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Derivation

The Tsiolkovsky Rocket Equation

D-034 Home PU-101 Threads force · matter Depends on Momentum Conservation from Newton's Third Law
Statement

For a body that propels itself by ejecting mass at a constant speed \(v_{\mathrm{ex}}\) relative to itself, in the absence of external forces, momentum conservation yields the velocity gained as a logarithmic function of the ratio of initial to final mass: \(\Delta v = v_{\mathrm{ex}} \ln\!\left(\dfrac{m_0}{m_f}\right)\).

Why it matters

The Tsiolkovsky equation is the master constraint of spaceflight. It says that the velocity budget of a rocket is set not by how much fuel it carries in absolute terms, but by the ratio of wet to dry mass, mediated by exhaust speed. Because the mass ratio enters only through a logarithm, doubling the payload's velocity gain requires squaring the mass ratio, which is why chemical rockets are overwhelmingly propellant by mass and why staging exists.

It is also the cleanest non-trivial application of Newton's second law to a variable-mass system, making it a touchstone for understanding what "the momentum of a system" really means when that system is exchanging mass with its surroundings.

Assumptions
No external forces act on the rocket-plus-exhaust system.If gravity or drag acts, the bare equation gives only the ideal \(\Delta v\); real velocity gain is reduced by gravity-loss and drag-loss integrals, \(\Delta v = v_{\mathrm{ex}}\ln(m_0/m_f) - \int g\sin\gamma\,dt - \int (D/m)\,dt\).
The exhaust speed \(v_{\mathrm{ex}}\) relative to the rocket is constant.If \(v_{\mathrm{ex}}\) varies (throttling, changing nozzle conditions), the logarithm is replaced by \(\Delta v = \int v_{\mathrm{ex}}(t)\,\dfrac{-dm}{m}\), evaluated along the actual burn.
All ejected mass leaves with the same relative velocity, directed opposite to the thrust axis (one-dimensional motion).If the exhaust has an angular spread or the motion is multi-dimensional, only the axial component contributes to thrust and an efficiency factor \(\cos\theta\) multiplies \(v_{\mathrm{ex}}\).
Momentum, not energy, is conserved for the closed system; the rocket carries no imposed energy budget.If you mistakenly impose energy conservation on the rocket alone, you get nonsense — the chemical energy released is an external supply to the mechanical problem, and only the total momentum of rocket + expelled gas is conserved.
Derivation
1
\[ p(t) = m\,v \]
At time \(t\) the rocket has mass \(m\) and velocity \(v\); this is its momentum in the inertial frame. A
2
\[ dm < 0, \qquad dm_{\mathrm{ex}} = -\,dm > 0 \]
In interval \(dt\) the rocket loses mass \(-dm\), which becomes the ejected parcel of mass \(dm_{\mathrm{ex}}\). Mass is conserved for the closed system. A
3
\[ v_{\mathrm{parcel}} = v - v_{\mathrm{ex}} \]
The parcel leaves at speed \(v_{\mathrm{ex}}\) relative to the rocket, directed backward, so in the inertial frame its velocity is the rocket velocity minus the exhaust speed. A
4
\[ p(t+dt) = (m+dm)(v+dv) + (-dm)\,(v - v_{\mathrm{ex}}) \]
Total momentum after \(dt\): the lighter rocket \((m+dm)\) at velocity \(v+dv\), plus the parcel of mass \(-dm\) at velocity \(v-v_{\mathrm{ex}}\). A
5
\[ p(t+dt) - p(t) = m\,dv + dm\,dv + v_{\mathrm{ex}}\,dm \]
Expand Step 4 and subtract Step 1. The terms \(v\,dm\) and \(-v\,dm\) cancel, and \(-dm\,(-v_{\mathrm{ex}})\) becomes \(+v_{\mathrm{ex}}\,dm\). B
6
\[ dp = m\,dv + v_{\mathrm{ex}}\,dm \]
The product \(dm\,dv\) is second-order in infinitesimals and is dropped; this is legitimate because it vanishes faster than the retained terms as \(dt\to 0\). C
7
\[ 0 = m\,dv + v_{\mathrm{ex}}\,dm \]
With no external force the total momentum of rocket + exhaust is conserved (momentum-conservation-from-third-law applied to the internal ejection), so \(dp = 0\). A
8
\[ dv = -\,v_{\mathrm{ex}}\,\frac{dm}{m} \]
Rearrange to isolate the velocity increment against the fractional mass change. Purely algebraic. A
9
\[ \int_{v_0}^{v_f} dv = -\,v_{\mathrm{ex}}\int_{m_0}^{m_f} \frac{dm}{m} \]
Integrate both sides over the burn; \(v_{\mathrm{ex}}\) is constant (assumption) so it comes outside the integral. B
10
\[ v_f - v_0 = -\,v_{\mathrm{ex}}\big[\ln m\big]_{m_0}^{m_f} = -\,v_{\mathrm{ex}}\big(\ln m_f - \ln m_0\big) \]
The antiderivative of \(1/m\) is \(\ln m\); evaluate at the limits. A
11
\[ \Delta v = v_{\mathrm{ex}}\,\ln\!\left(\frac{m_0}{m_f}\right) \]
Flip the sign into the logarithm: \(-(\ln m_f - \ln m_0) = \ln(m_0/m_f)\), and write \(\Delta v = v_f - v_0\). A
Result
\[ \Delta v = v_{\mathrm{ex}}\,\ln\!\left(\frac{m_0}{m_f}\right) \]

