Visual walkthrough — Gravity turn trajectory — pitch rate from aerodynamic angle of attack = 0

This page is the picture-first companion to the parent topic inside Rocket Flight Mechanics.

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Step 1 — Draw the rocket as a moving dot with an arrow

WHAT. Forget the tall shiny cylinder. A rocket, for our physics, is just a point (its mass, all squished to one spot) with a velocity arrow attached — an arrow that says "this is the direction I am moving, and its length is how fast."

WHY. Turning is a change in direction of that arrow. If we want to talk about "how fast the rocket turns," we must first have a single clean arrow whose swinging we can measure. A whole rocket shape would distract us.

PICTURE. Look at the burnt-orange arrow. Its tail is the rocket's current position; its length is the speed we will call $v$ (metres per second). The angle it makes above the flat horizon is drawn in teal — that angle is the star of the whole show.

Because the rocket flies with ==angle of attack $\alpha =0$== (nose exactly along velocity), the direction the nose points ($\theta$) equals the direction it moves ($\gamma$). So "how fast the nose pitches" is "how fast $\gamma$ changes." One arrow tells us everything.

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Step 2 — What does "turning" even mean? Watch the arrow swing

WHAT. Let a tiny sliver of time pass — call it $dt$ ("a little bit of time," so small the numbers barely change). In that instant the arrow swings by a tiny angle, which we call $d\gamma$ ("a little bit of change in $\gamma$").

WHY. We want a rate: change in angle per second. That is exactly $\frac{d\gamma }{dt}$ — read aloud as "d-gamma d-t," meaning "how much $\gamma$ changes for each second that passes." This fraction is the whole quantity we are hunting.

PICTURE. The old arrow (faint) and the new arrow (solid) sit nearly on top of each other; the plum wedge between them is $d\gamma$. Notice it swings downward — toward the horizon. That downward swing is why we will later see a minus sign.

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Step 3 — The tip of the arrow travels on a curve; find its sideways nudge

WHAT. If the arrow keeps swinging, the rocket's path is a curve, not a straight line. To bend a path, something must push the rocket sideways — perpendicular to where it's already going. Call that sideways acceleration $a_{\perp }$ ("a-perp," perpendicular acceleration).

WHY. Newton says a force is needed to change velocity's direction, not just its speed. The sideways push is what curves the path. We need a formula for how big a sideways push produces how fast a turn.

PICTURE. The rocket moves a little distance $vdt$ along the curve (orange). In that step the direction swung by $d\gamma$. Geometry of a small arc says the sideways displacement is $vdt\times d\gamma$, so the sideways acceleration is speed times turn-rate:

Read it backwards: if we know the sideways acceleration, we know the turn rate — $\dot{\gamma }=a_{\perp }/v$. So the hunt narrows to: what provides the sideways acceleration?

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Step 4 — Only gravity can push sideways (because $\alpha =0$)

WHAT. List every force. Thrust $T$ pushes forward. Drag $D$ pushes backward. Gravity $mg$ pulls straight down. Because the nose is exactly along velocity ($\alpha =0$), thrust and drag are purely along the arrow — they can speed us up or slow us down but they cannot push sideways.

WHY. Turning needs a sideways force. Thrust and drag have exactly zero sideways part here, so they are ruled out. That leaves gravity as the only candidate to bend the path. This is the deep reason the manoeuvre is called a gravity turn.

PICTURE. The orange (thrust) and its opposing drag lie flat along the arrow — no sideways reach. Only the grey gravity arrow, pointing straight down, leans across the direction of motion.

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Step 5 — Split gravity into "along" and "sideways" using a right triangle

WHAT. Gravity points straight down with size $mg$. We break it into two arrows: one along the velocity (backward when climbing) and one perpendicular (the sideways one we care about). This splitting is called resolving into components.

WHY. Only the perpendicular part turns the rocket. To use it we must measure exactly how big it is. A right triangle turns "straight-down gravity + tilted velocity" into two clean numbers.

