Third-body perturbations: Worked examples — Hinglish deep dive

3.2.36 · D3 · Physics › Orbital Mechanics & Astrodynamics › Third-body perturbations

Yeh page ek drill floor hai. Parent note ne physics build ki thi; yahan hum har woh case hit karte hain jo yeh topic tumhare saamne rakh sakta hai — satellite ki har direction third body ke relative, do limiting geometries (in-line vs sideways), degenerate cases ( $r \to 0$ , $r$ comparable to $d$ ), ek real stationkeeping word problem, aur ek exam twist jo tumhe mass ke saath trick karne ki koshish karta hai.

Shuru karne se pehle, parent se do workhorses ka ek reminder, taaki neeche har symbol already samjha hua ho:

Scenario matrix

Is topic ka har problem in cells mein se ek (ya mix) hota hai. Neeche har worked example us cell ke saath tagged hai jo woh cover karta hai.

Cell	Kya vary karta hai	Kaun sa sawaal answer karta hai	Example
A. Along-line, near side	$r ∥ \hat{d}$	Max stretch, accel ka sign	Ex 1
B. Along-line, far side	$r ∥ - \hat{d}$	Kya far side bhi stretch hota hai?	Ex 2
C. Perpendicular	$r ⊥ \hat{d}$	Squeeze case, sign flip	Ex 3
D. General angle	$\hat{d} \cdot r$ arbitrary	Dono components ek saath	Ex 4
E. Sun vs Moon	mass vs distance	Kaun sa body dominate karta hai, cube kyun	Ex 5
F. Degenerate $r \to 0$	satellite at Earth centre	Limiting behaviour, kya yeh vanish hota hai?	Ex 6
G. Exact vs tidal	$r$ not $≪ d$	Approximation kab break hoti hai?	Ex 7
H. Real-world word problem	GEO stationkeeping	Accel ko $Δ v$ budget mein convert karo	Ex 8
I. Exam twist	"zyada massive ⇒ stronger?"	$μ$ vs $μ / d^{3}$ par trap	Ex 9

Geometry cells (A, B, C, D) sab ek picture par rehte hain. Examples padhne se pehle ise dekho:

Red arrow $\hat{d}$ hai (third body ki taraf). Teen coloured satellites $\hat{d}$ ke along, $\hat{d}$ ke opposite, aur uske perpendicular baithte hain. Dekho ki har chhoti tidal arrow kis taraf point karti hai — wahi cells A, B, C ka answer hai.

Worked examples

Poore note mein hum standard numbers use karte hain:

Moon: $μ_{M} = 4.90 \times 1 0^{12} m^{3} / s^{2}$ , $d_{M} = 3.84 \times 1 0^{8}$ m.
Sun: $μ_{S} = 1.327 \times 1 0^{20} m^{3} / s^{2}$ , $d_{S} = 1.496 \times 1 0^{11}$ m.
GEO radius $r = 4.22 \times 1 0^{7}$ m.

Convenience ke liye tidal coefficient $k \equiv μ_{M} / d_{M}^{3}$ define karo. Ise ek baar compute karte hain aur reuse karte hain.

Worked example Ex 1 — Cell A: satellite Earth–Moon line PAR, near side

Ek GEO satellite seedha Earth aur Moon ke beech baitha hai, to $r = r \hat{d}$ with $r = 4.22 \times 1 0^{7}$ m. Uska tidal acceleration nikalo.

Forecast: direction guess karo — Moon ki taraf, door, ya sideways? Aur kya magnitude $k r$ hai ya $2 k r$ ?

$\hat{d} \cdot r$ compute karo. Kyunki $r = r \hat{d}$ aur $\hat{d} \cdot \hat{d} = 1$ , hume $\hat{d} \cdot r = r$ milta hai. Yeh step kyun? Tidal formula ko $r$ ka $\hat{d}$ par projection chahiye; yahan woh poori length hai.
Tidal form mein plug in karo. $a_{3 b} = k (3 (\hat{d} \cdot r) \hat{d} - r) = k (3 r \hat{d} - r \hat{d}) = 2 k r \hat{d} .$ Yeh step kyun? Dono terms $\hat{d}$ par line up hote hain; $3 - 1 = 2$ — stretch coefficient.
Number. $2 k r = 2 (8.65 \times 1 0^{- 14}) (4.22 \times 1 0^{7}) = 7.3 \times 1 0^{- 6} m/s^{2}$ , Moon ki taraf ( $+ \hat{d}$ ) point karta hai.

