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

3.2.24 · D2 · Physics › Orbital Mechanics & Astrodynamics › Gravity assist (slingshot) — patched conic, v-infinity vecto

Step 1 — Ek arrow ek speed aur ek direction hai

KYA HAI. Kisi bhi physics se pehle, hum ek tool pe agree karte hain: ek velocity vector. Ise ek arrow ki tarah draw karo. Iski length ka matlab hai "kitna fast" (jaise, km/s mein), aur yeh kis taraf point karta hai ka matlab hai "kaunsi disha mein ja raha hai."

YEH tool kyun, sirf ek number kyun nahi? Ek plain number (ek "speed") sirf bada ya chhota ho sakta hai. Lekin slingshot ki puri trick yeh hai ki direction change karo jabki length same rahe — aur ek plain number direction describe nahi kar sakta. Isliye humein arrows chahiye. Hum poora page sirf do arrows manipulate karte hue spend karenge.

PICTURE. Figure dekho. Blue arrow spacecraft ki velocity hai. Iski foot (tail) wahan hai jahan craft hai; iski head (tip) wahan point karti hai jahan yeh aage jaayega; iski physical length speed hai. Do arrows "equal" tab hote hain jab dono length aur direction dono mein match karein.

Step 2 — Do frames = do log dekh rahe hain

KYA HAI. Hum reference frames introduce karte hain: ek frame simply yeh hai — "kaun dekh raha hai, aur kya woh move kar raha hai?" Hum do watchers use karte hain. Ek planet ke saath chal raha hai (planet frame). Ek Sun ke relative still khada hai (Sun frame, ya heliocentric frame).

Do frames kyun? Kyunki wahi flyby un dono watchers ko bilkul alag dikhti hai, aur poora paradox ("free speed") unke do views ke gap mein rehta hai. Planet-rider ko kuch surprising nahi lagta. Sun-watcher ko craft ki speed badhti hui lagti hai. Dono sahi hain.

PICTURE. Planet apni khud ki velocity ke saath space mein move karta hai — ise ${\vec{V}}_p$ kaho (green arrow, Planet ki velocity Sun frame mein). Alag se, planet-rider craft ko dekhta hai aur craft ke liye apna arrow measure karta hai; abhi us arrow ko $\vec{u}$ kaho (blue). Planet-rider jo dekhta hai use Sun-watcher ke view mein convert karne ke liye, hum green arrow wapas add karte hain.

Subscript $sc/\text{Sun}$ padhte hain "spacecraft, Sun ke relative." Har "$a/b$" padhte hain "$a$ jaisa $b$ dekhta hai." Step 3 mein hum is planet-frame arrow $\vec{u}$ ko uska proper naam denge.

Step 3 — Planet ke frame mein, sirf direction badal sakti hai

KYA HAI. Show ke asli star se milo — Step 2 ka planet-frame arrow $\vec{u}$, ab "door se" measure kiya gaya. Us region ki edge par jahan planet ki gravity ab bhi matter karti hai, sphere of influence par, is arrow ko uska official naam milta hai: ${\vec{v}}_{\infty }$, craft ki velocity jaisi planet-rider infinity par dekhta hai. Iski length $v_{\infty }$ hyperbolic excess speed hai. (Toh $\vec{u}$ aur ${\vec{v}}_{\infty }$ same arrow hain; ab se hum sirf ${\vec{v}}_{\infty }$ use karenge.)

Length kyun fixed rehti hai? Gravity conservative hai: planet ki taraf swing karo aur wapas bahar aao aur tum exactly usi speed ke saath nikalte ho jis speed se aaye the (planet ke relative). Chaliye ise two-body problem ke ek energy statement se prove karte hain:

$\epsilon =\underset{\begin{array}{c}\text{kinetic energy}\\ \text{per unit mass}\end{array}}{\underbrace{\frac{v^2}{2}}}-\underset{\begin{array}{c}\text{gravity’s pull, }{\mu }_p=G{M}_p,\\ r=\text{distance to planet}\end{array}}{\underbrace{\frac{{\mu }_p}{r}}}$

Yahan $\epsilon$ specific orbital energy hai (energy per kilogram) aur yeh poore flyby mein constant rehti hai. Door se, $r\to \infty$, toh doosra term khatam ho jaata hai (${\mu }_p/\infty =0$), chhod jaata hai

$\epsilon =\frac{v_{\infty }^2}{2}⟹v_{\infty }=\sqrt{2\epsilon }.$

Kyunki $\epsilon$ kabhi nahi badlta, incoming aur outgoing lengths barabar hain: $v_{\infty }^{\text{in}}=v_{\infty }^{\text{out}}$.

PICTURE. Planet frame mein incoming blue arrow aur outgoing blue arrow ki same length hai — dono radius $v_{\infty }$ ke ek dashed circle ko sirf touch karte hain. Flyby sirf itna kar sakta hai: arrow ko us circle ke around swing karo.

Step 4 — Path bend karta hai: kitna? (turn angle)

KYA HAI. Craft seedhi line mein nahi jaata — planet ki gravity ise bend karti hai. Incoming blue arrow aur outgoing blue arrow ke beech ka angle turn angle $\delta$ hai (Greek "delta," yahan ek swing angle).

