Case Study: Designing a Fully Reusable Vehicle to Beat the Cost Floor

"You cannot amortize your way past the parts you keep throwing away." — the lesson of the Falcon 9 cost floor (Chapter 22, §38.4)

Executive Summary

Case Study 1 audited a partially reusable rocket and found its cost bottoming out on a floor set by the expended second stage. This case study takes the next step and designs the vehicle that removes the floor: a fully reusable, two-stage, stainless-steel, methalox super-heavy in the Starship class. We will size both stages under the rocket equation with reuse penalties on each, discover that full reuse leaves the upper stage nearly empty on reaching LEO — forcing on-orbit refueling — and then estimate the tanker campaign and the cost floor that full reuse achieves. We will find that the architecture is economically coherent only at high flight rate and high reuse count, where it undercuts Falcon 9's floor by an order of magnitude and makes the aspirational "hundreds of dollars per kilogram" numbers arithmetically possible — while remaining, as Appendix H warns, firmly Tier 3 until it flies. This is a design study, one tier harder than the audit: you build the vehicle, then stress it.

Skills applied

• Sizing a two-stage vehicle with reuse penalties on both stages (§38.3, §38.5; Chapters 3, 22).
• Discovering an architectural driver (refueling) from a mass-ratio result rather than assuming it (§38.5).
• Modeling full-reuse economics and locating the new cost floor (§38.4).
• Stress-testing a design against dry-mass growth — "mass is the enemy" made quantitative (Chapter 23).

Background

Every number in this study is a design goal (Tier 3), not a specification. Starship is under active development; the point is the reasoning, which is durable, not the tonnages, which change with every test flight. Our clean-sheet vehicle:

Design requirement: deliver $\sim 100\ \text{t}$ to LEO with both stages recovered, and reach beyond-LEO destinations through on-orbit refueling. LEO ascent needs about $9{,}400\ \text{m/s}$.

Phase 1: Size the ascent, both stages recovered

Work bottom-up, and charge each stage its recovery reserve — the propellant it must keep to come home.

Booster. It reserves ~$8\%$ of its propellant (~$272\ \text{t}$) for boostback and the tower catch. Its "payload" is the fully fueled Ship plus the mission payload: $100 + 120 + 1{,}200 = 1{,}420\ \text{t}$. $$m_{0,1} = 1{,}420 + 200 + 3{,}400 = 5{,}020\ \text{t}, \quad m_{f,1} = 1{,}420 + 200 + 272 = 1{,}892\ \text{t},$$ $$\Delta v_1 = 3{,}200\,\ln\!\left(\frac{5{,}020}{1{,}892}\right) = 3{,}200 \times 0.976 = 3{,}123\ \text{m/s}.$$

Ship. It reserves ~$30\ \text{t}$ for its own landing burn (the belly-flop descent is unpowered; only the final flip-and-land burns). Its payload is the $100\ \text{t}$ cargo. $$m_{0,2} = 100 + 120 + 1{,}200 = 1{,}420\ \text{t}, \quad m_{f,2} = 100 + 120 + 30 = 250\ \text{t},$$ $$\Delta v_2 = 3{,}700\,\ln\!\left(\frac{1{,}420}{250}\right) = 3{,}700 \times 1.737 = 6{,}427\ \text{m/s}.$$

Total: $\Delta v_{\text{total}} = 3{,}123 + 6{,}427 = 9{,}550\ \text{m/s}$, just above the $9{,}400\ \text{m/s}$ LEO requirement. The vehicle delivers $\sim 100\ \text{t}$ to LEO with both stages recovered — with a thin margin, exactly as a real launcher lives.

🔧 Sanity check. A thin positive margin (~$150\ \text{m/s}$) is realistic — launchers are optimized to the edge. If it had come out hugely positive we would suspect a units slip; hugely negative and the design would not close. Landing near the requirement is the expected result.

