Variable Mass Systems — Rockets

In short

In a variable-mass system like a rocket, the thrust force is F_{\text{thrust}} = v_e \lvert dm/dt \rvert, where v_e is the exhaust speed relative to the rocket and \lvert dm/dt \rvert is the rate of mass ejection. Integrating over the entire burn gives the Tsiolkovsky rocket equation: \Delta v = v_e \ln(m_0/m_f). The velocity gain depends on the exhaust speed and the natural logarithm of the mass ratio — which is why rockets are almost entirely fuel.

A Diwali skyrocket launches from an empty bottle. In the first instant the tube is packed with propellant and heavy — the thrust barely overcomes gravity. But as the powder burns away, the tube gets lighter. The same push from the exhaust now accelerates a lighter body, so it climbs faster and faster. By the time the charge is spent, the nearly-empty tube is moving at its peak speed.

Scale that picture up by a factor of a million. ISRO's PSLV sits on the launch pad at 320 tonnes. The payload — the satellite it carries to orbit — is about 1.7 tonnes. More than 99% of the rocket's mass is propellant, thrown backward as hot exhaust gas so that the tiny fraction at the top can reach orbital speed. A rocket is not a vehicle that carries fuel. It is fuel that carries a vehicle.

The physics that governs both the ₹10 Diwali patakha and the ₹300 crore PSLV is the same: conservation of momentum applied to a system whose mass is changing. This article derives that physics from scratch.

Why momentum conservation applies

When you first learn Newton's second law, F = ma, the mass m is constant. A block on a table, a ball in free fall, a car on a highway — the object does not gain or lose mass as it moves. But a rocket burns fuel and ejects it as hot gas. The rocket is getting lighter every instant. How do you apply F = ma when m itself is changing?

The answer: stop looking at the rocket alone. Look at the system — the rocket plus all the exhaust gas it has ever ejected. That system has a fixed total mass. Nothing enters or leaves it. If no external force acts on the system (as in deep space), the total momentum of the system is conserved. The rocket speeds up in one direction, the exhaust gas moves in the other, and their momenta always sum to the same value.

Near Earth's surface, gravity is an external force, so you account for it separately. But the core logic remains: the internal exchange of momentum between rocket and exhaust is governed entirely by conservation of momentum, just as it is in any collision.

The cricket-ball thought experiment

Stand on a frozen lake — no friction beneath your shoes. You hold a bag of ten cricket balls, each 160 g. Your mass (including the bag) is 80 kg.

Throw one ball backward at 20 m/s relative to you. By conservation of momentum, you slide forward:

v_1 = \frac{0.16 \times 20}{80} = 0.04 \text{ m/s}

Why: the ball carries backward momentum 0.16 \times 20 = 3.2 kg·m/s, so you gain 3.2 kg·m/s forward. Dividing by your mass gives your recoil speed.

Now throw a second ball. Here is the key: you are now 0.16 kg lighter (79.84 kg), and you are already moving at 0.04 m/s. The second ball, thrown at 20 m/s relative to your current velocity, gives you a slightly larger speed boost — because you are lighter.

By the tenth ball, you have shed 1.6 kg and each throw accelerates you a little more than the previous one. If you could keep throwing until your "fuel" (the balls) was exhausted, your final speed would not be 10 times 0.04 m/s = 0.4 m/s. It would be more — because each throw acts on a lighter you.

This is exactly how a rocket works. The exhaust gas is the cricket balls. The exhaust speed v_e is how fast you throw them. And the mass ratio — how much fuel you start with versus how much dry mass remains — determines your final speed. The frozen-lake thought experiment is a rocket in slow motion.

Deriving the thrust force

Set up the problem precisely. At time t, the rocket (including unburned fuel) has mass m and velocity v, both measured in the ground frame. Take the direction of travel as positive.

At time $t$, the rocket has mass $m$ and velocity $v$. A tiny time later, it has ejected exhaust mass $-dm$ (positive, since $dm < 0$) at velocity $v - v_e$ in the ground frame. The dashed boundary encloses the closed system whose total momentum is conserved.

In the next instant dt:

The rocket ejects a small mass \lvert dm \rvert of exhaust gas. Define dm as the change in rocket mass, so dm < 0 — the rocket is losing mass.
The exhaust leaves at speed v_e relative to the rocket, in the backward direction. In the ground frame, the exhaust velocity is v - v_e.
The rocket's mass is now m + dm (smaller, since dm < 0) and its velocity is v + dv.

