Staging events — separation dynamics, thrust tail-off: Exercises

Before you start, here is the whole toolkit on one card so no symbol appears "unearned" below.

Here is the geometry you will lean on for the separation problems.

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

Level 1 — Recognition

Goal: name the tool and plug in. No traps yet.

Exercise 1.1

An engine holds thrust $F_{0} = 40 kN$ and after the cutoff command its thrust decays with tail-off time $τ = 0.25 s$ . What is the tail-off impulse $J_{t ai l}$ ?

Recall Solution

WHAT tool: tail-off impulse is the area under the decaying-thrust curve, which equals a rectangle of height $F_{0}$ and width $τ$ . $J_{t ai l} = F_{0} τ = 40000 \times 0.25 = 10000 N\cdots = 10 kN\cdots .$ WHY $F_{0} τ$ and not something fancier: the exponential $e^{- t / τ}$ integrated from $0$ to $\infty$ gives exactly $τ$ , so all that survives is $F_{0} \times τ$ .

Exercise 1.2

Two stages: spent lower $m_{1} = 6000 kg$ , upper $m_{2} = 3000 kg$ . Compute the reduced mass $μ$ .

Recall Solution

WHY reduced mass: the two stages push on each other (internal force), so their relative motion behaves like one particle of mass $μ$ . $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{6000 \times 3000}{6000 + 3000} = \frac{1.8 \times 1 0 ^{7}}{9000} = 2000 kg .$ Notice $μ = 2000$ is smaller than either mass — always true. It sits nearer the lighter body.

Level 2 — Application

Goal: chain two formulas.

Exercise 2.1

An upper stage of mass $m = 4000 kg$ suffers tail-off with $F_{0} = 50 kN$ , $τ = 0.3 s$ . Find the uncommanded velocity kick $Δ v_{t ai l}$ .

Recall Solution

Step 1 — impulse (WHAT/WHY): the area under the tail is $J_{t ai l} = F_{0} τ = 50000 \times 0.3 = 15000 N\cdots$ . Step 2 — turn impulse into speed: by the Impulse-Momentum Theorem, $J = m Δ v$ , so $Δ v_{t ai l} = \frac{J _{t ai l}}{m} = \frac{15000}{4000} = 3.75 m/s .$ This 3.75 m/s is free velocity guidance must predict, or burnout speed is wrong.

Exercise 2.2

Springs deliver $J_{se p} = 2400 N\cdots$ between spent stage $m_{1} = 6000 kg$ and upper $m_{2} = 3000 kg$ . Find $v_{r e l}$ and the gap after $t = 2.0 s$ .

Recall Solution

Step 1: reuse $μ = 2000 kg$ from Exercise 1.2. Step 2 — relative speed: $v_{r e l} = \frac{J _{se p}}{μ} = \frac{2400}{2000} = 1.2 m/s .$ Step 3 — gap (constant drift, WHAT IT LOOKS LIKE: the orange arrow in the figure just gets longer): $d = v_{r e l} t = 1.2 \times 2.0 = 2.4 m .$

Level 3 — Analysis

Goal: split effects, check conservation, reason about "enough or not."

Exercise 3.1

Same masses ( $m_{1} = 6000$ , $m_{2} = 3000$ kg) and $J_{se p} = 2400 N\cdots$ . Find each stage's own velocity change, then verify total momentum is conserved.

Recall Solution

Newton's third law: each body gets equal-and-opposite impulse.

Upper (gets forward push): $Δ v_{2} = + \frac{J _{se p}}{m _{2}} = \frac{2400}{3000} = + 0.8 m/s .$
Lower (recoils backward): $Δ v_{1} = - \frac{J _{se p}}{m _{1}} = - \frac{2400}{6000} = - 0.4 m/s .$
Relative check: $Δ v_{2} - Δ v_{1} = 0.8 - (- 0.4) = 1.2 m/s = v_{r e l}$ ✓.
Momentum check (Conservation of Momentum): $m_{2} Δ v_{2} + m_{1} Δ v_{1} = 3000 (0.8) + 6000 (- 0.4) = 2400 - 2400 = 0$ ✓. Internal forces move nothing's centre of mass.

Exercise 3.2

Mission rule: the stages must open a gap of $d = 3.0 m$ within $t_{c l e a r} = 1.5 s$ before the upper engine lights. With $μ = 2000 kg$ , what is the minimum $J_{se p}$ the springs must deliver?

Recall Solution

Step 1 — required relative speed: $v_{r e l} \geq \frac{d}{t _{c l e a r}} = \frac{3.0}{1.5} = 2.0 m/s .$ Step 2 — invert the separation formula: since $v_{r e l} = J_{se p} / μ$ , $J_{se p} \geq μ v_{r e l} = 2000 \times 2.0 = 4000 N\cdots .$ Anything less and the plume can strike the not-yet-cleared stage.

Level 4 — Synthesis

Goal: combine tail-off + separation, or separation + the rocket equation.

Exercise 4.1

An upper stage of mass $m = 4000 kg$ first experiences tail-off ( $F_{0} = 50 kN$ , $τ = 0.3 s$ ) which adds forward speed, then the springs ( $J_{se p} = 3600 N\cdots$ ) shove it forward off the spent stage $m_{1} = 6000 kg$ . Compute (a) the upper stage's total velocity gain from these two events, and (b) $v_{r e l}$ .

