Case Study 2: Designing a Reusable Launcher — and Deciding If Reuse Pays

"The first flight of a reusable rocket costs more than an expendable one. Everything after that is the whole point."

Executive Summary

In Case Study 1 we analyzed an existing vehicle; here we design one, and then make the business decision that separates a demonstration from a company. We size a two-stage launcher to put $10\ \text{t}$ into low Earth orbit, first as a conventional expendable rocket, then modify its first stage for reuse and measure the payload penalty that recovery hardware and landing propellant impose. Finally we run the reuse cost model of §22.6 on a per-kilogram basis — the honest comparison, since the reusable rocket carries less — and find the break-even flight count. The conclusion is the one that reshaped the launch industry: reuse costs you payload and costs you more on flight one, but if you fly the vehicle even a few times, it wins decisively.

Skills applied: - Top-down stage sizing with $m_{\text{stage}} = m_{\text{above}}(R-1)/(1-\varepsilon R)$ (§22.2; Chapter 3). - Modeling a reuse payload penalty as a rise in structural coefficient (§22.1, §22.5). - Building and solving the reuse cost model $C(N) = M/N + R + F$ on a cost-per-kg basis (§22.6). - Turning the analysis into a mission decision (§22.6; the Mission Design Checkpoint).

Cost figures are illustrative (Tier 3); the structure of the argument, not the exact dollars, is the lesson.

Background

Our requirement: deliver $m_L = 10{,}000\ \text{kg}$ to low Earth orbit, total $\Delta v_{\text{total}} = 9.4\ \text{km/s}$. We choose a two-stage serial vehicle (per §22.1's diminishing-returns argument). The first stage burns kerosene ($v_e = 2{,}950\ \text{m/s}$, structural coefficient $\varepsilon_1 = 0.06$); the second burns something more energetic ($v_e = 3{,}400\ \text{m/s}$, $\varepsilon_2 = 0.05$). Because the stages differ in $v_e$, §22.2 tells us the split should not be exactly even and the more efficient upper stage should do more. We adopt a split of $\Delta v_1 = 3.4\ \text{km/s}$ (first stage) and $\Delta v_2 = 6.0\ \text{km/s}$ (second stage), which leans the work upward as the rule prescribes.

Phase 1: Size the expendable baseline

Each stage's mass ratio is $R = e^{\Delta v/v_e}$:

$$ R_1 = e^{3400/2950} = e^{1.153} = 3.166, \qquad R_2 = e^{6000/3400} = e^{1.765} = 5.840. $$

Both sit below their ceilings ($1/\varepsilon_1 = 16.7$, $1/\varepsilon_2 = 20.0$), so both are buildable. We size top-down with the stage-mass formula, starting from the payload.

Second stage (carries the payload): $$ m_{\text{stage 2}} = 10{,}000 \times \frac{5.840 - 1}{1 - 0.05\times5.840} = 10{,}000 \times \frac{4.840}{0.708} = 68{,}360\ \text{kg}. $$ Its gross mass is $10{,}000 + 68{,}360 = 78{,}360\ \text{kg}$.

First stage (carries the whole second-stage stack): $$ m_{\text{stage 1}} = 78{,}360 \times \frac{3.166 - 1}{1 - 0.06\times3.166} = 78{,}360 \times \frac{2.166}{0.810} = 209{,}550\ \text{kg}. $$

Lift-off mass $= 78{,}360 + 209{,}550 = 287{,}900\ \text{kg} \approx 288\ \text{t}$. The payload fraction is $10{,}000/287{,}900 = 3.5\%$ — right in the range §22.1 predicted for a two-stage vehicle.

Phase 2: Make the first stage reusable — and pay for it

To recover the first stage propulsively (§22.5), we add landing legs, grid fins, and structure rugged enough to fly again, and we must reserve propellant for the boostback, entry, and landing burns. Both effects act like added dead mass on the ascent: the reserved landing propellant is mass the first stage lifts but does not spend on the payload, and the recovery hardware is extra structure. We model the combined effect as a rise in the first stage's effective structural coefficient from $\varepsilon_1 = 0.06$ to $\varepsilon_1' = 0.11$ — a realistic penalty for a recoverable stage.

Hold the design fixed (same split, same $288\ \text{t}$ lift-off mass) and ask what payload it now delivers. The per-stage payload ratios are $\pi_i = (1 - \varepsilon_i R_i)/[R_i(1-\varepsilon_i)]$:

$$ \pi_2 = \frac{1 - 0.05\times5.840}{5.840\times0.95} = 0.128, \qquad \pi_1' = \frac{1 - 0.11\times3.166}{3.166\times0.89} = 0.231. $$

Overall payload fraction $= \pi_1' \pi_2 = 0.231 \times 0.128 = 0.0295$, so the payload on the same $288\ \text{t}$ rocket falls to

$$ m_L' = 287{,}900 \times 0.0295 \approx 8{,}500\ \text{kg}. $$

Reuse cost us $10{,}000 - 8{,}500 = 1{,}500\ \text{kg}$, a $15\%$ payload penalty — squarely in the range real droneship-recovered vehicles accept. (Equivalently, to keep the full $10\ \text{t}$ payload we would have to grow the rocket to about $339\ \text{t}$ at lift-off. Either way, the rocket equation extracts its tax for the mass we chose to bring home.)