Step 1. Write conservation of momentum for the closed system.

\underbrace{m\,v}_{\text{before}} = \underbrace{(m + dm)(v + dv)}_{\text{rocket after}} + \underbrace{(-dm)(v - v_e)}_{\text{exhaust gas}}

Why: the total momentum of the closed system (rocket + just-ejected gas) cannot change. The exhaust mass is -dm (positive, since dm is negative), and its ground-frame velocity is v - v_e.

Step 2. Expand the right side.

m\,v = m\,v + m\,dv + dm\,v + dm\,dv - dm\,v + dm\,v_e

The m\,v terms cancel. The dm\,v terms cancel:

0 = m\,dv + dm\,dv + dm\,v_e

Why: the cancellations happen because we are comparing the same system at two nearby instants. The dm\,dv term is the product of two infinitesimals — negligibly small compared to the others.

Step 3. Drop the second-order term dm\,dv and rearrange.

\boxed{m\,dv = -v_e\,dm} \tag{1}

Why: this is the fundamental equation of rocket propulsion. The left side is the rocket's momentum change. The right side is the momentum carried away by the exhaust. Since dm < 0, the right side is positive — the rocket accelerates forward.

Step 4. Divide both sides by dt to get the thrust force.

m\,\frac{dv}{dt} = -v_e\,\frac{dm}{dt}

The left side is mass times acceleration — the net force on the rocket. The right side is the thrust:

\boxed{F_{\text{thrust}} = v_e \left\lvert\frac{dm}{dt}\right\rvert} \tag{2}

Why: the thrust depends on two things — how fast you eject the exhaust (v_e) and how much mass you eject per second (\lvert dm/dt \rvert). Double either one and the thrust doubles.

ISRO's Vikas engine, used in the PSLV second stage, ejects exhaust at about 2800 m/s and burns roughly 260 kg of propellant per second. Thrust: 2800 \times 260 = 728{,}000 N — approximately 74 tonnes of force from a single engine.

The Tsiolkovsky rocket equation

Equation (1) tells you the thrust at any instant. But what you really want to know is: after all the fuel is burned, how fast is the rocket going?

Step 5. Separate variables in equation (1).

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

Why: divide both sides of m\,dv = -v_e\,dm by m. Now the left side has only v and the right side has only m — the equation is ready to integrate.

Step 6. Integrate. The velocity goes from 0 to \Delta v (starting from rest). The mass goes from m_0 (total mass at launch) to m_f (dry mass after all fuel is burned):

\int_0^{\Delta v} dv = -v_e \int_{m_0}^{m_f} \frac{dm}{m}

\Delta v = -v_e \bigl[\ln m\bigr]_{m_0}^{m_f} = -v_e\,(\ln m_f - \ln m_0)

\boxed{\Delta v = v_e \ln\!\left(\frac{m_0}{m_f}\right)} \tag{3}

Why: the natural logarithm appears because you are integrating dm/m. This is a calculus step — you add up infinitely many tiny velocity increments, one for each tiny mass ejection. The sum is v_e \ln(m_0/m_f). Since m_0 > m_f (you started with more mass than you end with), the logarithm is positive — the rocket speeds up.

This is the Tsiolkovsky rocket equation. It encodes three facts that govern all of rocketry:

The velocity gain depends logarithmically on the mass ratio. Doubling the fuel does not double \Delta v. Going from a mass ratio of 2 to 4 adds v_e \ln 2 \approx 0.69\,v_e — the same gain as going from 1 to 2. The logarithm imposes brutal diminishing returns.
Exhaust speed is a direct multiplier. Every improvement in v_e multiplies the entire \Delta v. This is why rocket engineers obsess over exhaust velocity — or equivalently, specific impulse.
The mass ratio is why rockets are mostly fuel. To reach low Earth orbit (\sim7.8 km/s) with an exhaust speed of 3 km/s, you need m_0/m_f = e^{7.8/3} = e^{2.6} \approx 13.5. For every kilogram that reaches orbit, you need 12.5 kg of propellant. That is before accounting for gravity losses and air drag, which make it worse.

Why variable mass matters — the rocket advantage

The simulation below makes this concrete. Two bodies start from rest in the absence of gravity. Both receive the same total impulse over 4 seconds from a 20 N thrust. The red body is a rocket: it ejects mass continuously, getting lighter as it burns, so each second of thrust accelerates it more. The dark body has constant mass — it gets the same push every second, but never gets lighter.