Reading. The velocity a rocket can gain equals its exhaust speed times the natural logarithm of the mass ratio (initial mass over final mass). Every factor-of-\(e\approx 2.718\) increase in mass ratio buys one more \(v_{\mathrm{ex}}\) of \(\Delta v\). The relation is independent of burn rate, thrust profile and time — only the endpoints \(m_0\), \(m_f\) and the exhaust speed matter.

Units check. \(v_{\mathrm{ex}}\) has units of \(\mathrm{m\,s^{-1}}\); the mass ratio \(m_0/m_f\) is dimensionless, so its logarithm is dimensionless. The product carries units of \(\mathrm{m\,s^{-1}}\), matching \(\Delta v\). Consistent.

Limiting cases
  • Small mass loss (\(m_f \to m_0\)): let \(m_f = m_0(1-\epsilon)\) with \(\epsilon\ll 1\); then \(\Delta v \approx v_{\mathrm{ex}}\,\epsilon\), the linear impulse-momentum result — throwing a small fraction of your mass backward gives proportional velocity.
  • Large mass ratio (\(m_0/m_f \to \infty\)): \(\Delta v\) grows without bound but only logarithmically, so \(\Delta v = 2\,v_{\mathrm{ex}}\) needs \(m_0/m_f = e^2 \approx 7.4\), and \(\Delta v = 3\,v_{\mathrm{ex}}\) needs \(\approx 20\). Diminishing returns.
  • Zero exhaust speed (\(v_{\mathrm{ex}} \to 0\)): \(\Delta v \to 0\) for any finite mass ratio — dropping mass with no relative velocity produces no thrust.
  • Payload-dominated (\(m_f\) mostly structure/payload): the usable mass ratio is capped by the structural coefficient, setting a hard ceiling on single-stage \(\Delta v\).
Breaks when
  • Relativistic exhaust or vehicle speeds. When \(v_{\mathrm{ex}}\) or \(\Delta v\) become comparable to \(c\), Newtonian momentum \(mv\) is wrong; the relativistic rocket equation replaces the logarithm with a rapidity relation, \(\dfrac{m_0}{m_f} = \left(\dfrac{1+\beta}{1-\beta}\right)^{c/(2 v_{\mathrm{ex}})}\).
  • External forces are not negligible. During launch through a gravity field and atmosphere, gravity-loss and drag-loss subtract from the ideal \(\Delta v\); the bare equation over-predicts performance, sometimes by \(1\text{–}2\ \mathrm{km\,s^{-1}}\) for a surface launch.
  • Mass is gained, not just lost, or ambient mass is ingested. Air-breathing engines (jets, ramjets) and interstellar-ram concepts violate the closed-ejection picture; momentum flux of ingested mass must be included and the sign structure changes.
  • Continuous thrust with variable \(v_{\mathrm{ex}}\). Electric propulsion that throttles specific impulse breaks the constant-\(v_{\mathrm{ex}}\) assumption; one must integrate \(\int v_{\mathrm{ex}}\,dm/m\) along the actual profile.
Failure modes
  • Sign flip in the log. Writing \(\ln(m_f/m_0)\) gives a negative \(\Delta v\); initial mass belongs on top because the rocket gets lighter.
  • Confusing \(v_{\mathrm{ex}}\) with \(g_0 I_{sp}\) carelessly. Effective exhaust velocity is \(v_{\mathrm{ex}} = g_0 I_{sp}\) with \(g_0 = 9.81\ \mathrm{m\,s^{-2}}\) a fixed reference constant, not local gravity; using local \(g\) is an error.
  • Keeping the \(dm\,dv\) term. Retaining the second-order product in Step 6 yields a spurious correction; it must be dropped in the infinitesimal limit.
  • Applying energy conservation to the rocket alone. Setting \(\tfrac12 m v^2\) constant gives the wrong law; only total momentum of rocket + exhaust is conserved, and chemical energy is an external input.
  • Using the exhaust velocity in the ground frame. \(v_{\mathrm{ex}}\) is the speed relative to the rocket; using the parcel's inertial velocity in \(dv = -v_{\mathrm{ex}}\,dm/m\) is inconsistent.
  • Forgetting staging. Treating a multistage vehicle with one mass ratio ignores that discarded stage structure resets \(m_f\); total \(\Delta v\) is the sum over stages, each with its own ratio.
Discussion