Sanity check. A $15\%$ payload hit from recovery is consistent with the widely reported figure that Falcon 9 gives up roughly $15$–$30\%$ of its expendable payload to land the booster downrange, and more for a return-to-launch-site profile. Our number is in the right neighborhood.

Phase 3: Does reuse pay? Compare cost per kilogram

Now the decision. Reuse is only worth a $15\%$ payload penalty if getting the booster back saves more than it costs. Use the §22.6 model with illustrative figures (Tier 3): reusable booster build $M = \$25\ \text{M}$, refurbishment $R = \$2\ \text{M}$/flight, and fixed cost (expended second stage + operations + propellant) $F = \$20\ \text{M}$/flight. An expendable version costs $C_{\text{expend}} = M + F = \$45\ \text{M}$. The reusable cost per flight is

$$ C_{\text{reuse}}(N) = \frac{25}{N} + 2 + 20 = 22 + \frac{25}{N}\ \ (\$\text{M}). $$

But the reusable rocket delivers only $8.5\ \text{t}$, not $10\ \text{t}$, so we must compare cost per kilogram, dividing each option's per-flight cost by the payload it actually carries:

Read the first row carefully: flown once, the reusable design is more expensive per kilogram ($\$5.53$ vs $\$4.50\ \text{M/t}$) — it cost more to build with recovery hardware, and it carried less. Reuse is a loss on flight one. The break-even, solving $(22 + 25/N)/8.5 = 4.50$, comes at $N \approx 1.5$: by the second flight of a booster, reuse is cheaper per kilogram, and by ten flights it is nearly $40\%$ cheaper. The payload penalty is real, but amortization overwhelms it fast.

# Cost per tonne vs. flight count (reuse) against the expendable baseline. Not executed.
def reuse_cost_per_tonne(N, payload_t=8.5):
    return (22 + 25 / N) / payload_t     # $M per tonne
for N in (1, 2, 5, 10):
    print(N, round(reuse_cost_per_tonne(N), 2), "vs expendable 4.50")
# Expected output:
# 1 5.53 vs expendable 4.50
# 2 4.06 vs expendable 4.50
# 5 3.18 vs expendable 4.50
# 10 2.88 vs expendable 4.50

Phase 4: The decision for your mission

The break-even is about cadence. Reuse wins only if you fly the booster several times, so the decision turns on how often your mission — and the market around it — launches.

• Track A (GEO comsat constellation), or any high-cadence business: reuse is a clear win. Dozens of flights drive the cost per kilogram toward the floor; the $15\%$ payload penalty is trivial next to a near-halving of price.
• A single, one-off deep-space probe (Track C/D): reuse may not pay for that mission alone — one flight sits on the wrong side of break-even, and you simply gave up $15\%$ of payload. You would fly expendable, or buy a ride on a reusable vehicle whose costs are already amortized across someone else's many flights (which is, in practice, exactly what deep-space missions now do).

The floor, again. Notice the cost per tonne bottoms out at $\$2.59\ \text{M/t}$, not zero, because the second stage is still expended every flight (that is what the $F = \$20\ \text{M}$ buys). To break that floor you must reuse the second stage too — the Starship bet of Chapter 38. Our two-stage design shows exactly why partial reuse plateaus and full reuse is the next revolution.

Discussion Questions

1. We modeled the reuse penalty as a jump in $\varepsilon_1$ from $0.06$ to $0.11$. Which physical items contribute to each part of that jump, and which would a return-to-launch-site profile make worse than a downrange droneship landing?
2. The break-even flight count fell at $N \approx 1.5$. What would push it higher (make reuse harder to justify)? Consider refurbishment cost $R$ and the size of the payload penalty.
3. Why is comparing cost per flight misleading here, and cost per kilogram the honest metric? When would cost per flight actually be the right comparison?

Your Turn: Extensions

• Option A (design). Re-optimize the split. Try $\Delta v_1 = 3.0$ and $\Delta v_2 = 6.4\ \text{km/s}$ and re-size the expendable rocket. Does leaning even more delta-v onto the efficient upper stage reduce lift-off mass? By how much?
• Option B (economics). Suppose refurbishment turns out to cost $R = \$10\ \text{M}$/flight (a Shuttle- like surprise) instead of $\$2\ \text{M}$. Rebuild the cost-per-tonne table. Does reuse still beat expendable, and after how many flights? What does this say about why the Shuttle's reusability failed to cut cost (Chapter 37)?
• Option C (mission). Run this whole analysis on your mission's launch. Estimate your realistic flight cadence, pick rough costs, and write the reuse verdict into your MDR (the Mission Design Checkpoint asks for exactly this).

Key Takeaways

• A two-stage, $10\ \text{t}$-to-LEO expendable launcher masses about $288\ \text{t}$ at lift-off — a $3.5\%$ payload fraction, as §22.1 predicts.
• Making the first stage reusable (modeled as $\varepsilon_1: 0.06 \to 0.11$) costs about $15\%$ of payload — the rocket equation's tax for bringing the booster home.
• On a cost-per-kilogram basis the reusable design loses on flight one but breaks even by flight two and is far cheaper by flight ten; reuse is an amortization bet that pays off with cadence.
• The cost floor is set by whatever you still throw away (here, the second stage), which is why full reuse — not partial — is the path to the next order-of-magnitude cost drop.