Staging events — separation dynamics, thrust tail-off: Worked examples

Before we compute anything, we restate the two formulas and the picture behind them so nothing is used unearned.

The scenario matrix

Every staging problem is one of these cells. The worked examples below are each tagged with the cell they cover.

Cell	What makes it special	Covered by
A Ordinary tail-off	plain $F_{0}, τ$ finite, positive	Ex 1
B Tail-off, degenerate $τ \to 0$	instant cutoff limit	Ex 2
C Tail-off, large- $τ$ / partial window	integrate only to finite $t$ , not $\infty$	Ex 3
D Separation, unequal masses	general $m_{1} \neq = m_{2}$	Ex 4
E Separation, equal masses $m_{1} = m_{2}$	symmetric limit, $μ = m /2$	Ex 5
F Separation, extreme mass ratio	$m_{1} ≫ m_{2}$ (tiny upper stage) → $μ \to m_{2}$	Ex 6
G Clearance / collision-risk word problem	solve for required $J_{se p}$ from a gap deadline	Ex 7
H Combined real-world / exam twist	tail-off and separation together; sign bookkeeping	Ex 8
I Zero-input sanity	$J_{se p} = 0$ → what happens?	inside Ex 5 verify

Signs matter: we take forward (direction of flight) as positive. The upper stage gets a $+$ push, the discarded lower stage a $-$ recoil.

Ex 1 — Cell A: ordinary tail-off

Worked example The plain case

A first-stage engine reads $F_{0} = 120 kN$ at the cutoff command. Tail-off time $τ = 0.6 s$ . The stage mass at that instant is $m = 20000 kg$ . Find the tail-off impulse $J_{t ai l}$ and the un-commanded velocity kick $Δ v_{t ai l}$ .

Forecast: guess — will the extra $Δ v$ be a few cm/s, a few m/s, or tens of m/s?

Convert units. $F_{0} = 120 kN = 120000 N$ . Why this step? The formula uses newtons; a stray factor of 1000 is the single most common numeric error.
Impulse = area under the fade curve. Using the restated profile $F (t) = F_{0} e^{- t / τ}$ , $J_{t ai l} = \int_{0}^{\infty} F_{0} e^{- t / τ} d t = F_{0} τ = 120000 \times 0.6 = 72000 N\cdots .$ Why this step? Impulse is the area under $F (t)$ ; the area of the whole decaying curve equals a rectangle of height $F_{0}$ and width $τ$ .
Velocity kick. $Δ v_{t ai l} = \frac{J _{t ai l}}{m} = \frac{72000}{20000} = 3.6 m/s$ . Why this step? Impulse is the change in momentum, $J = m Δ v$ , so $Δ v = J / m$ .

Verify: Units — $\frac{N\cdots}{kg} = \frac{kg\cdotm/s ^{2} \cdot s}{kg} = m/s$ ✓. Magnitude $3.6$ m/s is "a few m/s" — the middle forecast. That is real velocity guidance must budget for.

Ex 2 — Cell B: instant-cutoff limit ( $τ \to 0$ )

Worked example The degenerate case people

assume is always true Same engine, $F_{0} = 120 kN$ , but now imagine an idealized valve that shuts perfectly instantly: $τ \to 0$ . What tail-off impulse does the model predict?

Forecast: if the fade takes zero time, is there still a "hidden" impulse?

Take the limit. $J_{t ai l} = F_{0} τ \to 120000 \times 0 = 0 N\cdots$ . Why this step? We are testing the degenerate input to make sure the formula behaves sanely at its boundary.
Interpret. With $τ = 0$ the thrust really is a switch — no leftover propellant burns, no extra impulse. Why this step? This confirms the impulse-momentum picture: no area under the $F (t)$ curve ⇒ no momentum change.

Verify: As $τ \to 0$ , $\int_{0}^{\infty} F_{0} e^{- t / τ} d t = F_{0} τ \to 0$ ✓. So the "thrust is a switch" belief is exactly the $τ = 0$ special case — real engines have $τ > 0$ , which is why the parent note warns you not to assume it.