Verify: direction $+ \hat{d}$ = Moon ki taraf, figure mein near-side arrow se match karta hai (stretch near satellite ko Earth se aur door Moon ki taraf kheenchta hai). Magnitude parent Ex 2 se match karti hai. ✓

Worked example Ex 2 — Cell B: satellite Earth ke FAR side par

Wahi GEO satellite lekin opposite side par: $r = - r \hat{d}$ . Kya yeh Moon ki taraf push hota hai (andar girnа) ya door?

Forecast: ek naive "Moon sab kuch apni taraf kheenchta hai" Moon ki taraf kehta hai. Kya yeh sahi hai?

Projection. $\hat{d} \cdot r = \hat{d} \cdot (- r \hat{d}) = - r$ . Yeh step kyun? Ab satellite $- \hat{d}$ side par hai, isliye projection negative hai — sign matter karta hai.
Plug in. $a_{3 b} = k (3 (- r) \hat{d} - (- r \hat{d})) = k (- 3 r + r) \hat{d} = - 2 k r \hat{d} .$ Yeh step kyun? Dono signs mil kar $- 2 k r \hat{d}$ dete hain — $- \hat{d}$ direction mein point karta hai, yaani Moon se door.
Number. Magnitude phir $7.3 \times 1 0^{- 6} m/s^{2}$ , lekin direction hai Moon se door (far side par bahar ki taraf).

Verify: isliye tides ke do bulges hote hain. Near side: Moon ki taraf khicha jaata hai. Far side: Moon se door dhakela jaata hai. Dono bulges Earth ke centre se bahar point karte hain — line ke along net stretch. Figure mein dono along-line arrows dekho: woh opposite absolute directions mein point karte hain lekin dono Earth se door. ✓

Worked example Ex 3 — Cell C: satellite Moon line ke PERPENDICULAR

Ab $r$ sideways point karta hai, $\hat{d}$ ke perpendicular, phir bhi $∣ r ∣ = r = 4.22 \times 1 0^{7}$ m. Tidal acceleration nikalo.

Forecast: stretch ya squeeze? Earth ki taraf ya door?

Projection. $\hat{d} \cdot r = 0$ kyunki perpendicular vectors ka dot product zero hota hai. Yeh step kyun? Perpendicular case ka yahi toh poora point hai — $3 (\hat{d} \cdot r) \hat{d}$ stretch term switch off ho jaata hai.
Plug in. $a_{3 b} = k (3 (0) \hat{d} - r) = - k r .$ Yeh step kyun? Sirf $- r$ bachta hai — yeh Earth ke centre ki taraf wapas point karta hai, ek squeeze hai.
Number. Magnitude $k r = (8.65 \times 1 0^{- 14}) (4.22 \times 1 0^{7}) = 3.65 \times 1 0^{- 6} m/s^{2}$ , andar ki taraf directed.

Verify: squeeze magnitude exactly stretch magnitude ki aadhi hai ( $k r$ vs $2 k r$ ) aur sense mein opposite (inward vs outward). Yeh classic tidal signature hai: stretch $+ 2$ , squeeze $- 1$ , teen axes par sum $= 2 - 1 - 1 = 0$ — ek trace-free (volume-preserving) field, jaise kisi bhi pure tidal field mein hona chahiye. ✓

Worked example Ex 4 — Cell D: general 45° angle (dono components ek saath)

Maano $r$ $\hat{d}$ ke saath $4 5^{\circ}$ angle banata hai, magnitude $r = 4.22 \times 1 0^{7}$ m. Ek plane mein $\hat{d} = (1, 0)$ ke saath kaam karo, to $r = r (cos 4 5^{\circ}, sin 4 5^{\circ})$ . $a_{3 b}$ nikalo.

Forecast: answer mein $\hat{d}$ ke along ek component hoga aur ek perpendicular — purely stretch ya squeeze nahi.

