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

Fixed: $v_{\infty }$ ki magnitude (planet ke relative speed) — kyunki planet frame mein energy conserved hoti hai aur $\epsilon =v_{\infty }^2/2$ arrival conditions se set hoti hai. Changes: ${\vec{v}}_{\infty }$ ki direction, jo turn angle $\delta$ se rotate hoti hai. Craft ke planet ke saath relation mein koi aur cheez permanently change nahi hoti.

Flyby mein positive specific energy hoti hai $\epsilon =v_{\infty }^2/2>0$. Positive energy $\Rightarrow$ eccentricity $e>1\Rightarrow$ hyperbolic. (Koi bhi nonzero $v_{\infty }$ isko guarantee karta hai — yahi wajah hai ki parent note planetocentric leg ko hyperbola kehta hai. Dekho Hyperbolic orbits & orbital eccentricity.)

False. Planet ke peechhe se guzro (uski wake se) taaki ${\vec{v}}_{\infty }$ ko $+{\vec{V}}_p$ ki taraf rotate kiya ja sake aur speed gain ho. Aage se guzarne pe woh $-{\vec{V}}_p$ ki taraf rotate hoti hai aur speed kho jaati hai.

$e=1+\frac{(2\times 10^5)(9^2)}{1.267\times 10^8}=1+\frac{1.62\times 10^7}{1.267\times 10^8}=1+0.12786=1.1279$ $\sin \frac{\delta }{2}=\frac{1}{1.1279}=0.88663\Rightarrow \frac{\delta }{2}=62.42^{\circ }\Rightarrow \delta =124.8^{\circ }$ Bahut bada bend hai, kyunki Jupiter ka huge ${\mu }_p$ term $r_p{v}_{\infty }^2$ ko dwarf kar deta hai.

$e=1+\frac{(7.15\times 10^4)(81)}{1.267\times 10^8}=1+0.045717=1.04572$ $\sin \frac{\delta }{2}=\frac{1}{1.04572}=0.95627\Rightarrow \frac{\delta }{2}=72.90^{\circ }\Rightarrow \delta =145.8^{\circ }$ Kareeb se fly karne pe $e$ shrink hokar 1 ke paas aa gaya, jisse $\sin (\delta /2)$ 1 ki taraf push hua aur turn $124.8^{\circ }$ se badhkar $145.8^{\circ }$ ho gaya. Closer = sharper turn.

$\Delta v_{\max }=2v_{\infty }=12$ km/s. Sirf ${\vec{v}}_{\infty }$ ki direction change ho sakti hai, isliye outgoing tip radius $v_{\infty }$ ke circle pe incoming tip se maximum ek diameter ($2v_{\infty }$) dur ho sakti hai. Planet ki speed $V_p=24$ km/s cap mein nahi aati — woh sirf set karti hai ki kaunsi direction "gain" hai.

Periapsis pe velocity radius ke perpendicular hoti hai, isliye $h=r_p{v}_p$. Vis-viva (dekho Two-body problem & vis-viva equation) at periapsis: $v_p^2=v_{\infty }^2+\frac{2{\mu }_p}{r_p}\text{(from }\epsilon =\frac{v_p^2}{2}-\frac{{\mu }_p}{r_p}=\frac{v_{\infty }^2}{2}\text{)}.$ Insert karo $h^2=r_p^2{v}_p^2=r_p^2\left(v_{\infty }^2+\frac{2{\mu }_p}{r_p}\right)$ aur $2\epsilon =v_{\infty }^2$:

Before: ${\vec{v}}_{\infty }^{in}=(-5,0)$, toh ${\vec{v}}_{sc}^{in}=(13-5,0)=(8,0)$, speed $=8$ km/s. ${\vec{v}}_{\infty }$ ko $90^{\circ }$ rotate karo: $(-5,0)\to (0,-5)$ (ek $90^{\circ }$ turn; perpendicular component ka sign flyby side pe depend karta hai — $-y$ lo). Phir ${\vec{v}}_{sc}^{out}=(13,-5)$. $|{\vec{v}}_{sc}^{out}|=\sqrt{13^2+5^2}=\sqrt{194}=13.928\text{km/s}.$ Speed mein change: $13.928-8=5.928$ km/s. Note karo yeh $2v_{\infty }=10$ cap se kam hai, kyunki humne sirf $90^{\circ }$ turn kiya, poora $180^{\circ }$ nahi.

Final speed maximum hoti hai jab ${\vec{v}}_{\infty }^{out}$ poori tarah $+{\vec{V}}_p$ along point kare, yaani ${\vec{v}}_{\infty }$ $(-5,0)$ se $(+5,0)$ tak rotate ho — ek $180^{\circ }$ turn. ${\vec{v}}_{sc}^{out}=(13+5,0)=(18,0),|{\vec{v}}_{sc}^{out}|=18\text{km/s}.$ Inbound $8$ km/s se compare karein toh yeh poora $2v_{\infty }=10$ km/s gain hai.