PICTURE. Build the right triangle: the downward gravity arrow $mg$ is the hypotenuse; the velocity direction sits at angle $\gamma$ from the horizon. Drop a perpendicular. The two legs are:

• along velocity: $mg\sin \gamma$
• perpendicular to velocity: $mg\cos \gamma$

Why does $\cos$ land on the perpendicular leg? Because the angle between "straight down" and "perpendicular-to-velocity" is $\gamma$ itself, and $\cos$ is "adjacent over hypotenuse" — the leg hugging that angle. Trace the plum leg in the figure: that is $mg\cos \gamma$, our sideways push.

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Step 6 — Set sideways force equal to sideways acceleration (Newton)

WHAT. Newton's law along the perpendicular direction: (sideways force) $=$ (mass) $\times$ (sideways acceleration). The sideways force is gravity's perpendicular leg pulling the arrow down — so it is negative (it reduces $\gamma$). The sideways acceleration is $v\dot{\gamma }$ from Step 3.

WHY. This is the one honest equation that connects the cause (gravity's perpendicular leg) to the effect (the turn). Everything before was building both sides; now we glue them.

PICTURE. Left side of the balance: the plum sideways-gravity leg $-mg\cos \gamma$. Right side: $m$ times the sideways acceleration $v\dot{\gamma }$. The scale balances.

Now cancel the mass $m$ from both sides (it appears in every term — a light rocket and a heavy rocket turn the same way, because gravity accelerates all masses equally), then divide by $v$:

Since $\alpha =0\Rightarrow \theta =\gamma$, this is also the pitch rate $\dot{\theta }$.

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Step 7 — The two edge cases the equation quietly handles

WHAT. Check the extremes so no scenario surprises us.

WHY. A formula you trust must survive its boundaries. Two matter: perfectly vertical ($\gamma =90^{\circ }$) and perfectly level ($\gamma =0^{\circ }$).

PICTURE. Two mini-scenes.

• Vertical, $\gamma =90^{\circ }$: gravity points exactly opposite the arrow. Its sideways leg has length $mg\cos 90^{\circ }=0$. No sideways push → $\dot{\gamma }=0$. A perfectly vertical rocket never turns on its own — that's why we give a tiny deliberate pitch-over to start things (see Ascent Guidance and Pitch Program).
• Level, $\gamma =0^{\circ }$: gravity is entirely perpendicular to the arrow; $mg\cos 0^{\circ }=mg$, the full weight bends the path. Turning is fastest here (for a given speed).

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The one-picture summary

Everything on one canvas: the arrow at angle $\gamma$, the down-gravity split into its along-leg $mg\sin \gamma$ and its all-important perpendicular leg $mg\cos \gamma$, that leg feeding the sideways acceleration $v\dot{\gamma }$, and the resulting curved path bending toward the horizon — labelled with the final boxed law.

Recall Feynman retelling — say it back in plain words

A rocket is a dot with a speed-arrow tilted at angle $\gamma$ above the horizon. To turn, its arrow must swing, and to swing it needs a sideways push. With the nose glued along the velocity ($\alpha =0$), the engine and the drag both point straight along the arrow — they can only change how fast, never which way. So the only thing left to steer with is gravity. Gravity points straight down; only the slice of it that lies across the arrow can turn the path, and a right triangle says that slice is $mg\cos \gamma$. Newton's sideways law, "sideways force $=$ mass times sideways acceleration," reads $mv\dot{\gamma }=-mg\cos \gamma$. Cancel the mass, divide by speed, and out drops $\dot{\gamma }=-g\cos \gamma /v$: the turn is downward (minus), driven by gravity ($g$), using only its across-slice ($\cos \gamma$), and slowed by speed ($1/v$). Straight up, $\cos 90^{\circ }=0$, so nothing turns — hence the gentle nudge at launch. Level flight, $\cos 0^{\circ }=1$, turns hardest.

Recall Quick self-test

Which force turns the rocket? ::: The perpendicular slice of gravity, $mg\cos \gamma$. Why don't thrust and drag turn it? ::: With $\alpha =0$ they lie along the velocity, so their sideways part is zero. What is the sideways acceleration of a turning body? ::: $v\dot{\gamma }$ (speed times turn rate). What makes $\dot{\gamma }=0$ at launch? ::: $\gamma =90^{\circ }$ gives $\cos 90^{\circ }=0$, no sideways gravity. Does doubling $v$ speed up or slow the turn? ::: Slows it — $\dot{\gamma }\propto 1/v$, so it halves.