Phase 2: The refueling driver falls out of the mass ratio

Look hard at the Ship's ascent. It burned almost all $1{,}200\ \text{t}$ of its propellant to reach LEO, arriving with only its ~$30\ \text{t}$ landing reserve. On reaching orbit, the Ship is essentially empty. That is the price of full reuse: the heat shield, flaps, and landing propellant it must carry ruin its mass ratio, so nothing is left for a beyond-LEO burn. We did not assume refueling — the rocket equation forced it on us.

How much delta-v could the Ship produce if it were topped back up to $1{,}200\ \text{t}$ in orbit, still carrying $100\ \text{t}$? $$\Delta v = 3{,}700\,\ln\!\left(\frac{120 + 1{,}200 + 100}{120 + 100}\right) = 3{,}700\,\ln\!\left(\frac{1{,}420}{220}\right) = 3{,}700 \times 1.865 = 6{,}900\ \text{m/s}.$$ That $6.9\ \text{km/s}$ from LEO covers trans-Mars injection ($\sim 3.6\ \text{km/s}$) with room to spare, or a trans-lunar injection plus a lunar landing on a tight budget. Refueling converts a vehicle that can barely reach LEO into one that can reach the Moon or Mars — which is why on-orbit refueling is not an add-on but the keystone of the whole architecture (and the enabler of the Mars mission in Chapter 34).

Phase 3: Size the tanker campaign

The Ship needs ~$1{,}170\ \text{t}$ of propellant added in LEO (from its ~$30\ \text{t}$ residual up to $1{,}200\ \text{t}$). Each tanker is itself a Ship that carries propellant as its payload, delivering about $120\ \text{t}$ of usable propellant per flight (Tier 3; boiloff and residuals eat the rest).

import math

def tanker_flights(prop_needed_t, per_tanker_t):
    """Number of tanker flights to deliver prop_needed_t of propellant to LEO."""
    return math.ceil(prop_needed_t / per_tanker_t)

need = 1200 - 30          # tonnes to add in orbit
print("tanker flights:", tanker_flights(need, 120))
print("total launches:", tanker_flights(need, 120) + 1)  # + the mission Ship itself
# Expected output:
# tanker flights: 10
# total launches: 11

Hand-trace: $\lceil 1170/120 \rceil = \lceil 9.75 \rceil = 10$ tanker flights, plus the mission Ship itself, for 11 launches per Mars mission — squarely in the widely cited "roughly 8–16" range. Now the critical observation: eleven launches to send one payload to Mars is only conceivable if each launch is cheap and the vehicles are reused. At Falcon 9's floor of ~$\$19\ \text{M}$/flight the campaign would cost ~$\$210\ \text{M}$; at a Shuttle-like $\$1.5\ \text{B}$/flight it would cost $\$16\ \text{billion}$ and be absurd. The refueling architecture is a child of the cost economics, not independent of them.

Phase 4: The new cost floor

Now the economics that justify the whole design. With both stages reused, the fixed cost $F$ that trapped Falcon 9 — the expended second stage — largely disappears. Model full reuse as $$C_{\text{full}}(N) = \frac{M_{\text{booster}} + M_{\text{ship}}}{N} + R + F_{\text{ops}},$$ with illustrative build costs $M_{\text{booster}} = \$100\ \text{M}$, $M_{\text{ship}} = \$50\ \text{M}$, refurbishment $R = \$2\ \text{M}$, and only propellant + operations left as $F_{\text{ops}} = \$2\ \text{M}$ (Tier 3):

The floor is now $R + F_{\text{ops}} = \$4\ \text{M}$/flight — an order of magnitude below Falcon 9's $\$19\ \text{M}$ floor, because there is no expended stage to keep buying. At high reuse count the cost per kilogram to LEO approaches the tens-of-dollars range, the aspirational "$\sim\$100/\text{kg}$" figure Appendix H flags as unproven. Two honest cautions preserve the audit spirit:

• The first flight is ruinous. At $N = 1$, full reuse costs $\$154\ \text{M}$ — worse than an expendable rocket — because you paid to build two recoverable stages and reused neither. Full reuse is a bet on a high flight rate that only pays off after many reflights; the ledger is red until then.
• These are Tier 3 aspirations. Refurbishment $R$ could balloon (as it did, catastrophically, for the Shuttle — Chapter 37), operations may not shrink to two people and a checklist, and the vehicle may never fly the hundreds of times the floor assumes. The arithmetic is coherent; whether reality obeys it is the experiment now underway.