Projection. $\hat{d} \cdot r = r cos 4 5^{\circ} = r / 2$ . Yeh step kyun? Intermediate angle par stretch term partially on hai — $cos 4 5^{\circ}$ se scale hua.
Dono terms assemble karo. $r = (r / 2, r / 2)$ aur $\hat{d} = (1, 0)$ ke saath: $a_{3 b} = k (3 \frac{r}{2} (1, 0) - (\frac{r}{2}, \frac{r}{2})) = k (\frac{2 r}{2}, - \frac{r}{2}) .$ Yeh step kyun? Component-by-component subtraction: $\hat{d}$ ke along hume $3 \frac{r}{2} - \frac{r}{2} = \frac{2 r}{2}$ milta hai (stretch), perpendicular mein $- \frac{r}{2}$ (squeeze).
Magnitude. $∣ a_{3 b} ∣ = k r (2)^{2} + (1/ 2)^{2} = k r 2 + 0.5 = k r 2.5 \approx 1.58 k r .$ Number: $1.58 \times (3.65 \times 1 0^{- 6}) \approx 5.8 \times 1 0^{- 6} m/s^{2}$ .

Verify: magnitude pure squeeze ( $1 k r$ ) aur pure stretch ( $2 k r$ ) ke beech hai — sanity confirm hua. Jaise angle $\to 0$ hota hai woh $2 k r$ approach karna chahiye; $4 5^{\circ}$ par yeh $1.58 k r$ hai, comfortably beech mein. ✓

Worked example Ex 5 — Cell E: Sun vs Moon, aur cube kyun jeetta hai

Kisi bhi Earth satellite par tidal strength ke hisaab se Sun aur Moon ko rank karo.

Forecast: Sun Moon se $\sim 27$ million times zyada massive hai. Surely woh jeetta hai?

Recognize karo $r$ cancel hota hai. Fixed satellite ke liye Tidal accel $\propto μ_{3} / d^{3}$ , to ratio $= \frac{μ _{S} / d _{S}^{3}}{μ _{M} / d _{M}^{3}}$ . Yeh step kyun? Hum bodies compare kar rahe hain, to common $r$ drop ho jaata hai — sirf $μ / d^{3}$ factor matter karta hai.
Har factor compute karo. $\frac{μ _{M}}{d _{M}^{3}} = 8.65 \times 1 0^{- 14}, \frac{μ _{S}}{d _{S}^{3}} = \frac{1.327 \times 1 0 ^{20}}{( 1.496 \times 1 0 ^{11} ) ^{3}} = 3.96 \times 1 0^{- 14} .$
Ratio (Moon ÷ Sun). $8.65/3.96 \approx 2.18$ . Yeh step kyun? Moon ka tide Sun ke tide se 2.2× zyada hai, mass mein tiny hone ke bavajood — inverse-cube distance ko brutally punish karta hai ( $d_{S} / d_{M} \approx 390$ , aur $39 0^{3} \approx 5.9 \times 1 0^{7}$ ).

Verify: mass ratio se cross-check karo. $μ_{S} / μ_{M} \approx 2.7 \times 1 0^{7}$ ; $39 0^{3} \approx 5.9 \times 1 0^{7}$ se divide karo → $\approx 0.46$ , yaani Sun/Moon $\approx 0.46$ , to Moon/Sun $\approx 2.2$ . Do independent routes agree karte hain. ✓

Worked example Ex 6 — Cell F: degenerate case

$r \to 0$ Jab satellite Earth ke centre ke paas jaata hai ( $r \to 0$ ) to $a_{3 b}$ ka kya hota hai? Safe rehne ke liye exact form use karo.

Forecast: ek common dar hai "Moon ka direct pull $μ_{3} / d^{2}$ blow up hoga ya finite rahega" — lekin kya perturbation finite hai? Kya yeh vanish hoti hai?

Exact formula mein $r = 0$ set karo. $a_{3 b} = μ_{3} (\frac{d - 0}{∣ d - 0 ∣ ^{3}} - \frac{d}{d ^{3}}) = μ_{3} (\frac{d}{d ^{3}} - \frac{d}{d ^{3}}) = 0 .$ Yeh step kyun? Centre par, satellite aur Earth ek hi point hain; Moon unhe identically pull karta hai, to difference exactly zero hai.
Tidal form se confirm karo. $3 (\hat{d} \cdot 0) \hat{d} - 0 = 0$ . Yeh bhi zero. Yeh step kyun? Dono forms ko limit mein agree karna chahiye — aur karte hain.

