Third-body perturbations: Exercises — Hinglish deep dive

3.2.36 · D4 · Physics › Orbital Mechanics & Astrodynamics › Third-body perturbations

Poore note mein, hum yeh constants reuse karenge (SI units, yaani metres, seconds):

Do formulas jis par neeche sab kuch depend karta hai (parent se):

Level 1 — Recognition

L1.1 — Perturbing quantity ko pehchano

Problem. Ek student likhta hai: "Moon mera satellite $μ_{M} / s^{2}$ acceleration se perturb karta hai, jahan $s$ Moon–satellite distance hai." Kya yeh perturbing acceleration hai? Haan/nahi mein jawab do aur woh quantity name karo jo aslaan orbit ko perturb karti hai.

Recall Solution

Nahi. $μ_{M} / s^{2}$ ek inertial frame mein satellite par Moon ka raw pull hai. Lekin hum orbit ko Earth-centred frame mein track karte hain, aur Moon Earth ko bhi pull karta hai. Jo quantity orbit ko perturb karti hai woh hai satellite par Moon ke pull aur Earth par uske pull ka difference — tidal (differential) acceleration $a_{3 b} = μ_{3} (\frac{d - r}{∣ d - r ∣ ^{3}} - \frac{d}{d ^{3}})$ . Common part cancel ho jaata hai kyunki Earth bhi Moon ki taraf free-fall kar raha hai bilkul satellite ki tarah.

L1.2 — Scaling law padho

Problem. Tidal strength $μ_{3} / d^{n}$ ke roop mein scale karti hai. $n$ kya hai, aur agar third body ko do baar door le jaao (same mass) toh perturbation badhegi ya ghateggi?

Recall Solution

$n = 3$ (inverse cube, inverse square nahi). $d$ ko double karne se tidal strength $1/ 2^{3} = 1/8$ se multiply hoti hai — yeh one-eighth reh jaati hai. Inverse cube isliye hai kyunki yeh inverse-square field ka gradient (rate of change) hai, aur $1/ d^{2}$ ko differentiate karne se $1/ d^{3}$ aata hai.

Level 2 — Application

L2.1 — Kaun dominate karta hai, Sun ya Moon?

Problem. Moon aur Sun ke liye tidal strength factor $μ_{3} / d^{3}$ compute karo, phir unka ratio. Kaun jeetega?

Recall Solution

Moon: $\frac{μ _{M}}{d _{M}^{3}} = \frac{4.90 \times 1 0 ^{12}}{( 3.84 \times 1 0 ^{8} ) ^{3}} = 8.66 \times 1 0^{- 14} s^{- 2}$ . Sun: $\frac{μ _{S}}{d _{S}^{3}} = \frac{1.327 \times 1 0 ^{20}}{( 1.496 \times 1 0 ^{11} ) ^{3}} = 3.96 \times 1 0^{- 14} s^{- 2}$ . Ratio $= 8.66/3.96 \approx 2.19$ . Moon lagbhag 2× se dominate karta hai. Hume sirf $μ_{3} / d^{3}$ chahiye kyunki tidal formula mein satellite distance $r$ dono bodies ko identically multiply karta hai aur ratio mein cancel ho jaata hai.

L2.2 — GEO par Moon se peak tidal acceleration

Problem. Ek GEO satellite Earth–Moon line ke along baitha hai ( $\hat{d} \cdot r = r$ ). Moon se peak (stretching) tidal acceleration nikalo.

Recall Solution

Line ke along, $3 (\hat{d} \cdot r) \hat{d} - r = 3 r \hat{d} - r \hat{d} = 2 r \hat{d}$ , toh magnitude hai stretch coefficient $\frac{2 μ _{M} r}{d _{M}^{3}}$ . $a = 2 (8.66 \times 1 0^{- 14}) (4.22 \times 1 0^{7}) = 7.3 \times 1 0^{- 6} m/s^{2} .$ GEO par Earth ke apne pull se kaafi chhota ( $μ / r^{2} \approx 0.224$ m/s²), lekin relentless — saalon ke dauran yeh orbit plane ko precess karta hai aur North–South drift drive karta hai jise GEO stationkeeping fight karta hai.

