Gravity assist (slingshot) — patched conic, v-infinity vectors: Worked examples

3.2.24 · D3 · Physics › Orbital Mechanics & Astrodynamics › Gravity assist (slingshot) — patched conic, v-infinity vecto

Yeh page slingshot topic ki drill-ground hai. Parent note ne teen master equations build ki thi; yahan hum unpe har tarah ki situation throw karte hain — har sign, har geometry, degenerate edges, ek real mission problem, aur ek exam twist. End tak tumhe koi bhi flyby scenario aisa nahi milna chahiye jo tumne pehle na dekha ho.

Shuru karne se pehle, toolkit ko plain words mein re-state karte hain taaki koi bhi symbol undefined na rahe.

Definition Paanch symbols jo hum use karenge, words mein

$V_{p}$ — planet ki velocity Sun se dekhi hui. Length $V_{p}$ . Ek mota arrow jo planet apne solar orbit mein kheenchta hai.
$v_{\infty}$ — spacecraft ki velocity planet se dekhi hui, Sphere of influence (SOI) ke edge par door se measure ki gayi jahan planet ki pakad khatam hoti hai. Iska length $v_{\infty}$ ("vee-infinity") flyby ke dauran kabhi nahi badlta.
$v_{sc}$ — spacecraft ki velocity Sun se dekhi hui (heliocentric). Yeh addition rule $v_{sc} = V_{p} + v_{\infty}$ follow karta hai (yeh sirf Galilean velocity addition hai). Hum $v_{sc}^{in}$ likhte hain flyby se theek pehle ki magnitude ke liye aur $v_{sc}^{o u t}$ flyby ke theek baad ki magnitude ke liye — superscripts "in"/"out" ka matlab hamesha pre-flyby aur post-flyby hota hai.
$e$ — planet-centred hyperbola ki eccentricity: flyby path kitna "open" hai. Genuine flyby ke liye $e > 1$ (value $e = 1$ ek parabolic limit hai, actually kabhi nahi pahuncha jaata).
$δ$ — turn angle: planet ki gravity $v_{\infty}$ ko aane aur jaane ke beech kitna ghuma deti hai.

Master trio: $v_{\infty} = const, e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}, sin \frac{δ}{2} = \frac{1}{e} .$ Yahan $r_{p}$ closest-approach distance hai aur $μ_{p}$ planet ki gravity strength hai ( $= G M_{p}$ ).

Scenario matrix

Har flyby problem in cells mein se ek (ya mix) hoti hai. Neeche ke worked examples un cells ke saath tagged hain jo woh hit karte hain, aur saath milke yeh sab ko cover karte hain.

#	Cell class	Kya cheez ise alag banati hai	Covered by
A	Speed-up geometry ( $v_{\infty}$ rotate hoti hai $+ V_{p}$ ki taraf)	planet ke peechhe se guzarna	Ex 1
B	Speed-down geometry ( $v_{\infty}$ rotate hoti hai $- V_{p}$ ki taraf)	aage se guzarna	Ex 2
C	General oblique angle (na aligned na anti-aligned)	full vector addition, law of cosines	Ex 3 (figure)
D	Limiting max case $δ \to 18 0^{\circ}$	perfect reversal, gain $\to 2 v_{\infty}$	Ex 4
E	Degenerate: grazing / tiny $r_{p}$ vs huge $r_{p}$	$e \to 1$ vs $e \to \infty$	Ex 5
F	Zero input edge $v_{\infty} = 0$ (co-moving)	koi boost possible nahi	Ex 6
G	Real mission word problem	numbers pick karo, side choose karo	Ex 7 (Voyager-style)
H	Exam twist: solve backwards	$Δ v$ diya hai, $r_{p}$ nikalo	Ex 8

Ex 1 — Cell A: peechhe se guzarna, speed gain karna

Forecast: pehle guess karo ki after-speed $10$ se badi hogi ya nahi, aur kabhi kabhi kitni, padne se pehle.

Flyby se pehle. $v_{\infty}$ anti-parallel hai $V_{p}$ se, isliye heliocentric speed seedha subtraction hai: $v_{sc}^{in} = ∣ V_{p} - v_{\infty} ∣ = 10 - 4 = 6 km/s$ . Yeh step kyun? Jab do vectors opposite direction mein point karte hain, unke sum ki length unki lengths ka difference hoti hai.
$9 0^{\circ}$ turn ke baad. Ab $v_{\infty}$ perpendicular hai $V_{p}$ se. Do arrows right angle banate hain, isliye Pythagoras use karo: $v_{sc}^{o u t} = V_{p}^{2} + v_{\infty}^{2} = 100 + 16 = 116 = 10.77 km/s$ . Yeh step kyun? $v_{sc} = V_{p} + v_{\infty}$ ; right angle ke liye sum ki length hypotenuse hoti hai.
Gain. $Δ v = 10.77 - 6 = 4.77 km/s$ .

