Exercises — Equations of motion — 3DOF point mass (trajectory analysis)

WHAT: We want a rate of change of speed, i.e. $\dot{V}$. WHY: speed is the length of the velocity arrow, and only the tangential (along-$V$) rule changes length. So we use $\dot{V}=\frac{T-D}{m}-g\sin \gamma .$ The gravity term against the climb is $-g\sin \gamma$. On the figure it is the shadow of the "down" arrow projected backwards along the velocity arrow.

• (a) bending downward → $\dot{\gamma }$ rule (direction changes): $\dot{\gamma }=\frac{L-mg\cos \gamma }{mV}$.
• (b) climb rate → $\dot{h}=V\sin \gamma$ (vertical component of the velocity arrow).
• (c) getting faster → $\dot{V}$ rule (length of arrow grows). Mnemonic from the parent: S-along, C-across — sine ties gravity to speed, cosine ties it to turning.

WHAT: compute the speed-rate. WHY: tangential rule. WHAT IT LOOKS LIKE: thrust stretches the arrow, drag + gravity's along-shadow shrink it. $\dot{V}=\frac{350000-45000}{12000}-9.81\sin 50^{\circ }=25.4167-7.5147=17.90m/{\mathbf{s}}^2.$

WHY the $\dot{\gamma }$ rule: we want how the arrow turns, not its length. Start from the full turning rule and set $L=0$: $\dot{\gamma }=\frac{L-mg\cos \gamma }{mV}=\frac{0-mg\cos \gamma }{mV}=\frac{-mg\cos \gamma }{mV}.$ Why does $m$ vanish? Every surviving term in the numerator carries one factor of $m$ (weight is $mg\cos \gamma$), and the denominator also carries one $m$. That common factor cancels top-and-bottom — heavier or lighter, gravity bends the direction at the same rate (only speed $V$ matters). This is exactly why we can write the lift-free turning rule with no mass at all: $\dot{\gamma }=-\frac{g\cos \gamma }{V}=-\frac{9.81\cos 50^{\circ }}{420}=-\frac{6.3057}{420}=-0.015014\text{rad/s}.$ Convert: $-0.015014\times \frac{180}{\pi }=-0.860deg/s$. Negative because gravity always drags the arrow's tip downward.

The velocity arrow of length $V=420$ splits by right-triangle trig onto the horizon and the vertical: $\dot{x}=420\cos 50^{\circ }=269.96m/s,\dot{h}=420\sin 50^{\circ }=321.75m/s.$

$\dot{\gamma }=-\frac{g\cos 0^{\circ }}{V}=-\frac{9.81(1)}{180}=-0.0545\text{rad/s}=-3.12^{\circ }/\text{s}.$ Reading it: at $\gamma =0$ cosine is maximal, so gravity's turning power is at its strongest here — the arrow tips over fastest exactly at the apex. This is why trajectories "round over" sharply at the top: the flat part is where gravity bends hardest.

$\dot{V}=\frac{0-8000}{5000}-9.81\sin (-30^{\circ })=-1.6-9.81(-0.5)=-1.6+4.905=+3.305m/{\mathbf{s}}^2.$ Reading it: $\gamma$ is negative, so $\sin \gamma <0$, so the gravity term flips sign and helps the motion — a dive gains speed. Drag alone would slow it ($-1.6$), but gravity's downhill pull ($+4.905$) wins. It speeds up.

Speed: $\dot{V}=\frac{900000-120000}{30000}-9.81\sin 75^{\circ }=26-9.4767=16.523m/{\mathbf{s}}^2.$ Turn: $\dot{\gamma }=-\frac{9.81\cos 75^{\circ }}{250}=-\frac{2.5389}{250}=-0.010156rad/s(=-0.582^{\circ }/\text{s}).$ Position: $\dot{x}=250\cos 75^{\circ }=64.70m/s,\dot{h}=250\sin 75^{\circ }=241.48m/s.$ Mass: $\dot{m}=-\frac{T}{I_{sp}{g}_0}=-\frac{900000}{290\times 9.81}=-316.4kg/s.$ WHY the mass rule: this is the parent mass-flow rule $\dot{m}=-T/(I_{sp}{g}_0)$ — it says the propellant leaves at a rate set by thrust divided by exhaust efficiency ($I_{sp}{g}_0$ is the effective exhaust speed). We include it because the other four rules quietly assume a value of $m$; as $m$ shrinks, that same thrust buys ever more acceleration, so we must track $m$ to march the trajectory forward honestly. Story: still climbing steeply and accelerating hard, barely turning (steep $\gamma$ ⇒ small $\cos \gamma$), burning propellant at ~316 kg every second.

