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

3.2.24 · D1 · Physics › Orbital Mechanics & Astrodynamics › Gravity assist (slingshot) — patched conic, v-infinity vecto

Yeh child page assume karta hai ki aap kuch nahi jaante. Hum har letter ko naam denge, har picture draw karenge, aur tabhi parent page ki equations ka matlab samajh aayega. Upar se neeche padhein — har idea uske upar wale idea pe build hota hai.

0. "Vector" kya hota hai? (woh arrow jo add hota hai)

Kisi bhi physics se pehle, yahan sabse important object hai velocity vector.

Hum vectors ko upar ek chhota sa arrow laga ke likhte hain: $v$ . Bina arrow ka plain letter, $v$ , sirf us arrow ki length (uski size) darshaata hai, jise hum magnitude kehte hain. Toh $v$ arrow hai aur $v = ∣ v ∣$ uski length hai.

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

Figure 1 (upar): left mein, ek yellow arrow $v$ label kiya hua — notice karein ki uski length ka label magnitude $v$ hai. Right mein, dekho blue "pehla" arrow aur pink "doosra" arrow tip-to-tail rakh ke; mota yellow "sum" arrow bilkul shuru se bilkul end tak jaata hai. Woh yellow arrow hi velocities add karne ka poora idea hai.

Topic ko yeh kyun chahiye: poora slingshot ek hi arrow-addition hai, $v_{sc / S u n} = V_{p} + v_{\infty}$ . Agar aap do arrows ko tip-to-tail add nahi kar sakte, toh koi aur equation samajh nahi aayegi. Deeper rules ke liye Reference frames & Galilean velocity addition note dekho.

1. Reference frame — "kaun dekh raha hai?"

Is poore topic mein do frames chalte hain:

Planet-centered frame — aap planet ke saath chalte hain. Yahan planet still baitta hai aur spacecraft uske paas se ghumaav lete hua guzarta hai.
Sun-centered (heliocentric) frame — aap Solar System ke upar float kar rahe hain. Yahan planet apni orbit mein tez daud raha hai aur spacecraft ka path Sun ke around ek bada curve hai.

Figure 2 (upar): left panel planet frame hai — blue planet still baitta hai aur craft uske paas se ek clean yellow hyperbola trace karta hai. Right panel Sun frame hai — ab yellow Sun still baitta hai, aur wahi encounter ek aisi planet (blue dot) par hota hai jo khud apni dashed orbit (blue arrow) mein move kar rahi hai. Dekho ki "kaun still hai" ka picture panels ke beech kaise flip hota hai; wahi flip slingshot kaam karne ki poori wajah hai.

Topic ko yeh kyun chahiye: "free speed boost" tabhi dikhta hai jab hum frames switch karte hain. Planet frame mein kuch tez nahi hota; Sun frame mein hota hai. Dono frames ko confuse karna gravity assist ke baare mein yeh sochne ka #1 tarika hai ki yeh energy conservation violate karta hai.

2. Subscript notation $v_{A / B}$ — "A ki velocity B ke relative"

Toh:

$v_{sc / S u n}$ = spacecraft ki velocity jaise Sun se dikhti hai.
$v_{sc / pl an e t}$ = spacecraft ki velocity jaise planet se dikhti hai (yeh important wali hai — yeh $v_{\infty}$ ban jaati hai).
$V_{p}$ = planet ki velocity jaise Sun se dikhti hai (capital $V$ bas flag karta hai "yeh bada tez planet hai").

Yeh simple arrow-adding se connect hote hain: $v_{sc / S u n} = v_{sc / pl an e t} + V_{p}$

Ise zor se padho: "spacecraft-from-Sun = spacecraft-from-planet + planet-from-Sun." Beech wala term (planet) aapके dimagh mein linking chains ki tarah cancel ho jaata hai. Yeh Galilean velocity addition hai.