Verify: Behind pass ke baad craft ko naive $6$ se tez hona chahiye — hai ( $10.77$ ), isliye geometry ( $+ V_{p}$ ki taraf rotate hona) Cell A se match karti hai. Units: sab km/s mein. ✓

Ex 2 — Cell B: aage se guzarna, speed lose karna

Forecast: kya craft pehle se tez ya dheema end hoga?

Pehle. $v_{\infty}$ parallel hai $V_{p}$ se: seedha addition, $v_{sc}^{in} = V_{p} + v_{\infty} = 10 + 4 = 14 km/s$ . Yeh step kyun? Same-direction vectors length mein add hote hain.
$9 0^{\circ}$ ke baad. $v_{\infty}$ ab perpendicular hai: $v_{sc}^{o u t} = V_{p}^{2} + v_{\infty}^{2} = 116 = 10.77 km/s$ . Yeh step kyun? Right angle ke saath vector sum ki magnitude hypotenuse hai — same formula Ex 1 ki tarah, lekin yahan yeh 14 se drop represent karta hai.
Change. $Δ v = 10.77 - 14 = - 3.23 km/s$ — ek loss.

Verify: Front-side pass ko craft slow karna chahiye (MESSENGER ne Mercury pahunchne ke liye use kiya). Humein negative $Δ v$ mila. ✓ Note: same $v_{\infty}$ , same $δ$ as Ex 1, lekin result ka opposite sign kyunki starting orientation aur side alag hai — yahi exactly kyun "which side" (Cell A vs B) poora game hai.

Ex 3 — Cell C: general oblique angle (figure)

Forecast: itni oblique geometry ke saath, kya gain upar ke tidy $9 0^{\circ}$ cases se bada ya chota hoga?

Figure — Gravity assist (slingshot) — patched conic, v-infinity vectors

Figure mein dono addition triangles same black $V_{p}$ base share karte hain. Red dashed arrow outgoing heliocentric velocity $v_{sc}^{o u t}$ hai (key object); black dashed arrow inbound $v_{sc}^{in}$ hai. Notice karo ki red arrow $V_{p}$ ki taraf black se zyada lean kar raha hai — yahi "gained speed" ka visual meaning hai.

Hum triangle $v_{sc} = V_{p} + v_{\infty}$ par law of cosines use karte hain. Agar do arrows ke beech angle $ϕ$ hai, toh $v_{sc}^{2} = V_{p}^{2} + v_{\infty}^{2} + 2 V_{p} v_{\infty} cos ϕ .$ Plus sign kyun? Kyunki hum vectors ko tip-to-tail add kar rahe hain; interior triangle angle $18 0^{\circ} - ϕ$ hai, aur $cos (18 0^{\circ} - ϕ) = - cos ϕ$ , jo usual minus ko plus mein flip kar deta hai.

Inbound, $ϕ_{in} = 15 0^{\circ}$ , $cos 15 0^{\circ} = - 0.8660$ : $v_{sc}^{in} = 169 + 36 + 2 (13) (6) (- 0.8660) = 205 - 135.1 = 69.9 = 8.36 km/s .$ Yeh step kyun? Obtuse angle ka matlab hai $v_{\infty}$ largely $V_{p}$ ka oppose karta hai, isliye hum low heliocentric speed expect karte hain.
Outbound, $ϕ_{o u t} = 8 0^{\circ}$ , $cos 8 0^{\circ} = 0.1736$ : $v_{sc}^{o u t} = 169 + 36 + 2 (13) (6) (0.1736) = 205 + 27.1 = 232.1 = 15.23 km/s .$ Yeh step kyun? Near-right angle ka matlab hai $v_{\infty}$ ab $V_{p}$ ke saath hai, isliye heliocentric speed jump karti hai.
Gain. $Δ v = 15.23 - 8.36 = 6.87 km/s$ .

Verify: Gain ko cap $2 v_{\infty} = 12 km/s$ (dekho Ex 4) respect karni chahiye — aur $6.87 < 12$ . ✓ Figure mein red outgoing vector $V_{p}$ ki taraf black incoming se zyada lean karta hai, speed increase se match karta hai.

