{r}_{2} = \dot{R} + \frac{m _{1}}{M} \dot{r}$ . Har ek square karo ( $a^{2} = a \cdot a$ use karke): $\dot{r}_{1}^{2} = \dot{R}^{2} - 2 \frac{m _{2}}{M} \dot{R} \cdot \dot{r} + \frac{m _{2}^{2}}{M ^{2}} \dot{r}^{2},$ $\dot{r}_{2}^{2} = \dot{R}^{2} + 2 \frac{m _{1}}{M} \dot{R} \cdot \dot{r} + \frac{m _{1}^{2}}{M ^{2}} \dot{r}^{2} .$ Ab $T = \frac{1}{2} m_{1} \dot{r}_{1}^{2} + \frac{1}{2} m_{2} \dot{r}_{2}^{2}$ build karo:

$\dot{R}^{2}$ term: $\frac{1}{2} (m_{1} + m_{2}) \dot{R}^{2} = \frac{1}{2} M \dot{R}^{2}$ . ✓
Cross term: $\frac{1}{2} (- 2 m_{1} \frac{m _{2}}{M} + 2 m_{2} \frac{m _{1}}{M}) \dot{R} \cdot \dot{r} = 0$ . Dono pieces cancel ho jaate hain — isliye coordinates decouple hote hain.
$\dot{r}^{2}$ term: $\frac{1}{2} (m_{1} \frac{m _{2}^{2}}{M ^{2}} + m_{2} \frac{m _{1}^{2}}{M ^{2}}) \dot{r}^{2} = \frac{1}{2} \frac{m _{1} m _{2} ( m _{2} + m _{1} )}{M ^{2}} \dot{r}^{2} = \frac{1}{2} \frac{m _{1} m _{2}}{M} \dot{r}^{2} = \frac{1}{2} μ_{red} \dot{r}^{2} .$ Sum karo: $T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}$ . ∎ Cross term ka vanish hona hi woh cheez hai jo orbit (energy) problem ko COM drift se independently solve karne deta hai.

L5.2 — Split par sanity numeric

$m_{1} = 2$ , $m_{2} = 6$ (units kg), $\dot{r}_{1} = (1, 0)$ , $\dot{r}_{2} = (0, 2)$ m/s ke liye, $T$ directly compute karo, phir split ke through, aur confirm karo ki dono match karte hain.

Recall Solution

Direct: $T = \frac{1}{2} (2) (1^{2}) + \frac{1}{2} (6) (2^{2}) = 1 + 12 = 13 J .$ Split: $M = 8$ . $\dot{R} = \frac{m _{1} r ˙ _{1} + m _{2} r ˙ _{2}}{M} = \frac{( 2 , 0 ) + ( 0 , 12 )}{8} = (0.25, 1.5)$ . $\dot{r} = \dot{r}_{2} - \dot{r}_{1} = (- 1, 2)$ . $μ_{red} = \frac{2 \cdot 6}{8} = 1.5$ . $\frac{1}{2} M \dot{R}^{2} = \frac{1}{2} (8) (0.2 5^{2} + 1. 5^{2}) = 4 (0.0625 + 2.25) = 9.25$ . $\frac{1}{2} μ_{red} \dot{r}^{2} = \frac{1}{2} (1.5) (1 + 4) = 0.75 \times 5 = 3.75$ . Sum $= 9.25 + 3.75 = 13 J$ ✓. Dono decoupled energies total ko exactly rebuild karte hain.

Recall Feynman recap

Har ladder rung ek idea hai: (1) rope $r$ ko naam do, (2) dono $μ$ 's mein masses plug karo, (3) prove karo ki middle point drift karta hai kyunki internal forces cancel hote hain, (4) sab kuch ek real orbit mein combine karo aur dono bodies recover karo, (5) prove karo ki energy cleanly split hoti hai. Agar tum L5 bina help ke kar sako, toh two bodies se one body reduction tumhara apna ho gaya.

Connections

Kepler's laws — L4.1 directly reduced third law use karta hai.
Conservation of angular momentum (central force) — guarantee karta hai ki binary orbits planar rahenge.
Vis-viva equation — L5 ka energy split orbital energy ko feed karta hai.
Center of mass frame — woh frame jisme L4.2 aur L5 sabse clean hain.
Reduced mass in molecular vibrations — diatomic molecules mein wohi $μ_{red}$ algebra.
Three-body problem — jahan yeh poori clean ladder collapse ho jaati hai.