Both bodies receive the same 20 N thrust for 4 seconds — same total impulse. The rocket (red) sheds mass as it burns fuel, so each second of thrust produces a bigger velocity increment. By burnout ($t = 4$ s), the rocket reaches 16 m/s while the constant-mass body reaches only 8 m/s. After that, both coast — but the rocket is already far ahead. Click replay to watch again.

At burnout, the rocket has reached 16 m/s while the constant-mass body has reached only 8 m/s. Both received the same total impulse (80 N·s). The difference is the logarithm at work: the rocket's mass ratio was 5:1, giving \Delta v = v_e \ln 5 \approx 1.6\,v_e instead of the 0.8\,v_e that constant mass would yield.

Explore the mass ratio yourself

Drag the red dot to explore how the mass ratio m_0/m_f determines the velocity gain. The curve flattens at high ratios — doubling the mass ratio from 10 to 20 adds less velocity than doubling from 2 to 4.

Drag the red dot to change the mass ratio $m_0/m_f$. At $m_0/m_f \approx 2.72$ (Euler's number $e$), $\Delta v$ equals exactly one $v_e$. The bottom readout uses $v_e = 3$ km/s — a typical value for kerosene-liquid oxygen engines — to show the velocity gain in real units. Reaching orbital speed (7.8 km/s) requires a mass ratio above 13.

Worked examples

Example 1: A sounding rocket reaches for the sky

A small sounding rocket has a total launch mass of 500 kg. Of this, 400 kg is propellant and 100 kg is structure plus payload. The exhaust speed is 2500 m/s relative to the rocket. The propellant burns uniformly over 40 seconds. Find (a) the thrust, (b) the ideal \Delta v in the absence of gravity, and (c) the actual \Delta v accounting for gravity during the burn.

Left: 500 kg rocket on the pad, mostly fuel. Right: after burnout, 100 kg of structure rockets upward at 3631 m/s, trailing exhaust.

Step 1. Compute the burn rate.

\left\lvert\frac{dm}{dt}\right\rvert = \frac{400}{40} = 10 \text{ kg/s}

Why: 400 kg of fuel burned uniformly over 40 seconds gives a constant burn rate of 10 kg per second.

Step 2. Compute the thrust.

F_{\text{thrust}} = v_e \times \left\lvert\frac{dm}{dt}\right\rvert = 2500 \times 10 = 25{,}000 \text{ N} = 25 \text{ kN}

Why: apply equation (2). The thrust is constant because both the exhaust speed and the burn rate are constant throughout the burn.

Step 3. Compute \Delta v in the absence of gravity.

\Delta v_{\text{ideal}} = v_e \ln\!\left(\frac{m_0}{m_f}\right) = 2500 \times \ln\!\left(\frac{500}{100}\right) = 2500 \times \ln 5 = 2500 \times 1.609

\Delta v_{\text{ideal}} = 4023 \text{ m/s} \approx 4.0 \text{ km/s}

Why: apply equation (3). The mass ratio is 5. In deep space, with no gravity or drag, this rocket would reach 4 km/s — about half of orbital speed.

Step 4. Account for gravity loss.

During the 40-second burn, gravity continuously opposes the rocket's ascent. The velocity lost to gravity is:

\Delta v_{\text{gravity}} = g \times t_{\text{burn}} = 9.8 \times 40 = 392 \text{ m/s}

\Delta v_{\text{actual}} = 4023 - 392 = 3631 \text{ m/s} \approx 3.6 \text{ km/s}

Why: gravity steals g \times t_{\text{burn}} of velocity (assuming vertical flight). This is called the gravity loss. The faster you burn your fuel, the smaller this loss — which is one reason rockets are designed with high thrust.

Result: Thrust = 25 kN. Ideal \Delta v = 4023 m/s. With gravity, actual \Delta v ≈ 3631 m/s (about 3.6 km/s).

What this shows: Even with a mass ratio of 5 and a high exhaust speed, this rocket reaches only 3.6 km/s — well short of orbital speed (7.8 km/s). A single stage cannot do the job. This is exactly why orbital rockets use staging: you throw away the empty first stage and start a fresh burn with a better mass ratio.

