Question bank — Δv = v_e · ln(m₀ - m_f) — understanding each term

False. Δv is a change in velocity — a budget you add to your starting velocity (and from which real gravity/drag losses are subtracted). A rocket already moving at 2000 m/s that earns Δv = 3000 m/s ends near 5000 m/s, not 3000.

False. Δv scales with the log of the mass ratio, not with propellant mass. Doubling Δv requires squaring the mass ratio ($\ln R^2=2\ln R$), which needs far more than double the fuel.

True. No propellant was burned, so the mass ratio is 1 and $\ln 1=0$, giving Δv = 0. There is nothing to throw backward, hence no kick — see Newton's Third Law.

True. Since $\Delta v=v_e\ln R$, Δv is directly proportional to $v_e$. Throwing exhaust faster gives a bigger forward kick per kilogram thrown — this is why engineers chase high $v_e$ (ion engines, hot gases).

False. The derivation assumes a vacuum with no external force; thrust comes purely from ejecting mass and Conservation of Momentum. Air actually hurts (drag), it does not help push.

False. It is a ratio of two masses, so the units cancel — $R$ is dimensionless. That is essential, because you can only take $\ln$ of a pure number.

True. Δv depends only on the ratio $m_0/m_f$; scaling both by the same factor keeps the ratio (and its log) fixed. A big rocket and a small one with the same proportions earn the same Δv.

False. It is the ideal case — inertial frame, no gravity, no drag, constant $v_e$. In reality you subtract gravity-loss and drag-loss terms from this ideal Δv.

Wrong. $m_f$ is the mass remaining after the burn — dry structure + engines + payload. The propellant burned is $m_0-m_f$, the mass that left as exhaust.

Wrong. It is a ratio inside the log: $\ln (m_0/m_f)$. A subtraction would give the wrong units (kg inside a log is meaningless) and the wrong physics — integrating $dv=-v_{e}dm/m$ yields a ratio, not a difference.

Wrong. Bigger tanks add dry mass too, raising $m_f$ and shrinking the ratio's benefit. Δv only grows logarithmically while structural mass grows fast — this is why we use Multistage Rockets.

Wrong. $dmdv$ is a product of two infinitesimals (second-order), vanishingly smaller than the first-order terms, so it is correctly dropped. Keeping it would add nothing to the final equation.

Wrong. $v_e$ is the exhaust speed relative to the rocket. In the derivation the ejected chunk moves at $v-v_e$ in the ground frame, precisely because $v_e$ is the relative speed.

Wrong. They are linked by $v_e=g_0{I}_{sp}$, where $g_0\approx 9.81{\text{ m/s}}^2$. The seconds of $I_{sp}$ become m/s once multiplied by $g_0$ — see Specific Impulse (Isp).

Wrong. The log is forced by the mathematics: velocity gain obeys $dv=-v_{e}dm/m$, and integrating $1/m$ must give $\ln m$. See Natural Logarithm and Integration of 1/x.

Because the same chunk of exhaust accelerates a lighter rocket more than a heavy one. Momentum conservation gives $dv=-v_{e}dm/m$, so what matters is mass thrown relative to current mass.

The fractional mass change $dm/m$ is smaller when $m$ is large, so each kilogram thrown produces less $dv$. Big rockets waste much of their fuel just moving the fuel that is still onboard.

Staging drops empty dry mass mid-flight, lowering $m_f$ for the remaining journey and boosting the effective mass ratio. This sidesteps the logarithm's punishing demand for exponentially more fuel — see Multistage Rockets.

Both come from the same momentum balance: thrust describes the instantaneous force from mass flow, while the rocket equation is that same relation integrated over the whole burn — see Thrust and Mass Flow Rate.

Because propellant is itself mass that must be carried and accelerated. Adding fuel raises $m_0$, but the log grows so slowly that returns collapse — you need to square the ratio to double Δv.

A mission is a sum of maneuvers (launch, transfers, landing), each demanding a chunk of speed change. Δv adds up cleanly across maneuvers regardless of direction, so planners track a total Δv budget.

Zero. No propellant is ejected, so $m_f=m_0$, the ratio is 1, and $\ln 1=0$. Unused fuel is just dead weight, not a speed source.

Δv $\to 0$, because $\ln R\to \ln 1=0$. For a small fuel fraction the equation is nearly linear ($\ln R\approx R-1$), so a little fuel gives a little Δv.

The ratio $R=m_0/m_f\to \infty$, so $\ln R\to \infty$ and Δv would grow without bound. This is unphysical — you can never reach $m_f=0$ because structure, engines, and payload always remain.

No. Since $m_0\ge m_f$ (you only lose mass), $R\ge 1$ and $\ln R\ge 0$, so Δv is always non-negative. Deceleration comes from pointing the thrust backward, not from a negative Δv.

Zero, because $\Delta v=v_e\ln R=0$ when $v_e=0$. Ejecting mass at zero relative speed carries no momentum away, so there is no forward kick — see Conservation of Momentum.

The forward Δv would shrink, because only the backward component of exhaust velocity produces forward thrust. The clean equation assumes exhaust is ejected directly opposite the motion.