3. $v_{\infty}$ — "velocity at infinity"

Magnitude $v_{\infty}$ (bina arrow ke) ek plain number hai — kilometres per second mein. Parent ka central claim hai: $v_{\infty}$ kabhi nahi badalta; sirf $v_{\infty}$ ki direction ghoomti hai.

Topic ko yeh kyun chahiye: $v_{\infty}$ woh ek arrow hai jo rotate hota hai. Ise rotate karna hi slingshot hai.

4. $r$ , $r_{p}$ , $G$ , $M_{p}$ , $μ_{p}$ — distance aur planet ki gravity strength

Topic ko yeh kyun chahiye: $r_{p}$ aur $μ_{p}$ set karte hain ki path kitna bend karta hai — closer approach aur stronger gravity matlab sharper turn. Turn-angle formula mein $v_{\infty}$ ke alaawa sirf yahi do "knobs" hain.

5. Specific energy $ε$ — energy per kilogram

Do pieces:

$\frac{v ^{2}}{2}$ kinetic energy per kilogram hai (standard $\frac{1}{2} m v^{2}$ mein $m$ divide ho ke), planet-frame speed $v$ use karke.
$- \frac{μ _{p}}{r}$ gravitational potential energy per kilogram hai. Yeh negative hai aur jaise $r$ badhta hai, kam negative hota jaata hai (zero ke closer), kyunki planet se door jaake aap uske well se bahar aa rahe hain.

Key move: door jaake $r \to \infty$ , toh $μ_{p} / r \to 0$ , aur bacha $ε = v_{\infty}^{2} /2$ . Kyunki $ε$ conserved hai (gravity planet frame mein koi net work nahi karti), $v_{\infty}$ lock ho jaata hai. Yahi parent ka poora proof hai ki $v_{\infty}$ constant hai. Dekho Two-body problem & vis-viva equation.

6. Angular momentum $h$ — "kitna sideways swing hai"

Closest point (periapsis) pe velocity purely sideways hoti hai, toh $ϕ = 90°$ aur $sin ϕ = 1$ , giving simple special case $h = r_{p} v_{p},$ jahan $v_{p}$ exactly periapsis pe speed hai. Hum yeh clean version use karte hain kyunki periapsis woh ek jagah hai jahan formula se angle nikal jaata hai.

Topic ko yeh kyun chahiye: $h$ aur $ε$ milke path ki shape fix karte hain (kitna curved, turn kitna sharp). Neeche eccentricity formula mein yeh doosra ingredient hai. Dekho Specific orbital energy & angular momentum.

7. Eccentricity $e$ aur path hyperbola kyun hota hai

Figure 3 (upar): same planet (white dot) ke saath teen possible paths — blue circle ( $e = 0$ ), pink ellipse ( $0 < e < 1$ , ek closed loop), aur mota yellow hyperbola ( $e > 1$ , ek open curve jo hamesha ke liye leave karta hai). Slingshot hamesha yellow wala hota hai. Notice karein ki jaise $e$ badhta hai curves progressively "more open" hoti jaati hain.

$e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}$ actually kahaan se aata hai

Hum yeh formula memorise nahi karne wale — hum ise do conserved quantities se build karenge jo humne abhi meet ki hain, $ε$ aur $h$ .

Step 1 — general shape formula KYA hai. Two-body problem solve karna (Two-body problem & vis-viva equation mein puri tarah kiya gaya hai) do conserved specific quantities ke terms mein eccentricity deta hai: $e = 1 + \frac{2 ε h ^{2}}{μ _{p}^{2}} .$ Yeh shape kyun? $ε$ "kitni energy" carry karta hai aur $h$ "kitna sideways swing" — milke woh exactly ek conic shape pin down karte hain. Zyada energy ya zyada swing → zyada open curve → bada $e$ .

