Rocket Equation — Variable Mass Problems

Rockets are the canonical example of variable-mass mechanics — and they show why $F = ma$ alone is not enough. Once you understand Tsiolkovsky’s equation, you understand why getting to orbit is hard, why staged rockets exist, and why JEE Advanced loves this topic.

This isn’t directly in CBSE Class 11 syllabus but appears in NCERT exemplar problems and is high-value for JEE Advanced. Variable-mass questions are nearly free marks if you know the derivation; nearly impossible if you do not.

Why Newton’s Second Law Needs an Update

The standard form $F = ma$ assumes mass is constant. For a rocket, mass changes as fuel is ejected. The general form of Newton’s law is:

F_{ext} = \frac{dp}{dt}

Momentum, not mass times acceleration. When mass varies, this distinction matters.

Deriving the Rocket Equation

Consider a rocket of mass $m$ moving at velocity $v$ . In time $dt$ , it ejects a small mass $dm_e$ of exhaust at velocity $u$ relative to the rocket (so velocity $v - u$ relative to the ground, with $u > 0$ being the exhaust speed).

Initial momentum: $p_i = mv$ .

After ejection, rocket mass is $m - dm_e$ at velocity $v + dv$ . Exhaust mass $dm_e$ at velocity $v - u$ .

Final momentum:

p_f = (m - dm_e)(v + dv) + dm_e (v - u)

Expanding and ignoring $dm_e \cdot dv$ (second order):

p_f = mv + m\, dv - v\, dm_e + v\, dm_e - u\, dm_e = mv + m\, dv - u\, dm_e

Change in momentum:

dp = m\, dv - u\, dm_e

If $F_{ext} = 0$ (no gravity, drag, etc.), $dp = 0$ :

m\, dv = u\, dm_e

Since the rocket loses mass at the rate at which exhaust is ejected, $dm_e = -dm$ (where $m$ is the rocket’s mass and decreases). So:

dv = -u \, \frac{dm}{m}

Integrating from $m_0$ (initial mass) to $m$ (final mass):

v - v_0 = u \ln\frac{m_0}{m}

This is the Tsiolkovsky rocket equation — the central result of variable-mass mechanics.

Including Gravity

If the rocket fires vertically against gravity, $F_{ext} = -mg$ (downward). The equation becomes:

m\, dv = u\, dm_e - mg\, dt

Or:

\frac{dv}{dt} = -\frac{u}{m}\frac{dm}{dt} - g

If the burn time is $T$ and exhaust is ejected at constant rate $\dot{m} = -dm/dt$ :

v(T) = u \ln\frac{m_0}{m_f} - gT

The $gT$ term is the “gravity loss” — fuel burned to fight gravity instead of accelerating.

Key Quantities

Specific impulse $I_{sp} = u/g$ (in seconds) — characterises engine efficiency. Chemical rockets have $I_{sp} \approx 300-450$ s. Ion engines reach $3000$ s but with very low thrust.

Mass ratio $m_0/m_f$ — fully fueled mass over empty mass. To reach orbital speed (~7.8 km/s) with chemical fuel ( $u \approx 4$ km/s), you need:

m_0/m_f = e^{7800/4000} = e^{1.95} \approx 7.0

So 86% of liftoff mass must be fuel — and that is for a single stage with no gravity loss.

Worked Example 1 (Easy)

A rocket has initial mass $1000$ kg including $800$ kg fuel. Exhaust speed $u = 2$ km/s. In a horizontal frictionless plane (no gravity), what is the final velocity after all fuel is burnt?

v = u \ln(m_0/m_f) = 2 \ln(1000/200) = 2 \ln 5 \approx 3.22 \, \text{km/s}

Worked Example 2 (Medium, JEE Main style)

A rocket of total mass $5000$ kg ejects gas at $1000$ kg/s with exhaust speed $u = 2$ km/s. Find the thrust force and the initial acceleration.

Thrust: $T = u \cdot \dot{m} = 2000 \times 1000 = 2 \times 10^6$ N.

Initial acceleration: $a = T/m_0 - g = 2 \times 10^6 / 5000 - 10 = 400 - 10 = 390$ m/s $^2$ .

Worked Example 3 (Hard, JEE Advanced)

A rocket with $m_0 = 1000$ kg, exhaust speed $u = 2$ km/s, and burn rate $\dot{m} = 10$ kg/s lifts off vertically. After 50 s, find its velocity. Take $g = 10$ m/s $^2$ .

Mass after 50 s: $m_f = 1000 - 500 = 500$ kg.

v = u\ln(m_0/m_f) - gT = 2000 \ln 2 - 500 \approx 1386 - 500 = 886 \, \text{m/s}

Why Multi-Stage Rockets

The mass ratio limit ( $m_0/m_f \approx 7$ for chemical fuel) makes single-stage-to-orbit impossible. Multi-staging discards empty fuel tanks during flight, effectively chaining several rocket equations:

\Delta v_{total} = \sum_i u_i \ln(m_{0,i}/m_{f,i})

Each stage has a fresh mass ratio. This is why the Saturn V had three stages — and why SpaceX’s Falcon 9 still uses two.

