In short

When a rocket ejects exhaust gas at speed v_e relative to itself, the thrust is F = v_e \times |dm/dt| — exhaust speed times the mass burn rate. The total velocity gained after burning all fuel is \Delta v = v_e \ln(m_i / m_f), the Tsiolkovsky rocket equation. The logarithm means you need exponentially more fuel for each additional unit of speed — which is why orbital rockets use multiple stages.

You light a Diwali rocket. It sits on the ground, fuse hissing, and then — whoosh — it tears into the sky, trailing sparks. A second later it is ten metres up, surrounded by nothing but air. What is it pushing against? The ground is gone. The stick is just dead weight. There is nothing solid beneath the rocket for it to push off. Yet it accelerates.

Now scale up. An ISRO PSLV sits on the launchpad at Sriharikota. The engines ignite and 320 tonnes of rocket rises into the sky — slowly at first, then faster and faster. Two minutes in, it is travelling at several kilometres per second, and the engine thrust has not changed. If the push is constant, why is it speeding up?

The answer to both questions is the same, and it has nothing to do with pushing against air or the ground. A rocket carries its own reaction mass — fuel. It hurls hot exhaust gas backward at enormous speed; Newton's third law hurls the rocket forward. And as fuel burns away, the rocket gets lighter, so the same thrust produces ever-increasing acceleration. The physics of rockets is the physics of a system whose mass changes with time.

The frozen-lake picture

Start with a thought experiment. You are standing on a perfectly smooth frozen lake — no friction at all. You hold a 10 kg bag of rice. You throw it away from you as hard as you can, and it leaves your hands at 5 m/s.

What happens to you? You slide backward. The total momentum of the system (you + bag) was zero before the throw, and no external horizontal force acted during it, so total momentum must stay zero. If the bag carries momentum to the left, you must carry equal momentum to the right.

m_{\text{you}}\,v_{\text{you}} + m_{\text{bag}}\,v_{\text{bag}} = 0

If you weigh 50 kg, you slide backward at 1 m/s. One throw, one push.

Person on a frozen lake throws a bag Before: person and bag together, at rest. After: bag moves left at 5 m/s, person slides right at 1 m/s. Momentum arrows show equal and opposite momenta. frictionless ice Before 50 kg 10 kg at rest After 1 m/s 5 m/s
You throw the bag left at 5 m/s. Conservation of momentum sends you sliding right at 1 m/s. Total momentum stays zero.

Now imagine you have ten bags, each 1 kg. You throw them one at a time, each at 5 m/s relative to yourself. After each throw you are a little lighter and a little faster. The first throw barely budges you, but by the tenth throw you weigh less, so the same impulse gives a larger speed boost. Each successive throw is more effective than the last.

That is a rocket. Replace the bags with a continuous stream of exhaust gas and the throws with combustion, and you have the physics of ISRO's PSLV. The key insight: a rocket does not push against anything external. It carries its own reaction mass and expels it at high speed. Newton's third law does the rest.

Deriving the thrust equation

Newton's second law F = ma assumes the mass stays constant. For a rocket that is shedding mass every second, you cannot use it directly. Instead, go back to the more fundamental principle: conservation of momentum.

Assumptions: The rocket is in deep space — no gravity, no air resistance. The exhaust leaves at a constant speed v_e relative to the rocket. This derivation applies to any burn rate; the rate does not need to be constant.

Momentum diagram at time t and time t + dt Left panel shows the rocket with mass m and velocity v. Right panel shows the rocket with mass m+dm and velocity v+dv, plus the exhaust gas with mass negative dm and velocity v minus v_e. Time $t$ Time $t + dt$ mass $m$ velocity $v$ $v$ $m + dm$ $v + dv$ $v + dv$ $-dm$ $v - v_e$
Left: the rocket at time $t$ — mass $m$, velocity $v$. Right: an instant later, the rocket is lighter ($m + dm$, where $dm < 0$) and faster ($v + dv$), while the exhaust carries mass $-dm$ backward at $v - v_e$.

At time t, the rocket (including unburned fuel) has mass m and velocity v. In the next tiny interval dt, it burns a sliver of fuel. The exhaust gas departs at speed v_e relative to the rocket — backward — which in the ground frame means it has velocity v - v_e.