Step 2 — apni do special values plug in kyun karein. Humne already $ε = \frac{v _{\infty}^{2}}{2}$ (far-away energy) find ki, aur periapsis pe, $h = r_{p} v_{p}$ . Substitute karte hain: $e = 1 + \frac{2 ( \frac{v _{\infty}^{2}}{2} ) ( r _{p} v _{p} ) ^{2}}{μ _{p}^{2}} = 1 + \frac{v _{\infty}^{2} r _{p}^{2} v _{p}^{2}}{μ _{p}^{2}} .$

Step 3 — periapsis pe speed evaluate kyun karein. Periapsis pe, energy conservation $v_{p}$ ko $v_{\infty}$ se tie karti hai: $\frac{v _{p}^{2}}{2} - \frac{μ _{p}}{r _{p}} = \frac{v _{\infty}^{2}}{2} \Rightarrow v_{p}^{2} = v_{\infty}^{2} + \frac{2 μ _{p}}{r _{p}} .$ Yeh bas "kinetic + potential door aur periapsis pe same hoti hai" hai. $v_{p}^{2}$ wapas daalne se aur algebra simplify karne se square root exactly is tidy result mein collapse ho jaata hai: $e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}$

Words mein WHAT kehta hai: badi leftover speed $v_{\infty}$ , ya door/tez pass (bada $r_{p} v_{\infty}^{2}$ ), zyada open (bada $e$ ) hyperbola deta hai jo mushkil se bend karta hai. Slow, close pass $e$ ko 1 ke paas rakhta hai — sabse sharp turn.

8. Asymptotes aur turn angle $δ$

Figure 4 (upar): yellow hyperbola blue planet ke paas se sweep karta hai. Pink arrow $v_{\infty}$ hai jo lower asymptote ke along andar aa raha hai; blue arrow $v_{\infty}$ hai jo upper asymptote ke along bahar ja raha hai. Unke beech ka white arc turn angle $δ$ hai — woh rotation jo encounter ne deliver ki. Dashed line closest approach $r_{p}$ mark karti hai. Slingshot mein sab kuch "woh white arc kitna bada hai?" hai.

$sin (\frac{δ}{2}) = \frac{1}{e}$ kahaan se aata hai

Yeh payoff hai, toh ise geometrically earn karte hain.

Step 1 — asymptote angle KYA hota hai. Hyperbola ke do straight asymptotes hote hain. Har ek hyperbola ke axis (planet aur periapsis ke through symmetry line) ke saath ek fixed angle $θ_{\infty}$ banata hai. Standard conic geometry (dekho Hyperbolic orbits & orbital eccentricity) us asymptote ki direction deta hai: $cos θ_{\infty} = - \frac{1}{e} .$ $- 1/ e$ kyun? Hyperbola pe spacecraft sirf $θ_{\infty}$ tak angles reach kar sakta hai infinitely door jaane se pehle; conic equation mein orbit ki radius ko infinity set karne se exactly yahi cosine force hota hai.

Step 2 — turn angle supplement kyun hai. Incoming aur outgoing $v_{\infty}$ do asymptotes ke along lie karte hain. Kyunki do asymptotes axis ke baare mein mirror images hain, velocity actually kitna rotate hoti hai, $δ$ , $θ_{\infty}$ se relate hota hai: $δ = π - 2 (π - θ_{\infty}) = 2 θ_{\infty} - π .$ Words mein: velocity ek asymptote ke along inward aim karte shuru hui aur doosre ke along outward aim karte khatam hoti hai; bacha hua rotation $δ$ hai.

Step 3 — yeh clean sine mein collapse kyun hota hai. Step 1 ka boxed relation lo, $cos θ_{\infty} = - 1/ e$ , aur half-turn rewrite karo. $δ = 2 θ_{\infty} - π$ se milta hai $θ_{\infty} = \frac{δ + π}{2}$ , toh $cos θ_{\infty} = cos (\frac{δ}{2} + \frac{π}{2}) = - sin (\frac{δ}{2}) .$ $cos θ_{\infty}$ ke liye dono expressions equal set karne se, $- sin (δ /2) = - 1/ e$ , minus signs cancel ho jaate hain aur hum land karte hain: $sin (\frac{δ}{2}) = \frac{1}{e}$

Words mein WHAT kehta hai: $e = 1$ ke close wala hyperbola (slow, close pass) mein $sin (δ /2)$ 1 ke near hota hai, toh $δ$ $180°$ ke near — ek bada turn. Bahut open hyperbola (bada $e$ ) mein $sin (δ /2)$ 0 ke near hota hai, toh $δ$ $0°$ ke near — barely deflect. Yahi parent ka "closer aur slower = bigger turn" hai.