$\sin (\delta /2)=1/e$ se jab $\delta /2=30^{\circ }$: $e=1/\sin 30^{\circ }=2$. $r_p$ ke liye eccentricity relation invert karo: $e=1+\frac{r_p{v}_{\infty }^2}{{\mu }_p}\Rightarrow r_p=\frac{(e-1){\mu }_p}{v_{\infty }^2}=\frac{(1)(3.793\times 10^7)}{5.5^2}=\frac{3.793\times 10^7}{30.25}=1.2539\times 10^6\text{km}.$ Yeh hai $r_p\approx 1.254\times 10^6$ km — Saturn ($R\approx 60,300$ km) se kaafi bahar, ek gentle high pass ek modest $60^{\circ }$ bend ke liye. (Chhota turn bada $r_p$ maangta hai; $e$ chhota, 1 ke paas.)

Before: ${\vec{v}}_{sc}^{in}={\vec{V}}_p+{\vec{v}}_{\infty }^{in}=(0+5.5,9.7+0)=(5.5,9.7)$, speed $=\sqrt{5.5^2+9.7^2}=\sqrt{30.25+94.09}=\sqrt{124.34}=11.151$ km/s. After: ${\vec{v}}_{sc}^{out}=(0+2.75,9.7+4.7631)=(2.75,14.4631)$, speed $=\sqrt{2.75^2+14.4631^2}=\sqrt{7.5625+209.181}=\sqrt{216.744}=14.723$ km/s. Heliocentric speed $11.151$ se $14.723$ km/s ho gayi, $3.572$ km/s ka gain — real energy, sirf ${\vec{v}}_{\infty }$ ko planet ki motion ki taraf bend karne se. (Yahi mechanism Voyager & Cassini mission trajectories ke peechhe hai.)

Sun frame mein momentum conservation: craft ${\vec{V}}_p$ ke along jo momentum gain karta hai woh $m_{sc}\Delta v$ hai. Jupiter equal-and-opposite momentum absorb karta hai: $M_p\Delta V_p=m_{sc}\Delta v\Rightarrow \Delta V_p=\frac{m_{sc}\Delta v}{M_p}=\frac{722\times 10^4\text{kg⋅m/s}}{1.898\times 10^{27}\text{kg}}=3.804\times 10^{-21}\text{m/s}.$ Fractional: $\frac{\Delta V_p}{V_p}=\frac{3.804\times 10^{-21}}{1.306\times 10^4}=2.91\times 10^{-25}.$ Bilkul negligible ($\sim 10^{-25}$), jo parent note ke claim se match karta hai. Energy conserved hai — Jupiter genuinely slow hota hai, ek aisi amount se jo koi bhi instrument detect nahi kar sakta. Koi free lunch nahi; sirf ek wildly lopsided trade kyunki $m_{sc}\ll M_p$.

(a) $e=1+\frac{(6300)(4^2)}{3.249\times 10^5}=1+\frac{100800}{324900}=1+0.31025=\mathrm{1.31025.}$ $\sin (\delta /2)=1/1.31025=0.76321\Rightarrow \delta /2=49.74^{\circ }\Rightarrow \delta =99.48^{\circ }.$ (b) ${\vec{v}}_{\infty }^{in}=(-4,0)$. Ek vector ko angle $\delta$ se $+{\vec{V}}_p$ ki taraf rotate karna (counter-clockwise from the $-x$ direction): incoming direction $180^{\circ }$ hai; $\delta =99.48^{\circ }$ add karne ke baad hum $279.48^{\circ }$ pe pahunche, yaani ${\vec{v}}_{\infty }^{out}=4(\cos 279.48^{\circ },\sin 279.48^{\circ })=4(0.16495,-0.98630)=(0.6598,-3.9452).$ Before: ${\vec{v}}_{sc}^{in}=(35.02-4,0)=(31.02,0)$, speed $=31.02$ km/s. After: ${\vec{v}}_{sc}^{out}=(35.02+0.6598,-3.9452)=(35.6798,-3.9452)$, speed $=\sqrt{35.6798^2+3.9452^2}=\sqrt{1273.05+15.565}=\sqrt{1288.62}=35.897$ km/s. Gain $=35.897-31.02=4.877$ km/s. ($2v_{\infty }=8$ cap se kam hai kyunki turn $99.48^{\circ }$ tha, $180^{\circ }$ nahi.)

$v_{\infty }$ conserve karo, $e$ nikalo $1+r_p{v}_{\infty }^2/{\mu }_p$ se, $\delta$ nikalo $\arcsin (1/e)$ se (times 2), ${\vec{v}}_{\infty }$ rotate karo, phir ${\vec{V}}_p$ ko vectors ki tarah add karo — heliocentric change kabhi bhi $2v_{\infty }$ se zyada nahi hoga.