Systems with variable mass — rocket equation derivation preview

WHY the ordinary $F=ma$ is not enough

The bare equation $F = m\frac{dv}{dt}$ assumes $m$ is constant. The honest statement of Newton's 2nd law is:

$\vec F_{\text{ext}} = \frac{d\vec p}{dt}$

but you must apply it to a system whose membership does not change during $dt$ . If you carelessly write $\frac{d(mv)}{dt} = m\dot v + \dot m v$ for "the rocket," you get the wrong sign/term, because that $v$ should be the velocity of the ejected mass, not the rocket. We fix this by being careful about what leaves and how fast it moves.

HOW to derive it — track a closed system over $dt$

Figure — Systems with variable mass — rocket equation derivation preview

Setup (all in one fixed inertial frame):

At time $t$ : rocket has mass $m$ , velocity $v$ . Total momentum $p(t) = m v$ .
During the interval $dt$ , the rocket ejects a tiny mass $dm_{\text{ej}} > 0$ of exhaust.
Let the rocket lose mass: $\dfrac{dm}{dt} < 0$ , so $dm_{\text{ej}} = -\,dm$ .
The exhaust leaves with speed $u$ relative to the rocket, pointing backward, so its velocity in the ground frame is $v - u$ .

At time $t + dt$ :

Rocket: mass $m + dm$ (note $dm<0$ ), velocity $v + dv$ .
Exhaust just released: mass $-dm$ , velocity $v - u$ .

Formula Momentum bookkeeping (the heart of it)

Closed system = rocket + the bit it ejects. Apply $\vec F_{\text{ext}}\,dt = dp_{\text{system}}$ .

$p(t) = m v$ $p(t+dt) = \underbrace{(m+dm)(v+dv)}_{\text{rocket}} + \underbrace{(-dm)(v-u)}_{\text{exhaust}}$

Why this step? We track the same total matter on both sides, so $F_{\text{ext}}\,dt$ equals the change in this combined momentum — a legitimate use of Newton's law.

Expand: $p(t+dt) = mv + m\,dv + v\,dm + dm\,dv - v\,dm + u\,dm$

Why this step? Multiply out and notice $v\,dm$ cancels with $-v\,dm$ — the ground-frame velocity terms of the lost mass cancel, leaving only the relative velocity $u$ to do real work.

Drop the second-order term $dm\,dv$ (product of two infinitesimals → negligible): $dp = p(t+dt)-p(t) = m\,dv + u\,dm$

Set $dp = F_{\text{ext}}\,dt$ : $\boxed{\,m\frac{dv}{dt} = F_{\text{ext}} - u\frac{dm}{dt}\,}$

The ideal rocket equation (Tsiolkovsky)

Take gravity-free, drag-free space: $F_{\text{ext}} = 0$ . Then

$m\,dv = -u\,dm \quad\Rightarrow\quad dv = -u\,\frac{dm}{m}$

The ratio $\dfrac{m_0}{m_f}$ is the mass ratio. Because it sits inside a $\ln$ , getting big $\Delta v$ is brutally expensive — you must carry exponentially more fuel. This is the tyranny of the rocket equation.

Worked examples

Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine you're on a frictionless skateboard holding a stack of heavy bricks. Every time you throw a brick backward, you roll forward a little — that's Newton's "push back" rule. A rocket is the same: it throws hot gas backward super fast, so it shoots forward. The twist is that the bricks (fuel) are part of you, so as you throw them you get lighter — and a lighter you speeds up more with each throw. That's why the speed adds up like a logarithm: the later throws count more because there's less of you left to push.

Flashcards

Why can't you apply

F=ma

directly to a rocket?

Because its mass changes;

F=\dot p

must be applied to a closed system (rocket + ejected gas), not to the shrinking rocket alone.

What is the general variable-mass equation?

m\dfrac{dv}{dt} = F_{\text{ext}} - u\dfrac{dm}{dt}

, where

u

is exhaust speed relative to the body.

What is the thrust of a rocket?

F_{\text{thrust}} = -u\dfrac{dm}{dt} = u|\dot m|

State the Tsiolkovsky rocket equation.

\Delta v = u\ln\dfrac{m_0}{m_f}

Why does a logarithm appear in

\Delta v

Each kg burned pushes the ever-smaller remaining mass, so gain per kg compounds — integrating

dm/m

gives

\ln

What does

u

physically mean in the rocket equation?

Exhaust speed relative to the rocket, not its ground speed.

In the derivation, which term cancels and why?

The ground-frame velocity terms

v\,dm

and

-v\,dm

of the ejected mass cancel, leaving only the relative speed

u

Condition for a rocket to lift off Earth?

Thrust > weight:

u|\dot m| > m_0 g

(thrust-to-weight ratio > 1).

Why is high

\Delta v

so costly?

Mass ratio is inside a

\ln

, so fuel needed grows exponentially with required

\Delta v

("tyranny of the rocket equation").

What is the mass ratio?

m_0/m_f

— initial (fuelled) mass over final (burnout) mass.

Connections

Conservation of Linear Momentum — the closed-system principle the whole derivation rests on.
Newton's Third Law — exhaust pushes rocket; rocket pushes exhaust.
Impulse–Momentum Theorem — $F_{\text{ext}}\,dt = dp$ used per interval.
Center of Mass Motion — CoM of (rocket+exhaust) obeys $F_{\text{ext}}$ even as the rocket accelerates.
Relative Velocity — why $u$ is measured w.r.t. the rocket.
Logarithms and Exponential Growth — the math behind mass ratio.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, rocket aage isliye jaata hai kyunki wo apne hot gas ko bahut tezi se peeche phenkta hai. Newton ka third law: gas peeche, rocket aage. Lekin ek twist hai — rocket ka apna mass kam hota jaata hai jaise-jaise fuel jalta hai. Isliye seedha $F=ma$ likhna galat hai, kyunki "rocket" cheez hi badal rahi hai.

Trick yeh hai: har chhote time $dt$ me ek box banao jisme rocket + jo gas abhi nikalne wala hai dono ho. Yeh ek closed system hai, iska total momentum conserve hota hai. Algebra karne par ground-frame ke $v$ wale terms cancel ho jaate hain, aur sirf $u$ (exhaust ka speed rocket ke relative) bachta hai. Result: $m\,dv = -u\,dm$ , jisse thrust nikalta hai $F = u|\dot m|$ .

Ise integrate karo to milti hai Tsiolkovsky equation: $\Delta v = u\ln(m_0/m_f)$ . Yahan $\ln$ isliye aata hai kyunki har kg fuel jab jalta hai to wo bache hue (chhote) mass ko push karta hai, to baad ke kilos zyada speed dete hain — yeh compounding effect log banata hai.

Important baat: kyunki mass ratio $\ln$ ke andar hai, thoda zyada $\Delta v$ chahiye to exponentially zyada fuel chahiye — isko "tyranny of the rocket equation" kehte hain. Yaad rakho: $u$ hamesha rocket ke relative hai, aur $\dot m$ negative hota hai, isliye thrust ka sign dhyaan se lagana.

Systems with variable mass — rocket equation derivation preview

WHY the ordinary $F=ma$ is not enough

HOW to derive it — track a closed system over $dt$

The ideal rocket equation (Tsiolkovsky)

Worked examples

Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Momentum & Collisions

Connections