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

3.2.24 · Physics › Orbital Mechanics & Astrodynamics

"Free" energy boost exist kyon karta hai?

Puzzle kya hai? Gravity conservative hai. Agar tum kisi planet ki taraf giro aur wapas nikal jao, to tum utni hi speed se niklo ge jitni speed se aaye the (planet ke relative). To slingshot speed kaise badha sakta hai?

Resolution — frames matter karte hain. Energy aur speed reference frame pe depend karti hai.

Planet-centered frame mein: flyby ek hyperbola hai. Speed in = speed out $= v_{\infty}$ . Sirf direction change hoti hai.
Sun-centered (heliocentric) frame mein: planet $V_{p}$ pe move kar raha hai. Jab hum rotated velocity ko Sun frame mein transform karte hain, magnitude change ho jaati hai.

Patched-Conic Method

Yeh kyon kaam karta hai: SOI ke andar planet ka pull Sun ka pull dominate karta hai; bahar, ulta. Kyunki SOI heliocentric orbit ke compare mein bahut chhota hai, hum poori flyby ko ek single point pe hoti maan lete hain — isliye spacecraft ki heliocentric position nahi badalti, sirf uski velocity rotate hoti hai.

v-infinity: key vector

Dono frames ke beech link ek simple vector addition hai patch point pe:

$v_{sc / S u n} = V_{p} + v_{\infty}$

jahan $V_{p}$ planet ki heliocentric velocity hai.

$v_{\infty}$ ka conservation first principles se derive karna

Planet frame mein energy (specific orbital energy): $ε = \frac{v ^{2}}{2} - \frac{μ _{p}}{r}$ Yeh constant hai (gravity conservative hai). Jaise $r \to \infty$ potential term $\to 0$ , to $ε = \frac{v _{\infty}^{2}}{2} \Rightarrow v_{\infty} = 2 ε .$ Kyunki $ε$ incoming conditions se fix hota hai aur change nahi hota, == $v_{\infty}^{in} = v_{\infty}^{o u t}$ ==. Sirf $v_{\infty}$ ki direction rotate hoti hai.

Turn (deflection) angle derive karna

SOI ke andar path ek hyperbola hai. Hyperbola kyun? Kyunki $ε > 0$ (excess speed), ek unbound orbit → eccentricity $e > 1$ .

Hyperbola ki asymptotes ek doosre ke saath angle banati hain. Incoming aur outgoing $v_{\infty}$ in asymptotes ke along point karte hain, isliye velocity turn angle $δ$ se rotate hoti hai:

$sin (\frac{δ}{2}) = \frac{1}{e}$

Derivation sketch: Hyperbola ke liye asymptote axis ke saath angle $θ_{\infty}$ banati hai jahan $cos θ_{\infty} = - 1/ e$ . Asymptotes ke beech total exterior turn deta hai $δ = π - 2 (π - θ_{\infty})$ ... zyada useful tarike se, orbit shape use karte hue:

$e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}$

jahan $r_{p}$ = closest approach (periapsis) radius. Yeh form kyun? Vis-viva/energy relation $e = 1 + \frac{2 ε h ^{2}}{μ _{p}^{2}}$ se, jahan $ε = v_{\infty}^{2} /2$ aur angular momentum $h = r_{p} v_{p}$ ; periapsis pe algebra ise boxed expression mein reduce kar deta hai. To:

$sin (\frac{δ}{2}) = \frac{1}{1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}}$

Closer & slower = bigger turn kyun: Chhota $r_{p}$ ya chhota $v_{\infty}$ → chhota $e$ → bada $δ$ . Bahut mod lene ke liye paas se aur dheere guzro; bahut tez aur door se guzro to muskil se mudoge.

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

Worked Example 1 — Kitna gain ho sakta hai?

Planet $V_{p} = 13 km/s$ pe move kar raha hai. Spacecraft $v_{\infty} = 5 km/s$ se arrive karta hai. Best-case flyby $v_{\infty}$ ko fully reverse karti hai ( $δ = 180°$ se turn, jo tabhi possible hai jab $v_{\infty}$ kaafi chhota ho / $r_{p}$ tiny ho) to yeh anti-parallel se parallel ho jaata hai $V_{p}$ ke saath.

Before: $v_{h e l i o} = V_{p} - v_{\infty} = 13 - 5 = 8 km/s$ Yeh step kyun? Incoming $v_{\infty}$ ko $V_{p}$ ke opposite choose karo (sabse kharab inbound speed).
After: $v_{h e l i o} = V_{p} + v_{\infty} = 13 + 5 = 18 km/s$ Yeh step kyun? Full reversal $v_{\infty}$ ko $V_{p}$ ke saath align kar deti hai.
Gain: $Δ v = 10 km/s = 2 v_{\infty}$ . Yeh step kyun? Maximum heliocentric change $2 v_{\infty}$ hai — ek flyby se kabhi bhi apni excess speed se zyada gain nahi ho sakta.

