Exercises — Systems with variable mass — rocket equation derivation preview

• $u=3000$ m/s (exhaust velocity is measured relative to the rocket).
• $|\dot{m}|=40$ kg/s (the burn rate).
• $m_0=5000$ kg ("wet mass" = fuelled = start).
• $m_f=2000$ kg ("dry mass" = empty = end).
• Mass ratio $=\frac{m_0}{m_f}=\frac{5000}{2000}=2.5$.

What we did: matched everyday engineering words to our symbols. Nothing computed yet — recognition only.

$F_{\text{thrust}}=u|\dot{m}|=3000\times 40=1.2\times 10^5\text{N}=120\text{kN}.$ Why this formula: thrust is the reaction force from throwing mass backward. Each second, $40$ kg leaves at $3000$ m/s relative to the rocket, so momentum leaves at $40\times 3000$ kg·m/s per second — and force is momentum per second (Impulse–Momentum Theorem).

No external force, so Tsiolkovsky applies directly: $\Delta v=u\ln \frac{m_0}{m_f}=2000\ln \frac{4000}{1000}=2000\ln 4.$ Why $\ln$: each kg burned pushes the remaining, smaller mass, so the payoff compounds — integrating $dm/m$ gives $\ln$. $=2000\times 1.3863\approx 2773\text{m/s}.$ Started at rest, so final speed $=2773$ m/s. We burned $3/4$ of the mass yet gained only $\approx 1.39u$ — the log is stingy.

$F_{\text{thrust}}=u|\dot{m}|=2000\times 25=5.0\times 10^4\text{N}.$ $a=\frac{F_{\text{thrust}}}{m}=\frac{5.0\times 10^4}{2000}=25{\text{m/s}}^2.$ Why current mass: acceleration always uses the mass right now. As $m$ falls, $a$ climbs even though thrust is constant.

Start from Tsiolkovsky and invert it — we know $\Delta v$ and want the ratio: $\Delta v=u\ln \frac{m_0}{m_f}\Rightarrow \frac{\Delta v}{u}=\ln \frac{m_0}{m_f}\Rightarrow \frac{m_0}{m_f}=e^{\Delta v/u}.$ Why exponentiate: $\ln$ and $e^{(\cdot )}$ undo each other — to get the ratio out of the log, raise $e$ to both sides (Logarithms and Exponential Growth). $\frac{m_0}{m_f}=e^{6000/3000}=e^2\approx \mathrm{7.389.}$ Fuel fraction $=\frac{m_0-m_f}{m_0}=1-\frac{m_f}{m_0}=1-e^{-2}\approx 1-0.1353=0.8647$. So about 86.5% of the rocket must be fuel — this is the "tyranny of the rocket equation" in action.

The 1-D near-Earth equation (from the parent) is $m\frac{dv}{dt}=u|\dot{m}|-mg.$ Hover means $\frac{dv}{dt}=0$, so thrust balances weight: $u|\dot{m}|=mg\Rightarrow |\dot{m}|=\frac{mg}{u}=\frac{8000\times 9.8}{2500}=\frac{78400}{2500}=31.36\text{kg/s}.$ For thrust-to-weight $=1.5$, thrust must be $1.5mg$: $u|\dot{m}|=1.5mg\Rightarrow |\dot{m}|=1.5\times 31.36=47.04\text{kg/s}.$ Why: thrust-to-weight ratio scales the burn rate linearly, since thrust $=u|\dot{m}|$ is linear in $|\dot{m}|$.

(a) Single stage: $\Delta v_a=3000\ln \frac{10000}{2000}=3000\ln 5\approx 3000\times 1.6094=4828\text{m/s}.$ (b) Stage 1 (burns $10000\to 4000$, so its final mass is $4000$ kg before jettison): $\Delta v_1=3000\ln \frac{10000}{4000}=3000\ln 2.5\approx 3000\times 0.9163=2749\text{m/s}.$ Then drop the $2000$ kg empty casing (no momentum change — it leaves at the rocket's speed). Stage 2 now starts at $2000$ kg: $\Delta v_2=3000\ln \frac{2000}{500}=3000\ln 4\approx 3000\times 1.3863=4159\text{m/s}.$ Total: $\Delta v_b=2749+4159=6908\text{m/s}.$ Why staging wins: in the single stage you drag the heavy empty tank the whole way. Dropping dead mass mid-flight means later fuel pushes a lighter vehicle — the log rewards you. Same total structure, $\approx 2080$ m/s more $\Delta v$.