Ex 3 — Cell C: only part of the tail counts

Worked example Finite window, not infinity

Sometimes separation happens before the tail fully dies. Same $F_{0} = 120 kN$ , $τ = 0.6 s$ , but you only care about the impulse delivered in the first $t_{1} = 0.6 s$ (one $τ$ ) after cutoff. Find that partial impulse $J (t_{1})$ .

Forecast: one full $τ$ has passed — do you think you've collected roughly half, two-thirds, or nearly all of the total $72000$ N·s?

Set up the integral with a finite top limit. $J (t_{1}) = \int_{0}^{t_{1}} F_{0} e^{- t / τ} d t .$ Why this step? The full-tail integral goes to $\infty$ ; here the physical window is cut short at $t_{1}$ , so the top limit is finite and the answer will keep a leftover term.
Do the integral, step by step. The antiderivative of $e^{- t / τ}$ is $- τ e^{- t / τ}$ (differentiate it back: $\frac{d}{d t} [- τ e^{- t / τ}] = e^{- t / τ}$ ✓). So $J (t_{1}) = F_{0} [- τ e^{- t / τ}]_{0}^{t_{1}} = F_{0} (- τ e^{- t_{1} / τ} - (- τ e^{0})) = F_{0} τ (1 - e^{- t_{1} / τ}) .$ Why this step? Plugging the two limits in and subtracting is exactly how a definite integral collects the area between $0$ and $t_{1}$ ; the constant $τ$ pulls out and the two exponentials give the $(1 - e^{- t_{1} / τ})$ shape.
Plug in $t_{1} = τ$ so $t_{1} / τ = 1$ . $J (τ) = 72000 (1 - e^{- 1}) = 72000 \times 0.63212 \approx 45513 N\cdots .$ Why this step? One time-constant always delivers the same fraction $1 - e^{- 1} \approx 63.2%$ — a number worth memorising.

Verify: Fraction $= 1 - e^{- 1} = 0.632$ ✓, matching the forecast "roughly two-thirds". The remaining $36.8%$ dribbles out after $t_{1}$ , confirming Ex 1's total is bigger.

Figure 1 below draws exactly this split. The horizontal axis is time $t$ (s) after cutoff, the vertical axis is thrust $F$ (kN). The blue shaded area from $0$ to $τ$ is the $63.2%$ we just computed; the pink faint tail to the right of the yellow $t = τ$ line is the remaining $36.8%$ .

Figure — Staging events — separation dynamics, thrust tail-off

Figure 1 — Thrust tail-off $F (t) = F_{0} e^{- t / τ}$ : impulse is the area under the curve, split at one time-constant.

Ex 4 — Cell D: separation, unequal masses

Worked example The general two-body shove

Spent lower stage $m_{1} = 12000 kg$ , upper stage $m_{2} = 3000 kg$ . Separation springs deliver $J_{se p} = 4500 N\cdots$ . Find the reduced mass $μ$ , the relative separation velocity $v_{r e l}$ , and each stage's own $Δ v$ .

Forecast: which stage moves faster after the push — the heavy one or the light one?

Reduced mass. $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{12000 \times 3000}{15000} = \frac{3.6 \times 1 0 ^{7}}{15000} = 2400 kg$ . Why this step? Relative motion under an internal force behaves like one particle of mass $μ$ — see Reduced Mass and Two-Body Problem.
Relative velocity. $v_{r e l} = \frac{J _{se p}}{μ} = \frac{4500}{2400} = 1.875 m/s$ . Why this step? This is the drift-apart speed, the thing that decides collision safety.
Individual kicks (forward $= +$ ): $Δ v_{2} = + \frac{J _{se p}}{m _{2}} = \frac{4500}{3000} = + 1.5 m/s$ (upper speeds up); $Δ v_{1} = - \frac{J _{se p}}{m _{1}} = - \frac{4500}{12000} = - 0.375 m/s$ (lower falls behind). Why this step? Newton's third law: equal-and-opposite impulse on each body.

Verify: $Δ v_{2} - Δ v_{1} = 1.5 - (- 0.375) = 1.875 = v_{r e l}$ ✓. Momentum check (Conservation of Momentum): $m_{2} Δ v_{2} + m_{1} Δ v_{1} = 3000 (1.5) + 12000 (- 0.375) = 4500 - 4500 = 0$ ✓. The lighter upper stage moves faster — forecast confirmed.

