Trajectory optimization — minimum gravity loss, minimum drag loss: Visual walkthrough

Step 0 — The one arrow everything hangs on

WHAT. A rocket at some instant is moving. Its motion is an arrow: it has a length (how fast — the speed $v$ , measured in metres per second) and a direction (which way it points).

WHY. Every loss we will compute is about the direction this arrow points relative to the ground. So we must name that direction first, before any formula.

PICTURE. Look at the figure. The magenta arrow is the velocity. The dashed orange line is the local horizon — flat ground directly under the rocket. The angle between the arrow and the horizon is the star of the whole show.

Two words we will lean on, defined on that same picture:

Step 1 — Split gravity into "slows you" and "bends you"

WHAT. Gravity always points straight down with strength $g$ (about $9.8$ metres-per-second-per-second near the ground). We split that down-arrow into two pieces: one along the velocity, one across it.

WHY. A force along your motion changes your speed. A force across your motion changes your direction but not your speed. Gravity does both, in a proportion set entirely by $γ$ . We must separate them because they cause two different effects (a speed loss, and a free turn).

PICTURE. In the figure the navy arrow is gravity pointing down. The violet arrow is its piece pointing backwards along the path — that is the part that slows you. The orange arrow is its piece pointing across the path — that part bends you.

Here is the piece of geometry the whole page rests on. Drop the down-arrow onto the two directions:

Step 2 — Write Newton's law along the path

WHAT. We add up every force acting along the velocity arrow and set the total equal to mass times how fast the speed changes.

WHY. Newton's second law says (net force along a direction) = (mass) × (acceleration in that direction). Along the path, the "acceleration" is just $\frac{d v}{d t}$ — the rate at which the speed number $v$ grows. That is the quantity we ultimately care about: it becomes orbital speed.

PICTURE. Three forces live on the velocity line: thrust $T$ pushing forward (magenta), drag $D$ pushing back (violet), and the gravity-along piece $m g sin γ$ pushing back (navy).

Step 3 — Divide by mass so each term is an acceleration

WHAT. Divide every term by the mass $m$ .

WHY. We want to end up with $Δ v$ (a speed budget), and speed change comes from acceleration × time. Dividing a force by mass turns it into an acceleration (Newton again: $a = F / m$ ). Now every term on the right is "how many m/s per second this effect gives or takes."

Step 4 — Add up over the whole burn (integrate)

WHAT. Sum each term over every instant from ignition ( $t = 0$ ) to burnout ( $t = t_{b}$ ). That summing-over-time operation is the integral $\int_{0}^{t_{b}} (\dots) d t$ .

WHY. Acceleration acting for one instant gives a tiny speed change; the total speed change is all those tiny bits added up over the burn. The integral is just "add up a continuously changing quantity." Each term integrates into a total- $Δ v$ contribution.

PICTURE. The figure shows each acceleration curve as a shaded area under it — the area is the integral, the $Δ v$ that term contributes.

Rearranged, this is the bookkeeping equation the parent note promised: $Δ v_{achieved} = Δ v_{ideal} - Δ v_{drag} - Δ v_{grav}$

Step 5 — Read the two integrands: why $γ$ can't win both

WHAT. Stare at the two loss integrands and ask: as I change the pitch $γ$ , which way does each move?

WHY. This is the entire optimization problem in one glance. The pitch $γ$ is a single knob sitting inside both integrals — pushing them in opposite directions.

PICTURE. Two curves against $γ$ : gravity loss falls as you go flat; drag loss rises as you go flat (flat means low and fast in thick air). Their sum is a bowl with a minimum in between.

Knob position	$sin γ$	Gravity loss	Drag loss
Vertical, $γ \to 9 0^{\circ}$	$\to 1$	big	small (leaves thick air fast)
Flat, $γ \to 0^{\circ}$	$\to 0$	small	big ( $ρ v^{2}$ huge down low)

Step 6 — Edge cases: check the formula at its extremes

WHAT. Push $γ$ to its two limits and one degenerate burn, and confirm the formula does something sane.

WHY. A derivation you trust must survive its corners. If any extreme gave nonsense, we'd have a sign or a sine wrong.

PICTURE. Three mini-rockets: straight up, flat, and a zero-length burn.

Straight up, $γ = 9 0^{\circ}$ : $sin 9 0^{\circ} = 1$ , so $Δ v_{grav} = \int g d t$ — the whole of gravity fights you the whole burn. Worst-case gravity loss. ✔ matches intuition.
Flat, $γ = 0^{\circ}$ : $sin 0^{\circ} = 0$ , so $Δ v_{grav} = 0$ . Gravity takes no speed — but Step 1's across piece $g cos 0^{\circ} = g$ is now at full strength, curving you into the ground. The formula correctly says gravity stopped slowing you and started only bending you. ✔
Zero burn time, $t_{b} = 0$ : every integral is over an empty interval, so all terms are $0$ . No burn, no gain, no loss. ✔ The bookkeeping is self-consistent even at the degenerate limit.

The one-picture summary

One velocity arrow, one down-pointing gravity, split by the angle $γ$ into a slow-you piece ( $sin$ ) and a bend-you piece ( $cos$ ); Newton along the path, divided by mass, summed over the burn — out drops the loss bookkeeping, and the single knob $γ$ trades the two losses on a see-saw whose bottom is the gravity turn.

Recall Feynman: the whole walkthrough in plain words

Draw the arrow the rocket moves along. Gravity points straight down; slice that down-arrow into the part lined up against the arrow and the part sideways to it. The against-part is $g sin γ$ and it steals speed; the sideways-part is $g cos γ$ and it just bends your path for free. Write Newton's law along the arrow: thrust forward, drag and gravity-along backward. Divide by weight so everything is "speed gained per second," then add up every second of the burn. The thrust total is the fuel's ideal $Δ v$ ; the drag total and the gravity total are the two losses you paid for but didn't keep. The angle $γ$ hides inside both losses and shoves them opposite ways — steep kills you on gravity, flat kills you on drag — so the cheapest climb tips over just enough to let gravity do the turning by itself.

Recall Quick self-test

Why is gravity loss a sine and not a cosine? ::: Because the along-path (speed-robbing) piece of gravity is $g sin γ$ — it's max when vertical ( $sin 9 0^{\circ} = 1$ ) and zero when flat ( $sin 0^{\circ} = 0$ ), exactly matching how gravity opposes motion. What does dividing Newton's law by $m$ accomplish? ::: It turns each force into an acceleration (m/s per s), so integrating over time directly gives a $Δ v$ budget. At $γ = 0^{\circ}$ gravity loss is zero — is gravity doing nothing? ::: No: its across-path piece $g cos γ$ is at full strength, bending the trajectory (curving you toward the ground). It just isn't slowing you.