(i) Constant burn, so mass falls linearly: $m(t)=m_0-|\dot{m}|t$. Burnout when $m=m_f$: $t_b=\frac{m_0-m_f}{|\dot{m}|}=\frac{6000-2000}{20}=200\text{s}.$ (ii) In deep space, $v(t)=u\ln \frac{m_0}{m(t)}$. Substituting $m(t)=m_0-|\dot{m}|t$: $v(t)=2500\ln \frac{6000}{6000-20t}.$ At burnout $t=200$ s: $m=2000$ kg, $v_f=2500\ln \frac{6000}{2000}=2500\ln 3\approx 2500\times 1.0986=2747\text{m/s}.$ Shape (see figure): $v$ rises slowly at first (heavy rocket) and steepens toward the end (light rocket, same thrust ⇒ bigger acceleration). It is not a straight line — the acceleration grows.

The 1-D equation with constant gravity is $m\frac{dv}{dt}=u|\dot{m}|-mg.$ Integrate over the burn. The thrust term integrates to the Tsiolkovsky log; the gravity term integrates to $-g{t}_b$ (a steady drain over the burn time). Result: $v_f=\underset{\text{ideal}}{\underbrace{u\ln \frac{m_0}{m_f}}}-\underset{\text{gravity loss}}{\underbrace{g{t}_b}}.$ Why the gravity term is just $g{t}_b$: in $mdv=u|\dot{m}|dt-mgdt$, dividing by $m$ leaves the gravity piece as $-gdt$, whose integral over the burn is simply $-g{t}_b$ — independent of how the mass changes. Burn time: $t_b=\frac{m_0-m_f}{|\dot{m}|}=\frac{500000-150000}{2500}=\frac{350000}{2500}=140\text{s}.$ Ideal $\Delta v$: $u\ln \frac{m_0}{m_f}=3000\ln \frac{500000}{150000}=3000\ln \frac{10}{3}\approx 3000\times 1.20397=3611.9\text{m/s}.$ Gravity loss: $g{t}_b=9.8\times 140=1372\text{m/s}.$ True burnout speed: $v_f\approx 3611.9-1372=2239.9\text{m/s}.$ Gravity stole $\approx 1372$ m/s — about 38% of the ideal $\Delta v$. This is why launches burn hard and fast (short $t_b$) and turn horizontal early: to shrink $g{t}_b$.

Invert Tsiolkovsky for $m_0$: $\frac{m_0}{m_f}=e^{\Delta v/u}=e^{9000/4000}=e^{2.25}\approx \mathrm{9.4877.}$ $m_0=m_f{e}^{2.25}=3000\times 9.4877\approx 28463\text{kg}.$ Fuel needed $=m_0-m_f\approx 28463-3000=25463\text{kg}.$ Reading it: to move a $3000$ kg payload+structure by $9000$ m/s you must carry $\approx 25.5$ tonnes of fuel — over $8\times$ the dry mass. Because the ratio is $e^{\Delta v/u}$, pushing $\Delta v$ up (or $u$ down) blows the fuel budget up exponentially. This is the tyranny, quantified.

L1 — $u$ is relative to the rocket ::: an engine property, ~2000–4500 m/s, not a ground speed. L2 — acceleration uses which mass ::: the current mass, so $a$ rises as fuel burns. L3 — invert Tsiolkovsky how ::: $m_0/m_f=e^{\Delta v/u}$ (exponentiate to undo the log). L4 — why staging beats a single stage ::: dropping dead mass mid-flight lets later fuel push a lighter vehicle. L5 — gravity loss equals ::: $g{t}_b$, subtracted once from the ideal $\Delta v$.