Case Study: Auditing Falcon 9's Reuse Payload Penalty and Cost per Kilogram

"The rocket that comes home is not free — it is bought with payload." — the thesis of this case study

Executive Summary

Falcon 9 is the first orbital rocket to make first-stage reuse routine, and the two claims most often attached to it — "reuse costs about a quarter of the payload" and "it flies for around $2{,}700$ per kilogram" — are exactly the kind of numbers this book insists you audit rather than repeat. In this case study we take the Chapter 3 Falcon 9 model and reconstruct both claims from the rocket equation and a simple cost model. We will confirm that recovering the booster costs roughly a quarter of the payload for a droneship landing and closer to forty percent for a return-to-launch-site (RTLS) profile, and we will find that the headline "$\$2{,}700/\text{kg}$" quietly assumes a full rocket — a typical recovered mission is nearer $\$3{,}600/\text{kg}$, still an order of magnitude below the Space Shuttle. The habit on display is the one the whole book trains: trust a number only after you have rebuilt it and labeled its tier.

Skills applied

• Reconstructing a staged ascent delta-v from real masses and specific impulses (§38.2, §38.3; Chapter 3).
• Placing reserved landing propellant in the final mass, and reading the payload consequence (§38.3).
• Auditing a cost-per-kilogram figure against the reuse cost model and the price-versus-cost distinction (§38.4).
• Sanity-checking every result against the real vehicle and labeling each number's tier.

Background

We use the same approximate Falcon 9 (Block 5 class) masses as Chapter 3 — Tier 2, rounded for legible arithmetic. Absolute tonnages here run a little low (our second stage is modeled conservatively), so read the percentages, which track reality closely, rather than the absolute payloads.

Reaching low Earth orbit demands about $9{,}400\ \text{m/s}$ of delta-v (orbital speed plus the gravity and drag losses of Chapter 4). Our audit fixes that requirement and asks: for each recovery mode, how much payload can the rocket deliver and still produce $9{,}400\ \text{m/s}$? The first stage always lifts the fueled second stage ($96{,}000\ \text{kg}$) plus the payload; reserved landing propellant leaves with the first stage at separation, so it counts in the first stage's burnout mass but never burdens the second stage.

Phase 1: The expendable baseline

Throw the booster away, and all $395{,}700\ \text{kg}$ of first-stage propellant goes to ascent. We hunt for the payload $m_L$ that makes the two stages sum to exactly $9{,}400\ \text{m/s}$.

For the first stage, $m_{0,1} = m_L + 96{,}000 + 25{,}600 + 395{,}700 = m_L + 517{,}300$ and (all propellant burned) $m_{f,1} = m_L + 121{,}600$. For the second stage, $m_{0,2} = m_L + 96{,}000$ and $m_{f,2} = m_L + 4{,}000$. Trying $m_L = 18{,}200\ \text{kg}$:

$$\Delta v_1 = 2{,}840\,\ln\!\left(\frac{535{,}500}{139{,}800}\right) = 2{,}840\times 1.343 = 3{,}814\ \text{m/s},$$ $$\Delta v_2 = 3{,}410\,\ln\!\left(\frac{114{,}200}{22{,}200}\right) = 3{,}410\times 1.638 = 5{,}586\ \text{m/s},$$ $$\Delta v_{\text{total}} = 3{,}814 + 5{,}586 = 9{,}400\ \text{m/s}. \checkmark$$

So the expendable baseline is about $18.2\ \text{t}$ to LEO in our model. (The real vehicle does ~22.8 t expended; our conservative stage 2 makes the absolute number low, which is fine — we care about the ratio as we add recovery.)

Phase 2: Droneship recovery

Now recover the booster on a downrange droneship, reserving about $10\%$ of first-stage propellant (~$39{,}570\ \text{kg}$) for the entry and landing burns. That reserve joins the burnout mass: $m_{f,1} = m_L + 121{,}600 + 39{,}570 = m_L + 161{,}170$. Re-solving for $9{,}400\ \text{m/s}$, try $m_L = 13{,}500\ \text{kg}$:

$$\Delta v_1 = 2{,}840\,\ln\!\left(\frac{530{,}800}{174{,}670}\right) = 2{,}840\times 1.111 = 3{,}155\ \text{m/s},$$ $$\Delta v_2 = 3{,}410\,\ln\!\left(\frac{109{,}500}{17{,}500}\right) = 3{,}410\times 1.834 = 6{,}254\ \text{m/s},$$ $$\Delta v_{\text{total}} = 3{,}155 + 6{,}254 = 9{,}409\ \text{m/s} \approx 9{,}400. \checkmark$$

The droneship payload is about $13.5\ \text{t}$ — a cut of $(18.2 - 13.5)/18.2 \approx 26\%$ from the expendable baseline.

🔧 Sanity check. The real vehicle drops from ~22.8 t (expended) to ~17 t (droneship), a $25\%$ cut (Appendix H). Our model gives $26\%$ — the percentage matches reality even though our absolute tonnages are low. That is the audit working: the physics reproduces the reported penalty from first principles.