The deep content of the equation is that thrust is the reaction to a momentum flux. Differentiating the momentum balance in time gives the thrust equation \(F = -v_{\mathrm{ex}}\,\dfrac{dm}{dt} = \dot{m}_{\mathrm{ex}}\,v_{\mathrm{ex}}\): force equals mass-ejection rate times exhaust speed. The Tsiolkovsky equation is simply the time-integral of \(m\,dv/dt = F\) once you recognize that \(F\) itself depends on the same \(dm\) that changes \(m\). This is why the result is time-independent: whether you burn fast or slow, the total impulse per unit mass ejected is fixed.

Notice the logarithm is the fingerprint of a multiplicative process rendered additive. Each infinitesimal slug of propellant multiplies the mass ratio by a factor, and velocity increments add; summing multiplicative factors of \((1+dm/m)\) is exactly what produces \(\ln\). This is the same mathematical structure as compound interest or radioactive decay, and it explains the brutal exponential cost of high \(\Delta v\): to reach \(\Delta v/v_{\mathrm{ex}} = N\) you must carry roughly \(e^N\) times the dry mass in propellant.

The equation also motivates the entire logic of propulsion technology. Since \(\Delta v\) scales linearly with \(v_{\mathrm{ex}}\) but only logarithmically with mass ratio, the highest-leverage engineering move is to raise exhaust speed. Chemical rockets top out near \(v_{\mathrm{ex}} \sim 4.5\ \mathrm{km\,s^{-1}}\) (limited by combustion temperature and the molecular weight of the products); ion thrusters reach \(30\text{–}50\ \mathrm{km\,s^{-1}}\) at the price of tiny thrust. Staging is the complementary trick: by shedding empty tankage mid-flight it keeps the mass ratio of each phase favourable, since carrying dead structure into the log is wasteful.

At a deeper level the variable-mass problem exposes a subtlety about Newton's second law. The naive form \(F = d(mv)/dt = m\dot v + \dot m v\) is frame-dependent and generally wrong for open systems, because \(\dot m v\) is not a physical force — it depends on the arbitrary choice of inertial frame. The correct, frame-independent statement is that the net external force equals the rate of change of total momentum of a fixed collection of particles (rocket plus the gas it has ejected). Writing instead \(F_{\mathrm{ext}} = m\dot v - v_{\mathrm{ex}}\dot m\) for the rocket alone, the \(-v_{\mathrm{ex}}\dot m\) is the thrust — a genuine force arising from Newton's third law acting between vehicle and exhaust, which is exactly what the prior result momentum-conservation-from-third-law supplies.

