Classical-Mechanics · Unit 13 · Video 7 · Interactive Practice

Pushing on Its Own Fuel: Deriving the Rocket Equation

IKey Formulas

Formula	Name	What you need
$\Delta m_r = -\Delta m_f$	Mass bookkeeping	One constant-mass system over $[t,\ t+\Delta t]$
$\vec{F}_{\text{ext}} = m_r(t)\dfrac{d\vec{v}_r}{dt} - \dfrac{dm_r}{dt}\,\vec{u}$	The rocket equation	$\vec{u}$ measured relative to the rocket
$F_{\text{thrust},x} = -\dfrac{dm_r}{dt}\,u = \dfrac{dm_f}{dt}\,u$	Thrust	Burn rate and exhaust speed, both positive
$m_r(t)\dfrac{dv_{r,x}}{dt} = F_{\text{ext},x} + F_{\text{thrust},x}$	Newton II form	External force plus thrust, draining mass

Key Insight: Thrust is not a new force — nothing in the environment touches the rocket. It is the momentum the rocket keeps for itself by throwing mass away, moved to the force side of the equation by rearrangement.

IIOne Exhaust, Two Frames

Riding with the rocket the exhaust leaves at $u$ ; from the ground it moves at $\vec{u} + \vec{v}_r$ .

Rocket speed in the ground frame (m/s) 1200

IIIThe Derivation, Step by Step

One constant-mass system across $[t,\ t+\Delta t]$ turns momentum bookkeeping into the rocket equation.

Step 1

Step 1 — Choose a constant-mass system

m_{\text{sys}} = m_r(t) = m_{r,d} + m_f(t)

Step 2 — The same matter at $t + \Delta t$

m_r(t) + \Delta m_r + \Delta m_f = m_r(t) \;\Longrightarrow\; \Delta m_r = -\Delta m_f

Step 3 — Momentum at both snapshots

\vec{p}_{\text{sys}}(t) = m_r(t)\,\vec{v}_r(t)

\vec{p}_{\text{sys}}(t{+}\Delta t) = \big(m_r(t) + \Delta m_r\big)\vec{v}_r(t{+}\Delta t) - \Delta m_r\big(\vec{u} + \vec{v}_r(t{+}\Delta t)\big)

Step 4 — Expand: the middle terms cancel exactly

m_r(t)\vec{v}_r(t{+}\Delta t) + \underbrace{\Delta m_r\vec{v}_r(t{+}\Delta t) - \Delta m_r\vec{v}_r(t{+}\Delta t)}_{=\;0} - \Delta m_r\vec{u}

\vec{p}_{\text{sys}}(t{+}\Delta t) = m_r(t)\,\vec{v}_r(t{+}\Delta t) - \Delta m_r\,\vec{u}

Step 5 — Divide by $\Delta t$ and let $\Delta t \to 0$

\vec{F}_{\text{ext}} = \lim_{\Delta t \to 0} \frac{\vec{p}_{\text{sys}}(t{+}\Delta t) - \vec{p}_{\text{sys}}(t)}{\Delta t} = m_r(t)\frac{d\vec{v}_r}{dt} - \frac{dm_r}{dt}\,\vec{u}

IVThrust as Recoil

Thrust is recoil: every kilogram hurled backward at $u$ kicks the rocket forward.

Time since ignition (s) 0.0 Engine

💡 The freight car leaking sand ran the same term to zero: its grains left at the car's own velocity, so the relative velocity was $0$ and no recoil appeared.

VQuiz Questions

Problem 1 · Thrust from Burn Rate and Exhaust Speed

Given: an engine burning fuel at $\dfrac{dm_f}{dt} = 250\ \text{kg/s}$ and ejecting it at $u = 2800\ \text{m/s}$ relative to the rocket, whose instantaneous mass is $m_r = 1.4\times10^{4}\ \text{kg}$ — find the thrust $F_{\text{thrust},x}$ .