Verify: yeh parent ke teesre "common mistake" ko khatam karta hai. Indirect term precisely wahi hai jo perturbation ko $r = 0$ par vanish karata hai. Naive "direct pull $μ_{3} / d^{2}$ " (Earth–Moon distance $d$ use karte hue, kyunki centre par satellite–Moon distance $d$ ke barabar hai) $μ_{3} / d^{2} \neq = 0$ dega — clearly galat hai ek satellite ke liye jo Earth ke centre par baitha hai, jo koi differential tug feel hi nahi karta. ✓

Worked example Ex 7 — Cell G: exact vs tidal, error kitna bada hai?

$r = r \hat{d}$ lo (near-side, along-line) to exact formula scalar hai. Exactly exact aur tidal answers ko do cases ke liye compare karo: GEO ( $r / d_{M} = 0.11$ ) aur ek deep case $r = 0.5 d_{M}$ .

Forecast: tidal form $O ((r / d)^{2})$ drop karta hai. $r / d = 0.11$ par error chhota hona chahiye; $r / d = 0.5$ par bada hona chahiye.

Along-line exact form. $r = r \hat{d}$ ke saath, $d - r = (d - r) \hat{d}$ : $a_{exact} = μ_{3} (\frac{1}{( d - r ) ^{2}} - \frac{1}{d ^{2}}) .$ Yeh step kyun? Dono terms $\hat{d}$ ke along point karte hain, to hum vectors drop kar ke signed scalars use kar sakte hain; note karo near-side satellite Moon ke zyada paas hai, to $(d - r)^{2} < d^{2}$ → positive (Moon ki taraf). ✓
GEO case ( $x \equiv r / d_{M} = 0.110$ ). Exact factor $μ_{M} [(d - r)^{- 2} - d^{- 2}]$ . Numerically exact along-line accel $\approx 8.7 \times 1 0^{- 6}$ m/s² hai; tidal ne $2 k r = 7.3 \times 1 0^{- 6}$ m/s² diya. Error $\approx 16%$ — pehla neglected term $O (x) \approx 0.11$ hai, aur next correction ise roughly one-sixth tak le jaata hai. Yeh step kyun? GEO par "small" parameter $r / d$ $0.11$ hai, isliye tidal form sirf $15%$ – $16%$ tak good hai — scaling arguments ke liye theek, high-precision propagation ke liye nahi.
Deep case $r = 0.5 d_{M}$ . Exact: $μ_{M} [(0.5 d)^{- 2} - d^{- 2}] = μ_{M} / d^{2} \cdot (4 - 1) = 3 μ_{M} / d^{2}$ . Tidal: $2 k r = 2 (μ_{M} / d^{3}) (0.5 d) = μ_{M} / d^{2}$ . Exact tidal estimate se 3× hai — approximation collapse ho gayi. Yeh step kyun? Jab $r$ no longer $≪ d$ hota hai to tumhe exact form use karni chahiye (yeh lunar-transfer trajectories ka regime hai).

Verify: tidal form exact form ka leading term hai — near side par yeh hamesha ek underestimate honi chahiye (kyunki $(d - r)^{- 2}$ super-linearly grow karta hai). Dono cases confirm karte hain exact $>$ tidal. ✓

Worked example Ex 8 — Cell H: real-world GEO stationkeeping

$Δ v$ Luni-solar tidal acceleration dheere dheere ek GEO orbit ke plane ko tilt kar deta hai, jo North–South stationkeeping burns force karta hai. Yearly $Δ v$ estimate karo agar effective out-of-plane tidal accel average $a_{⊥} \approx 3.6 \times 1 0^{- 6} m/s^{2}$ ho (Ex 3 se Moon ke squeeze ka order).

Forecast: guess karo kya yeh grams-per-year hai ya tens of m/s per year.

Ek saal mein seconds. $T = 365.25 \times 24 \times 3600 = 3.156 \times 1 0^{7}$ s. Yeh step kyun? Ek continuous acceleration velocity accumulate karta hai $a \cdot t$ ki tarah; hume $t$ SI mein chahiye.
Accumulated velocity change. $Δ v \sim a_{⊥} T = 3.6 \times 1 0^{- 6} \times 3.156 \times 1 0^{7} \approx 114 m/s per year .$ Yeh step kyun? Tiny accel ko seconds ki huge number se multiply karo — aise $1 0^{- 6}$ operationally huge ban jaata hai.
Interpret karo. Real GEO N–S stationkeeping $\approx 45$ – $55$ m/s per year cost karta hai. Hamara crude estimate (114) sahi order of magnitude hai — true value chhoti hai kyunki accel oscillate karta hai aur sirf uska slow-secular part cancel karna padta hai.

