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

This page assumes nothing. The parent topic ends at one boxed equation, but that equation is written in a language of symbols — $\gamma$, $t$, $g$, $v$, $\cos$, and a strange fraction $\frac{d\gamma }{dt}$ — that we have not learned yet. So we will not write that equation here at the top. Instead we build every letter of it from scratch, one at a time, each on top of the last, each anchored to a picture. Only on the very last line, once every symbol is earned, will the full equation appear.

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1. Angle above the horizon — the picture of "how steep"

Everything in this topic is about angles measured up from the flat ground.

Look at the figure: the gray line is the horizon, and the angle is the little wedge between it and the arrow. A tiny wedge means "almost flat"; a wedge near a quarter-turn means "almost straight up". This is the only kind of angle in this topic, so hold onto the picture: wedge measured up from flat ground.

Why do we need this? Because a rocket's whole job is to go from straight up (to escape the ground and the thick air) to nearly sideways (to go fast enough to orbit). That journey is just this wedge shrinking from $90^{\circ }$ toward $0^{\circ }$.

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2. Two different arrows: where the nose points vs. where it moves

Here is the subtlety that the whole topic hinges on. A rocket has two directions that look like they should be the same but aren't always:

• The symbol $\theta$ is the Greek letter "theta". The symbol $\gamma$ is "gamma". They are just names, like $x$ and $y$, chosen by tradition.

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3. Angle of attack $\alpha$ — the gap between the two arrows

The single most important idea of the whole parent topic is the special case:

Why does the topic care so much about $\alpha =0$? Because the air only pushes sideways on a rocket when the nose is tilted away from the airflow — that sideways push is called lift, and it bends and stresses the structure. If $\alpha =0$, the air flows straight down the body and there is no sideways air-push. Keep this fact in your pocket — in Section 5 it will kill two whole forces. See Aerodynamic Drag and Max-Q for why that sideways push is so dangerous in thick air.

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4. Speed $v$ — how fast, ignoring direction

Picture the velocity arrow in Figure 2: its length is $v$, its tilt is $\gamma$. That split — length vs. tilt — is exactly why the final equation will carry both symbols.

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5. The forces: $T$, $D$, $m$, $g$ — and which ones can turn the rocket

We now name the four things that push and pull the rocket.

The crucial geometry, shown in the figure: thrust and drag point along the velocity line (because $\alpha =0$ makes nose = velocity), but gravity points stubbornly straight down, at an angle to that line. That mismatch is the entire story.

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6. Splitting a force into "along" and "across" — trigonometry earned

Gravity points down; the rocket moves at tilt $\gamma$. To use gravity we must ask: how much of it pushes backward along the path, and how much pushes sideways across the path? Splitting one arrow into two perpendicular arrows is called resolving into components.

Applying this to gravity $mg$ (the hypotenuse), reading the triangle in Figure 4:

• Along-the-path part $mg\sin \gamma$: slows the climb. When vertical ($\gamma =90^{\circ }$), $\sin 90^{\circ }=1$, so gravity fights the climb head-on — makes sense.
• Across-the-path part $mg\cos \gamma$: this is the turning force. When vertical, $\cos 90^{\circ }=0$, so there is no sideways tug — a vertical rocket cannot start turning. When flat ($\gamma =0$), $\cos 0^{\circ }=1$, gravity pulls fully sideways-to-motion.

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7. Rate of change $\frac{d()}{dt}$ — "how fast is it changing right now"

We have $\gamma$ (the tilt) and gravity's sideways pull. But the topic wants to know how fast $\gamma$ shrinks — a rate. We need a symbol for "how fast a quantity changes per second".

• Likewise $\frac{dv}{dt}$ is how fast the speed changes (acceleration along the path).
• The dot notation $\dot{\gamma }$ means exactly the same thing as $\frac{d\gamma }{dt}$ — a dot on top is just shorthand for "rate of change per second". We only use $\dot{\gamma }$ after this definition, never before.
• Units caution: when this rate sits inside the physics formula, $\dot{\gamma }$ comes out in radians per second. We may report it in degrees per second for feel, but only after computing in radians.

