Tsiolkovsky rocket equation — full first-principles derivation from momentum

WHAT are we deriving?

WHY it must be true (before the algebra)

The only external tool is Newton's law in momentum form: $\vec F_{ext} = \dfrac{d\vec p}{dt}$. In free space $\vec F_{ext}=0$, so total momentum of (rocket + already-ejected gas) is conserved. Any speed the rocket gains must be paid for by momentum carried away in the exhaust. That single bookkeeping fact forces the log.

HOW — the derivation, step by step

Consider a snapshot at time $t$: rocket has mass $m$ and velocity $v$ (all in one dimension, ground frame).

Step 1 — Momentum now. $p(t) = m v$ Why this step? We track the whole system. Right now the fuel about to be burned is still inside, moving at $v$, so it's part of $mv$.

Step 2 — A tiny time $dt$ later. The rocket ejects a small mass $dm_{gas}>0$ backwards. The rocket's mass drops by that much, so if we write the rocket mass as $m$, then $dm$ (the change in rocket mass) is negative, and $dm_{gas} = -dm$.

• Rocket now: mass $m+dm$, velocity $v+dv$.
• Ejected gas: mass $-dm$, moving at velocity $(v - v_e)$ in the ground frame.

Why $v-v_e$? The gas leaves at speed $v_e$ relative to the rocket, pointing backward. In the ground frame that's the rocket's speed minus $v_e$.

Step 3 — Momentum then. $p(t+dt) = \underbrace{(m+dm)(v+dv)}_{\text{rocket}} + \underbrace{(-dm)(v-v_e)}_{\text{gas}}$

Why this step? Total momentum = rocket's new momentum + the momentum the gas carried off.

Step 4 — Expand and cancel. $p(t+dt) = mv + m\,dv + v\,dm + dm\,dv - v\,dm + v_e\,dm$ The $+v\,dm$ and $-v\,dm$ cancel. Drop the second-order term $dm\,dv$ (product of two tiny quantities → negligible). $p(t+dt) = mv + m\,dv + v_e\,dm$

Step 5 — Apply conservation. Free space ⇒ $p(t+dt)=p(t)=mv$. Subtract: $0 = m\,dv + v_e\,dm$

Why this step? No external force means the change in total momentum over $dt$ is zero. This is the whole physics — everything after is calculus.

Step 6 — Separate variables. $m\,dv = -v_e\,dm \quad\Rightarrow\quad dv = -v_e\,\frac{dm}{m}$

Step 7 — Integrate from start ($v=0$... or $v_i$, $m=m_0$) to finish ($v=v_f$, $m=m_f$): $\int_{v_i}^{v_f} dv = -v_e \int_{m_0}^{m_f} \frac{dm}{m}$ $v_f - v_i = -v_e\big[\ln m\big]_{m_0}^{m_f} = -v_e(\ln m_f - \ln m_0)$

The thrust connection (bonus HOW)

From Step 5, divide by $dt$: $\; m\dfrac{dv}{dt} = -v_e\dfrac{dm}{dt}$. Since $\dfrac{dm}{dt}<0$, define mass flow rate $\dot m = -\dfrac{dm}{dt}>0$. Then the force on the rocket: $F_{thrust} = v_e\,\dot m$ Why this matters: thrust depends on how fast you throw mass ($v_e$) times how much per second ($\dot m$). The rocket equation is the time-integrated version of this.

Worked Examples

Common Mistakes (Steel-manned)

Assumptions baked in (the fine print)

• No external forces (no gravity loss, no drag). Real launches add a $-\int g\,dt$ gravity loss term.
• Constant $v_e$.
• Ejection is continuous (calculus limit), not discrete lumps.

Active Recall Flashcards

Connections

• Conservation of Momentum — the sole physical principle used.
• Newton's Second Law (momentum form) — $F = dp/dt$, source of Step 5.
• Thrust and Mass Flow Rate — the differential ($F=v_e\dot m$) form.
• Multistage Rockets — why staging beats the log's diminishing returns.
• Specific Impulse — $I_{sp} = v_e/g_0$, engineering measure of $v_e$.
• Gravity Loss and Drag Loss — corrections to ideal $\Delta v$.
• Variable Mass Systems — general framework this is a special case of.

Concept Map

Hinglish (regional understanding)