Variable-mass systems

Σ Fₑₓₜ = dP/dt comes from adding F = dp/dt over a set of particles and letting the internal forces cancel in third-law pairs, so it presumes a membership that does not change. A rocket's hull and remaining propellant lose members every instant, a belt under a hopper gains them, and a leaking tank sheds them. So do not apply the law to “the rocket” with a mass that happens to vary. Choose a system you can follow for one short interval dt: the object at time t together with the parcel dm that will cross its boundary during dt. That system has fixed membership, and the push between engine and exhaust, or between belt and gravel, is internal to it and drops out. External forces are only those exerted from outside the enlarged boundary: gravity, drag, the ground, the drive motor. Before writing any equation, run two checks: everything counted at t is still counted at t + dt, and no force on your list is exerted by one part of the system on another.

At time t a rocket of mass m moves at v, so the system's momentum is mv. At t + dt the rocket has mass m + dm, with dm negative, and velocity v + dv, while the ejected parcel of mass −dm moves at v − u, where u is the exhaust speed relative to the rocket. Setting the change equal to the external impulse gives (m + dm)(v + dv) + (−dm)(v − u) − mv = m dv + u dm + dm dv = Fₑₓₜ dt. Two things happen. The v dm terms cancel, so the rocket's speed in whichever frame you chose drops out and only u survives. And dm dv is a product of two small changes: for a 2000 kg rocket ejecting 30 kg s⁻¹ at 2500 m s⁻¹ in free space, a 0.010 s interval gives dm = −0.30 kg and m dv = 750 kg m s⁻¹, so dv = 0.375 m s⁻¹ and dm dv is only 0.11 kg m s⁻¹, 0.015% of m dv, vanishing as dt → 0. Dividing by dt gives m dv/dt = Fₑₓₜ − u dm/dt.

Write w for the velocity of whatever crosses the boundary and the same ledger gives m dv/dt = Fₑₓₜ + (w − v) dm/dt. The last term is a momentum flux: kilograms crossing per second times their velocity relative to the object, which is kg m s⁻², a newton. Its direction follows from the signs rather than from intuition. For a rocket, w − v = −u points backwards and dm/dt < 0, so the product points forwards: a thrust u|dm/dt|. An engine burning 220 kg s⁻¹ with u = 2.9 km s⁻¹ receives 6.4 × 10⁵ N, whatever its own speed. For material joining from rest, w = 0 and dm/dt > 0, so the term is −v dm/dt, a drag the drive must match to hold speed. For water dribbling out through a moving tank's floor, w = v and the term vanishes. The same bookkeeping works for a boundary drawn round a wall: a jet delivering 0.40 kg s⁻¹ at 15 m s⁻¹ that is brought to rest pushes with 6.0 N, and one that rebounds straight back at the same speed pushes with 12 N.

Let gravel fall from rest onto a belt running at 2.0 m s⁻¹, landing at 30 kg s⁻¹. The flux term says the drive must push with F = v dm/dt = 60 N to hold the belt's speed, so it delivers P = Fv = 120 W. The gravel's kinetic energy, though, grows at only ½v² dm/dt = 60 W. The other 60 W is dissipated while each grain skids on the belt until friction brings it up to speed: a continuous, completely inelastic collision. The split is exactly half at every belt speed, because the drive supplies v² dm/dt and the load keeps ½v² dm/dt. Solve the problem by energy conservation instead and you would predict a 30 N drive force, wrong by a factor of two, with nothing in the energy equation to warn you. Whenever matter joins or leaves with a velocity different from the object's, kinetic energy is created or destroyed in the transfer, and only the momentum ledger gives the force.

With Fₑₓₜ = 0 the ledger m dv = −u dm separates into dv = −u dm/m. Integrating from ignition mass mᵢ to burnout mass m_f gives Δv = −u ∫ dm/m = −u ln(m_f/mᵢ) = u ln(mᵢ/m_f). The logarithm has a physical reading. Burning 1 kg adds dv = u(1 kg)/m, so with u = 3000 m s⁻¹ it buys 0.30 m s⁻¹ while the vehicle still has 10 000 kg, and 3.0 m s⁻¹ once it is down to 1000 kg: early propellant is spent mostly on accelerating propellant that has not burned yet. The price is steep. With u = 3.0 km s⁻¹ a mass ratio of 4 gives 4.16 km s⁻¹, and doubling it to 8 adds only 3.0 ln 2 = 2.08 km s⁻¹. Inverting, mᵢ/m_f = e^(Δv/u), so 9.0 km s⁻¹ from the same exhaust needs e³ = 20.1: 95.0% of the ignition mass must be propellant, leaving 5.0% for tanks, engines and payload. That is the argument for staging, and the reason exhaust speed is the number engine designers fight hardest to raise.

Fire vertically and gravity joins as an external force: m dv/dt = u|dm/dt| − mg. Nothing moves unless the thrust beats the ignition weight, u|dm/dt| > mᵢ g; below that the rocket sits on the pad burning propellant for nothing. At a constant burn rate b the mass is m = mᵢ − bt and the equation integrates to v(t) = u ln(mᵢ/m) − gt, so the burnout speed is the ideal Δv minus a gravity loss g t_b. Take mᵢ = 60 000 kg with 45 000 kg of propellant, u = 2.8 km s⁻¹ and b = 300 kg s⁻¹. The thrust, 8.40 × 10⁵ N, beats the weight, 5.89 × 10⁵ N; the burn lasts 150 s; the ideal Δv is 2.8 ln 4 = 3.88 km s⁻¹ and gravity takes 1.47 km s⁻¹, leaving 2.41 km s⁻¹. Burn the same propellant at 450 kg s⁻¹ and the loss falls to 0.98 km s⁻¹, for 2.90 km s⁻¹ at burnout, but the acceleration just before burnout climbs from 46 to 74 m s⁻². Drag, the fall of g with height and pressure thrust at the nozzle all lie outside this model.