Gravity turn trajectory — pitch rate from aerodynamic angle of attack = 0: Worked examples

Before we start, a two-second refresher on the symbols, so nothing is used unexplained:

The scenario matrix

Every question this topic can pose falls into one of these cells. The Example column tells you which worked example nails it.

Cell	Case class	What makes it special	Example
A	Near-vertical ( $γ \approx 9 0^{\circ}$ )	$cos γ \approx 0$ → tiny turn rate	Ex 1
B	Mid-lean ( $γ \approx 4 5^{\circ}$ )	$cos γ$ moderate → brisk turn	Ex 2
C	Near-horizontal ( $γ \approx 0^{\circ}$ )	$cos γ \approx 1$ → fastest turn	Ex 3
D	Degenerate: exactly vertical ( $γ = 9 0^{\circ}$ )	$cos 9 0^{\circ} = 0$ → $\overset{γ}{˙} = 0$ , no turn	Ex 4
E	Degenerate: exactly horizontal ( $γ = 0^{\circ}$ )	$cos 0^{\circ} = 1$ → max possible $\overset{γ}{˙}$	Ex 4
F	Speed sensitivity (vary $v$ , fix $γ$ )	tests $\overset{γ}{˙} \propto 1/ v$	Ex 5
G	Limiting behaviour ( $v \to \infty$ , $v \to 0$ )	asymptotes of the equation	Ex 6
H	Real-world word problem (find turn radius)	connect to Centripetal / Normal Acceleration	Ex 7
I	Exam twist ( $α \neq = 0$ , so $θ \neq = γ$ )	pitch angle ≠ path angle	Ex 8

We take $g = 9.81 m/s^{2}$ everywhere unless told otherwise.

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

Example 1 — Near-vertical (Cell A)

Worked example Just after pitch-over

A rocket has $v = 100 m/s$ and $γ = 8 9^{\circ}$ . Find the turn rate $\overset{γ}{˙}$ in $rad/s$ and in $^{\circ} / s$ .

Forecast: the nose is almost straight up. Will it bend fast or barely at all? (Guess before reading.)

Step 1 — Find $cos γ$ . $cos 8 9^{\circ} = 0.01745$ . Why this step? The turning force is the perpendicular slice of gravity, and that slice is $m g cos γ$ . Near vertical, gravity points almost exactly backward along the path, leaving almost nothing perpendicular — so $cos γ$ is tiny.

Step 2 — Plug into the boxed formula. $\overset{γ}{˙} = - \frac{9.81 \times 0.01745}{100} = - 1.712 \times 1 0^{- 3} rad/s$ Why this step? Direct substitution — the formula is already solved for $\overset{γ}{˙}$ .

Step 3 — Convert to degrees. $- 1.712 \times 1 0^{- 3} \times 57.30 = - 0.098 1^{\circ} / s$ . Why this step? A number like $0.098$ degrees per second feels slow; the radian value doesn't.

Verify: The rate is negative (nose tips down, correct) and minuscule — matching the Forecast that a near-vertical rocket barely turns. At this rate it would take about $1^{\circ} /0.098 \approx 10$ seconds to drop just one degree. ✅

Example 2 — Mid-lean (Cell B)

Example 3 — Near-horizontal (Cell C)

Example 4 — The two degenerate endpoints (Cells D & E)

Worked example Exactly vertical and exactly horizontal

Take $v = 100 m/s$ . Compute $\overset{γ}{˙}$ at (a) $γ = 9 0^{\circ}$ and (b) $γ = 0^{\circ}$ .

Forecast: one of these gives zero turn. Which one, and why? (Guess.)

Step 1 — Case (a), $γ = 9 0^{\circ}$ . $cos 9 0^{\circ} = 0$ , so $\overset{γ}{˙} = - \frac{9.81 \times 0}{100} = 0 rad/s .$ Why this step? Straight up, gravity is exactly anti-parallel to velocity — pure braking, no perpendicular slice. A perfectly vertical rocket never starts turning. This is why launches deliberately inject a tiny initial pitch-over to break the symmetry.

Step 2 — Case (b), $γ = 0^{\circ}$ . $cos 0^{\circ} = 1$ , so $\overset{γ}{˙} = - \frac{9.81 \times 1}{100} = - 0.0981 rad/s = - 5.62 2^{\circ} / s .$ Why this step? Flying flat, gravity is entirely perpendicular — the maximum possible turn rate at this speed.

Verify: Case (a) $= 0$ (no-turn endpoint, matches the Forecast) and case (b) $= - g / v = - 9.81/100$ exactly (the largest magnitude, since $cos γ \leq 1$ ). Both endpoints behave as the physics demands. ✅

Example 5 — Speed sensitivity (Cell F)

Example 6 — Limiting behaviour (Cell G)

Worked example What happens at the extremes of speed?

With $γ = 4 5^{\circ}$ fixed, describe $\overset{γ}{˙}$ as $v \to \infty$ and as $v \to 0^{+}$ .

Forecast: for very fast $v$ , does the turn rate approach a big number or vanish? (Guess.)

Step 1 — Large $v$ . $\overset{γ}{˙} = - g cos γ / v$ . As $v \to \infty$ the fraction $\to 0$ . Why this step? The faster you fly, the less time gravity has to nudge each metre of path — the trajectory becomes nearly straight. Mathematically $1/ v \to 0$ .