Verify: units: $(m/s^{2}) (s) = m/s$ . ✓ Order of magnitude (tens of m/s/yr) us well-known fact se match karta hai ki N–S keeping GEO fuel budget dominate karta hai — parent note ke claim ko confirm karta hai ki stationkeeping "luni-solar effects se dominated" hai. ✓

Worked example Ex 9 — Cell I: exam trap

Exam statement: "Jupiter $\sim 318$ Earth masses ka hai aur Moon se vastly zyada massive hai. Isliye Jupiter ka tidal effect ek Earth satellite par Moon ke effect se zyada hai. Sach ya jhooth?"

Forecast: "vastly more massive" bait hai. $μ / d^{3}$ pakdo, kabhi sirf $μ$ mat pakdo.

$μ_{J}$ aur $d_{J}$ nikalo. $μ_{J} \approx 1.267 \times 1 0^{17} m^{3} / s^{2}$ ; closest approach $d_{J} \approx 5.9 \times 1 0^{11}$ m. Yeh step kyun? Tidal strength kabhi akele mass use nahi karti — hamesha $μ / d^{3}$ , isliye hume dono numbers chahiye.
Jupiter ka factor compute karo. $\frac{μ _{J}}{d _{J}^{3}} = \frac{1.267 \times 1 0 ^{17}}{( 5.9 \times 1 0 ^{11} ) ^{3}} = \frac{1.267 \times 1 0 ^{17}}{2.05 \times 1 0 ^{35}} \approx 6.2 \times 1 0^{- 19} s^{- 2} .$ Yeh step kyun? Yeh Jupiter ka tidal coefficient hai, directly Moon ke $k = 8.65 \times 1 0^{- 14}$ se comparable.
Moon se ratio. $\frac{6.2 \times 1 0 ^{- 19}}{8.65 \times 1 0 ^{- 14}} \approx 7 \times 1 0^{- 6}$ — Jupiter ka tide Moon ke tide ka ek lakh-wanh hissa hai. Yeh step kyun? Jupiter ka mass advantage ( $\sim 26000 \times$ Moon) $\sim 1500 \times$ door hone se annihilate ho jaata hai: $150 0^{3} \approx 3 \times 1 0^{9}$ .

Answer: Jhooth, aur bahut bade margin se — Jupiter Earth satellites ke liye bilkul negligible hai.

Verify: yeh trap exactly parent ka mistake "Sun dominate karta hai kyunki woh massive hai" hai, escalated. Annihilation directly sanity check karo: $(μ_{J} / μ_{M}) / (d_{J} / d_{M})^{3} = (1.267 \times 1 0^{17} /4.90 \times 1 0^{12}) / (5.9 \times 1 0^{11} /3.84 \times 1 0^{8})^{3} \approx 2.6 \times 1 0^{4} /3.6 \times 1 0^{9} \approx 7 \times 1 0^{- 6}$ — step 3 se match karta hai. Hamesha $μ / d^{3}$ se rank karo. ✓

Recall

Recall Near side vs far side — tidal accel kis taraf point karta hai?

Near side ::: Moon ki taraf (Earth se bahar); far side ::: Moon se door (Earth se bahar bhi). Dono bulges bahar point karte hain — do tidal bulges.

Recall Stretch vs squeeze coefficients

$\hat{d}$ ke along ::: $+ 2 μ_{3} r / d^{3}$ (stretch); perpendicular ::: $- μ_{3} r / d^{3}$ (squeeze); trace zero hai.

Recall

$a_{3 b} = 0$ Earth ke centre par kyun hai? Direct aur indirect terms identical ho jaate hain ::: unka difference vanish ho jaata hai; satellite aur Earth same pull feel karte hain.

Recall Do third bodies ko tidal strength se kaise rank karte ho?

$μ / d^{3}$ compare karo ::: kabhi akele mass se nahi — inverse cube nearness ko dominate karvaata hai.

Yeh bhi dekho: Tidal forces, J2 perturbation and oblateness, GEO stationkeeping, Gauss and Lagrange planetary equations, Restricted three-body problem, Two-body problem.