Ex 5 — Cell E: equal masses (and Cell I: zero impulse)

Worked example The symmetric limit — plus a zero-input sanity check

Two stages of equal mass $m_{1} = m_{2} = 5000 kg$ . (a) Springs deliver $J_{se p} = 4000 N\cdots$ : find $μ$ and $v_{r e l}$ . (b) Then check the degenerate case $J_{se p} = 0$ .

Forecast: with equal masses, how does $μ$ relate to a single stage mass?

Reduced mass for equal masses. $μ = \frac{m \cdot m}{m + m} = \frac{m}{2} = \frac{5000}{2} = 2500 kg$ . Why this step? Equal masses is a clean special case worth knowing: $μ$ is exactly half of each mass.
Relative velocity (a). $v_{r e l} = \frac{4000}{2500} = 1.6 m/s$ . Why this step? Same formula, checking it works at the symmetric point.
Individual kicks. $Δ v_{2} = + \frac{4000}{5000} = + 0.8$ , $Δ v_{1} = - 0.8 m/s$ — symmetric, as expected. Why this step? Equal masses ⇒ equal-and-opposite speeds, each exactly half of $v_{r e l}$ .
Zero-input case (b). $J_{se p} = 0 \Rightarrow v_{r e l} = 0/2500 = 0 m/s$ . Why this step? No shove ⇒ no drift ⇒ no separation. The stages ride together and will collide at ignition — the danger the parent note warns about.

Verify: $v_{r e l} = 0.8 - (- 0.8) = 1.6$ ✓. Momentum $5000 (0.8) + 5000 (- 0.8) = 0$ ✓. And $J_{se p} = 0 \Rightarrow v_{r e l} = 0$ ✓ (no impulse, no motion — the formula degrades gracefully).

Ex 6 — Cell F: extreme mass ratio ( $m_{1} ≫ m_{2}$ )

Worked example A giant spent stage kicks off a tiny upper stage

A huge spent booster $m_{1} = 90000 kg$ separates from a small upper stage $m_{2} = 1000 kg$ , with $J_{se p} = 2000 N\cdots$ . Find $μ$ , $v_{r e l}$ , and compare $μ$ to $m_{2}$ .

Forecast: when one body is 90× heavier, does the reduced mass sit near the light mass or near the heavy mass?

Reduced mass. $μ = \frac{90000 \times 1000}{91000} = \frac{9.0 \times 1 0 ^{7}}{91000} = 989.01 kg$ . Why this step? Testing the limiting behaviour $m_{1} ≫ m_{2}$ .
Compare to $m_{2}$ . $μ = 989 kg$ is within $1%$ of $m_{2} = 1000 kg$ . Why this step? This confirms the limit $μ \to m_{2}$ when $m_{1} \to \infty$ — the heavy body barely recoils, so the relative motion is almost entirely the light body's motion.
Relative velocity. $v_{r e l} = \frac{2000}{989.01} = 2.0222 m/s$ . Why this step? Nearly all the drift is the upper stage's own $Δ v_{2} = 2000/1000 = 2.0$ m/s; the heavy stage's $Δ v_{1} = - 2000/90000 = - 0.0222$ m/s is tiny.

Verify: $Δ v_{2} - Δ v_{1} = 2.0 - (- 0.02222) = 2.02222 = v_{r e l}$ ✓. And $μ / m_{2} = 989.01/1000 = 0.989$ ≈ 1, so $μ \to m_{2}$ — forecast (near the light mass) confirmed.

Ex 7 — Cell G: collision-risk word problem (solve backwards)

Worked example "How hard must the springs push?"

Mission rule: the upper stage must open a gap of $d = 4 m$ before its engine ignites at $t_{c l e a r} = 2.0 s$ after bolt-fire, or the plume will hit the spent stage. Masses: $m_{1} = 8000 kg$ , $m_{2} = 2000 kg$ . What minimum separation impulse $J_{se p}$ is required?

Forecast: will the needed impulse be hundreds or thousands of N·s?