✅ Correct! Burn rate times exhaust speed:

(250)(2800) = 7.0\times10^{5}\ \text{N}

, and the instantaneous mass never enters the thrust — only the acceleration it produces,

a = 50\ \text{m/s}^2

❌ That is $m_r u$ , not $\dot{m}_f u$ . The thrust comes from the mass leaving each second, so the factor is

250\ \text{kg/s}

, not the

1.4\times10^{4}\ \text{kg}

still aboard.

❌ You counted the third-law pair twice. The rocket's push on the fuel acts on the fuel; only its partner, the fuel's push on the rocket, belongs in the rocket's equation.

❌ Check the operation.

F_{\text{thrust},x} = (dm_f/dt)\,u

is a product; dividing gives

\text{m/s} \div (\text{kg/s})

, which is not a newton.

Show solution

The thrust term is the mass rate multiplied by the exhaust speed relative to the rocket:

F_{\text{thrust},x} = -\frac{dm_r}{dt}\,u = \frac{dm_f}{dt}\,u = (250\ \text{kg/s})(2800\ \text{m/s}) = 7.0\times10^{5}\ \text{N}

Sign check: $dm_f/dt > 0$ and $u > 0$ , so $F_{\text{thrust},x} > 0$ — the thrust points forward, opposite to the direction the fuel is thrown.

What the mass is for: in free space it fixes the acceleration, not the thrust:

a_x = \frac{F_{\text{thrust},x}}{m_r} = \frac{7.0\times10^{5}\ \text{N}}{1.4\times10^{4}\ \text{kg}} = 50\ \text{m/s}^2

Problem 2 · Which Velocity Belongs in the Thrust

Given: a rocket streaking through the ground frame at $v_{r,x} = 4000\ \text{m/s}$ , its engine ejecting $200\ \text{kg/s}$ at $u = 3000\ \text{m/s}$ relative to the rocket — find the thrust.

✅ Correct!

(200)(3000) = 6.0\times10^{5}\ \text{N}

— and note this rocket outruns its own exhaust, which drifts forward at

4000 - 3000 = 1000\ \text{m/s}

in the ground frame, with no effect at all on the thrust.

❌ That is the lazy product rule. Differentiating

m_r\vec{v}_r

carelessly puts

\vec{v}_r

where

\vec{u}

belongs — an answer that would change from one inertial frame to another, which no real force can do.

❌ That is the exhaust's ground-frame velocity.

4000 - 3000 = 1000\ \text{m/s}

describes where the batch drifts, but the recoil is set by how fast the engine throws it,

u = 3000\ \text{m/s}

❌ The speeds are not added. The exhaust's ground velocity is the vector sum

\vec{u} + \vec{v}_r

, and with

\vec{u}

pointing backward that is

-3000 + 4000

, not

3000 + 4000

— and neither sum belongs in the thrust.

Show solution

In the rocket equation the mass rate multiplies $\vec{u}$ , the exhaust velocity relative to the rocket:

F_{\text{thrust},x} = \frac{dm_f}{dt}\,u = (200\ \text{kg/s})(3000\ \text{m/s}) = 6.0\times10^{5}\ \text{N}

Exhaust speed is a property of the engine — the chemistry of the burn and the shape of the nozzle — so the same engine gives the same thrust whether the rocket crawls or streaks.

Where the ground frame does show up: the batch's ground-frame velocity is

\vec{u} + \vec{v}_r = (-3000 + 4000)\,\hat{i} = +1000\ \hat{i}\ \text{m/s}

so the exhaust actually drifts forward here, just $3000\ \text{m/s}$ slower than the rocket that dropped it. The algebra never asks which way it drifts.

Problem 3 · Launch Against Gravity

Given: a rocket of instantaneous mass $m_r = 3.0\times10^{5}\ \text{kg}$ lifting straight up, burning $1500\ \text{kg/s}$ at $u = 2500\ \text{m/s}$ , with $g = 9.8\ \text{m/s}^2$ and $+x$ taken upward — find the thrust and the acceleration.

What is the thrust?

What is the acceleration?

✅ Correct! The thrust

3.75\times10^{6}\ \text{N}

beats the weight

2.94\times10^{6}\ \text{N}

by only

8.1\times10^{5}\ \text{N}

, so this heavy rocket leaves the pad at a gentle

2.7\ \text{m/s}^2

❌ You used the mass, not the mass rate.