Common misconceptions. A rocket does not "push against the ground" or "against the air" — it works perfectly in vacuum because thrust is the reaction to expelled mass, not to any external medium. And the mass ratio, not the total propellant mass, sets performance: a small rocket and a giant rocket with the same mass ratio and \(v_{\mathrm{ex}}\) reach the same \(\Delta v\).

Worked examples
1
\[ \text{Given } v_{\mathrm{ex}} = 4500\ \mathrm{m\,s^{-1}},\quad m_0 = 5.0\times10^{5}\ \mathrm{kg},\quad m_f = 1.0\times10^{5}\ \mathrm{kg} \]
A single stage: find \(\Delta v\). Work symbolically first. A
2
\[ \Delta v = v_{\mathrm{ex}}\ln\!\left(\frac{m_0}{m_f}\right) \]
Direct application of the result. A
3
\[ \frac{m_0}{m_f} = \frac{5.0\times10^{5}}{1.0\times10^{5}} = 5.0, \qquad \ln 5.0 = 1.609 \]
Evaluate the dimensionless mass ratio and its logarithm. A
4
\[ \Delta v = 4500\ \mathrm{m\,s^{-1}} \times 1.609 = 7.24\times10^{3}\ \mathrm{m\,s^{-1}} \]
Multiply. Units: \(\mathrm{m\,s^{-1}}\times(\text{dimensionless})=\mathrm{m\,s^{-1}}\). A
\[ \Delta v \approx 7.24\ \mathrm{km\,s^{-1}} \]

Reading. With a 5:1 mass ratio and chemical-grade exhaust, this single stage delivers about \(7.2\ \mathrm{km\,s^{-1}}\) of ideal velocity — short of the \(\sim 9.4\ \mathrm{km\,s^{-1}}\) needed for low Earth orbit once gravity and drag losses are added, which is precisely why real launchers stage.

1
\[ \text{Required } \Delta v = 6.0\ \mathrm{km\,s^{-1}},\quad I_{sp} = 320\ \mathrm{s},\quad g_0 = 9.81\ \mathrm{m\,s^{-2}} \]
Find the propellant fraction needed. First convert specific impulse to exhaust speed. A
2
\[ v_{\mathrm{ex}} = g_0 I_{sp} = 9.81 \times 320 = 3139\ \mathrm{m\,s^{-1}} \]
Definition of effective exhaust velocity from specific impulse. Units: \(\mathrm{m\,s^{-2}}\times\mathrm{s}=\mathrm{m\,s^{-1}}\). A
3
\[ \frac{m_0}{m_f} = \exp\!\left(\frac{\Delta v}{v_{\mathrm{ex}}}\right) \]
Invert the rocket equation by exponentiating both sides. B
4
\[ \frac{m_0}{m_f} = \exp\!\left(\frac{6000}{3139}\right) = \exp(1.911) = 6.76 \]
Evaluate the dimensionless exponent, then exponentiate. A
5
\[ \frac{m_{\mathrm{prop}}}{m_0} = 1 - \frac{m_f}{m_0} = 1 - \frac{1}{6.76} = 0.852 \]
Propellant mass fraction is one minus the reciprocal of the mass ratio. A
\[ \frac{m_{\mathrm{prop}}}{m_0} \approx 0.85 \]

Reading. Roughly 85% of the launch mass must be propellant to achieve \(6\ \mathrm{km\,s^{-1}}\) with this engine. Only 15% is left for structure plus payload — a vivid illustration of the tyranny of the rocket equation.

