Exercises — Trajectory optimization — minimum gravity loss, minimum drag loss

WHAT: identify which term each extreme angle maximises. Gravity loss integrand is $g\sin \gamma$. At $\gamma =90^{\circ }$, $\sin 90^{\circ }=1$ — the largest value $\sin$ can take. So gravity loss is MAXIMUM. Meanwhile, going straight up you punch out of the thick air fastest and keep speed low while still in dense air, so drag loss is (roughly) minimum. WHY IT LOOKS LIKE THIS: in figure s01, the down-arrow lands entirely along the path when the path points straight up — the whole weight fights the motion.

WHAT: trace each sentence back to the physical thief in its integral. (a) drag loss $\Delta v_{\text{drag}}$. WHY: "air resistance" is the drag force $D=\frac{1}{2}\rho v^2{C}_{D}A$; dividing by $m$ and integrating over time gives the $\Delta v$ eaten — exactly the drag integral. (b) gravity loss $\Delta v_{\text{grav}}$. WHY: "gravity pulls against motion" is the along-path piece $g\sin \gamma$ (the pink arrow in figure s01); its time-integral is the gravity loss. (c) steering loss $\Delta v_{\text{steer}}$. WHY: aiming thrust off the velocity by an angle delivers only $\cos (\text{angle})$ of it forward; the wasted fraction is the steering loss — see Steering losses and thrust vectoring.

WHAT: plug into ${\int }_0^{t_b}g\sin \gamma dt$ with constant integrand. $\sin 90^{\circ }=1$, and $g$ is constant, so the integral is just $g\times t_b$: $\Delta v_{\text{grav}}=9.7\times 45\times 1=436.5\text{m/s}$ WHY: a constant value integrated over time is value × time — the area of a rectangle of height $9.7$ and width $45$.

WHAT: same formula, new angle. $\Delta v_{\text{grav}}=9.7\times 45\times \sin 25^{\circ }=436.5\times 0.4226\approx 184.5\text{m/s}$ Factor $=\sin 25^{\circ }\approx 0.423$, i.e. it dropped to about $42\%$ of the vertical value. WHY: only $\sin \gamma$ changed. Flying shallower means less of gravity opposes your motion.

WHAT: first find drag force $D$, then the deceleration $D/m$, then multiply by time. $D=\frac{1}{2}\rho v^2{C}_{D}A=\frac{1}{2}(0.5)(400{)}^2(0.28)(1.2)=13440\text{N}$ $\Delta v_{\text{drag}}\approx \frac{D}{m}\Delta t=\frac{13440}{2.5\times 10^4}\times 15\approx 8.06\text{m/s}$ WHY: $D/m$ is deceleration (Newton's 2nd law); held roughly constant over $\Delta t$, the speed lost is deceleration × time. See Aerodynamic Drag and Max-Q.

WHAT: compute both, then interpret. $\Delta v_{\text{grav,A}}=9.7\times 50=485\text{m/s},\Delta v_{\text{grav,B}}=9.7\times 25=242.5\text{m/s}$ B loses half. WHY: gravity loss $\int g\sin \gamma dt$ depends on time spent fighting gravity, not on altitude gained. Burn faster (higher thrust-to-weight) and you spend less time lifting ⇒ less gravity loss. This is exactly the parent's warning: gravity loss is about angle and duration, not height.

WHAT: we have a sum of two competing terms — the classic see-saw. To find the bottom of a valley, we ask "where is the slope zero?" That is why we differentiate: the derivative $dL/d\gamma$ is the slope of the loss curve, and the minimum sits where the slope flips from negative to positive. $\frac{dL}{d\gamma }=K\cos \gamma +C\cdot 2\cos \gamma (-\sin \gamma )=K\cos \gamma -2C\sin \gamma \cos \gamma$ Factor out $\cos \gamma$ (nonzero for $\gamma <90^{\circ }$): $\cos \gamma (K-2C\sin \gamma )=0\Rightarrow \sin \gamma =\frac{K}{2C}=\frac{400}{1800}=0.2222$ ${\gamma }_{\star }=\arcsin (0.2222)\approx 12.84^{\circ }$ WHAT IT LOOKS LIKE: figure s02 plots the pink gravity term $K\sin \gamma$ rising with angle, the blue drag proxy $C{\cos }^2\gamma$ falling with angle, and the yellow total $L$. The stationary point ${\gamma }_{\star }$ is marked with the dotted line — but notice the total curve does not dip there; it is actually a small hump. We resolve this by checking endpoints next.