(3.0\times10^{5}\ \text{kg})(2500\ \text{m/s})

is a momentum, not a force; the thrust needs

dm_f/dt = 1500\ \text{kg/s}

❌ That is the net force, not the thrust.

8.1\times10^{5}\ \text{N}

is what is left after gravity takes its share — the thrust itself is the full

(1500)(2500)

❌ Check the thrust.

F_{\text{thrust},x} = (dm_f/dt)\,u

— burn rate times exhaust speed, with

g

playing no part in it.

❌ Gravity is still acting.

12.5\ \text{m/s}^2

F_{\text{thrust},x}/m_r

alone; the launch equation is

m_r a_x = F_{\text{ext},x} + F_{\text{thrust},x}

with

F_{\text{ext},x} = -m_r g

❌ Check which one is bigger. The thrust exceeds the weight,

3.75\times10^{6} > 2.94\times10^{6}\ \text{N}

, so the net force — and the acceleration — points upward.

❌ Check the acceleration. Subtract the weight from the thrust before dividing by the mass.

Show solution

Step 1 — Thrust. It depends only on the engine:

F_{\text{thrust},x} = \frac{dm_f}{dt}\,u = (1500\ \text{kg/s})(2500\ \text{m/s}) = 3.75\times10^{6}\ \text{N}

Step 2 — External force. Gravity is the only thing the environment applies:

F_{\text{ext},x} = -m_r g = -(3.0\times10^{5}\ \text{kg})(9.8\ \text{m/s}^2) = -2.94\times10^{6}\ \text{N}

Step 3 — Newton II form of the rocket equation.

m_r(t)\frac{dv_{r,x}}{dt} = F_{\text{ext},x} + F_{\text{thrust},x} = -2.94\times10^{6} + 3.75\times10^{6} = 8.1\times10^{5}\ \text{N}

a_x = \frac{8.1\times10^{5}\ \text{N}}{3.0\times10^{5}\ \text{kg}} = 2.7\ \text{m/s}^2 \ \text{(upward)}

Note: as the tanks drain, $m_r$ falls while $F_{\text{thrust},x}$ stays fixed, so $a_x$ climbs throughout the burn.

Problem 4 · The Same Term on a Leaking Hopper Car

Given: a hopper car rolling at $15\ \text{m/s}$ on level frictionless track leaks sand through a hole in its floor at $20\ \text{kg/s}$ ; each grain leaves with the car's own velocity, so $u = 0$ — find the thrust the escaping sand delivers.

✅ Correct!

(20\ \text{kg/s})(0) = 0

— the sand is dropped, not thrown, so there is no recoil and the car rolls on at

15\ \text{m/s}

while getting lighter.

❌ Wrong velocity in the term.

15\ \text{m/s}

is the car's velocity in the ground frame; the thrust term needs the exhaust velocity relative to the vehicle, which here is zero.

❌ That is a weight, not a thrust.

(20\ \text{kg/s})(9.8\ \text{m/s}^2)

has the units of force, but the sand's weight acts vertically and the track carries it — it never pushes the car along

x

❌ Losing mass is not a brake. Nothing pushes backward on the car: the departing grains keep the velocity they had, so they take away momentum and mass in exactly the ratio that leaves

v_x

unchanged.

Show solution

The thrust term is always mass rate times the ejection velocity relative to the vehicle:

F_{\text{thrust},x} = \frac{dm_f}{dt}\,u = (20\ \text{kg/s})(0\ \text{m/s}) = 0

With no external horizontal force either, the rocket equation reduces to

m(t)\frac{dv_x}{dt} = 0 + 0 \quad\Longrightarrow\quad v_x = 15\ \text{m/s}\ \text{throughout}

The contrast: replace the hole with an engine that throws the same $20\ \text{kg/s}$ backward at $u = 3000\ \text{m/s}$ , and the identical term becomes

F_{\text{thrust},x} = (20)(3000) = 6.0\times10^{4}\ \text{N}

Same mass rate, same vehicle — the relative velocity is the entire difference between coasting and accelerating.

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