Question bank — Pitch program — open-loop pitch-over

False — the ideal gravity turn keeps the nose aligned with velocity ($\alpha \approx 0$); gravity bends the velocity vector and the nose merely follows, so almost no active steering happens after the kick.

False — an attitude loop actively holds the vehicle on the commanded angle; what is open-loop is only that the ==command ${\theta }_c(t)$== is a fixed time schedule, not computed from measured state.

False — since $\dot{\gamma }=-g\cos \gamma /v$, turn rate $\propto 1/v$; more speed means more inertia, so gravity bends the velocity vector less, not more.

False — at $\gamma =90^{\circ }$, $\cos \gamma =0$ so $\dot{\gamma }=0$; the turn rate is exactly zero and stays zero forever. You need the initial kick to seed a nonzero $\cos \gamma$.

False — the pitch-over exists to redirect thrust toward the horizontal for orbital speed, which is required regardless of drag; drag only changes how gently and when you may turn.

False — too large a kick over-rotates: the vehicle turns toward horizontal too early, dives into denser air with high dynamic pressure, and suffers large loads and losses.

True — thrust lies along the body axis, and the body axis lies along velocity ($\alpha =0$), so its normal component is zero; only gravity's normal component $mg\cos \gamma$ turns the path.

True (roughly) — open-loop is used in the atmosphere precisely because active closed-loop maneuvering there would risk building dangerous $\alpha$; a gentle fixed schedule keeps $\alpha \approx 0$ through max-Q.

Error: in the ideal turn the engine points straight along the body axis and gravity does the turning; continuous gimballing would create $\alpha$ and side loads. Gimbal is used only for the brief kick and for stabilization.

Error: turning early happens at low $v$ and high dynamic pressure, producing large $\alpha$ and aerodynamic loads → structural risk and steering loss. You pitch gently and low to keep $\alpha \approx 0$ through max-Q.

Error: the correct law is $\dot{\gamma }=-g\cos \gamma /v$ (divide, not multiply). Speed sits in the denominator, so higher $v$ turns you slower.

Error: the minus sign means $\gamma$ decreases — it is exactly what bends the path from vertical toward horizontal. Without it there would be no turn at all.

Error: they are used in different phases — open-loop schedule through the dense atmosphere, then closed-loop guidance on upper stages where aero loads are gone and precise orbit targeting matters.

Error: pointing thrust partly against gravity while climbing still incurs gravity loss; the gravity turn minimizes steering loss (by keeping $\alpha \approx 0$), not all losses.

Because the densest air is near the pad; flying straight along the body axis with $\alpha \approx 0$ minimizes aerodynamic load, and a pre-tilt would create angle of attack right where dynamic pressure is climbing.

Because $\dot{\gamma }$ depends only on the current state ($g,v,\gamma$); once you set ${\gamma }_0$ (and the thrust-to-weight $n$), integrating the equation from there fixes the whole $v(\gamma )$ curve — the ascent is self-determined after the nudge.

Because $\dot{\gamma }\propto 1/v$ and $v$ grows through the flight; while $v$ is still small the turn rate is largest, so the vehicle sweeps most of its angle change early.

A slow vehicle has small $v$, so $\dot{\gamma }=-g\cos \gamma /v$ is large — it gravity-turns quickly on its own; a big kick on top would over-rotate it, so it needs a gentler nudge.

Aerodynamic side force scales like $\alpha \cdot q$; at high dynamic pressure even a modest $\alpha$ produces loads that can break the vehicle, so structural survival dominates until the air thins.

Tsiolkovsky gives total achievable $\Delta v$ from mass ratio, but says nothing about direction over time; the pitch program decides how that $\Delta v$ is split between fighting gravity and building horizontal speed.

$\cos \gamma$ stays near zero, so $\dot{\gamma }$ stays tiny and the vehicle never turns enough — at burnout it is still climbing steeply, wasting $\Delta v$ against gravity and failing to reach the near-horizontal orbital condition.

$\cos \gamma \to 1$, so the turn rate reaches its maximum magnitude $\dot{\gamma }\to -g/v$; but by then $v$ is large, so in practice the actual angular change per second is still small. It is also where you want the turn to nearly stop.

$\cos \gamma$ becomes negative, so $\dot{\gamma }=-g\cos \gamma /v$ becomes positive — gravity would push $\gamma$ back up toward $90^{\circ }$; the vertical point is an unstable balance the kick must decisively move you off of.

No — the kick itself is an active maneuver (small gimbal deflection) that creates a brief nonzero $\alpha$; the pure gravity turn ($\alpha \approx 0$) begins immediately after, once velocity and nose realign.

The law $\dot{\gamma }=-g\cos \gamma /v$ diverges (division by zero), which just signals the model is invalid at rest — you cannot bend a velocity vector that has no length. The gravity turn only makes sense once the vehicle has real speed.

Not necessarily — without aero loads there is no penalty for $\alpha$, so closed-loop guidance that optimally steers thrust becomes preferable; the gravity turn is a response to the atmospheric constraint.