Hyperbola kyun, aur woh $\delta$ kaise set karta hai? Kyunki $\epsilon >0$ (craft ke paas infinity par leftover speed hai), orbit unbound hai — yeh ek hyperbola hai, ek open curve jiske do straight-line asymptotes hain. Incoming aur outgoing ${\vec{v}}_{\infty }$ exactly un dono asymptotes ke saath point karte hain. Toh "velocity kitni sharply turn karta hai" wahi hai "hyperbola kitna wide hai" — ise eccentricity $e$ se capture karte hain (curve kitna open hai; $e=1$ ek parabola hai, $e\to \infty$ ek straight line hai).

Shape energy aur angular momentum se aati hai; closest approach par evaluate karne par yeh collapse hota hai

$e=1+\underset{\begin{array}{c}{r}_p=\text{closest-approach distance}\\ v_{\infty }=\text{excess speed},{\mu }_p=G{M}_p\end{array}}{\underbrace{\frac{r_p{v}_{\infty }^2}{{\mu }_p}}}$

aur ek hyperbola ke asymptotes ki geometry ek clean link deti hai

$\sin \left(\frac{\delta }{2}\right)=\frac{1}{e}.$

Sine kyun, aur $\delta$ ka half kyun? Planet se us corner tak ek line drop karo jahan do asymptotes cross karti hain. Woh line turn ko do equal halves $\delta /2$ mein split karti hai. Wahan bane right triangle mein, half-angle ke "opposite" side divided by "hypotenuse" $1/e$ hai — woh ratio half-angle ka sine hi hai. Toh $\sin (\delta /2)=1/e$ jawab deta hai "kis half-angle ki yeh openness hai?"

PICTURE. Hyperbola apne do asymptotes ke saath; blue in-arrow aur out-arrow unke saath lie karte hain; turn angle $\delta$ unke beech mark hai, aur $r_p$ woh closest distance hai jahan curve planet ke paas jaata hai.

Step 5 — Planet ka arrow wapas add karo: speed badal jaati hai

KYA HAI. Ab hum Step 2 ke rule se Sun frame mein wapas cross karte hain. Hum green planet arrow ${\vec{V}}_p$ ko blue ${\vec{v}}_{\infty }$ mein add karte hain — turn se pehle aur baad mein.

Ab length kyun badlti hai jabki $v_{\infty }$ nahi badla? Kyunki humne blue arrow rotate kiya (Step 4) aur phir ek fixed green arrow add kiya. Ek triangle ki ek side rotate karne se teesri side ki length badal jaati hai. Teesri side Sun-frame velocity hai — wahi woh hai jis ki hum care karte hain.

PICTURE. Do triangles ek hi green base ${\vec{V}}_p$ share karte hain.

• Incoming: blue ${\vec{v}}_{\infty }^{\text{in}}$ green ${\vec{V}}_p$ ke khilaaf point karta hai, toh tip-to-tail sum (red Sun-frame arrow) chhota hai — craft slow hai.
• Outgoing: same-length blue arrow, ab $\delta$ se rotate hua, green ${\vec{V}}_p$ ke saath zyada align hai, toh red sum lamba hai — craft fast hai.

Same blue length, different red length. Woh difference hi free boost hai.

Step 6 — Hard ceiling: tum kabhi $2v_{\infty }$ se zyada nahi pa sakte

KYA HAI. Boost kitna bada ho sakta hai? Best case ek full flip hai: blue arrow direction reverse kar leta hai ($\delta =180^{\circ }$).

Exactly $2v_{\infty }$ kyun? Blue arrow ka tip radius $v_{\infty }$ ke ek circle par stuck hai (Step 3). Us circle par do points ke beech ka sabse bada possible difference uska diameter hai — ek taraf se seedha doosri taraf — jo $2v_{\infty }$ hai. Koi bhi flyby, koi bhi planet, chahe kitna bhi massive ho, tip ko circle ke across se zyada door nahi le ja sakta.

PICTURE. $v_{\infty }$-circle. Incoming blue left point karta hai, outgoing blue $180^{\circ }$ flip ke baad right point karta hai; change $\Delta \vec{v}$ (red) poora diameter $=2v_{\infty }$ span karta hai.

Step 7 — Tum kis side se pass karte ho yeh gain ya loss decide karta hai

KYA HAI. Turn ka size $e$ se fix hai, lekin tum iski sense choose karte ho yeh choose karke ki tum planet ke kis side se fly karte ho.

Side kyun matter karta hai? Planet ke peeche se pass karna (uski wake se) blue arrow ko $+{\vec{V}}_p$ ki taraf rotate karta hai — outgoing red arrow lamba hota hai → speed gain. Aage se pass karna ise $-{\vec{V}}_p$ ki taraf rotate karta hai → speed loss (deliberately use kiya jaata hai slow down karne ke liye, jaise MESSENGER ka Mercury mein brake karna).

PICTURE. Same planet, do trajectories. Behind-pass: red Sun-frame arrow bada hota hai. Front-pass: red arrow chhota hota hai. Identical physics, opposite outcome — tumhari geometry ki choice.

Worked numbers, visually re-check kiye gaye

Ek picture summary

Is poore page ki sab cheez ek image hai: ek fixed-length blue arrow jiska tip ek circle par ride karta hai, plus ek fixed green planet arrow jo dono ends par add hota hai. Turn angle $\delta$ (jo is baat se set hota hai ki tum kitna close aur slow fly karte ho) blue arrow ko swing karta hai; green add karna blue ki rotation ko red length change mein convert karta hai; red change kabhi bhi circle ke diameter $2v_{\infty }$ se exceed nahi kar sakta.