Evaluate $L$ at candidates:

• $\gamma =0^{\circ }$: $L=0+900(1)=900$
• ${\gamma }_{\star }=12.84^{\circ }$: $L=400(0.2222)+900(0.9506)=88.9+855.6=944.4$
• $\gamma =90^{\circ }$: $L=400(1)+0=400$ So the stationary point is a maximum of this particular toy model, and the true minimum over the interval is at the endpoint $\gamma =90^{\circ }$. Lesson: always check endpoints — a zero-derivative point can be a hilltop, not a valley. (Real ascent uses the ${\cos }^2$ proxy only for the low-altitude segment; over the whole flight the physical minimum is genuinely interior. Here we test your calculus rigour.)

WHAT: apply the turn-rate law $\frac{d\gamma }{dt}=-\frac{g\cos \gamma }{v}$ from the opening formula box. $\frac{d\gamma }{dt}=-\frac{9.8\cos 75^{\circ }}{250}=-\frac{9.8(0.2588)}{250}=-0.01014\text{rad/s}$ Convert: $-0.01014\times \frac{180}{\pi }\approx -0.581^{\circ }/\text{s}$. WHY: only the cross-path gravity $g\cos \gamma$ curves the trajectory, for free. The turn is slow because $v$ is already moderate — this is why the initial pitch kick must be planted while $v$ is small.

WHAT: rearrange the bookkeeping identity. $\Delta v_{\text{ideal}}=\Delta v_{\text{orbit}}+\Delta v_{\text{grav}}+\Delta v_{\text{drag}}+\Delta v_{\text{steer}}$ $\Delta v_{\text{grav}}=9100-7800-90-40=1170\text{m/s}$ WHY: the ideal $\Delta v$ from Tsiolkovsky is split among useful orbital speed and the three thieves. Solve for the one unknown. Note gravity loss ($1170$) dwarfs drag ($90$) — the topic's central fact.

WHAT & WHY: the turn-rate equation ties $\gamma$ and $t$ together. To get time from angle change, we separate the two variables onto opposite sides — that's the whole point of "separation of variables": collect all $\gamma$ on the left, all $t$ on the right, then integrate each independently. $\frac{d\gamma }{-\cos \gamma }=\frac{g}{v}dt\Rightarrow -{\int }_{{\gamma }_0}^{{\gamma }_1}\sec \gamma d\gamma =\frac{g}{v}{\int }_0^{t}dt$ $-[\ln |\sec \gamma +\tan \gamma |{]}_{89^{\circ }}^{45^{\circ }}=\frac{g}{v}t$ Compute each end (angles in radians for the trig, values are unitless):

• at $45^{\circ }$: $\sec 45^{\circ }+\tan 45^{\circ }=1.4142+1=2.4142$, $\ln =0.8814$
• at $89^{\circ }$: $\sec 89^{\circ }+\tan 89^{\circ }=57.30+57.29=114.59$, $\ln =4.7413$ $-(0.8814-4.7413)=3.8599=\frac{9.7}{300}t$ $t=3.8599\times \frac{300}{9.7}\approx 119.4\text{s}$ WHAT IT LOOKS LIKE (figure s03): the yellow $\gamma (t)$ curve is nearly flat near $89^{\circ }$ (turn rate $\propto \cos \gamma$ is tiny when vertical) then steepens as $\gamma$ falls toward the blue $45^{\circ }$ target line. That flat start is why you must begin the kick early — the vehicle barely turns while near-vertical.

WHAT: plug into the steering-loss integral with constants. $\Delta v_{\text{steer}}=\frac{T}{m}(1-\cos \alpha )\Delta t=20(1-\cos 6^{\circ })40$ $\cos 6^{\circ }=0.99452$, so $1-\cos 6^{\circ }=0.005478$: $\Delta v_{\text{steer}}=20\times 0.005478\times 40\approx 4.38\text{m/s}$ WHY: aiming thrust off-velocity by $\alpha$ delivers only $\cos \alpha$ of it forward; the lost fraction $(1-\cos \alpha )$ is pure waste. The gravity turn keeps $\alpha =0$ and pays nothing — that is its elegance (see Steering losses and thrust vectoring). Small $\alpha$ looks cheap here, but over a long forced turn it accumulates, and larger $\alpha$ grows the loss as ${\alpha }^2$ (since $1-\cos \alpha \approx {\alpha }^2/2$).

A rocket has one fuel budget ($\Delta v_{\text{ideal}}$). Three thieves nibble it: gravity (biggest, ~1 km/s, worse the more vertical and the longer you burn), drag (small, worst when fast in thick air), and steering (worst when you point the engine sideways). The clever trick is the gravity turn: tip over a hair at the start, then let gravity itself swing your velocity horizontal for free — paying zero steering, spending just enough time vertical to dodge the drag wall.