Rocket staging — series staging, parallel staging: Exercises

Before we start, one tool appears in almost every solution, so let us earn it here.

Level 1 — Recognition

L1.1 — Reading a mass ratio

A single stage starts at $m_{0} = 20, 000$ kg and ends at $m_{f} = 5, 000$ kg after burning fuel. Its exhaust velocity is $v_{e} = 3000$ m/s. What is its mass ratio, and its $Δ v$ ?

Recall Solution

What we do: plug numbers into Tsiolkovsky. Mass ratio $= \frac{m _{0}}{m _{f}} = \frac{20000}{5000} = 4$ . $Δ v = 3000 \times ln (4) = 3000 \times 1.3863 \approx 4159 m/s .$ What it means: the rocket was $4 \times$ heavier at ignition; that ratio (not the raw masses) is what sets the speed gain.

L1.2 — Series or parallel?

A rocket lights its bottom stage, burns it dry, drops it, then lights the next stage on top. No two stages ever burn at the same time. Is this series or parallel staging?

Recall Solution

Series (sequential). The defining feature of series staging is that stages ignite one at a time and each is dropped before the next lights. Parallel staging would have the core and boosters burning together from liftoff.

Level 2 — Application

L2.1 — One stage from fractions

A stage has hardware mass (structure + propellant) $S = 10, 000$ kg with propellant fraction $λ = 0.9$ , and it carries a payload of $2, 000$ kg. Its $v_{e} = 3200$ m/s. Find $Δ v$ .

Recall Solution

What we do: split hardware, then build $m_{0}$ and $m_{f}$ .

Propellant $m_{p} = 0.9 \times 10000 = 9000$ kg.
Structure $m_{s} = 0.1 \times 10000 = 1000$ kg.
$m_{0} = S + payload = 10000 + 2000 = 12000$ kg (full).
$m_{f} = m_{s} + payload = 1000 + 2000 = 3000$ kg (fuel burnt, tank still attached). $Δ v = 3200 \times ln (\frac{12000}{3000}) = 3200 \times ln (4) \approx 4436 m/s .$

L2.2 — Effective exhaust velocity from thrust-weighting

Two engine groups fire together: the core gives $F_{c} = 6$ MN at $v_{e, c} = 4400$ m/s, and boosters give combined $F_{b} = 24$ MN at $v_{e, b} = 2600$ m/s. What single "effective" exhaust velocity $\overset{v}{ˉ}_{e}$ describes the pair?

Recall Solution

= \frac{6 \times 4400 + 24 \times 2600}{6 + 24} = \frac{26400 + 62400}{30} = \frac{88800}{30} = 2960 \text{ m/s}.$$ The answer sits close to the booster's $2600$ m/s because boosters supply $80\%$ of the thrust.

Level 3 — Analysis

L3.1 — Two-stage series, worked backward

Payload $= 1000$ kg. Stage 2: $S_{2} = 6000$ kg, $λ_{2} = 0.88$ , $v_{e, 2} = 3600$ m/s. Stage 1: $S_{1} = 60000$ kg, $λ_{1} = 0.86$ , $v_{e, 1} = 2900$ m/s. Find total $Δ v$ .

Figure — Rocket staging — series staging, parallel staging

Recall Solution

Why work top-down? Each stage's "payload" is everything above it, so we must know the upper masses first. Look at the figure: stage 2 sits on stage 1, and both sit under the capsule.

Stage 2 (carries only the 1000 kg capsule):

$m_{p, 2} = 0.88 \times 6000 = 5280$ kg, $m_{s, 2} = 0.12 \times 6000 = 720$ kg.
$m_{0} = 6000 + 1000 = 7000$ kg; $m_{f} = 720 + 1000 = 1720$ kg. $Δ v_{2} = 3600 ln (\frac{7000}{1720}) = 3600 \times 1.4036 \approx 5053 m/s .$

Stage 1 (carries all of stage 2's hardware + capsule $= 6000 + 1000 = 7000$ kg):

$m_{p, 1} = 0.86 \times 60000 = 51600$ kg, $m_{s, 1} = 0.14 \times 60000 = 8400$ kg.
$m_{0} = 60000 + 7000 = 67000$ kg; $m_{f} = 8400 + 7000 = 15400$ kg. $Δ v_{1} = 2900 ln (\frac{67000}{15400}) = 2900 \times 1.4708 \approx 4265 m/s .$

Total: $Δ v = 5053 + 4265 \approx 9318$ m/s — near the ~9.4 km/s needed for low-Earth orbit once you add gravity losses.