Worked Example 2 — Turn angle compute karo

Jupiter: $μ_{p} = 1.267 \times 1 0^{8} km^{3} / s^{2}$ . Flyby $r_{p} = 100, 000$ km pe $v_{\infty} = 6$ km/s ke saath.

$e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}} = 1 + \frac{( 1 0 ^{5} ) ( 36 )}{1.267 \times 1 0 ^{8}} = 1 + 0.0284 = 1.0284$ Yeh step kyun? Periapsis aur excess speed ko eccentricity relation mein plug karo.
$sin (δ /2) = 1/1.0284 = 0.9724 \Rightarrow δ /2 = 76.5° \Rightarrow δ = 153°$ . Yeh step kyun? Itna bada turn isliye kyunki Jupiter bahut massive hai aur $v_{\infty}$ modest hai. Yahi wajah hai ki Jupiter gravity assists ka king hai (Voyager, Cassini, Juno, Pioneer).

Worked Example 3 — Kis side se guzrein?

Tum heliocentrically speed up karna chahte ho.

Planet ke peeche se guzro (planet tumse door move kar raha hai jab tum uske wake ko cross karte ho) → $v_{\infty}$ rotate hokar $+ V_{p}$ ki taraf jaata hai → speed gain.
Aage se guzro → $- V_{p}$ ki taraf rotate hota hai → speed loss (slow down karne ke liye use hota hai, jaise MESSENGER ka Mercury ke liye).

Yeh step kyun? Kaunsi asymptote se exit karte ho iski geometry decide karti hai ki $v_{\infty}^{o u t}$ ka $V_{p}$ ke along component bada hai ya nahi.

Active Recall

Recall Gravity assist mein konsi single quantity conserved hoti hai, aur kya change hota hai?

$v_{\infty}$ (planet ke relative speed) conserved hoti hai; $v_{\infty}$ ki direction turn angle $δ$ se rotate hoti hai. Sun frame mein, velocity ka magnitude change hota hai $v_{sc} = V_{p} + v_{\infty}$ ke zariye.

Recall Ek flyby se maximum heliocentric velocity change kitna hoga?

$Δ v_{ma x} = 2 v_{\infty}$ , jab $v_{\infty}$ fully reverse ho jaaye.

Recall Feynman: ek 12-saal ke bacche ko slingshot explain karo

Socho tum ek marble ko ek bade truck par maarte ho. Agar tum use moving truck ke front par maaro, to woh tumhare throw se zyada tez wapas aata hai — truck ki motion ne use push kiya. Agar tum use door jaate truck ke back par maaro, to woh dheerey wapas aata hai. Spacecraft yehi karte hain planets ke saath: woh touch nahi karte, lekin planet ki gravity truck ke front ki tarah kaam karti hai, ship ko Sun ke around zyada tez (ya dheema) phenkti hai. Planet ko practically kuch farak nahi padta, kyunki woh ek zillion guna zyada bhaari hai.

Flashcards

Patched-conic method ek trajectory ko kaise pieces mein todta hai?

SOI ke bahar ek heliocentric conic, aur SOI ke andar ek planetocentric hyperbola, SOI boundary pe patch kiye hue.

v_{\infty}

kya hai?

Hyperbolic excess velocity — SOI boundary pe spacecraft ki planet ke relative speed;

v_{\infty} = 2 ε

v_{\infty}

flyby mein conserved kyun hoti hai?

Planet frame mein gravity conservative hai, isliye specific energy

ε = v^{2} /2 - μ / r

constant hai;

r \to \infty

pe yeh

v_{\infty}^{2} /2

deta hai, jo fixed hai.

Turn angle formula?

sin (δ /2) = 1/ e

jahan

e = 1 + r_{p} v_{\infty}^{2} / μ_{p}

Flyby hyperbola ki eccentricity periapsis ke terms mein?

e = 1 + \frac{r _{p} v _{\infty}^{2}}{μ _{p}}

Turn angle maximize kaise karein?

Chhota periapsis

r_{p}

aur chhota

v_{\infty}

(aur bada

μ_{p}

) →

e

near 1 →

δ

near

180°

Ek assist se max heliocentric speed change?

2 v_{\infty}

(

v_{\infty}

ka full reversal).

Planet aur Sun frames ke beech relation?

v_{sc / S u n} = V_{p} + v_{\infty}

Planet ke peeche vs aage se guzrna?

Peeche → speed up (heliocentric speed gain); aage → slow down.

Kya energy conservation violate hoti hai?

Nahi — planet exactly woh energy lose karta hai jo craft gain karta hai; planet par effect negligible hai kyunki

m_{sc} ≪ M_{p}

Jupiter gravity-assist ke liye best planet kyun hai?

Bada

μ_{p}

ek bada deflection

δ

deta hai, aur high

V_{p}

bade heliocentric changes allow karta hai.

Connections

Two-body problem & vis-viva equation
Hyperbolic orbits & orbital eccentricity
Sphere of influence (SOI)
Specific orbital energy & angular momentum
Interplanetary transfer & Hohmann transfer
Reference frames & Galilean velocity addition
Oberth effect (powered flybys)
Voyager & Cassini mission trajectories