Phase 3: Return to launch site (RTLS)

Bring the booster all the way back to the pad, and it must add a large boostback burn to cancel its downrange velocity — reserve about $18\%$ of first-stage propellant (~$71{,}230\ \text{kg}$). Now $m_{f,1} = m_L + 121{,}600 + 71{,}230 = m_L + 192{,}830$. Re-solving, try $m_L = 11{,}000\ \text{kg}$:

$$\Delta v_1 = 2{,}840\,\ln\!\left(\frac{528{,}300}{203{,}830}\right) = 2{,}840\times 0.952 = 2{,}705\ \text{m/s},$$ $$\Delta v_2 = 3{,}410\,\ln\!\left(\frac{107{,}000}{15{,}000}\right) = 3{,}410\times 1.965 = 6{,}701\ \text{m/s},$$ $$\Delta v_{\text{total}} = 2{,}705 + 6{,}701 = 9{,}406\ \text{m/s} \approx 9{,}400. \checkmark$$

The RTLS payload is about $11.0\ \text{t}$ — a cut of about $40\%$ from expendable. Putting the three modes side by side:

The pattern is exactly what §38.3 argued: the more delta-v you reserve to come home, the more payload you surrender. RTLS costs the most because the boostback burn is pure overhead. Why ever choose it? Because a booster back at the pad is faster and cheaper to turn around than one that must sail home on a barge — an operations saving ($F$ in the cost model) that can outweigh the payload it costs for a light mission.

Phase 4: Auditing "$2{,}700 per kilogram"

Now the cost claim. Cost per kilogram is price divided by payload, so it depends on which payload you use. Let us compute it honestly, in Python, and compare with the Shuttle.

def cost_per_kg(price_musd, payload_kg):
    """Dollars per kg from list price ($M) and delivered payload (kg)."""
    return price_musd * 1e6 / payload_kg

# Falcon 9 at a ~$62M list price, three ways to count the payload:
print("expended max :", round(cost_per_kg(62, 22800)))   # full rocket, expendable capacity
print("recovered    :", round(cost_per_kg(62, 17000)))   # typical droneship payload
# Space Shuttle, full program cost per flight over its payload:
print("shuttle      :", round(cost_per_kg(1500, 27500)))
# Expected output:
# expended max : 2719
# recovered    : 3647
# shuttle      : 54545

Hand-trace: $62\times10^{6}/22{,}800 = \$2{,}719/\text{kg}$; $62\times10^{6}/17{,}000 = \$3{,}647/\text{kg}$; $1.5\times10^{9}/27{,}500 = \$54{,}545/\text{kg}$. Audit finding: the widely quoted "$\sim\$2{,}700/\text{kg}$" uses the maximum (expendable-capacity) payload — it flatters the rocket by assuming it is filled to the brim. A representative recovered mission is closer to $\$3{,}600/\text{kg}$. Either way the comparison to the Shuttle's $\$54{,}500/\text{kg}$ is the real headline: a factor of $15$–$20$, a genuine order of magnitude. The exact multiple is soft; the order of magnitude is not.

⚠️ Price is not cost. The $\$62\ \text{M}$ is a list price, what a customer pays, including margin; SpaceX's internal cost per flight is reportedly lower. And the Shuttle's $\$54{,}500$ is a full-program accounting (development plus a large standing workforce spread over a low flight rate); its marginal cost per flight was lower. Auditing a per-kilogram figure means asking three questions every time: whose number — price or cost? what payload — full or typical? what accounting — marginal or fully burdened?

Phase 5: Where the cost stops falling

Finally, reuse only helps if you fly the booster many times. Apply the Chapter 22 model $C_{\text{reuse}}(N) = M/N + R + F$ with illustrative Falcon 9-class figures (Tier 3): booster build $M = \$30\ \text{M}$, refurbishment $R = \$1\ \text{M}$, expended-upper-stage-plus-ops $F = \$18\ \text{M}$:

The cost per flight falls fast at first and then flattens onto a floor of $R + F = \$19\ \text{M}$, set by the still-expended second stage and fixed operations. No amount of first-stage reuse crosses that floor — which is precisely the finding that motivates Starship's full reuse (§38.4, §38.5). Reuse of one stage is a revolution with a ceiling; the way through the ceiling is to stop expending the other stage.

Discussion Questions

1. Our model reproduces the real ~25% droneship penalty as a percentage despite low absolute tonnages. Why is the percentage robust to our conservative stage-2 numbers while the absolute payload is not?
2. RTLS costs more payload than a droneship landing, yet SpaceX flies RTLS for many missions. What operations cost ($F$) might justify surrendering payload?
3. The "$\$2{,}700/\text{kg}$" figure is technically defensible but misleading. Rewrite it as a single honest sentence a journalist could use.
4. The cost floor in Phase 5 is $R + F$. Which of the two terms does Starship attack with full reuse, and which does it attack with fast, few-person operations?

Your Turn: Extensions

• Option A (analysis). Recompute Phase 2 with a heavier reserve (say $14\%$, for a marginal-weather landing). Find the new droneship payload and penalty, and confirm the trend.
• Option B (computation). Write a function orbit_payload(reserve_frac, dv_target=9400) that solves for the payload giving dv_target, by scanning candidate payloads and returning the one whose staged delta-v is closest. Reproduce the Phase 1–3 table. (Do not run it — hand-trace one row and add # Expected output:.)
• Option C (design). Suppose densified propellant (§38.2) adds $165\ \text{m/s}$ of ascent delta-v. Roughly how much of the droneship payload penalty does that recover? Estimate the new payload and comment.

Key Takeaways

1. Reuse costs payload, and the amount is recoverable from first principles. Reserved landing propellant raises burnout mass and cuts ascent delta-v — ~26% payload for a droneship, ~40% for RTLS in our model, matching the real ~25% droneship figure.
2. The recovery mode is a delta-v budget choice. More reserve (RTLS) means more payload surrendered; you choose it when operations savings outweigh the payload cost.
3. "Cost per kilogram" must be audited, not quoted. The headline number assumes a full rocket and a list price; a typical recovered mission is dearer, but still ~15–20× cheaper than the Shuttle.
4. First-stage reuse has a floor at the cost of the expended second stage plus operations — the finding that makes full reuse the next move.