# Rocket Propulsion and Variable Mass Systems

> Rocket propulsion and variable mass systems for JEE Physics: thrust v_r(dm/dt), Tsiolkovsky v = u + v_r ln(m0/m), multistage design and conveyor-belt problems.

- Canonical URL: https://prepelephant.com/topics/jee/physics/rocket-propulsion-variable-mass
- Exam / course: JEE · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Rocket Propulsion and Variable Mass Systems", PrepElephant, https://prepelephant.com/topics/jee/physics/rocket-propulsion-variable-mass

## Direct answer

Newton's second law in the form F = ma quietly fails when mass itself changes, and a rocket is the standard case: it accelerates by throwing its own mass backwards. The thrust is F = v_r × |dm/dt| — exhaust speed times burn rate — and integrating momentum conservation across the continually shrinking rocket gives the Tsiolkovsky equation v = u + v_r ln(m0/m), where u is the initial speed, v_r the exhaust speed relative to the rocket and m0/m the mass ratio. The logarithm is the sobering part: doubling the fuel fraction does not double the speed, which is precisely why real launchers stack stages and discard empty tanks. The same bookkeeping answers the conveyor-belt classic — F = v dm/dt to keep a belt moving as sand lands — and its energy paradox, where half the supplied power vanishes into sliding.

## What you must remember

- **Thrust:** F = v_r|dm/dt|, depending on both how fast the gas leaves and how fast it is burnt; a heavy slow exhaust can match a light fast one.
- **Tsiolkovsky equation:** v = u + v_r ln(m0/m); in gravity, subtract gt: v = u + v_r ln(m0/m) − gt over the burn.
- **Logarithmic price:** speed grows only as the logarithm of the mass ratio — 3v_r demands e³ ≈ 20 kg launched per kilogram delivered.
- **Sign care:** dm/dt is negative for the rocket; thrust magnitude uses the positive burn rate, and the exhaust velocity is measured relative to the rocket, not the ground.
- **Multistage logic:** dropping empty stages improves the effective mass ratio, since dead tank mass would otherwise ride along as payload; this is engineering's answer to the logarithm.
- **Conveyor-belt result:** F = v(dm/dt) to keep a belt at speed v while mass lands on it at rate dm/dt; the motor supplies power Fv = v²(dm/dt) while the sand gains only ½v²(dm/dt) — the other half is lost to sliding at contact.
- **Momentum conservation is the tool:** in every variable-mass problem, apply conservation of momentum to the system just before and just after a small interval, never F = ma directly.

## Deriving the rocket equation

At some instant the rocket has mass m and upward speed v. In the next dt it ejects mass −dm (> 0) backwards at speed v_r relative to the rocket, while the rocket gains speed dv. Momentum before is mv; after, the rocket carries (m + dm)(v + dv) and the gas (−dm)(v − v_r). Discarding the second-order dm·dv term leaves m dv = −v_r dm; integrating gives Δv = v_r ln(m0/m) — algebra JEE Advanced has asked students to reproduce. With gravity, each burn second steals g from the velocity budget.

Now the numbers that make the logarithm real. At exhaust speed 2000 m/s, a mass ratio of e gives Δv = 2000 m/s; 4000 m/s demands burning down to m0/e², and 6000 m/s a ratio of e³ ≈ 20. Since structure itself weighs something, a single stage cannot reach orbit — hence stacking. The conveyor case runs the same arithmetic in the opposite direction: sand landing at 2 kg/s on a belt moving 3 m/s needs F = v(dm/dt) = 6 N; the motor supplies 18 W, the sand gains 9 W of kinetic energy, and the missing 9 W dissipates in slipping — a half-loss theorem at every rate.

## Traps in variable-mass problems

The exhaust speed must be relative to the rocket; students who read "gases ejected at 2 km/s" as ground-frame speed break the derivation at its first line. Second, thrust is not the rocket's net force: gravity and drag still act, so the acceleration is (F_thrust − mg)/m, which grows during the burn as m falls at roughly constant thrust — why launches feel gentle at liftoff and violent at burnout. Third, the conveyor energy split: F = v(dm/dt) is a momentum result, and computing the sand's kinetic energy as if the belt force acted without slipping is the designed error; honest bookkeeping gives motor power double the sand's gain. Fourth, "the rocket burns 100 kg/s" is |dm/dt|, ready for the thrust formula without conversion.

## Frequently asked questions

### Why does F = ma fail for a rocket?

The rocket's mass keeps changing, so acceleration comes from momentum conservation applied over a small time interval; the correct statement is thrust = v_r|dm/dt| and v = u + v_r ln(m0/m).

### What is the Tsiolkovsky rocket equation?

v = u + v_r ln(m0/m): the speed gain equals exhaust speed times the natural log of the mass ratio, minus any gravity losses during the burn.

### Why do rockets use multiple stages?

Because speed grows only logarithmically with mass ratio; discarding empty stages removes dead mass and multiplies the effective mass ratio, which no amount of single-stage fuel can achieve.

### What force keeps a conveyor belt moving as sand falls onto it?

F = v(dm/dt) in the direction of motion, since each second's sand must be brought from rest to the belt speed v — a momentum-rate force.

### Where does half the conveyor motor's power go?

The motor supplies v²(dm/dt) while the sand gains only ½v²(dm/dt) of kinetic energy — exactly half is lost to relative sliding.