Ex 4 — Cell D: limiting maximum, $δ \to 18 0^{\circ}$

Forecast: kya ek flyby kabhi $2 v_{\infty}$ se zyada de sakti hai? Haan ya nahi commit karo.

Geometry. Sirf $v_{\infty}$ ki direction badal sakti hai; uski tip $v_{\infty}$ radius ke circle par hoti hai. Us circle par do points ke beech sabse bada difference uska diameter, $2 v_{\infty}$ hai. Yeh step kyun? $Δ v = v_{\infty}^{o u t} - v_{\infty}^{in}$ ; circle ki sabse lambi chord uska diameter hoti hai.
Best case realize hota hai $δ \to 18 0^{\circ}$ par: $v_{\infty}$ anti-parallel se parallel $V_{p}$ ke saath flip hoti hai. Before $= V_{p} - v_{\infty} = 13 - 6 = 7$ ; after $= V_{p} + v_{\infty} = 13 + 6 = 19 km/s$ . Yeh step kyun? Full reversal along- $V_{p}$ component swing ko maximize karti hai.
Change. $Δ v = 19 - 7 = 12 = 2 v_{\infty}$ . Exactly the cap.

Verify: $2 v_{\infty} = 2 (6) = 12 = Δ v$ . ✓ Full $18 0^{\circ}$ turn ke liye $sin (δ /2) = sin 9 0^{\circ} = 1 = 1/ e \Rightarrow e = 1$ chahiye, parabolic limit — koi achievable hyperbola nahi. $e = 1 + r_{p} v_{\infty}^{2} / μ_{p}$ se, $e \to 1$ reach karne ka matlab $r_{p} \to 0$ ; lekin physically $r_{p}$ kabhi planet ki surface/atmosphere se neeche nahi ja sakta ( $r_{p} \geq R_{m i n}$ ). Isliye actual best turn woh smallest safe $r_{p}$ set karta hai, aur $2 v_{\infty}$ ek unattainable upper bound hai, near pahuncha ja sakta hai lekin kabhi meet nahi hota. ✓

Ex 5 — Cell E: degenerate radii (grazing vs distant)

Forecast: kaunsa pass trajectory ko zyada bend karta hai — close ya far?

(a) grazing. $e = 1 + \frac{( 71 , 500 ) ( 36 )}{1.267 \times 1 0 ^{8}} = 1 + 0.02032 = 1.02032$ . Phir $sin (δ /2) = 1/1.02032 = 0.9801 \Rightarrow δ /2 = 78.5 5^{\circ} \Rightarrow δ = 157. 1^{\circ} .$ Yeh step kyun? Chota $r_{p}$ → $e$ near $1$ → $sin (δ /2)$ near $1$ → bada turn.
(b) distant. $e = 1 + \frac{( 5 \times 1 0 ^{6} ) ( 36 )}{1.267 \times 1 0 ^{8}} = 1 + 1.4207 = 2.4207$ . Phir $sin (δ /2) = 1/2.4207 = 0.4131 \Rightarrow δ /2 = 24.4 0^{\circ} \Rightarrow δ = 48. 8^{\circ} .$ Yeh step kyun? Bada $r_{p}$ → $e$ bada → chota turn.

Verify: Close pass bahut zyada turn karta hai ( $15 7^{\circ}$ vs $4 9^{\circ}$ ), rule "fly close to bend a lot" se match karta hai. Jab $r_{p} \to \infty$ , $e \to \infty$ aur $δ \to 0$ (koi bend nahi, straight line) — yeh sahi degenerate limit hai. ✓

Ex 6 — Cell F: zero input, $v_{\infty} = 0$

Forecast: kya ek bahut close, bahut massive planet yahan bhi help karti hai?

Eccentricity aur turn. $e = 1 + \frac{r _{p} \cdot 0}{μ _{p}} = 1$ , isliye $sin (δ /2) = 1/ e = 1 \Rightarrow δ /2 = 9 0^{\circ} \Rightarrow δ = 18 0^{\circ}$ . Yeh step kyun? Formally turn full reversal hai — lekin yeh ek zero-length vector ki reversal hai, isliye kuch move nahi hota. Equations real hyperbola ki jagah degenerate parabolic limit indicate karti hain.
Boost cap. Maximum change $= 2 v_{\infty} = 2 (0) = 0 km/s$ . Yeh step kyun? $v_{\infty}$ ki tip zero-radius circle ke center par hai — koi chord exist nahi karti, isliye $δ = 18 0^{\circ}$ turning nothing yields nothing.