Level 3 — Analysis

L3.1 — Stretch vs squeeze directions

Problem. $\hat{d} = (1, 0, 0)$ lo (Moon $x$ -axis ke along). Tidal acceleration likho ek satellite ke liye (a) exactly Moon ki taraf, $r = (r, 0, 0)$ ; aur (b) exactly perpendicular, $r = (0, r, 0)$ . Signs interpret karo.

Neeche diya figure woh geometry fix karta hai jo hum use karenge. Earth origin par hai (black dot). Black horizontal arrow $\hat{d}$ hai, Earth se Moon ki taraf direction (hamara $+ x$ -axis). Teen satellites draw kiye gaye hain: ek line ke along near side par ( $x = + r$ ) jiska red acceleration arrow outward, Earth se door point karta hai (stretch); ek far side par ( $x = - r$ ) jiska red arrow bhi outward point karta hai (stretch, opposite direction); aur ek perpendicular ( $y = + r$ ) jiska black arrow inward, Earth ki taraf point karta hai (squeeze). Picture se dekho ki do along-line arrows perpendicular wale se do baar lambe hain — yeh 2:1 stretch-to-squeeze ratio hai jo tum ab algebraically confirm karoge.

Recall Solution

Use karo $a_{3 b} = \frac{μ _{3}}{d ^{3}} (3 (\hat{d} \cdot r) \hat{d} - r)$ .

(a) $\hat{d} \cdot r = r$ , toh $a = \frac{μ _{3}}{d ^{3}} (3 r \hat{d} - r \hat{d}) = \frac{μ _{3}}{d ^{3}} (2 r, 0, 0)$ . Positive $x$ = Earth se door, Moon side ki taraf point karta hai → stretching, coefficient $+ 2 μ_{3} r / d^{3}$ . Yeh figure mein red near-side arrow hai.

(b) $\hat{d} \cdot r = 0$ , toh $a = \frac{μ _{3}}{d ^{3}} (0 - r \overset{y}{^}) = \frac{μ _{3}}{d ^{3}} (0, - r, 0)$ . Negative $y$ = Earth ki taraf point karta hai → squeezing, coefficient $- μ_{3} r / d^{3}$ . Yeh figure mein black inward arrow hai.

Stretch (line ke along) exactly squeeze magnitude (perpendicular) se do baar hai. Yeh 2:1 ratio har tidal field ka fingerprint hai.

L3.2 — Near side aur far side dono stretch hote hain

Problem. Kya Moon se far side par baitha ek satellite ( $r = (- r, 0, 0)$ , still with $\hat{d} = (1, 0, 0)$ ) Moon ki taraf khicha jaata hai ya Earth se door? Dikhao, aur explain karo ki "dono sides outward kyun bulge karti hain."

Recall Solution

$\hat{d} \cdot r = - r$ , toh $a = \frac{μ _{3}}{d ^{3}} (3 (- r) \hat{d} - (- r \hat{d})) = \frac{μ _{3}}{d ^{3}} (- 3 r + r, 0, 0) = \frac{μ _{3}}{d ^{3}} (- 2 r, 0, 0)$ . Acceleration $- x$ mein point karta hai = far side par bhi Earth se door. Toh dono near side (Moon ki taraf khicha, Earth se door) aur far side (Earth Moon ki taraf satellite se zyada khicha jaata hai, satellite ko "peeche" chhod deta hai = Earth ke relative outward) outward stretch hote hain. Yeh two-bulge tidal pattern hai — issi liye Earth mein din mein do ocean tides aate hain. Dekho Tidal forces.

