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

False. With $\alpha =0$ thrust points exactly along velocity, so its component perpendicular to the path is zero — only gravity's normal component $mg\cos \gamma$ bends the trajectory.

True. Lift is a force perpendicular to velocity generated by $\alpha \ne 0$; since $\alpha =0$ the airflow hits the vehicle head-on, so lift vanishes and only drag (tangential) remains.

False. At $\gamma =90^{\circ }$, $\cos 90^{\circ }=0$, so $\dot{\gamma }=0$. Gravity is anti-parallel to velocity and has no sideways component to start the turn.

False. $\dot{\gamma }\propto 1/v$, so doubling $v$ halves $|\dot{\gamma }|$ — gravity gets less time to bend each metre of path.

True for this trajectory only, because $\theta =\gamma +\alpha$ and $\alpha =0$. If a gust or manoeuvre made $\alpha \ne 0$, the two would separate.

True. $\dot{\gamma }\propto \cos \gamma$, and $\cos \gamma$ is largest ($=1$) at $\gamma =0^{\circ }$, so the horizontal-flight instant has the maximum turn rate for a given speed.

False. Drag acts along the velocity line (opposing motion), so it is purely tangential — it slows the rocket but has zero normal component to curve the path.

False. High $v$ makes $\dot{\gamma }$ tiny, so a fast upper stage barely bends — most of the pitching must be done earlier, while slow and low.

Error: that describes thrust-vector steering, not a gravity turn. A gravity turn keeps thrust along velocity ($\alpha =0$); the sideways force is gravity's normal component, requiring no steering effort.

Error: the minus sign only means $\gamma$ decreases (nose tips from vertical toward horizontal). $\gamma$ approaches $0^{\circ }$ (horizontal), not negative territory, during powered ascent.

Error: $\alpha =0$ only sets the angle between nose and velocity to zero. Thrust, drag, and gravity are all still present and generally do not cancel; the vehicle accelerates.

No mistake: for curved motion the radius satisfies $r=v/\dot{\gamma }$, so $v^2/r=v^2\cdot (\dot{\gamma }/v)=v\dot{\gamma }$. The two expressions are identical.

Error: you launch vertically to clear the tower and cross the dense low atmosphere quickly, minimising drag and structural loads. A tiny pitch-over then seeds the turn; an angled launch raises drag and bending stress. (See Aerodynamic Drag and Max-Q.)

Error: the $\sin \gamma$ term is tangential — it slows the climb, changing speed, not direction. The turning term is the normal component $-mg\cos \gamma$.

Error: $\gamma$ is the velocity's angle above the horizon; $\alpha$ is the angle between the nose and the velocity. They are related through $\theta =\gamma +\alpha$, not equal.

Bending loads come from lift, a side force that scales with $\alpha$ and dynamic pressure. With $\alpha =0$ lift is zero, so the airframe feels no aerodynamic side-load even in dense air.

Because $\dot{\gamma }\propto 1/v$, low speed gives the largest turn rate per second, so a small nudge early sets up the whole trajectory efficiently. (See Ascent Guidance and Pitch Program.)

Both the required force ($mv\dot{\gamma }$) and the available turning force ($mg\cos \gamma$) carry a factor $m$. Like a free-falling body, the acceleration is mass-independent, so $\dot{\gamma }$ depends only on $g$, $\gamma$, and $v$.

Gravity already has a component perpendicular to a tilted velocity, so it bends the path with no propellant cost. Making thrust turn the vehicle would require $\alpha \ne 0$, which wastes energy on lift and drag.

Because $\alpha =0$ eliminates aerodynamic lift entirely; the only curving agent left is gravity, hence "gravity turn" and "zero-lift" name the same condition.

As $\gamma$ drops from $90^{\circ }$, $\cos \gamma$ grows, boosting $\dot{\gamma }$; but the climbing speed $v$ also grows, and past some point the $1/v$ effect dominates and the turn rate falls. It is a tug-of-war between $\cos \gamma$ rising and $v$ rising.

The instantaneous radius of the ascent path is hard to state directly, but the direction rate $\dot{\gamma }$ is exactly what we want. Writing $a_{\perp }=v\dot{\gamma }$ puts the equation directly in terms of the angle we are solving for. (See Centripetal / Normal Acceleration.)

$\cos 90^{\circ }=0$, so $\dot{\gamma }=0$. The rocket stays vertical forever — this degenerate case is why a deliberate initial pitch-over is needed to break symmetry.

$\cos \gamma \to 1$, giving the maximum turn rate $|\dot{\gamma }|=g/v$ for that speed — gravity is now fully perpendicular to velocity and bends it most effectively.

$\dot{\gamma }=-g\cos \gamma /v$ blows up as $v\to 0$, predicting an infinitely fast turn. Physically this is why the rocket must gain some vertical speed before pitching over, and why the flat-plane model is only valid once $v$ is nonzero.

$\cos (\text{small negative})\approx 1>0$, so $\dot{\gamma }$ stays negative — the path would keep bending downward toward the ground, exactly the diving behaviour you'd expect; the model remains consistent.

No. With $\alpha \ne 0$ thrust and lift gain normal components, adding terms to the turning equation, and $\theta =\gamma +\alpha$ so pitch rate and flight-path rate differ. The boxed formula is a zero-$\alpha$ special case.

$\dot{\gamma }\to 0$: with no gravity there is no normal force at $\alpha =0$, so the path cannot bend on its own. Turning then genuinely requires thrust vectoring ($\alpha \ne 0$), unlike ascent through a planet's gravity.