Example 2: Sand falling on a conveyor belt

Sand pours from a hopper onto a horizontal conveyor belt at a steady rate of 5 kg/s. The belt moves at a constant speed of 2 m/s. What horizontal force must the motor exert to keep the belt at constant speed?

Sand drops vertically from the hopper onto a belt moving at 2 m/s. The motor must exert a force $F$ to keep the belt at constant speed as it continuously accelerates new sand from rest to belt speed.

Step 1. Identify the variable-mass system.

The system is the belt plus the sand already on it. Fresh sand arrives with zero horizontal velocity and must be accelerated to the belt speed v = 2 m/s. This is the reverse of a rocket — mass is added to the system, not ejected from it.

Why: in a rocket, mass leaves at some relative velocity, providing thrust forward. Here, mass arrives at zero horizontal speed and must be dragged up to belt speed — requiring a continuous force from the motor.

Step 2. Apply the thrust equation.

The force needed equals the rate of horizontal momentum delivered to the new sand:

F = v \times \frac{dm}{dt} = 2 \times 5 = 10 \text{ N}

Why: each kilogram of sand that lands must gain 2 m/s of horizontal velocity. At 5 kg per second, the belt must supply 5 \times 2 = 10 kg·m/s of momentum every second — and that is 10 N.

Step 3. Compute the power delivered by the motor.

P = F \times v = 10 \times 2 = 20 \text{ W}

Why: power is force times velocity. The motor exerts 10 N on a belt moving at 2 m/s, so the power input is 20 W.

Step 4. Check against the kinetic energy gained by the sand.

\frac{dK}{dt} = \frac{1}{2}\,\frac{dm}{dt}\,v^2 = \frac{1}{2} \times 5 \times 4 = 10 \text{ W}

Why: the motor delivers 20 W, but the sand gains only 10 W of kinetic energy. The other 10 W goes into heat — the sand slides on the belt before reaching belt speed, and friction converts that kinetic energy difference into thermal energy. Exactly half the motor's energy is "wasted" as heat, regardless of the belt speed. This 50% ratio is a fundamental result for any system where mass arrives at rest and is accelerated to a constant velocity.

Result: The motor must exert a constant 10 N. It delivers 20 W of power, but only 10 W ends up as kinetic energy of the sand; the other 10 W dissipates as frictional heat.

What this shows: Variable-mass physics is not just about rockets. Any system where mass enters or leaves — sand on a belt, a chain being lifted off a table, rain falling into a moving cart — obeys the same momentum principle. Here the mass arrives instead of leaving, and the force resists the motion instead of assisting it.

Common confusions

"Rockets push against the air." No — rockets work perfectly in vacuum, and in fact work better there because there is no air drag. The thrust comes entirely from ejecting mass backward. Newton's third law: the rocket pushes the exhaust backward, the exhaust pushes the rocket forward. No air, no ground, no external surface needed. ISRO's upper-stage engines fire in the vacuum of space and produce thrust just fine.
"More fuel always means more speed." Only logarithmically. Going from a mass ratio of 3 to 6 adds v_e \ln 2 \approx 0.69\,v_e — the same gain as going from 1.5 to 3. The Tsiolkovsky equation has punishing diminishing returns: the first doubling of fuel gives you a big speed boost, but the fifth doubling gives almost nothing. At some point, the extra fuel is mostly just accelerating other fuel.
"The thrust changes as fuel burns." Only if the burn rate or exhaust speed changes. If both are constant (as in many solid-fuel rockets), the thrust F = v_e \lvert dm/dt \rvert stays constant. What does change is the acceleration: a = F/m increases as m drops. This is why you feel pushed harder into your seat as a rocket's burn progresses — the force is constant, but the mass it acts on is shrinking.
"Variable mass means F = d(mv)/dt = m\,dv/dt + v\,dm/dt." This is a common textbook trap. The product rule d(mv)/dt applies to a fixed set of particles whose identity does not change. A rocket's identity does change — it is a different collection of particles at every instant. The correct approach is to apply momentum conservation to the closed system (rocket + all exhaust), as done in the derivation above. The result is m\,dv/dt = v_e\,\lvert dm/dt \rvert + F_{\text{ext}}, which must be derived, not assumed from a naive product rule.
"A heavier rocket is always slower." Not necessarily. A rocket that starts heavier because it carries more fuel can end up faster — as long as the extra fuel improves the mass ratio m_0/m_f. But a rocket that starts heavier because of a massive structure (thick tank walls, heavy engines) ends up slower. The mass ratio is what matters, and the structural mass is dead weight that hurts both m_0 and m_f.