Required relative velocity. During the short window (gravity/drag neglected, as flagged up top), drift is steady, so $d = v_{r e l} t_{c l e a r}$ . Thus $v_{r e l} \geq \frac{d}{t _{c l e a r}} = \frac{4}{2.0} = 2.0 m/s .$ Why this step? The clearance condition converts a distance deadline into a minimum speed.
Reduced mass. $μ = \frac{8000 \times 2000}{10000} = 1600 kg$ . Why this step? We need $μ$ to turn required speed into required impulse.
Solve for impulse. From $v_{r e l} = J_{se p} / μ$ : $J_{se p} = μ v_{r e l} = 1600 \times 2.0 = 3200 N\cdots$ . Why this step? This is the design number the spring/retro-rocket must deliver — at minimum.

Verify: Plug back: $v_{r e l} = 3200/1600 = 2.0$ m/s, gap $= 2.0 \times 2.0 = 4.0$ m ✓ = $d$ . Units: $kg \times m/s = N\cdots$ ✓. Any smaller impulse ⇒ gap $< 4$ m ⇒ collision — so $3200$ N·s is the safety floor.

Ex 8 — Cell H: exam twist, tail-off and separation together

Worked example The combined scenario with sign bookkeeping

Just before staging the whole vehicle (both stages) moves at $v_{0} = 2500 m/s$ . The upper-stage engine has already cut, but its tail-off adds a kick. Data: $F_{0} = 30 kN$ , $τ = 0.5 s$ ; masses at separation $m_{1} = 6000 kg$ (spent), $m_{2} = 4000 kg$ (upper); separation impulse $J_{se p} = 5000 N\cdots$ . Find the upper stage's final velocity after both the tail-off and the separation push.

Forecast: tail-off and separation both push the upper stage forward — will its final speed be above or below $2500$ m/s, and by how much?

Convert units first. $F_{0} = 30 kN = 30000 N$ (same care as Ex 1 — kilonewtons must become newtons before multiplying by seconds). Why this step? Consistent SI units are what make $J = F_{0} τ$ come out in N·s.
Tail-off kick on the upper stage. The tail-off thrust acts on the upper stage (mass $m_{2}$ ): $Δ v_{t ai l} = \frac{F _{0} τ}{m _{2}} = \frac{30000 \times 0.5}{4000} = \frac{15000}{4000} = 3.75 m/s .$ Why this step? This is Cell A's tool applied to the upper stage's own mass — a forward $(+)$ kick.
Separation kick on the upper stage. $Δ v_{2} = + \frac{J _{se p}}{m _{2}} = \frac{5000}{4000} = + 1.25 m/s$ . Why this step? The upper stage receives the forward half of the equal-opposite spring impulse.
Add signed contributions. Both are forward, so $v_{f ina l} = v_{0} + Δ v_{t ai l} + Δ v_{2} = 2500 + 3.75 + 1.25 = 2505.0 m/s .$ Why this step? Velocities from separate impulses on the same body simply add (each is $J / m$ ); careful sign tracking keeps both as $+$ .

Verify: $2500 + 3.75 + 1.25 = 2505.0$ m/s ✓ — above the start, as forecast. Sanity: the spent stage's separation recoil $Δ v_{1} = - 5000/6000 = - 0.8333$ m/s is not added to the upper stage — it belongs to the lower body. Momentum of the separation alone: $4000 (1.25) + 6000 (- 0.8333) = 5000 - 5000 = 0$ ✓.

Recall Quick self-test across the matrix

Which cell has $μ = m /2$ ? ::: Equal masses (Cell E). As $m_{1} \to \infty$ , $μ \to$ ? ::: $m_{2}$ (Cell F, the light stage). As $τ \to 0$ , $J_{t ai l} \to$ ? ::: $0$ (Cell B, instant-cutoff limit). One time-constant delivers what fraction of the total tail-off impulse? ::: $1 - e^{- 1} \approx 63.2%$ (Cell C). To turn a gap deadline into a minimum spring impulse, use ::: $J_{se p} = μ d / t_{c l e a r}$ (Cell G). What three effects do we neglect during the short staging window? ::: gravity, drag, and continued mass loss (constant-mass impulse-momentum model).

See also: Tsiolkovsky Rocket Equation, Multistage Rocket Optimization, Impulse-Momentum Theorem, Gravity Losses and Ascent Trajectory.