Case Study: Reconstructing Falcon 9's Ascent Losses

"The rocket equation tells you what you have. The atmosphere and gravity tell you what you get to keep."

Executive Summary

In Chapter 3 we fed Falcon 9's real masses into the rocket equation and found that its two stages supply roughly $9.9\ \text{km/s}$ of ideal delta-v. In this chapter we learned that a real ascent does not keep all of it: gravity and drag take a cut. Now we close the loop. We will follow a Falcon 9 from the pad to orbit, milestone by milestone, and reconstruct where the delta-v goes — showing concretely that the first stage spends about $1.5\ \text{km/s}$ of its output fighting gravity, that drag takes only about a tenth of that, and that when the accounting is done, the $\sim 9.9\ \text{km/s}$ on tap comfortably covers the $\sim 9.4\ \text{km/s}$ that orbit actually demands. This is the anchor example of the book — Tsiolkovsky × Falcon 9 — advanced from "how much delta-v does it have?" to "how much does it get to keep, and why?"

Skills applied

• Reading an ascent timeline and identifying the flight-path angle regime (§4.2).
• Computing gravity loss from the loss integral and from a milestone difference (§4.3).
• Estimating drag loss and showing it is the small term (§4.4).
• Reconciling ideal delta-v (rocket equation), required delta-v (orbit + losses), and achieved velocity — the habit of labeling every delta-v (§4.3, §4.6).

Background

The vehicle and its ascent

Falcon 9 is a two-stage kerosene/LOX launcher: nine Merlin engines on the first stage, one vacuum-optimized Merlin on the second. The masses we used in Chapter 3 (approximate, Tier 2) gave a first-stage ideal delta-v of $\Delta v_1 \approx 3.86\ \text{km/s}$ and a second-stage ideal delta-v of $\Delta v_2 \approx 6.02\ \text{km/s}$, for a total of about $9.9\ \text{km/s}$ carrying a $15\ \text{t}$ payload.

Here is a representative ascent timeline to low Earth orbit. All figures are approximate (Tier 2) and vary with mission, payload, and vehicle block; they are here to establish the shape of the trajectory, not to certify any single flight.

Notice the trajectory doing exactly what §4.2 described: vertical at lift-off, bending through the gravity turn, and arriving essentially horizontal at orbital insertion.

Why this matters

Chapter 3 left a question hanging, and Chapter 4 promised to answer it: the vehicle produces $\sim 9.9\ \text{km/s}$ of delta-v, but orbital speed is only $\sim 7.7$–$7.8\ \text{km/s}$. Where does the rest go? The timeline above holds the answer, and the milestone that reveals it most cleanly is MECO.

Phase 1: The delta-v on tap

From Chapter 3, the two stages supply:

$$ \Delta v_1 \approx 3.86\ \text{km/s} \quad(\text{first stage}), \qquad \Delta v_2 \approx 6.02\ \text{km/s} \quad(\text{second stage}), $$ $$ \Delta v_{\text{ideal}} = 3.86 + 6.02 = 9.88 \approx 9.9\ \text{km/s}. $$

This is what the rocket equation promises in gravity-free, airless space. It is the "bank balance." Now we watch the vehicle spend it against a real planet.

Phase 2: The first stage pays the gravity bill

Here is the reconstruction that pays off the anchor. The first stage delivers an ideal $3.86\ \text{km/s}$. Yet at MECO — when that stage shuts down, about $2.5$ minutes into flight — the vehicle is moving at only about $2.3\ \text{km/s}$. The stage spent $3.86\ \text{km/s}$ of delta-v and the speedometer gained only $2.3$. The difference is the losses the first stage paid:

$$ \Delta v_{\text{loss, stage 1}} = \Delta v_1 - v_{\text{MECO}} \approx 3.86 - 2.3 = 1.56\ \text{km/s}. $$

Almost all of that is gravity loss. We can check it against the loss integral of §4.3. The first stage burns for about $155\ \text{s}$, starting vertical ($\sin\gamma = 1$) and ending well tilted ($\gamma \approx 25^\circ$, $\sin\gamma \approx 0.42$). Taking an average $\langle\sin\gamma\rangle \approx 0.6$:

$$ \Delta v_{\text{grav, stage 1}} \approx g\,\langle\sin\gamma\rangle\,t_b = 9.81 \times 0.6 \times 155 \approx 912\ \text{m/s} \approx 0.9\ \text{km/s}. $$

That accounts for the bulk of the loss; the remainder (a few hundred m/s) is the higher gravity loss of the very vertical early seconds (our flat average of $0.6$ understates the first-stage-heavy weighting), plus a small drag contribution. The headline is robust and beautiful: the mighty first stage spends somewhere around $1.5\ \text{km/s}$ of its delta-v just holding the rocket up against gravity while it climbs. That is the concrete meaning of the "extra" delta-v to orbit.