9. Infinity symbol, Greek letters, aur "delta" of change

Ek quick glossary taaki koi squiggle aapko trip na kare:

$\infty$ — "infinity", matlab "arbitrarily door."
$μ$ — Greek "mu", gravity parameter ke liye use hota hai.
$ε$ — Greek "epsilon", specific energy ke liye use hota hai.
$δ$ — Greek "delta", turn angle.
$Δ$ (capital delta) — "mein change." Toh $Δ v = v_{o u t} - v_{in}$ hai "final arrow minus initial arrow," net kick.

Sab kuch topic ko kaise feed karta hai

Equipment checklist

Khud ko test karo — right side cover karo aur zor se jawab do.

v

pe arrow ka kya matlab hai aur plain

v

ka kya matlab hai?

v

poora arrow hai (size + direction); plain

v

sirf uski length hai (magnitude).

Do velocity vectors kaise add karte hain?

Tip-to-tail: doosre arrow ki tail ko pehle ki tip pe rakho; sum bilkul shuru se bilkul end tak jaata hai.

v_{sc / pl an e t}

zor se padho.

"Velocity of the spacecraft relative to (as seen from) the planet."

Slingshot mein kaun se do reference frames use hote hain?

Planet-centered frame aur Sun-centered (heliocentric) frame.

Ise

v_{\infty}

("at infinity") kyun kehte hain?

Yeh planet-relative speed hai jo door se measure ki jaati hai, jahan gravity chhod deti hai — ek clean constant value, planet ke paas ki messy speed se alag.

G

kya hai aur

μ_{p}

se kaise alag hai?

G

universal gravitational constant hai (har jagah same);

μ_{p} = G M_{p}

ise ek planet ke mass ke saath combine karta hai.

μ_{p}

kya bundle karta hai aur kyun?

μ_{p} = G M_{p}

; dono hamesha multiply hote hain aur product

M_{p}

akele se zyada precisely measure hota hai.

ε = v^{2} /2 - μ_{p} / r

mein

v

kis frame ki speed hai?

Planet frame:

v = ∣ v_{sc / pl an e t} ∣

h

ki general definition likho aur yeh conserved kyun hai.

h = ∣ r \times v ∣ = r v sin ϕ

; conserved hai kyunki gravity

r

ke along pull karti hai aur sideways swing ko twist nahi kar sakti.

e = 1 + r_{p} v_{\infty}^{2} / μ_{p}

kaise aata hai?

e = 1 + 2 ε h^{2} / μ_{p}^{2}

se shuru karo,

ε = v_{\infty}^{2} /2

aur

h = r_{p} v_{p}

ke saath

v_{p}^{2} = v_{\infty}^{2} + 2 μ_{p} / r_{p}

substitute karo, phir simplify karo.

sin (δ /2) = 1/ e

kahaan se aata hai?

Asymptote angle

cos θ_{\infty} = - 1/ e

banata hai;

δ = 2 θ_{\infty} - π

use karke woh cosine

sin (δ /2) = 1/ e

ban jaata hai.

e

ki kaun si value hyperbola mean karti hai, aur slingshot hamesha ek kyun hota hai?

e > 1

; kyunki leftover speed

ε > 0

banati hai (unbound), path ek open hyperbola hoti hai.

Turn angle

δ

kya hai?

Incoming aur outgoing

v_{\infty}

ke beech ka angle — velocity arrow kitna rotate hua.

Δ v

mein capital

Δ

ka kya matlab hai?

"Mein change": final arrow minus initial arrow.