Systems with variable mass — rocket equation derivation preview: Question bank

Throughout, recall the two headline results we lean on:

General variable-mass law: $m \frac{d v}{d t} = F_{ext} - u \frac{d m}{d t}$ .
Ideal (force-free) rocket equation: $Δ v = u ln \frac{m _{0}}{m _{f}}$ .

Here $m$ is the rocket's current mass, $\overset{m}{˙} = \frac{d m}{d t} < 0$ (mass is leaving), and $u$ is the exhaust speed relative to the rocket. See Conservation of Linear Momentum, Newton's Third Law, and Impulse–Momentum Theorem for the machinery these rest on.

True or false — justify

A rocket in deep space needs something (air, ground) to push against to accelerate.

False — it pushes against its own ejected gas via Newton's Third Law; the gas goes back, the rocket goes forward. No external medium is involved, which is exactly why rockets work in vacuum.

If a rocket ejected its fuel at zero speed relative to itself (

u = 0

), it would still speed up because it is losing mass.

False — with

u = 0

the thrust term

- u \overset{m}{˙}

vanishes, so

m \overset{v}{˙} = F_{ext}

; merely getting lighter gives no push. You need the fuel to actually leave you behind at nonzero relative speed.

Doubling the exhaust speed

u

doubles the final

Δ v

(all else equal).

True — in

Δ v = u ln (m_{0} / m_{f})

the factor

u

is out front and linear, so doubling

u

doubles

Δ v

. Contrast this with the mass ratio, which only helps through a stingy logarithm.

Doubling the mass ratio

m_{0} / m_{f}

doubles

Δ v

False —

Δ v

depends on

ln (m_{0} / m_{f})

, not on the ratio itself. Doubling the ratio only adds

u ln 2

(about

0.69 u

), it does not double the answer.

For a force-free rocket,

Δ v

depends on how fast you burn the fuel.

False — the burn rate

\overset{m}{˙}

cancels when you integrate

d v = - u d m / m

; only the start and end masses survive. Fast or slow, the same fuel gives the same

Δ v

(in the ideal, gravity-free case).

The velocity terms

v d m

and

- v d m

cancelling in the derivation is a coincidence of algebra.

False — it is physically meaningful: the ground-frame speed at which mass leaves cancels out, leaving only the relative speed

u

to produce thrust. That's why

u

, not ground speed, is what matters.

A raindrop growing as it falls through mist is not a variable-mass problem because nothing is being thrown out.

False — variable mass covers both loss and gain of matter; an accreting drop uses the same

F_{ext} = d p / d t

bookkeeping, just with

\overset{m}{˙} > 0

and incoming relative velocity.

Spot the error

F = \frac{d ( m v )}{d t} = m \overset{v}{˙} + v \overset{m}{˙}

gives the rocket's motion directly."

The

v \overset{m}{˙}

term secretly assumes the lost mass departs at the rocket's own velocity

v

(zero relative speed), so it predicts no thrust.

F = \overset{p}{˙}

is only valid for a closed system, so you must track rocket plus ejected gas.

"Thrust

= u \frac{d m}{d t}

Missing minus sign: since

\overset{m}{˙} < 0

this would make thrust negative. The correct form is

F_{thrust} = - u \frac{d m}{d t} = u ∣ \overset{m}{˙} ∣ > 0

— keep the minus and let the sign of

\overset{m}{˙}

do its work.

u

is the exhaust's speed measured from the ground."

u

is the exhaust speed relative to the rocket (an engine property, ~2000–4500 m/s). The ground-frame exhaust velocity is

v - u

, which even changes sign once the rocket is faster than

u

. See Relative Velocity.

"Because

F_{thrust} = u ∣ \overset{m}{˙} ∣

is constant, the acceleration is constant too."

Acceleration is

a = F_{thrust} / m

with the current

m

, which keeps shrinking. So even at constant thrust,

a

rises as the rocket lightens.

"On the launch pad, if thrust equals weight the rocket rises slowly."

u ∣ \overset{m}{˙} ∣ = m g

exactly then

a = 0

: the rocket hovers, it does not rise. You need thrust-to-weight strictly greater than 1 (

u ∣ \overset{m}{˙} ∣ > m g

) to lift off.