Step 2 — Tiny $v$ . As $v \to 0^{+}$ , $\overset{γ}{˙} = - g cos γ / v \to - \infty$ . Why this step? At vanishing speed, gravity has "forever" per metre — the direction whips around. In reality the rocket never sits at $v = 0$ pointed sideways, but the equation warns that low-speed turning is violently fast, which is exactly why the pitch-over is a tiny gentle kick, not a sharp one.

Verify: Check with numbers at $γ = 4 5^{\circ}$ : at $v = 1000$ , $\overset{γ}{˙} = - 9.81 (0.7071) /1000 = - 0.006936 rad/s$ (small, → 0 trend); at $v = 5$ , $\overset{γ}{˙} = - 9.81 (0.7071) /5 = - 1.387 rad/s$ (huge, → −∞ trend). Both confirm the limits. ✅

Example 7 — Real-world word problem: turn radius (Cell H)

Worked example How tight is the curve the rocket is carving?

A rocket flies at $v = 200 m/s$ , $γ = 3 0^{\circ}$ . Treat its curving path locally as an arc of a circle. Find (a) the turn rate $\overset{γ}{˙}$ , and (b) the radius of curvature $R$ of the flight path at this instant.

Forecast: will the radius be a few metres, or kilometres? (Guess — a fast rocket bends gently.)

Step 1 — Turn rate. $cos 3 0^{\circ} = 0.86603$ , $\overset{γ}{˙} = - \frac{9.81 \times 0.86603}{200} = - 0.042479 rad/s .$

Step 2 — Relate to radius. A point moving at speed $v$ along a circle of radius $R$ sweeps angle at rate $∣ \overset{γ}{˙} ∣ = v / R$ (this is Centripetal / Normal Acceleration: $a_{⊥} = v \overset{γ}{˙} = v^{2} / R$ ). So $R = \frac{v}{∣ γ ˙ ∣} .$ Why this step? The parent note noted $r = v / \overset{γ}{˙}$ ; the perpendicular gravity slice supplies exactly the centripetal turning, so the instantaneous circle has radius $v /∣ \overset{γ}{˙} ∣$ .

Step 3 — Compute. $R = \frac{200}{0.042479} = 4708 m \approx 4.71 km .$

Verify (two ways).

Substitute the formula directly: $R = v /∣ \overset{γ}{˙} ∣ = v \cdot v / (g cos γ) = v^{2} / (g cos γ) = 20 0^{2} / (9.81 \times 0.86603) = 4708 m$ . ✅
Units: $\frac{m/s}{rad/s} = m$ (radians are dimensionless). ✅ The multi-kilometre radius matches the Forecast that fast rockets bend gently.

Example 8 — Exam twist: pitch angle when $α \neq = 0$ (Cell I)

Worked example The trap: nose ≠ velocity

A gust briefly gives the rocket an angle of attack $α = 4^{\circ}$ while its flight-path angle is $γ = 5 5^{\circ}$ . What is the pitch angle $θ$ (the direction the nose points)? Is this still a pure gravity turn?

Forecast: in the ideal gravity turn $θ = γ$ . With a gust, is $θ$ bigger or smaller than $γ$ ? (Guess.)

Step 1 — Recall the general relation. From Flight-path angle and pitch angle, the pitch angle, flight-path angle and angle of attack always satisfy $θ = γ + α .$ Why this step? $α$ is defined as the angle between where the nose points ( $θ$ ) and where the velocity points ( $γ$ ). So the nose is $α$ above the velocity.

Step 2 — Plug in. $θ = 5 5^{\circ} + 4^{\circ} = 5 9^{\circ}$ .

Step 3 — Is it still a gravity turn? No. A gravity turn requires $α = 0$ . With $α = 4^{\circ}$ the wings/body generate lift, thrust now has a small normal component, and extra structural side-loads appear near Aerodynamic Drag and Max-Q. The pitch-rate equation $\overset{γ}{˙} = - g cos γ / v$ no longer captures the full turning — there's now an aerodynamic contribution.

Verify: Sanity check the special case: if the gust vanishes ( $α \to 0$ ), $θ = γ + 0 = γ = 5 5^{\circ}$ , recovering the gravity-turn condition $θ = γ$ . The formula degrades gracefully to the ideal case. ✅

Active recall

Recall Test yourself (hide the answers)

Near vertical ( $γ \to 9 0^{\circ}$ ), why is the turn rate tiny? ::: $cos γ \to 0$ , so gravity's perpendicular slice vanishes. Which endpoint gives $\overset{γ}{˙} = 0$ exactly? ::: $γ = 9 0^{\circ}$ (straight up). Doubling $v$ at fixed $γ$ does what to $∣ \overset{γ}{˙} ∣$ ? ::: Halves it, since $\overset{γ}{˙} \propto 1/ v$ . Formula for the instantaneous turn radius? ::: $R = v /∣ \overset{γ}{˙} ∣ = v^{2} / (g cos γ)$ . If $α = 4^{\circ}$ and $γ = 5 5^{\circ}$ , what is $θ$ ? ::: $θ = γ + α = 5 9^{\circ}$ . As $v \to \infty$ at fixed $γ$ , what happens to $\overset{γ}{˙}$ ? ::: It $\to 0$ ; the path straightens out.