After the burn:

  • The rocket has mass m + dm (where dm < 0, since mass is leaving) and velocity v + dv.
  • The exhaust chunk has mass -dm (positive) and velocity v - v_e.

Step 1. Write conservation of momentum for the system (rocket + exhaust).

\underbrace{m\,v}_{\text{before}} = \underbrace{(m + dm)(v + dv)}_{\text{rocket after}} + \underbrace{(-dm)(v - v_e)}_{\text{exhaust after}} \tag{1}

Why: over the interval dt, no external forces act. The total momentum of the closed system — rocket plus the gas it just expelled — must be the same before and after.

Step 2. Expand the right side.

(m + dm)(v + dv) + (-dm)(v - v_e)
= mv + m\,dv + dm\,v + dm\,dv - dm\,v + dm\,v_e

Why: multiply out both products. Notice the +dm\,v from the first product and -dm\,v from the second — they cancel.

Step 3. Cancel and simplify. Setting the right side equal to mv:

\cancel{mv} = \cancel{mv} + m\,dv + dm\,dv + dm\,v_e
0 = m\,dv + dm\,dv + dm\,v_e

Why: mv cancels from both sides. The term dm\,dv is the product of two infinitesimals — vanishingly small compared to the other terms. Drop it.

Step 4. Drop dm\,dv (second-order infinitesimal) and rearrange.

\boxed{m\,dv = -v_e\,dm} \tag{2}

Why: this is the fundamental variable-mass equation. Since dm < 0 (the rocket is losing mass), the right side -v_e\,dm is positive, confirming dv > 0 — the rocket speeds up. Physically: the momentum gained by the rocket equals the momentum carried away by the exhaust.

Divide both sides by dt to get the force form:

m\frac{dv}{dt} = -v_e\frac{dm}{dt} = v_e\left|\frac{dm}{dt}\right|

The right side is the thrust — the force the exhaust exerts on the rocket:

\boxed{F_{\text{thrust}} = v_e \times \left|\frac{dm}{dt}\right|} \tag{3}

Why: thrust equals exhaust speed times the rate of mass ejection. A higher exhaust speed or a faster burn rate gives more thrust. This is why rocket engines heat exhaust to extreme temperatures — hotter gas moves faster, and faster exhaust means more thrust per kilogram of fuel.

For perspective: ISRO's PSLV first stage burns about 500 kg of solid propellant per second with an exhaust speed around 2500 m/s, producing roughly 4500 kN of thrust — enough to lift 320 tonnes off the launchpad.

The Tsiolkovsky rocket equation

Equation (2) describes what happens in a single instant. To find the total velocity gain over the entire burn, integrate.

Step 5. Separate variables — divide both sides of equation (2) by m.

dv = -v_e\,\frac{dm}{m} \tag{4}

Why: put dv on one side and dm/m on the other so each can be integrated independently. The exhaust speed v_e is constant (it depends on the engine design and propellant chemistry, not on how much fuel remains).

Step 6. Integrate from the initial state (mass m_i, velocity v_i) to the final state (mass m_f, velocity v_f).

\int_{v_i}^{v_f} dv = -v_e \int_{m_i}^{m_f} \frac{dm}{m}
v_f - v_i = -v_e\,\Big[\ln m\Big]_{m_i}^{m_f} = -v_e\,(\ln m_f - \ln m_i) = v_e\,\ln\frac{m_i}{m_f}

Why: the integral of 1/m is \ln m. The negative sign flips the fraction inside the logarithm: -(\ln m_f - \ln m_i) = \ln(m_i/m_f). Since m_i > m_f (you started with more mass), the logarithm is positive and \Delta v is positive.

Tsiolkovsky rocket equation

\boxed{\Delta v = v_e \,\ln\!\frac{m_i}{m_f}} \tag{5}

where \Delta v is the total velocity change, v_e is the exhaust speed relative to the rocket, m_i is the initial mass (rocket + fuel), and m_f is the final mass (empty rocket + payload).

