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

Fixed: the magnitude $v_{\infty }$ (speed relative to the planet) — because energy is conserved in the planet frame and $\epsilon =v_{\infty }^2/2$ is set by arrival conditions. Changes: the direction of ${\vec{v}}_{\infty }$, which rotates by the turn angle $\delta$. Nothing else about the craft's relation to the planet changes permanently.

A flyby has positive specific energy $\epsilon =v_{\infty }^2/2>0$. Positive energy $\Rightarrow$ eccentricity $e>1\Rightarrow$ hyperbolic. (Any nonzero $v_{\infty }$ guarantees this — that is why the parent note calls the planetocentric leg a hyperbola. See Hyperbolic orbits & orbital eccentricity.)

False. Fly behind the planet (through its wake) to rotate ${\vec{v}}_{\infty }$ toward $+{\vec{V}}_p$ and gain speed. Flying in front rotates it toward $-{\vec{V}}_p$ and loses speed.

$e=1+\frac{(2\times 10^5)(9^2)}{1.267\times 10^8}=1+\frac{1.62\times 10^7}{1.267\times 10^8}=1+0.12786=1.1279$ $\sin \frac{\delta }{2}=\frac{1}{1.1279}=0.88663\Rightarrow \frac{\delta }{2}=62.42^{\circ }\Rightarrow \delta =124.8^{\circ }$ A big bend, because Jupiter's huge ${\mu }_p$ dwarfs the $r_p{v}_{\infty }^2$ term.

$e=1+\frac{(7.15\times 10^4)(81)}{1.267\times 10^8}=1+0.045717=1.04572$ $\sin \frac{\delta }{2}=\frac{1}{1.04572}=0.95627\Rightarrow \frac{\delta }{2}=72.90^{\circ }\Rightarrow \delta =145.8^{\circ }$ Flying closer shrank $e$ toward 1, which pushed $\sin (\delta /2)$ toward 1 and the turn from $124.8^{\circ }$ up to $145.8^{\circ }$. Closer = sharper turn.

$\Delta v_{\max }=2v_{\infty }=12$ km/s. Only the direction of ${\vec{v}}_{\infty }$ can change, so the outgoing tip can sit at most a diameter ($2v_{\infty }$) from the incoming tip on the circle of radius $v_{\infty }$. The planet's speed $V_p=24$ km/s does not enter the cap — it only sets which direction is "gain."

At periapsis the velocity is perpendicular to the radius, so $h=r_p{v}_p$. Vis-viva (see 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 $h^2=r_p^2{v}_p^2=r_p^2\left(v_{\infty }^2+\frac{2{\mu }_p}{r_p}\right)$ and $2\epsilon =v_{\infty }^2$:

Before: ${\vec{v}}_{\infty }^{in}=(-5,0)$, so ${\vec{v}}_{sc}^{in}=(13-5,0)=(8,0)$, speed $=8$ km/s. Rotate ${\vec{v}}_{\infty }$ by $90^{\circ }$: $(-5,0)\to (0,-5)$ (a $90^{\circ }$ turn; sign of the perpendicular component depends on flyby side — take $-y$). Then ${\vec{v}}_{sc}^{out}=(13,-5)$. $|{\vec{v}}_{sc}^{out}|=\sqrt{13^2+5^2}=\sqrt{194}=13.928\text{km/s}.$ Change in speed: $13.928-8=5.928$ km/s. Note this is less than the $2v_{\infty }=10$ cap, because we only turned $90^{\circ }$, not the full $180^{\circ }$.

Final speed is maximal when ${\vec{v}}_{\infty }^{out}$ points fully along $+{\vec{V}}_p$, i.e. ${\vec{v}}_{\infty }$ has been rotated from $(-5,0)$ to $(+5,0)$ — a $180^{\circ }$ turn. ${\vec{v}}_{sc}^{out}=(13+5,0)=(18,0),|{\vec{v}}_{sc}^{out}|=18\text{km/s}.$ Compared with the $8$ km/s inbound, that's the full $2v_{\infty }=10$ km/s gain.

From $\sin (\delta /2)=1/e$ with $\delta /2=30^{\circ }$: $e=1/\sin 30^{\circ }=2$. Invert the eccentricity relation for $r_p$: $e=1+\frac{r_p{v}_{\infty }^2}{{\mu }_p}\Rightarrow r_p=\frac{(e-1){\mu }_p}{v_{\infty }^2}=\frac{(1)(3.793\times 10^7)}{5.5^2}=\frac{3.793\times 10^7}{30.25}=1.2539\times 10^6\text{km}.$ That's $r_p\approx 1.254\times 10^6$ km — well outside Saturn ($R\approx 60,300$ km), a gentle high pass for a modest $60^{\circ }$ bend. (A smaller turn wants a larger $r_p$; a smaller $e$ closer to 1.)

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 rose from $11.151$ to $14.723$ km/s, a gain of $3.572$ km/s — real energy, all from bending ${\vec{v}}_{\infty }$ toward the planet's motion. (This is the mechanism behind Voyager & Cassini mission trajectories.)

Momentum conservation in the Sun frame: the momentum the craft gains along ${\vec{V}}_p$ is $m_{sc}\Delta v$. Jupiter absorbs the equal-and-opposite momentum: $M_p\Delta V_p=m_{sc}\Delta v\Rightarrow \Delta V_p=\frac{m_{sc}\Delta v}{M_p}=\frac{722\times 10^4\text{kg⋅m/s}}{1.898\times 10^{27}\text{kg}}=3.804\times 10^{-21}\text{m/s}.$ Fractional: $\frac{\Delta V_p}{V_p}=\frac{3.804\times 10^{-21}}{1.306\times 10^4}=2.91\times 10^{-25}.$ Utterly negligible ($\sim 10^{-25}$), matching the parent note's claim. Energy is conserved — Jupiter genuinely slows, by an amount no instrument could ever detect. No free lunch; just a wildly lopsided trade because $m_{sc}\ll M_p$.

(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) ${\vec{v}}_{\infty }^{in}=(-4,0)$. Rotating a vector by angle $\delta$ toward $+{\vec{V}}_p$ (counter-clockwise from the $-x$ direction): the incoming direction is $180^{\circ }$; after adding $\delta =99.48^{\circ }$ we land at $279.48^{\circ }$, i.e. ${\vec{v}}_{\infty }^{out}=4(\cos 279.48^{\circ },\sin 279.48^{\circ })=4(0.16495,-0.98630)=(0.6598,-3.9452).$ 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. (Below the $2v_{\infty }=8$ cap because the turn was $99.48^{\circ }$, not $180^{\circ }$.)

Conserve $v_{\infty }$, get $e$ from $1+r_p{v}_{\infty }^2/{\mu }_p$, get $\delta$ from $\arcsin (1/e)$ (times 2), rotate ${\vec{v}}_{\infty }$, then add ${\vec{V}}_p$ as vectors — the heliocentric change never exceeds $2v_{\infty }$.