JEE Advanced 2018 asked about the relative effectiveness of single-stage vs two-stage rockets given equal total fuel. Two-stage wins because the second stage doesn’t have to lift the first stage’s empty tank.

Common Mistakes

Mistake 1: Using $F = ma$ with constant $m$ — gives wrong thrust.

Mistake 2: Confusing exhaust speed (relative to rocket) with exhaust speed (relative to ground). The Tsiolkovsky equation uses relative-to-rocket.

Mistake 3: Forgetting gravity loss when the problem says “vertical takeoff”.

Mistake 4: Assuming fuel burns linearly with time — only true if $\dot{m}$ is constant.

Mistake 5: Mixing units — exhaust speed in km/s, mass in kg, time in seconds. Stick to SI units throughout.

Conveyor Belt and Falling Chain Variants

Variable-mass problems also appear in non-rocket contexts:

Sand falling on a conveyor belt — Sand drops at rate $\lambda$ (kg/s) onto a belt moving at speed $v$ . The motor must supply force $F = \lambda v$ to keep the belt moving. Work done by motor splits 50-50 between sand’s KE and heat from sand-belt friction.

Chain falling onto a table — A chain hangs above a table and falls. The force on the table equals the weight of chain already on the table plus the impulse from incoming chain: $F = \mu g x + \mu v^2$ where $\mu$ is mass per unit length and $x$ is length already piled.

These are JEE Advanced classics that use the same $dp/dt$ idea.

Practice Questions

Q1. A rocket exhausts fuel at $3$ km/s. To reach $9$ km/s in space, what mass ratio is needed?

$m_0/m_f = e^{9/3} = e^3 \approx 20$ .

Q2. A rocket of mass $2000$ kg burns fuel at $20$ kg/s with $u = 1.5$ km/s. Find the thrust.

Thrust $= u\dot{m} = 1500 \times 20 = 30000$ N $= 30$ kN.

Q3. Why is liquid hydrogen preferred as rocket fuel despite its low density?

High exhaust speed ( $u \approx 4.4$ km/s) gives the best $I_{sp}$ . The low density forces large tanks, but the velocity gain is worth it.

Q4. Sand falls onto a belt at $5$ kg/s. The belt moves at $2$ m/s. What power does the motor supply?

$P = F v = (\lambda v) v = 5 \times 2^2 = 20$ W. Half goes to sand’s KE, half to heat.

Q5. A rocket fires vertically with $u = 2$ km/s. After 30 s, half the initial mass has been ejected. Find velocity ( $g = 10$ ).

$v = 2000 \ln 2 - 10 \times 30 = 1386 - 300 = 1086$ m/s.

Q6. A water tank ejects water from a hole at $5$ m/s. The tank has mass $10$ kg and ejects $1$ kg/s. Find initial thrust.

Thrust $= u\dot{m} = 5 \times 1 = 5$ N.

Q7. Why can’t a rocket exceed its exhaust speed?

False — it can. From Tsiolkovsky, $v = u \ln(m_0/m_f)$ , and $\ln(m_0/m_f) > 1$ when $m_0/m_f > e \approx 2.72$ . So with mass ratio above $e$ , the rocket exceeds exhaust speed.

Q8. A 1 kg/s chain falls vertically and lands on a scale. The end of the chain has fallen $5$ m before hitting. What does the scale read?

Speed at landing $v = \sqrt{2gh} = \sqrt{100} = 10$ m/s. Force from impact $= \dot{m} v = 1 \times 10 = 10$ N. Plus weight of chain already on scale.

FAQs

Why is the rocket equation logarithmic? Because each kg of exhaust pushes the remaining rocket, which is lighter than the original. Diminishing returns lead to the logarithm.

Can a rocket work in vacuum? Yes, and better than in atmosphere. Air resistance is the only thing reducing efficiency on Earth.

Is the rocket equation relativistic? A relativistic version exists for fuel ejected near the speed of light. For chemical rockets ( $u \ll c$ ), the classical version is exact.

What is delta-v budget? The total $\Delta v$ a mission requires — sum of all required velocity changes (launch, transfer, landing). Mission designers sum required $\Delta v$ , then size the rocket.

Why doesn’t $F = ma$ work for rockets? Mass is changing. Use $F = dp/dt$ to handle variable mass correctly.

What is the Oberth effect? Burning fuel at high speed (e.g., near periapsis of an orbit) gives more $\Delta v$ than burning at low speed, because kinetic energy goes as $v^2$ .