Rocket Propulsion and Variable Mass Systems
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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.