Euler = "assume the rate holds constant for the little step, walk that far." $V\approx 250+16.523(2)=283.05m/s.$ $\gamma \approx 75^{\circ }+(-0.582^{\circ }/\text{s})(2)=73.836^{\circ }.$ $h\approx h_0+241.48(2)=h_0+482.96\mathbf{m}.$ The arrow got longer, tipped over slightly, and rose ~483 m in 2 s. The figure below draws this exact step: the magenta arrow is the velocity direction at the start ($\gamma =75^{\circ }$), and the orange segment is where a single Euler step carries the point mass over those 2 seconds — notice it moves far more up ($h$) than along ($x$) because $\gamma$ is steep, matching $\dot{h}\gg \dot{x}$.

Put $\gamma =90^{\circ }$: $\cos 90^{\circ }=0$, so $\dot{\gamma }=-\frac{g\cos 90^{\circ }}{V}=-\frac{g\cdot 0}{V}=0.$ With $\dot{\gamma }=0$, $\gamma$ never leaves $90^{\circ }$ — vertical stays vertical. And $\sin 90^{\circ }=1$ gives $\dot{V}=\frac{T-D}{m}-g\cdot 1=\frac{T-D}{m}-g,$ exactly the ideal one-dimensional rocket. Why it makes sense: straight up, gravity's turning shadow ($\cos$) vanishes and its braking shadow ($\sin$) is total.

$\dot{\gamma }=-\frac{g\cos \gamma }{V}⟶-\infty \text{as }V\to 0^{+}.$ WHAT this means: at near-zero speed the tiniest off-vertical tilt would swing the arrow wildly — the model blows up. WHY reality doesn't: at lift-off the rocket is held vertical ($\gamma =90^{\circ }$, so $\cos \gamma =0$ cancels the tiny $V$), and only after speed builds is a small pitch-over introduced. Then $\cos \gamma$ is still small and $V$ is now large, so $\dot{\gamma }$ is gentle. That controlled hand-off is precisely the Gravity Turn Trajectory: pitch over only once you have speed, and let gravity do the bending for free.

WHY the full rule: here direction is changed not only by gravity but by an aerodynamic force $L$, so we must keep the $L$ term: $\dot{\gamma }=\frac{L-mg\cos \gamma }{mV}=\frac{80000-6000\cdot 9.81\cdot \cos 10^{\circ }}{6000\cdot 200}.$ Numerator: $80000-6000(9.81)(0.98481)=80000-57967.3=22032.7$. Denominator: $1,200,000$. $\dot{\gamma }=\frac{22032.7}{1200000}=+0.018361\text{rad/s}=+1.052^{\circ }/\text{s}.$ Reading it: $\dot{\gamma }$ is now positive — the path bends upward. Lift ($80$ kN) beat gravity's perpendicular shadow ($mg\cos \gamma \approx 58$ kN), so the net sideways force curls the velocity arrow up instead of down. This is an aerodynamic (lift-driven) turn, the opposite of the gravity turn: wings, not gravity, steer the arrow. Set $L=0$ and you recover the familiar $\dot{\gamma }=-g\cos \gamma /V<0$.

• (a) Speed loss $\propto \sin \gamma$. $\sin 70^{\circ }>\sin 20^{\circ }$, so the steeper ($70^{\circ }$) rocket bleeds more speed to gravity — climbing costs energy.
• (b) Turn rate $\propto \cos \gamma$. $\cos 20^{\circ }>\cos 70^{\circ }$, so the shallower ($20^{\circ }$) rocket bends faster — nearer the horizon gravity's sideways pull dominates. Picture: as the arrow tips from vertical toward horizontal, the "along" shadow shrinks and the "across" shadow grows — sine and cosine trade roles. This single trade-off is the heart of every ascent-profile design.

length of the velocity arrow, and which changes its direction? Length (speed) → $\dot{V}=(T-D)/m-g\sin \gamma$. Direction (turning) → $\dot{\gamma }=(L-mg\cos \gamma )/(mV)$.

$\gamma <0$) speed a rocket up even with drag present? Because $-g\sin \gamma$ becomes positive when $\gamma <0$ (gravity now pulls along motion); if it beats drag, $\dot{V}>0$.