Level 4 — Synthesis

L4.1 — Exact vs tidal error at GEO

Problem. Earth–Moon line ke along ek GEO satellite ke liye, (a) ratio $r_{GE O} / d_{M}$ compute karo aur (b) exact ke bajaye tidal (first-order) formula use karne se hone wali fractional error estimate karo. Leading dropped term $O ((r / d)^{2})$ order ka hai, toh error size roughly $r / d$ estimate karo.

Recall Solution

(a) $r_{GE O} / d_{M} = \frac{4.22 \times 1 0 ^{7}}{3.84 \times 1 0 ^{8}} = 0.110$ . (b) Tidal formula $r / d$ mein linear terms rakhta hai aur $(r / d)^{2}$ order ke terms drop karta hai. Pehle dropped term ki relative size $r / d \approx 0.11$ ke order par hai, yaani GEO par roughly 11% error. Quick estimates ke liye acceptable; precision propagation ke liye exact form rakhna hoga ya next (quadrupole) term add karna hoga. Yahi small parameter hai jo J2 perturbation and oblateness-style expansions ko bhi govern karta hai.

L4.2 — GEO par Sun aur Moon combine karo

Problem. Maano Moon aur Sun momentarily satellite ke same line ke along aligned hain (dono ke liye $\hat{d} \cdot r = r$ ). GEO par combined peak stretching acceleration nikalo, aur L2.2 se Moon-only value se compare karo.

Recall Solution

Combined stretch coefficient $= 2 r (\frac{μ _{M}}{d _{M}^{3}} + \frac{μ _{S}}{d _{S}^{3}})$ . $a = 2 (4.22 \times 1 0^{7}) (8.66 \times 1 0^{- 14} + 3.96 \times 1 0^{- 14}) = 2 (4.22 \times 1 0^{7}) (1.262 \times 1 0^{- 13}) = 1.065 \times 1 0^{- 5} m/s^{2} .$ Yeh $\approx 1.07 \times 1 0^{- 5}$ m/s² hai, Moon-only $7.3 \times 1 0^{- 6}$ m/s² se lagbhag 1.46×. Jab aligned hote hain toh reinforce karte hain; jab perpendicular hote hain toh partly cancel karte hain — isisi wajah se luni-solar stationkeeping demand saal bhar vary karti hai. Dekho GEO stationkeeping.

Level 5 — Mastery

L5.1 — Third-body $J_{2}$ ko kahan beat karta hai?

Problem. Earth ki oblateness ( $J_{2}$ ) ek perturbing acceleration produce karti hai jo altitude ke saath roughly $a_{J 2} \approx \frac{3}{2} J_{2} \frac{μ R _{E}^{2}}{r ^{4}}$ ke roop mein fall off karti hai ( $1/ r^{4}$ scale karti hai), jabki third-body tidal term $a_{3 b} \approx \frac{2 μ _{M} r}{d _{M}^{3}}$ ke roop mein grow karta hai ( $r$ scale karta hai). $J_{2} = 1.083 \times 1 0^{- 3}$ aur $R_{E} = 6.378 \times 1 0^{6}$ m use karke crossover radius $r_{\times}$ nikalo jahan Moon ki tidal acceleration $J_{2}$ ki acceleration ke barabar ho. Interpret karo.

Recall Solution

Dono ko equal set karo: $\frac{3}{2} J_{2} \frac{μ R _{E}^{2}}{r ^{4}} = \frac{2 μ _{M} r}{d _{M}^{3}} \Rightarrow r^{5} = \frac{3 J _{2} μ R _{E}^{2} d _{M}^{3}}{4 μ _{M}} .$ Numerator: $3 (1.083 \times 1 0^{- 3}) (3.986 \times 1 0^{14}) (6.378 \times 1 0^{6})^{2} (3.84 \times 1 0^{8})^{3}$ .