"In

Δ v = u ln (m_{0} / m_{f})

we plug

m_{f} = 0

for a fully burnt rocket to get max speed."

m_{f}

is the burnout (empty structure + payload) mass, never zero — engines, tanks, and payload remain. As

m_{f} \to 0

the log blows up to infinity, which is unphysical: you can never burn all of yourself.

"The

ln

appears because we chose to integrate; a different method would avoid it."

The

ln

is forced by the physics: each kg burned pushes the ever-smaller remaining mass, so

d v = - u d m / m

has

m

in the denominator, and

\int d m / m = ln m

inevitably. See Logarithms and Exponential Growth.

Why questions

Why must we draw the box around rocket plus the fuel it's about to eject, rather than the rocket alone?

Because Newton's law in the form

F_{ext} = d p / d t

is only valid for a system whose membership is fixed over

d t

; the rocket alone loses members, so its momentum changes for a reason that isn't an external force.

Why does the second-order term

d m d v

get dropped?

It is the product of two infinitesimally small quantities, so it is vanishingly tiny compared to the first-order terms

m d v

and

u d m

— in the

d t \to 0

limit it contributes nothing.

Why is achieving high

Δ v

so brutally expensive in fuel?

Because the mass ratio sits inside a logarithm, so to raise

Δ v

linearly you must raise

m_{0} / m_{f}

exponentially — the "tyranny of the rocket equation."

Why does the same rocket accelerate harder near burnout than at launch (deep space, constant thrust)?

Thrust

u ∣ \overset{m}{˙} ∣

stays fixed but the mass

m

a = F_{thrust} / m

has shrunk, so the same push moves less stuff faster.

Why doesn't the burn rate

\overset{m}{˙}

appear in the final Tsiolkovsky formula, even though it clearly sets the thrust?

Thrust and time both depend on

\overset{m}{˙}

, but when you integrate to get total velocity change they cancel;

Δ v

only cares about total mass expelled, not how quickly it left.

Why is it the relative speed

u

and not the rocket's speed that determines thrust?

Because momentum you hand to the exhaust is set by how fast it leaves you; the ground-frame velocity terms cancel in the bookkeeping, leaving only the relative kick

u

as the source of thrust.

Edge cases

What happens to the rocket equation if

F_{ext} \neq = 0

(say gravity)?

You can no longer separate variables cleanly into

d v = - u d m / m

; the term

F_{ext}

stays in

m \overset{v}{˙} = F_{ext} - u \overset{m}{˙}

and you must integrate the full equation, giving a smaller

Δ v

than Tsiolkovsky.

What is

Δ v

if the rocket burns no fuel, so

m_{f} = m_{0}

Δ v = u ln (m_{0} / m_{0}) = u ln 1 = 0

— no mass expelled means no thrust and no speed change, exactly as intuition demands.

Can a variable-mass body ever slow down due to mass change alone?

Yes — for an accreting body (

\overset{m}{˙} > 0

) picking up mass at rest relative to the ground, the incoming mass drags it down; the sign of the relative velocity flips the thrust into a retarding force.

What if the exhaust is somehow ejected forward (same direction as motion,

u < 0

Then the thrust

- u \overset{m}{˙}

reverses sign and pushes the rocket backward (decelerates it) — this is exactly how retro-rockets and reverse thrusters brake a craft.

In the ground frame, what is the exhaust's velocity once the rocket exceeds its own exhaust speed,

v > u

The exhaust velocity

v - u

becomes positive — the gas is still moving forward relative to the ground, even though it left the rocket going backward relative to it. Thrust is unaffected because only the relative speed matters. See Relative Velocity.

What limits

Δ v

if you could make the mass ratio arbitrarily large?

In the ideal equation nothing caps it mathematically (

ln \to \infty

), but physically you cannot have infinite fuel and zero structure — real tanks, engines, and payload fix a finite

m_{f}

, so

Δ v

is bounded well below the mathematical limit.

Recall One-line self-test

If you can answer "why $u$ and not $v$ ?", "why $ln$ and not the raw ratio?", and "why the closed box?" in one sentence each, you own this topic.

Parent: topic note · builds on Conservation of Linear Momentum, Impulse–Momentum Theorem, Newton's Third Law, Logarithms and Exponential Growth.