The logarithm is the physical punchline. Velocity gain grows only logarithmically with the mass ratio m_i/m_f. To double your \Delta v, you do not double the fuel — you square the mass ratio. Going from \Delta v = v_e (mass ratio e \approx 2.72, fuel is 63% of total mass) to \Delta v = 2v_e (mass ratio e^2 \approx 7.39, fuel is 86%) demands nearly three times the mass ratio. This logarithmic penalty is the fundamental reason why rockets are mostly fuel, payloads are tiny, and space launch is expensive.

Seeing the acceleration increase

Animated: rocket with constant thrust but increasing acceleration A single body moves left to right. Ghost markers at t equals 1, 2, and 3 seconds show increasing spacing — the rocket covers more distance each second because its mass is shrinking while thrust stays constant.
The thrust is constant throughout the 3-second burn. But watch the ghost markers at each second — they get farther and farther apart. At $t = 1$ s the rocket has covered 1.4 m; by $t = 2$ s the second-second gap is 4.8 m; by $t = 3$ s the third-second gap is 10.0 m. Same push, less mass to push, more acceleration. Click replay to watch again.

Exploring the mass ratio

Interactive: Tsiolkovsky equation — Δv/v_e versus mass ratio A logarithmic curve showing how delta-v over exhaust speed depends on the mass ratio m_i over m_f. A draggable point lets you explore different mass ratios and see the corresponding delta-v and fuel fraction. mass ratio $m_i\!/m_f$ $\Delta v\;/\;v_e$ 0 1 2 3 1 5 10 15 $\Delta v/v_e = \ln(m_i/m_f)$ drag the red point along the axis
Drag the red point to change the mass ratio. At $e \approx 2.72$ (63% fuel), you gain one exhaust speed. At $e^2 \approx 7.39$ (86% fuel), you gain two. The curve bends ever more gently — each additional unit of $\Delta v$ demands exponentially more fuel.

Worked examples

Example 1: A sounding rocket in space

An ISRO sounding rocket has a total liftoff mass of 500 kg, of which 400 kg is solid propellant. The exhaust speed is 2500 m/s. Assuming the rocket fires in deep space (no gravity losses), find the speed the rocket attains after all fuel is burned.

Before and after diagram for sounding rocket Left: rocket at rest, total mass 500 kg. Right: empty rocket of 100 kg moving right at 4.02 km/s, with exhaust plume trailing behind. Before (at rest) 500 kg $v = 0$ After (fuel spent) 100 kg 4.0 km/s exhaust
The full rocket (500 kg) starts at rest. After burning 400 kg of fuel, the 100 kg shell reaches 4.0 km/s.

Step 1. Identify the knowns.

m_i = 500 kg (total), m_f = 500 - 400 = 100 kg (structure + payload), v_e = 2500 m/s.

Mass ratio: m_i / m_f = 500/100 = 5.

Why: the mass ratio tells you how many times heavier the loaded rocket is compared to the empty one. A ratio of 5 means 80% of the initial mass is fuel.

Step 2. Apply the Tsiolkovsky equation.

\Delta v = v_e \ln\frac{m_i}{m_f} = 2500 \times \ln 5

Why: no gravity, no drag — the full \Delta v from the equation goes into speed.

Step 3. Compute.

\ln 5 = \ln\frac{10}{2} = \ln 10 - \ln 2 \approx 2.303 - 0.693 = 1.610
\Delta v = 2500 \times 1.610 = 4024 \text{ m/s} \approx 4.0 \text{ km/s}

Why: breaking \ln 5 into \ln 10 - \ln 2 uses two values worth memorising — \ln 2 \approx 0.693 and \ln 10 \approx 2.303. These let you estimate most mass-ratio logarithms in your head.

Step 4. Put this in perspective.

4.0 km/s is about Mach 12. But reaching low Earth orbit requires \Delta v \approx 9.4 km/s. With the same exhaust speed, the mass ratio needed would be e^{9400/2500} = e^{3.76} \approx 43. That means 97.7% of the rocket would have to be fuel, leaving only 2.3% for structure and payload — essentially impossible for a single stage.

Why: this is the practical consequence of the logarithmic penalty: orbital speed demands a mass ratio so extreme that no single-stage chemical rocket can achieve it. Multi-stage rockets solve this by dropping empty structure.

