Question bank — Trajectory optimization — minimum gravity loss, minimum drag loss

False. Gravity loss $\int g\sin \gamma dt$ does vanish at $\gamma =0^{\circ }$, but flying flat and fast in dense low air makes $\frac{1}{2}\rho v^2$ enormous, so drag loss explodes — you must optimize the sum, not one term.

False. Gravity loss is $\int g\sin \gamma dt$ — it depends on flight-path angle and burn time, not on altitude reached. Confusing it with potential energy $mgh$ is the trap; a slow rocket loitering vertically loses far more than a fast one reaching the same height.

True. In a gravity turn thrust is kept along the velocity vector ($\alpha =0$), so no thrust is wasted sideways; the trajectory is bent only by gravity's free perpendicular component.

False. Drag loss is typically 30–150 m/s, while gravity loss runs 1–2 km/s. Gravity loss dominates the optimization — that is why the pitch program cares mostly about turning horizontal in time.

False. Off-axis thrust at angle $\alpha$ delivers only $\cos \alpha$ forward; the shortfall $\int \frac{T}{m}(1-\cos \alpha )dt$ is steering loss (see Steering losses and thrust vectoring). The gravity turn avoids it by keeping $\alpha =0$.

True. Higher thrust-to-weight means a shorter burn, and gravity loss $\int g\sin \gamma dt$ shrinks when the burn time $t_b$ shrinks — you spend less time fighting gravity.

True. With $\sin 90^{\circ }=1$ the full $g$ opposes the velocity, so every second of vertical burn costs the maximum possible gravity loss $g\Delta t$.

False. Gravity still acts — it just does no work along the path. Its full perpendicular component $g\cos \gamma$ now bends the trajectory (curves the velocity), which is exactly the turning force the gravity turn exploits.

False. $L(\gamma )=\Delta v_{\text{grav}}+\Delta v_{\text{drag}}$ is convex: high at $\gamma =90^{\circ }$ (gravity) and high at $\gamma =0^{\circ }$ (drag), so the minimum lies in between — that interior compromise is the gravity turn.

The sine, not cosine, projects gravity onto the velocity. Gravity opposes motion by $g\sin \gamma$; the $g\cos \gamma$ piece is the perpendicular part that turns the path, not the part that costs speed.

Wrong trig function. Only the perpendicular gravity component turns the velocity, so $\frac{d\gamma }{dt}=-\frac{g\cos \gamma }{v}$. Using $\sin$ would (wrongly) make the turn fastest when already vertical.

Backwards. The turn rate is inversely proportional to $v$: at high speed the velocity vector is hard to bend, so you must apply the pitch kick while $v$ is still small.

Sign error. Losses add on the ideal side: $\Delta v_{\text{ideal}}=\Delta v_{\text{orbit}}+\Delta v_{\text{grav}}+\Delta v_{\text{drag}}+\Delta v_{\text{steer}}$. The ideal $\Delta v$ from Tsiolkovsky must cover orbit plus all losses.

That integrand is a force, not an acceleration. Drag loss is $\int \frac{D}{m}dt$ — you must divide by mass to get the deceleration $D/m$ before integrating over time (see Aerodynamic Drag and Max-Q).

Throttling up raises $v$ down low where $\rho$ is large, making $\frac{1}{2}\rho v^2$ worse. Real vehicles throttle down through max-Q to keep $\rho v^2$ (dynamic pressure) manageable.

It scales as $1-\cos \alpha$, which grows like ${\alpha }^2$ for small angles — a $2^{\circ }$ offset costs about four times a $1^{\circ }$ offset, not twice.

To punch through the dense low atmosphere quickly before speed builds, keeping $\frac{1}{2}\rho v^2$ (and heating and structural load) low. Only after clearing thick air can it afford to pitch over.

Because thrust stays aligned with velocity and no aerodynamic lift (or side thrust) is used to bend the path — the only turning force is gravity's perpendicular component, which is free.

The turn rate $g\cos \gamma /v$ is inversely proportional to $v$. While $v$ is small the rate is large, so a small early nudge seeds a turn that grows naturally; wait too long and $v$ is high, the path is stiff, and you can't reach horizontal before fuel runs out.

Both are controlled by the same knob, the pitch angle $\gamma$, and they respond in opposite directions. Pushing $\gamma$ to help one always hurts the other, so only the sum has a meaningful minimum.

Gravity loss $\int g\sin \gamma dt$ falls because the burn is shorter in time; but a shorter, harder burn pushes $v$ up faster low down, which can raise drag loss. The benefit is on the time-integral of gravity, not on drag.

Orbit requires large horizontal velocity to keep falling around the planet; vertical speed just goes up and comes back. So thrust must eventually be redirected sideways — which is precisely the job the gravity turn accomplishes gradually.

The along-track (tangential) equation only sees force components parallel to velocity. Gravity's parallel part is $g\sin \gamma$; its perpendicular part $g\cos \gamma$ lives in the normal equation, where it turns the path instead of slowing it.

$\Delta v_{\text{grav}}=\int gdt$ (since $\sin 90^{\circ }=1$) — the maximum possible, because the entire weight opposes motion every instant. No orbital (horizontal) speed is built at all.

Gravity loss vanishes ($\int g\sin \gamma dt=0$) and the turn rate $g\cos \gamma /v\to 0$, so the velocity vector never rotates on its own — there is no gravity turn, and any turning must be paid for with steering thrust.

$\frac{d\gamma }{dt}=-\frac{g\cos 90^{\circ }}{v}=0$ — a perfectly vertical rocket will never start turning by itself. A deliberate small pitch kick off vertical is required to break the symmetry and let gravity take over.

No — at liftoff the rocket is held near vertical ($\cos \gamma \approx 0$) and is under active control, so the product stays finite. The formula warns that turning is easiest when slow, not that it happens uncontrollably.

Drag loss $\int \frac{\frac{1}{2}\rho v^2{C}_{D}A}{m}dt\to 0$ because $\rho \to 0$, even though $v$ is now huge. This is why the whole drag-vs-gravity trade only matters low down; once out of the air, only gravity loss remains to fight.

Drag depends on the product $\rho v^2$; past max-Q, $\rho$ drops faster than $v^2$ rises, so the dynamic pressure — and thus the drag deceleration $D/m$ — falls despite increasing speed.

$\Delta v_{\text{grav}}={\int }_0^{t_b}g\sin \gamma dt\to 0$ — with no time spent burning, gravity gets no chance to act against the motion. This is the idealized limit that shows why high thrust-to-weight is prized.