If you came here to derive the thrust equation and the Tsiolkovsky rocket equation, you have everything you need. What follows is for readers who want to understand multi-stage rockets, gravity losses, and why every orbital launch vehicle has staging.

Multi-stage rockets — why staging works

The Tsiolkovsky equation has a practical problem. To reach low Earth orbit, you need \Delta v \approx 9.5 km/s (including gravity and air-drag losses). With a typical exhaust speed of 3 km/s:

\frac{m_0}{m_f} = e^{9.5/3} = e^{3.17} \approx 23.7

For every kilogram in orbit, you need 22.7 kg of propellant on the pad. But there is a catch: the tanks, engines, and plumbing that hold and burn that propellant have mass too. Call it structural mass. A typical structural fraction is 8–12% of the stage's propellant-plus-structure mass. So a stage carrying 100 kg of fuel has about 10 kg of tankage. That 10 kg is dead weight that the fuel must also lift and accelerate.

In a single-stage rocket, you carry all that structural mass to orbit — the empty tanks, the spent first-stage engine, everything. In a multi-stage rocket, you throw away each stage's structure once its fuel is exhausted. The next stage starts its burn free of the dead weight.

Why the gain is so large. Split the required \Delta v into two equal burns: 4750 m/s each. Each stage needs a mass ratio of e^{4750/3000} \approx 4.87 — far more achievable than 23.7 for a single stage. After the first stage burns out, you drop its empty tanks and engines. The second stage's initial mass is dramatically lower.

This is why every orbital rocket — ISRO's PSLV (4 stages), GSLV (3 stages), SpaceX's Falcon 9 (2 stages) — uses staging. The PSLV's solid first stage provides raw thrust through the dense lower atmosphere. Its liquid upper stages provide the efficiency and precision needed for orbital insertion. Each stage is optimised for its portion of the flight.

Gravity loss and the thrust-to-weight ratio

The Tsiolkovsky derivation assumed no external forces. Near Earth's surface, gravity acts as a continuous drag:

\Delta v_{\text{actual}} = v_e \ln\!\left(\frac{m_0}{m_f}\right) - g\,t_{\text{burn}}

The gravity loss g \cdot t_{\text{burn}} means you want to burn your fuel as quickly as possible — a short, intense burn minimises the time gravity has to slow you down. This is why rockets have enormous thrust-to-weight ratios. The PSLV's thrust-to-weight ratio at liftoff is about 1.5 — the thrust exceeds the rocket's weight by 50%, giving it a strong initial acceleration.

But there is a trade-off: higher thrust requires larger, heavier engines, which increases structural mass and worsens the mass ratio. Rocket design is the art of optimising between gravity loss (which favours high thrust and short burns) and structural efficiency (which favours smaller, lighter engines). This optimisation is one reason rocket engineering is considered among the hardest problems in aerospace.

Effective exhaust velocity and specific impulse

Rocket engineers rarely quote exhaust velocity directly. Instead, they use specific impulse I_{sp}, defined as:

I_{sp} = \frac{v_e}{g_0}

where g_0 = 9.8 m/s² is standard gravity. The units of I_{sp} are seconds. A Vikas engine with v_e = 2800 m/s has I_{sp} = 2800/9.8 \approx 286 s. The physical meaning: each kilogram of propellant provides thrust equivalent to supporting 1 kg against gravity for 286 seconds. Higher I_{sp} means more efficient use of fuel.

Solid rockets: I_{sp} \approx 250 s. Kerosene-LOX: \approx 300 s. Hydrogen-LOX: \approx 450 s. Ion thrusters: > 3000 s but with tiny thrust — useful only in space, where there is no gravity loss.

Where this leads next

Centre of Mass — Definition and Calculation — the point whose motion obeys F_{\text{ext}} = Ma_{\text{cm}} even when individual parts of the system move in complex ways.
Conservation of Linear Momentum — the foundational law that makes the thrust equation derivable.
Impulse and Momentum — the force-time integral connecting Newton's second law to momentum change.
Elastic Collisions — another application of momentum conservation where the closed-system approach is essential.
Collisions in Two Dimensions — when the momentum exchange happens at an angle, requiring vector decomposition.