$(6.378 \times 1 0^{6})^{2} = 4.068 \times 1 0^{13}$
$(3.84 \times 1 0^{8})^{3} = 5.662 \times 1 0^{25}$
Product $= 3 (1.083 \times 1 0^{- 3}) (3.986 \times 1 0^{14}) (4.068 \times 1 0^{13}) (5.662 \times 1 0^{25}) = 2.983 \times 1 0^{51}$ $4 μ_{M} = 1.96 \times 1 0^{13}$ se divide karo: $r^{5} = 1.522 \times 1 0^{38}$ . $r_{\times} = (1.522 \times 1 0^{38})^{1/5} = 4.33 \times 1 0^{7} m \approx 6.8 R_{E} .$ Interpretation: $\approx 4.3 \times 1 0^{7}$ m se neeche, $J_{2}$ oblateness perturbation budget dominate karti hai (LEO ko $J_{2}$ ki chinta hai, Moon ki nahi). Uske upar — GEO bilkul is crossover ke paas baitha hai, jabki GTO apogee, Molniya apogee, aur lunar transfers kaafi aage hain — third-body luni-solar effects dominate karne lagte hain. Yeh crossover hi wajah hai ki mission designers orbit altitude ke hisaab se decide karte hain ki pehle kaun sa perturbation model karein. Gauss and Lagrange planetary equations se compare karo ki yeh accelerations slow orbital-element drift mein kaise feed karte hain, aur Restricted three-body problem full nonlinear regime ke liye jab $r$ ab $d$ se $≪$ nahi raha.

L5.2 — Tidal limit ko exact form ke against sanity check karo

Problem. $\hat{d} = (1, 0, 0)$ , $r = (r, 0, 0)$ ke liye, algebraically dikhao ki exact $a_{3 b}$ limit $r / d \to 0$ mein $+ 2 μ_{3} r / d^{3}$ reduce hoti hai, tidal coefficient confirm karte hue. Exact expression do aur $r / d = 0.11$ par uski first-order value do.

Recall Solution

Line ke along, $∣ d - r ∣ = d - r$ (dono $+ x$ ke along), toh exact $x$ -component hai $a_{x} = μ_{3} (\frac{d - r}{( d - r ) ^{3}} - \frac{d}{d ^{3}}) = μ_{3} (\frac{1}{( d - r ) ^{2}} - \frac{1}{d ^{2}}) .$ Expand karo $\frac{1}{( d - r ) ^{2}} = \frac{1}{d ^{2}} (1 - r / d)^{- 2} \approx \frac{1}{d ^{2}} (1 + 2 r / d)$ . Tab $a_{x} \approx μ_{3} (\frac{1}{d ^{2}} + \frac{2 r}{d ^{3}} - \frac{1}{d ^{2}}) = \frac{2 μ _{3} r}{d ^{3}},$ bilkul tidal stretch coefficient. $r / d = 0.11$ par exact bracket ( $μ_{3} / d^{2}$ ke units mein) hai $(1 - 0.11)^{- 2} - 1 = 0.263$ jabki linear estimate $2 (0.11) = 0.22$ hai — relative discrepancy $(0.263 - 0.22) /0.263 \approx 16%$ (linear estimate ke against measure karne par roughly $20%$ ). Yeh confirm karta hai ki first-order approximation self-consistent hai aur uski error $r / d$ ke saath badhti hai, L4.1 mein order-of-magnitude estimate se match karti hai.

Recall check

Recall Quick self-quiz

Perturbation quantity — raw pull ya difference? ::: Difference (tidal/differential acceleration). Tidal strength $d$ ke saath kaise scale karti hai? ::: Inverse cube, $μ_{3} / d^{3}$ . $μ / d^{3}$ ka Sun-vs-Moon ratio? ::: Lagbhag 2.19, Moon jeetata hai. Tidal field mein stretch-to-squeeze ratio? ::: 2 to 1. Roughly kahan third-body Earth orbits ke liye $J_{2}$ se aage nikal jaata hai? ::: $r \approx 6.8 R_{E}$ ke paas ( $\approx 4.3 \times 1 0^{7}$ m, GEO ke close).