L3.2 — Parallel: two-phase burn

A parallel vehicle at liftoff: core $m_{c} = 30000$ kg (of which $m_{p, c} = 27000$ kg propellant, $m_{s, c} = 3000$ kg structure), two boosters each $m_{b} = 15000$ kg ( $m_{p, b} = 13000$ , $m_{s, b} = 2000$ ), upper stages + payload $m_{u pp er} = 4000$ kg. During phase 1 the boosters burn dry and the core burns $m_{p, c}^{(1)} = 9000$ kg. Effective $\overset{v}{ˉ}_{e} = 3000$ m/s in phase 1; core $v_{e, c} = 4200$ m/s in phase 2. Find $Δ v_{1}$ , $Δ v_{2}$ , and the total.

Recall Solution

Phase 1 — everything burns together.

$m_{0} = m_{c} + 2 m_{b} + m_{u pp er} = 30000 + 2 (15000) + 4000 = 64000$ kg.
Mass removed $= 2 m_{p, b} + m_{p, c}^{(1)} = 2 (13000) + 9000 = 35000$ kg.
$m_{1} = 64000 - 35000 = 29000$ kg. $Δ v_{1} = 3000 ln (\frac{64000}{29000}) = 3000 \times 0.7916 \approx 2375 m/s .$

Booster jettison: drop $2 m_{s, b} = 2 (2000) = 4000$ kg of empty booster shells.

$m_{2} = m_{1} - 4000 = 29000 - 4000 = 25000$ kg.

Phase 2 — core burns its leftover fuel.

Remaining core propellant $m_{p, c}^{(2)} = 27000 - 9000 = 18000$ kg.
$m_{2, f} = m_{s, c} + m_{u pp er} = 3000 + 4000 = 7000$ kg.
Check: $m_{2} = m_{s, c} + m_{p, c}^{(2)} + m_{u pp er} = 3000 + 18000 + 4000 = 25000$ kg. ✓ $Δ v_{2} = 4200 ln (\frac{25000}{7000}) = 4200 \times 1.2730 \approx 5347 m/s .$

Total: $Δ v = 2375 + 5347 \approx 7722$ m/s.

Level 4 — Synthesis

L4.1 — Where does single-stage break down?

A single stage must deliver $Δ v = 9400$ m/s with $v_{e} = 3500$ m/s. What mass ratio does it need? If the structural fraction can be no better than $ϵ = 0.08$ (8% empty tank), show that no single stage carrying any payload can reach this — this is why we stage.

Recall Solution

Required mass ratio: invert Tsiolkovsky. Since $Δ v = v_{e} ln R$ , we solve for $R = m_{0} / m_{f}$ : $R = e^{Δ v / v_{e}} = e^{9400/3500} = e^{2.6857} \approx 14.67.$ What this demands: $m_{f}$ must be at most $m_{0} /14.67 \approx 0.0682 m_{0}$ . So the empty mass (structure + payload) can be at most $6.82%$ of liftoff mass. The contradiction: the structure alone already eats $ϵ = 8%$ of the stage hardware, and the payload adds even more on top. Since $6.82% < 8%$ , the empty mass alone exceeds the budget before any payload is added. Impossible. Conclusion: you cannot both carry a payload and keep $m_{f}$ small enough — dropping dead structure mid-flight (staging) is the only escape. This connects to Payload fraction optimization.

L4.2 — Equal-split optimum (a compact synthesis)

Two identical series stages each have $v_{e} = 3000$ m/s and each achieves mass ratio $R = 6$ . What is the total $Δ v$ ? Compare to a single stage that somehow reached mass ratio $R = 6$ : how much does the second stage buy you?