Result: \Delta v = 4.0 km/s.

What this shows: Even with 80% of its mass as fuel, the rocket gains only 1.6 times the exhaust speed. The Tsiolkovsky logarithm is a harsh master — orbital speed requires either much higher exhaust velocities (ion engines, nuclear thermal) or multi-stage designs.

Example 2: Sand on a conveyor belt

At a construction site, sand falls vertically from a hopper onto a horizontal conveyor belt at a rate of 5 kg/s. The belt moves at a constant 3 m/s. Find the horizontal force the motor must exert to keep the belt at constant speed, and the power required.

Sand falling onto a moving conveyor belt Side view: a horizontal belt moves to the right at 3 m/s. Sand falls from a hopper above onto the belt. A force arrow shows the motor force pushing the belt to the right. 3 m/s hopper 5 kg/s $F$
Sand falls from the hopper onto the belt at 5 kg/s. The belt must accelerate each grain from zero horizontal speed to 3 m/s.

Step 1. Identify the variable-mass aspect.

The belt-plus-sand system is gaining mass. Each second, 5 kg of sand arrives with zero horizontal velocity. The belt must accelerate this new mass to 3 m/s.

Why: this is the reverse of a rocket — mass is being added to the system instead of ejected. The same momentum-transfer physics applies: new mass arrives with different velocity, so a force is needed.

Step 2. Compute the force.

F = v \times \frac{dm}{dt} = 3 \times 5 = 15 \text{ N}

Why: every second, 5 kg must go from 0 to 3 m/s horizontally. That requires 5 \times 3 = 15 kg·m/s of horizontal impulse per second — which is 15 N of force.

Step 3. Compute the power the motor delivers.

P = F \times v = 15 \times 3 = 45 \text{ W}

Why: power equals force times the velocity at which the force acts. The belt moves at 3 m/s under a 15 N load.

Step 4. Check where the energy goes.

Kinetic energy gained by sand per second: \frac{1}{2} \times 5 \times 3^2 = 22.5 W.

The motor delivers 45 W, but only 22.5 W appears as kinetic energy. The other 22.5 W is dissipated as heat — when a grain lands on the belt, it slides briefly before reaching belt speed, and kinetic friction during that sliding converts energy to heat.

Why: this is a general result for any "sticking" process. When mass is absorbed at a different velocity, exactly half the input power goes to heat. The physics is the same as a perfectly inelastic collision: kinetic energy is always lost when two objects reach a common velocity.

Result: F = 15 N, P = 45 W (of which 22.5 W is useful kinetic energy and 22.5 W is heat).

What this shows: Variable-mass problems are not just about rockets — any situation where mass enters or leaves a system at a different velocity requires a force. The sand-on-belt problem is the "inverse rocket": instead of mass leaving fast and pushing you forward, mass arrives slow and you must push it to match your speed.

Common confusions

  • "A rocket pushes against the air." No. Rockets work in vacuum — in fact, they are more efficient in vacuum, because there is no atmospheric back-pressure slowing the exhaust. The exhaust pushes the rocket forward by Newton's third law, not by pushing on anything external.

  • "Constant thrust means constant acceleration." Wrong, and this is the central point of variable-mass physics. Thrust may be constant (v_e |dm/dt| doesn't change if v_e and the burn rate are constant), but the mass is dropping. Since a = F/m and m is shrinking, the acceleration increases as the rocket burns fuel. The simulation above shows this — the ghost markers get farther apart even though the thrust never changes.

  • "More fuel always means more speed." Only logarithmically. Doubling the fuel does not double the \Delta v. Going from 50% fuel (mass ratio 2, \Delta v = 0.69\,v_e) to 90% fuel (mass ratio 10, \Delta v = 2.30\,v_e) requires nine times the fuel fraction — but you gain only 1.61\,v_e more speed.

  • "Exhaust speed doesn't matter much — just burn fuel faster." Burning faster increases thrust but not \Delta v. The Tsiolkovsky equation contains only v_e and the mass ratio. A higher burn rate gets you to your final speed faster (more thrust, more acceleration), but the final speed itself depends only on exhaust speed and the mass ratio. This is why ion engines — with their tiny thrust but enormous exhaust speeds (30,000–50,000 m/s) — achieve far higher \Delta v than chemical rockets.