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8. Normal (turning) acceleration $v\dot{\gamma }$ — derived, not asserted

When something moving turns, its direction changes even if its speed doesn't. Changing direction is itself an acceleration, and by Newton it needs a sideways force. Let us build the formula rather than quote it.

So the faster you're going ($v$) or the faster the arrow swings ($\dot{\gamma }$), the bigger the sideways acceleration. This is the curved-motion cousin of the familiar $v^2/r$; here the turn radius is $r=v/\dot{\gamma }$. Deeper detail lives in Centripetal / Normal Acceleration.

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Putting it together (preview of the parent result)

Now — and only now — every symbol is earned, so we may finally write the parent equation. Newton across the path says: (mass)$\times$(sideways acceleration) = (sideways force). The sideways acceleration is $v\dot{\gamma }$ (Section 8); the sideways force is gravity's across-path part, $-mg\cos \gamma$ (Section 6), negative because it pulls $\gamma$ down: $mv\dot{\gamma }=-mg\cos \gamma .$ Cancel the mass $m$ from both sides: $\boxed{\frac{d\gamma }{dt}=-\frac{g\cos \gamma }{v}}$ Read it back with your new eyes: the sideways fraction of gravity ($\cos \gamma$), scaled by gravity's strength ($g$), shared out over how fast you're already going ($v$) — that's your turn rate. (And remember: put $\gamma$ in radians to compute, and the answer $\dot{\gamma }$ comes out in radians per second.) The full derivation is the parent note the parent topic within Rocket Flight Mechanics; see also Ascent Guidance and Pitch Program and Tsiolkovsky Rocket Equation for what happens to $v$ and $m$ over the flight.

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Prerequisite map

The diagram below shows how each foundation you just built feeds into the final pitch-rate equation. Read arrows as "is needed for". Notice the two streams: the left stream (angles → components → sideways gravity pull) supplies the force; the right stream (speed and rate of change → normal acceleration) supplies the turning. They meet at the boxed equation.

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Equipment checklist

Test yourself: cover the answer after the dash and say it aloud. If any answer is fuzzy, reread its section.

• What does "angle above the horizon" measure, and what sign is "below"? — The upward tilt from the flat ground line; $0^{\circ }$ = sideways, $+90^{\circ }$ = straight up, negative = pointing below the horizon (descending).
• Why must $\gamma$ be in radians inside the formula? — Because $\sin$ and $\cos$ in physics equations expect radians; degrees give wrong numbers unless converted ($180^{\circ }=\pi$).
• What is the pitch angle $\theta$? — The direction the rocket's nose (and thrust) points, measured above the horizon.
• What is the flight-path angle $\gamma$? — The direction the rocket actually travels (its velocity), measured above the horizon.
• What is the angle of attack $\alpha$? — The signed gap between nose and travel directions: $\alpha =\theta -\gamma$.
• What does $\alpha =0$ force to be true, and which forces does it disable? — $\theta =\gamma$; and thrust and drag then have no across-path component, so they cannot turn the rocket.
• What does the symbol $v$ stand for? — Speed — the length of the velocity arrow, with direction stripped away.
• On the gravity triangle, which part turns the rocket? — The across-path part $mg\cos \gamma$; the along-path part $mg\sin \gamma$ only slows the climb.
• Why is there no sideways tug when vertical? — $\cos 90^{\circ }=0$, so gravity's across-path component vanishes.
• What does $\frac{d\gamma }{dt}$ (same as $\dot{\gamma }$) mean? — How fast the flight-path angle is changing per second right now (negative = tipping over).
• Where does the normal acceleration $v\dot{\gamma }$ come from? — The velocity arrow (length $v$) swinging by $d\gamma$ moves its tip by $vd\gamma$; divide by $dt$ to get $v\dot{\gamma }$.