Phase 5: Stress test — mass is the enemy

Reusable upper stages live and die by dry mass, so stress the design. Suppose the Ship's dry mass grows from $120\ \text{t}$ to $160\ \text{t}$ — heat-shield tiles heavier than hoped, structure reinforced after a test failure. Re-run Phase 1 with $100\ \text{t}$ payload: $$\Delta v_2 = 3{,}700\,\ln\!\left(\frac{100 + 160 + 1{,}200}{100 + 160 + 30}\right) = 3{,}700\,\ln\!\left(\frac{1{,}460}{290}\right) = 3{,}700 \times 1.616 = 5{,}979\ \text{m/s},$$ and the booster leg falls slightly to $\sim 3{,}081\ \text{m/s}$, for a total of about $9{,}060\ \text{m/s}$ — now below the $9{,}400\ \text{m/s}$ needed. The design no longer closes at $100\ \text{t}$. Restoring orbit means offloading payload: re-solving, the vehicle carries only about $65\ \text{t}$. A $40\ \text{t}$ growth in dry mass cost about $35\ \text{t}$ of payload — nearly tonne-for-tonne. That brutal exchange rate is why a reusable vehicle's dry mass is the program's central battle (Chapter 23's mass is the enemy), and why every gram of heat shield, every flap hinge, and every landing leg deleted (the tower catch of §38.5) is fought over. On a fully reusable upper stage, the rocket equation is at its least forgiving.

Discussion Questions

1. We did not assume refueling; it emerged from the Ship's mass ratio. Explain in your own words why full reuse forces orbital refueling for beyond-LEO missions.
2. Full reuse costs more than expendable at $N = 1$. What has to be true about the mission model for full reuse to be the right choice anyway?
3. The Phase 5 stress test showed nearly tonne-for-tonne payload loss with dry-mass growth. Why is this exchange rate so much worse on a fully reusable upper stage than on an expendable one?
4. The tanker campaign needs ~11 launches per Mars mission. What single failure or delay in that chain would you most want redundancy against, and why?

Your Turn: Extensions

• Option A (design). Re-size the vehicle for a lunar cargo mission (Track B) instead of Mars: the Ship must land ~50 t on the Moon (no atmosphere for braking). Estimate the LEO departure delta-v needed and how many tanker flights it implies.
• Option B (computation). Write closes(dry_ship, payload, dv_req=9400) that returns whether the two-stage stack meets dv_req for a given Ship dry mass and payload. Use it to map the payload the vehicle can carry as dry mass grows from 120 t to 180 t. (Hand-trace one point; add # Expected output:.)
• Option C (economics). Suppose refurbishment $R$ turns out to be $\$8\ \text{M}$, not $\$2\ \text{M}$ (Shuttle-like). Recompute the cost floor and the cost per kilogram at $N = 10$ and $N = 100$. How much of the revolution survives a heavy $R$?

Key Takeaways

1. Full reuse can beat Falcon 9's floor by an order of magnitude — but only at high reuse count; the first flights are more expensive than expendable.
2. Refueling is not optional. Full reuse leaves the upper stage nearly empty at LEO, so beyond-LEO missions require on-orbit refueling — an architectural driver that falls straight out of the rocket equation.
3. The tanker campaign is a child of cheap launch: ~11 launches per Mars mission is only sane when each launch is cheap and reusable.
4. Dry mass is decisive. On a fully reusable upper stage, dry-mass growth costs payload nearly tonne-for-tonne — the sharpest form of "mass is the enemy" in the book.
5. The arithmetic is coherent; the flight rate is the gamble. Every headline Starship cost number is a Tier 3 aspiration contingent on flying often and refurbishing cheaply.