Recall Solution

(a) Two forward pushes on the same body add:

Tail-off kick: $Δ v_{t ai l} = F_{0} τ / m = 15000/4000 = 3.75 m/s$ (from Ex 2.1).
Separation push on the upper stage: $Δ v_{2} = J_{se p} / m_{2} = 3600/4000 = 0.9 m/s$ .
Total upper-stage gain $= 3.75 + 0.9 = 4.65 m/s .$ (b) Relative speed uses reduced mass with $m_{1} = 6000$ , $m_{2} = 4000$ : $μ = \frac{6000 \times 4000}{6000 + 4000} = \frac{2.4 \times 1 0 ^{7}}{1 0 ^{4}} = 2400 kg, v_{r e l} = \frac{3600}{2400} = 1.5 m/s .$ WHY they differ: $Δ v_{2} = 0.9$ m/s is how much the upper stage alone speeds up; $v_{r e l} = 1.5$ m/s is how fast the two separate (it also counts the lower stage recoiling backward).

Exercise 4.2

An upper stage burns from wet mass $m_{0} = 9000 kg$ to dry mass $m_{f} = 3000 kg$ with $I_{s p} = 340 s$ (take $g_{0} = 9.81 m/s^{2}$ ). Find its exhaust speed $v_{e}$ and the stage's $Δ v$ from the Tsiolkovsky Rocket Equation.

Recall Solution

Step 1 — exhaust speed: $v_{e} = I_{s p} g_{0} = 340 \times 9.81 = 3335.4 m/s .$ Step 2 — rocket equation (WHY the log: it counts how the mass ratio shrinks continuously as fuel leaves): $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 3335.4 \times ln \frac{9000}{3000} = 3335.4 \times ln 3.$ With $ln 3 = 1.0986$ : $Δ v = 3335.4 \times 1.0986 \approx 3664.4 m/s .$

Level 5 — Mastery

Goal: full reasoning, edge cases, design decisions.

Exercise 5.1 (design margin)

Springs are certified to deliver $J_{se p} = 4000 N\cdots$ . Stages: $m_{1} = 8000 kg$ , $m_{2} = 2000 kg$ . Ignition must wait until a gap of $d = 3.5 m$ exists, and the flight plan allows only $t_{c l e a r} = 1.8 s$ . Does the design clear? State the margin.

Recall Solution

Step 1 — $μ$ : $μ = \frac{8000 \times 2000}{10000} = 1600 kg .$ Step 2 — achieved relative speed: $v_{r e l} = \frac{4000}{1600} = 2.5 m/s .$ Step 3 — gap achieved in the allowed time: $d_{a c h} = v_{r e l} t_{c l e a r} = 2.5 \times 1.8 = 4.5 m .$ Step 4 — verdict: required $3.5$ m, achieved $4.5$ m → clears, with margin $4.5 - 3.5 = 1.0 m$ (about $29%$ over the requirement). ✓

Exercise 5.2 (limiting / degenerate case)

Take the separation formula $v_{r e l} = J_{se p} / μ$ with $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ . Suppose the spent lower stage is enormous compared to the upper: $m_{1} \to \infty$ with $m_{2}$ fixed and $J_{se p}$ fixed. What does $v_{r e l}$ approach, and what does it physically mean?

Recall Solution

Take the limit: $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{m _{2}}{1 + m _{2} / m _{1}} m_{1} \to \infty m_{2} .$ So $v_{r e l} \to \frac{J _{se p}}{m _{2}} = Δ v_{2}$ . Physical meaning: an infinitely heavy lower stage does not recoil at all ( $Δ v_{1} = - J_{se p} / m_{1} \to 0$ ). Then all of the relative motion is the upper stage moving; separating speed equals the upper stage's own speed change. Check with numbers: $m_{2} = 2000$ , $J_{se p} = 4000$ → $v_{r e l} \to 2.0 m/s$ . This is the same "recoil vanishes for a huge partner" idea as firing a bullet from a battleship: the ship barely moves.

Exercise 5.3 (cutoff timing correction)

A stage must hit a precise burnout speed. Guidance predicts a tail-off kick $Δ v_{t ai l} = F_{0} τ / m$ . With $F_{0} = 60 kN$ , $τ = 0.4 s$ , $m = 5000 kg$ , by how much must the commanded cutoff be moved earlier in velocity terms, and if the stage is accelerating at $a = 25 m/s^{2}$ at that instant, roughly how much earlier in time should cutoff be commanded to cancel the tail-off kick?

Recall Solution

Step 1 — the unwanted kick: $Δ v_{t ai l} = \frac{F _{0} τ}{m} = \frac{60000 \times 0.4}{5000} = 4.8 m/s .$ Guidance must subtract this from the target. Step 2 — convert speed error to a time-advance: near cutoff the stage gains speed at rate $a$ , so cutting off earlier by $δ t$ removes $a δ t$ of speed. Set $a δ t = Δ v_{t ai l}$ : $δ t = \frac{Δ v _{t ai l}}{a} = \frac{4.8}{25} = 0.192 s .$ Interpretation: command shutdown about $0.19$ s early so the tail-off dribble fills in the last $4.8$ m/s exactly. Notice $δ t = 0.192 s$ is smaller than $τ = 0.4$ s — you don't advance by the whole tail-off duration, only by however long it takes normal thrust to build the same $4.8$ m/s.

Recall Quick self-test (cloze)

The tail-off impulse equals == $F_{0} τ$ == because the area under $F_{0} e^{- t / τ}$ from $0$ to $\infty$ is == $τ$ . Separation relative speed uses the reduced mass== $μ = m_{1} m_{2} / (m_{1} + m_{2})$ , which is always smaller than either mass. To cancel a tail-off kick $Δ v_{t ai l}$ you advance cutoff by == $Δ v_{t ai l} / a$ ==, not by $τ$ .