Verify: Planet ke saath drift kar raha craft ke paas rotate karne ke liye kuch nahi hai, isliye chahe planet kitna bhi massive ho, $Δ v = 0$ . ✓ Yahi reason hai kyun tum kisi planet ke saath saath drift karte hue usse slingshot nahi kar sakte — tumhe genuine relative speed chahiye. $μ_{p}$ aur $r_{p}$ se independent, correctly.

Ex 7 — Cell G: real mission word problem (Voyager-style)

Forecast: kya ek Jupiter pass Solar-System escape reach karne ke liye enough hoga (heliocentric $≳ 18.5$ km/s Jupiter ki distance par)?

Eccentricity. $e = 1 + \frac{( 2 \times 1 0 ^{5} ) ( 100 )}{1.267 \times 1 0 ^{8}} = 1 + 0.15785 = 1.15785.$ Yeh step kyun? Mission ke $r_{p}$ aur $v_{\infty}$ ko eccentricity relation mein feed karo.
Turn angle. $sin (δ /2) = 1/1.15785 = 0.8636 \Rightarrow δ /2 = 59.7 2^{\circ} \Rightarrow δ = 119. 4^{\circ} .$ Yeh step kyun? Yeh real (idealized $18 0^{\circ}$ nahi) rotation hai jo geometry allow karti hai.
Heliocentric speed. Inbound $v_{\infty}$ anti-parallel hai $V_{p}$ se (angle $18 0^{\circ}$ ); $δ = 119. 4^{\circ}$ $V_{p}$ ki taraf turn karne ke baad, outbound angle $ϕ = 18 0^{\circ} - 119. 4^{\circ} = 60. 6^{\circ}$ hai. Law of cosines: $v_{sc}^{o u t} = 1 3^{2} + 1 0^{2} + 2 (13) (10) cos 60. 6^{\circ} = 169 + 100 + 260 (0.4909) = 396.6 = 19.91 km/s .$ Yeh step kyun? Rotated $v_{\infty}$ ko addition triangle ke zariye Sun frame mein wapas convert karo.

Verify: Inbound heliocentric $13 - 10 = 3 km/s$ thi; outbound $19.91 km/s$ — gain of $16.9$ , cap $2 v_{\infty} = 20$ se neeche. ✓ Aur $19.91 > 18.5$ , isliye haan, ek single Jupiter assist ek probe ko Solar-System escape tak fling kar sakta hai — Voyagers ke liye historically true. ✓

Ex 8 — Cell H: exam twist, solve backwards

Forecast: required $r_{p}$ tight (planet ke paas) hoga ya loose?

$δ$ se $e$ . $sin (δ /2) = sin 3 0^{\circ} = 0.5 = 1/ e \Rightarrow e = 2.$ Yeh step kyun? $sin (δ /2) = 1/ e$ ko invert karo; hum master trio ko reverse mein work karte hain.
$e$ se $r_{p}$ . Rearrange $e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}$ : $r_{p} = \frac{( e - 1 ) μ _{p}}{v _{\infty}^{2}} = \frac{( 1 ) ( 3.79 \times 1 0 ^{7} )}{5. 5 ^{2}} = \frac{3.79 \times 1 0 ^{7}}{30.25} = 1.253 \times 1 0^{6} km .$ Yeh step kyun? Ek unknown ke liye algebraically solve karo.

Verify: Plug back karo: $e = 1 + \frac{( 1.253 \times 1 0 ^{6} ) ( 30.25 )}{3.79 \times 1 0 ^{7}} = 1 + 1.000 = 2.00$ , giving $sin (δ /2) = 0.5$ , $δ = 6 0^{\circ}$ . ✓ Round-trip consistent. Woh $r_{p} \approx 1.253$ million km lagbhag $21$ Saturn radii hai — ek loose pass, sahi hai ek modest $6 0^{\circ}$ bend ke liye. ✓

Recall Kaunsa cell heliocentric change ka

sign decide karta hai, aur kaunsa uski size? Sign: tum kis side se fly karte ho (behind → gain, Cell A; front → loss, Cell B) aur tumhari inbound orientation. Size: turn angle $δ$ ( $r_{p}$ , $v_{\infty}$ , $μ_{p}$ se set hota hai), $2 v_{\infty}$ par capped (Cell D).

Recall Kyun koi

$v_{\infty} = 0$ flyby kabhi help nahi kar sakti, chahe planet kitni bhi massive ho? Heliocentric change $2 v_{\infty} = 0$ se bounded hai; koi relative speed nahi hai toh rotate karne ke liye koi vector nahi hai. Mass aur $r_{p}$ irrelevant hain.