Recall Solution

Two stages: each gives $Δ v_{i} = 3000 ln 6$ , and gains add: $Δ v = 2 \times 3000 ln 6 = 6000 \times 1.7918 \approx 10751 m/s .$ Single stage at same $R = 6$ : only $3000 ln 6 \approx 5375$ m/s. What the second stage buys: it doubles the $Δ v$ , because it starts fresh with its own favourable mass ratio instead of dragging a spent tank. This is the compounding effect: adding the logs of equal ratios beats trying to force one giant ratio (which structure forbids, per L4.1).

Level 5 — Mastery

L5.1 — Design to a $Δ v$ target

You need a two-stage series rocket delivering exactly $Δ v_{total} = 9000$ m/s to a $1000$ kg payload. Both stages use $v_{e} = 3400$ m/s. You choose to split the job equally: $Δ v_{1} = Δ v_{2} = 4500$ m/s. Each stage has $ϵ = 0.10$ . Find the required mass ratio per stage, then the stage-2 hardware mass $S_{2}$ needed.

Recall Solution

Step 1 — mass ratio per stage (invert Tsiolkovsky): $R = e^{Δ v_{i} / v_{e}} = e^{4500/3400} = e^{1.3235} \approx 3.757.$

Step 2 — relate $R$ to hardware. For a stage of hardware mass $S$ carrying payload $P$ : $R = \frac{m _{0}}{m _{f}} = \frac{S + P}{ϵ S + P}, ϵ = 0.10.$ Solve for $S$ : cross-multiply $R (ϵ S + P) = S + P$ , giving $S (1 - R ϵ) = P (R - 1) \Rightarrow S = \frac{P ( R - 1 )}{1 - R ϵ} .$

Step 3 — stage 2 carries $P = 1000$ kg:

= \frac{2757}{0.6243} \approx 4416 \text{ kg}.$$ **Sanity check:** $m_0 = 4416 + 1000 = 5416$, $m_f = 0.1(4416)+1000 = 1442$, ratio $= 5416/1442 \approx 3.756$ ✓ — matches the target $R$. (Stage 1 would then carry $S_2 + P = 5416$ kg as its payload and be sized the same way.)

L5.2 — When does the " $1 - R ϵ$ " denominator explode?

Using $S = \frac{P ( R - 1 )}{1 - R ϵ}$ from L5.1, explain physically what happens as $R \to 1/ ϵ$ , and compute the maximum achievable $Δ v$ per stage when $v_{e} = 3400$ m/s and $ϵ = 0.10$ .

Recall Solution

The blow-up: as $R$ climbs toward $1/ ϵ = 10$ , the denominator $1 - R ϵ \to 0^{+}$ , so the required hardware mass $S \to \infty$ . Physically, a mass ratio of $1/ ϵ$ would mean the whole stage is propellant with the empty structure alone as final mass — no room for any payload. You would need infinite fuel to lift infinite tanks: the vicious cycle from the parent note, made precise. Ceiling on $R$ : since $R$ must stay below $1/ ϵ = 10$ , the largest single-stage $Δ v$ is

= 3400 \times 2.3026 \approx 7829 \text{ m/s}.$$ **Meaning:** even a payload-free, perfectly-fuelled single stage with $\epsilon = 0.10$ caps out near $7.8$ km/s — below orbital needs. Staging (resetting $\epsilon$ each time you drop a spent stage) is the only way past this wall. See [[Payload fraction optimization]] and [[Falcon Heavy]].

Recall Quick self-check (cloze)

Series staging fires stages ::: one at a time, dropping each before the next lights Parallel staging fires core and boosters ::: simultaneously from liftoff The effective exhaust velocity in a parallel burn is weighted by ::: thrust The mass ratio a single stage can never exceed is ::: $1/ ϵ$ (one over its structural fraction) To hit a $Δ v$ target you invert Tsiolkovsky as ::: $R = e^{Δ v / v_{e}}$

Related: Specific impulse · Saturn V · Tsiolkovsky rocket equation.