  • "The sand-on-belt problem is unrelated to rockets." They are the same physics. A rocket ejects mass at high relative velocity; a belt absorbs mass at a different velocity. Both require a force proportional to v_{\text{rel}} \times |dm/dt|. One speeds up, the other maintains speed — but the variable-mass equation governs both.

If you came here to derive the thrust equation and the Tsiolkovsky rocket equation, you have everything you need. What follows is for readers curious about why real rockets use multiple stages — and the quantitative proof that staging beats a single stage.

Multi-stage rockets — why dropping empty tanks matters

The Tsiolkovsky equation has a sobering consequence for single-stage rockets. To reach low Earth orbit, you need approximately \Delta v \approx 9.4 km/s (including gravity and drag losses). A good chemical rocket achieves v_e \approx 3000 m/s. The required mass ratio:

\frac{m_i}{m_f} = e^{9400/3000} = e^{3.13} \approx 22.9

Only 1/22.9 \approx 4.4\% of the total mass can be structure plus payload. With current materials, the empty structure of a rocket stage (tanks, engines, plumbing) is typically 8–10% of the fuel mass. That alone exhausts the 4.4% budget — leaving nothing for payload. A single-stage-to-orbit chemical rocket is, for all practical purposes, impossible.

The solution: staging. Burn the first stage's fuel, then drop its empty tank. The second stage does not have to accelerate dead weight.

A numerical comparison. Take 1000 kg total mass, 100 kg payload, and v_e = 3000 m/s for both designs.

Single stage: 800 kg fuel, 100 kg structure, 100 kg payload.

\Delta v = 3000 \times \ln\frac{1000}{200} = 3000 \times \ln 5 = 3000 \times 1.609 = 4828 \text{ m/s}

Two stages (same total fuel, same total structure, same payload):

Fuel Structure Running total
Stage 1 500 kg 50 kg 1000 kg at ignition
Stage 2 300 kg 50 kg 450 kg at stage-2 ignition
Payload — — 100 kg delivered

Stage 1 burns 500 kg of fuel:

\Delta v_1 = 3000 \times \ln\frac{1000}{500} = 3000 \times \ln 2 = 3000 \times 0.693 = 2079 \text{ m/s}

Drop the 50 kg empty first-stage structure. Stage 2 starts at 1000 - 500 - 50 = 450 kg.

\Delta v_2 = 3000 \times \ln\frac{450}{150} = 3000 \times \ln 3 = 3000 \times 1.099 = 3296 \text{ m/s}
\Delta v_{\text{total}} = 2079 + 3296 = 5375 \text{ m/s}

The two-stage rocket gains 5375 m/s versus 4828 m/s for the single stage — an 11% improvement with the exact same fuel, structure, and payload. The only difference is that the second stage did not carry the first stage's empty 50 kg tank.

Real rockets push this further. ISRO's PSLV uses four stages plus six strap-on boosters. Each staging event drops dead weight, and the cumulative \Delta v gains are enough to place 1750 kg into a 600 km polar orbit — a feat that would be physically impossible with a single stage using the same propellants.

The general principle: for a given total fuel and exhaust speed, the more stages you use, the higher the total \Delta v — with diminishing returns for each additional stage. Three or four stages capture most of the benefit; beyond that, the added complexity and inter-stage structure outweigh the marginal \Delta v gains.

Where this leads next

  • Conservation of Linear Momentum — the foundation beneath every result in this article: momentum is conserved when no external forces act.
  • Impulse and Momentum — the force–time connection: how a finite push over a finite time changes a body's momentum.
  • Centre of Mass — Definition and Calculation — the point that moves as if all external force were applied there, even when mass is distributed or changing.
  • Inelastic Collisions — the sand-on-belt problem is an inelastic collision in continuous form. The energy-loss result (\frac{1}{2} goes to heat) generalises.
  • Elastic Collisions — the other extreme: collisions where kinetic energy is fully conserved.