Problems
  1. A rocket has exhaust velocity \(v_{\mathrm{ex}} = 3.0\ \mathrm{km\,s^{-1}}\) and a mass ratio of \(m_0/m_f = 4.0\). Find its ideal \(\Delta v\).
    Solution\(\Delta v = v_{\mathrm{ex}}\ln(m_0/m_f) = 3000 \times \ln 4.0 = 3000 \times 1.386 = 4.16\times10^{3}\ \mathrm{m\,s^{-1}} \approx 4.2\ \mathrm{km\,s^{-1}}.\)
  2. A probe must gain \(\Delta v = 10\ \mathrm{km\,s^{-1}}\) using an ion engine with \(v_{\mathrm{ex}} = 30\ \mathrm{km\,s^{-1}}\). What mass ratio is required, and what is the propellant fraction?
    Solution\(m_0/m_f = e^{\Delta v/v_{\mathrm{ex}}} = e^{10/30} = e^{0.333} = 1.396.\) Propellant fraction \(= 1 - 1/1.396 = 1 - 0.716 = 0.284\), about 28%. Contrast with a chemical engine at \(v_{\mathrm{ex}}=3\ \mathrm{km\,s^{-1}}\), which would need \(m_0/m_f = e^{3.33}\approx 28\) — the leverage of high exhaust speed.
  3. A two-stage rocket has stage-1 mass ratio \(3.0\) and stage-2 mass ratio \(2.5\), both with \(v_{\mathrm{ex}} = 3.2\ \mathrm{km\,s^{-1}}\). Find the total ideal \(\Delta v\).
    SolutionTotal \(\Delta v\) is the sum over stages: \(\Delta v = v_{\mathrm{ex}}(\ln 3.0 + \ln 2.5) = 3200(1.099 + 0.916) = 3200 \times 2.015 = 6.45\times10^{3}\ \mathrm{m\,s^{-1}} \approx 6.4\ \mathrm{km\,s^{-1}}.\) Equivalently \(v_{\mathrm{ex}}\ln(3.0\times2.5)=3200\ln 7.5\).
  4. During a vertical burn straight up, a rocket with \(v_{\mathrm{ex}} = 2.8\ \mathrm{km\,s^{-1}}\) and mass ratio \(5.0\) burns for \(t_b = 120\ \mathrm{s}\) in a uniform gravity field \(g = 9.81\ \mathrm{m\,s^{-2}}\). Estimate the actual \(\Delta v\) including gravity loss (take the burn as entirely vertical, \(\gamma = 90^\circ\)).
    SolutionIdeal \(\Delta v_{\mathrm{ideal}} = 2800\ln 5.0 = 2800\times1.609 = 4.51\times10^{3}\ \mathrm{m\,s^{-1}}.\) Gravity loss for a vertical burn \(= \int g\sin\gamma\,dt = g\,t_b = 9.81\times120 = 1.18\times10^{3}\ \mathrm{m\,s^{-1}}.\) Actual \(\Delta v \approx 4510 - 1177 = 3.33\times10^{3}\ \mathrm{m\,s^{-1}} \approx 3.3\ \mathrm{km\,s^{-1}}.\) Gravity loss costs about 26% here — real launches pitch over to reduce it.
  5. An engineer wants \(\Delta v = 9.4\ \mathrm{km\,s^{-1}}\) (LEO-class) from a single stage with structural coefficient \(\varepsilon = m_{\mathrm{struct}}/(m_{\mathrm{struct}}+m_{\mathrm{prop}}) = 0.08\) and no payload. What exhaust velocity would be required, and comment on feasibility.
    SolutionWith no payload \(m_f = m_{\mathrm{struct}}\) and \(m_0 = m_{\mathrm{struct}}+m_{\mathrm{prop}}\), so \(m_f/m_0 = \varepsilon = 0.08\) and \(m_0/m_f = 12.5.\) Then \(v_{\mathrm{ex}} = \Delta v / \ln(m_0/m_f) = 9400 / \ln 12.5 = 9400 / 2.526 = 3.72\times10^{3}\ \mathrm{m\,s^{-1}}.\) So \(v_{\mathrm{ex}}\approx 3.7\ \mathrm{km\,s^{-1}}\) (\(I_{sp}\approx 379\ \mathrm{s}\)) — achievable only by good cryogenic engines, and this leaves zero payload while ignoring gravity and drag losses. In practice single-stage-to-orbit is marginal, confirming why staging dominates real designs.