🔧 Engineering Reality: This is also why the first stage's ideal delta-v ($3.86\ \text{km/s}$) looks disappointingly small next to the second stage's ($6.02\ \text{km/s}$), even though the first stage burns far more propellant. The first stage does its work deep in the gravity-loss regime — steep flight-path angles, low down — where a large fraction of every second's thrust is eaten by $g\sin\gamma$. The second stage flies nearly horizontal in vacuum, where gravity loss nearly vanishes and almost all of its delta-v becomes speed. Staging does not just shed dead mass (Chapter 3); it also moves the upper stage's work into the cheap, low-loss part of the trajectory.

Phase 3: Drag takes only a sliver

The second thief barely shows up. Using the §4.4 estimate at max-Q — $q \approx 30\ \text{kPa}$, $A = 10.75\ \text{m}^2$, $C_d \approx 0.4$, vehicle mass $\approx 360\ \text{t}$ — the drag deceleration peaks at only

$$ \frac{D}{m} = \frac{q\,C_d\,A}{m} = \frac{30{,}000 \times 0.4 \times 10.75}{3.6\times10^5} \approx 0.36\ \text{m/s}^2, $$

against gravity's $9.81\ \text{m/s}^2$. Significant drag lasts only from roughly $T+40\ \text{s}$ to $T+100\ \text{s}$, and only near max-Q is it even this large. Integrated, the whole-ascent drag loss is on the order of $0.1\ \text{km/s}$ — small enough that Falcon 9's decision to throttle down through max-Q is made to protect the structure, not to save delta-v. Drag's delta-v bite is a rounding error next to gravity's.

Phase 4: The books balance

Now assemble the required budget and compare it with what the vehicle supplies.

The vehicle supplies about $9.9\ \text{km/s}$; orbit, after crediting the eastward launch, requires about $9.0\ \text{km/s}$; the roughly $0.9\ \text{km/s}$ left over is the performance reserve that pays for a heavier payload, a higher orbit, or — famously — flying the first stage back and landing it. The accounting closes.

🔧 Engineering Reality: Do not expect this budget to balance to the last $10\ \text{m/s}$ from the round numbers here — it will not, and it should not. Real gravity loss depends on the exact pitch program; the rotation credit depends on launch azimuth and latitude (Chapter 30); and our masses are approximate. What is robust is the structure of the answer: orbital speed is the biggest line, gravity loss is a firm second at $\sim 1.5\ \text{km/s}$, drag is a distant third at $\sim 0.1\ \text{km/s}$, and the sum is the $\sim 9.4\ \text{km/s}$ (before rotation credit) that Chapters 1 and 3 quoted. That is the lesson: always know whether a delta-v number is ideal, required, or achieved.

Discussion Questions

1. The first stage burns far more propellant than the second but produces less delta-v and keeps less of what it produces. Give both reasons (one from Chapter 3, one from Chapter 4).
2. At MECO the vehicle is moving $\sim 2.3\ \text{km/s}$ but is only $\sim 65\ \text{km}$ high. Explain why it is nowhere near orbital velocity yet, and what the second stage must still accomplish.
3. If Falcon 9 flew a steeper trajectory (stayed vertical longer), what would happen to its gravity loss, its drag loss, and its max-Q? Would the trade be worth it?
4. The margin in Phase 4 is what allows first-stage recovery. In delta-v terms, explain why landing the booster "costs payload."

Your Turn: Extensions

• Option A (analysis). Recompute the first-stage gravity loss assuming $\langle\sin\gamma\rangle = 0.7$ instead of $0.6$. How much does the estimate change, and is it still consistent with the $\sim 1.5\ \text{km/s}$ from the MECO difference?
• Option B (computation). Write a function ascent_losses(dv_stage, v_milestone) that returns the loss a stage paid as dv_stage - v_milestone, and apply it to the first stage ($3.86, 2.3$) and to a hypothetical stage that reached $3.0\ \text{km/s}$ at cutoff. Add # Expected output: comments by hand; do not run it.
• Option C (design). Suppose a customer wants $\sim 0.5\ \text{km/s}$ more performance. List three levers from Chapters 3–4 (e.g., higher upper-stage $I_{sp}$, an easterly equatorial launch site, a flatter trajectory) and estimate which gives the most delta-v for the least disruption.

Key Takeaways

1. The first stage pays the gravity bill. Its ideal $3.86\ \text{km/s}$ becomes only $\sim 2.3\ \text{km/s}$ at MECO; the missing $\sim 1.5\ \text{km/s}$ is gravity loss, the concrete "extra" beyond orbital speed.
2. Drag is the small thief (~$0.1\ \text{km/s}$); throttling through max-Q protects the structure, not the delta-v budget.
3. The budget closes: orbital speed (~7.8) + gravity (~1.5) + drag (~0.1) ≈ 9.4 km/s required, covered by the ~9.9 km/s the vehicle supplies — with the surplus enabling booster recovery.
4. Always label a delta-v as ideal (rocket equation), required (orbit + losses), or achieved (the speedometer), or you will compare the wrong numbers.