Rocket equation

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

Two initially stationary rockets exchange a fuel load $M$: A ejects it with constant rearward relative speed $u_0$ and B captures all of it. A has final velocity $v_{Af}=u_0\ln [(m_A+M)/m_A]$. Since final momentum is zero, $(m_B+M)v_{Bf}=-m_A{v}_{Af}$. No elastic-capture assumption is needed. With negligible flight time, transferred mass $q$ gives the same balance with instantaneous masses $m_A+M-q$ and $m_B+q$; their opposite speeds are equal when those masses agree. Final complete collection is a condition on the model, not a guarantee that every ballistic exhaust trajectory intercepts B.

For exhaust moving freely from a rocket A toward a collector B behind it along a common axis, the exhaust must overtake B in the rearward direction. At negligible separation and flight time this gives the displayed condition. In the ideal simultaneous-collection solution, write $a=m_A+M-q$, $b=m_B+q$ and $v_A=u_0\ln [(m_A+M)/a]$, $v_B=-a{v}_A/b$. The criterion becomes $[(a+b)/b]\ln [(m_A+M)/a]<1$. A formal endpoint obtained by assuming all fuel is collected need not satisfy this interception test throughout a large burn.

A rocket discards exhausted stages, reducing the mass accelerated by later burns. Apply the rocket equation separately to each burn and add the signed velocity increments. Ideal jettison with no relative impulse leaves velocity continuous; it changes the mass ratio available to subsequent burns. Dry-stage mass must not be counted as propellant expelled at the rocket exhaust speed.

With downward speed $v$, downward relative exhaust speed $v_E>0$ and mass $M=M_0-\alpha t$, the ideal vertical equation is $\dot{v}=g-\alpha v_E/M$. Integration gives $v=v_0+gt+v_E\log (M/M_0)$. Height follows from $\dot{h}=-v$; its touchdown condition must be imposed together with the final fuel fraction.

The rocket exhaust speed is the magnitude of the emitted material's velocity relative to the instantaneous rocket. In the ideal rearward-ejection convention of the rocket equation, a rocket with inertial velocity $v$ emits material with inertial velocity $v-u$ to first order. A constant $u$ does not make $v-u$ constant.

Rocket propellant is the onboard material expelled to produce rocket thrust. A chemical rocket normally carries both fuel and oxidizer; both contribute to its mass and to the expelled material. In the ideal rocket equation, $M(t)=M_{\mathrm{dry}}+M_{\mathrm{propellant}}(t)$, with fixed rocket dry mass and decreasing remaining rocket propellant. ^[ref]

In an ideal rocket equation calculation where all expelled rocket propellant is called fuel, the rocket fuel fraction is its initial mass divided by the total initial mass, including payload and structure. Complete consumption leaves rocket dry mass $m_f=(1-f)m_0$ and rocket mass ratio $m_0/m_f=1/(1-f)$. The payload is one component of $m_f$, not an additional mass to add to an already inclusive $m_0$.

In this ideal rocket equation model, rocket dry mass is the mass remaining after all usable rocket propellant has been expelled: structure, engines and payload. It is the final mass in the worked example; if rocket propellant remains at shutdown, the final mass is larger than the rocket dry mass.

The rocket mass ratio compares the initial and final total rocket masses for a burn. It is dimensionless, and the ideal rocket equation becomes $\Delta v=u\ln R$. For the same fixed rocket exhaust speed, two successive burns with rocket mass ratios $R_1$ and $R_2$ have combined rocket mass ratio $R_1{R}_2$ and total velocity gain $u\ln R_1+u\ln R_2$.

Rocket thrust is the propulsive force associated with ejecting rocket propellant. In the ideal rocket equation with no pressure correction, $T=u(-\dot{M})$, where $\dot{M}=dM/dt<0$ is the signed rate of change of rocket mass and $-\dot{M}$ is the positive mass loss rate. For $u=2000m/s$ and $\dot{M}=-5kg/s$, $T=10000\mathrm{N}$. With no other force, the acceleration is $T/M$: initially $10m/{\mathrm{s}}^2$ for $M=1000\mathrm{kg}$, rising as mass falls if this rocket thrust stays constant.

A real exhaust outlet can also contribute a pressure term: $T=u(-\dot{M})+A_e(p_e-p_a)$, with exit area $A_e$, exit pressure $p_e$ and ambient pressure $p_a$. The ideal formula here assumes that term vanishes; it must not silently absorb this correction while still interpreting $u$ as the actual relative rocket exhaust speed. ^[ref]

The instantaneous acceleration contribution from rocket thrust $T$ acting on remaining rocket mass $M$ is $a_T=T/M$. Gravity and other external forces contribute separately. A constant rocket thrust therefore produces an increasing thrust acceleration as rocket propellant is consumed.

For constant gravity and downward rocket exhaust speed $u$, hovering requires rocket thrust $u\alpha =Mg$ with $\alpha =-\dot{M}$. Hence $M=M_0{e}^{-gt/u}$ and the required mass flow rate decays exponentially. A nonzero rocket dry mass ends the hover at finite time, even though this mass equation tends to zero only asymptotically.

For fuel mass rate $s>0$ and backward relative exhaust speed $u$, a rocket sliding on a horizontal plane has remaining mass $q(t)$ and satisfies the displayed rocket equation with kinetic friction. Integration over a sliding interval gives $v(t)-v(t_0)=u\log [q(t_0)/q(t)]-\mu g(t-t_0)$. From initial rest, immediate forward sliding additionally requires thrust to overcome static friction. If both friction coefficients equal $\mu$, motion starts only when $su>\mu gq(t)$. An algebraically negative sliding speed is not the physical solution at rest.

A body ejecting mass at constant rate $Q>0$ and relative exhaust speed $V>0$, against quadratic drag $\alpha u^2$, has remaining mass $m=m_0-Qt$. A short-interval momentum balance gives $m\dot{u}=QV-\alpha u^2$. From rest, separation of variables givesFor final mass $m_f>0$, this formula applies until $t=(m_0-m_f)/Q$. The zero-drag limit is the rocket equation $u_f=V\log (m_0/m_f)$. Two dimensionless variables are $\alpha V/Q$ and $m_0/m_f$.

When matter leaves a moving body with zero instantaneous relative velocity, it carries the body's current momentum per unit mass but creates no additional recoil term. A short-interval momentum balance including the newly released material gives $q\dot{v}=F_{\mathrm{ext}}$ for the remaining body's mass $q(t)$. The expression $\frac{d}{dt}(qv)=F_{\mathrm{ext}}$ omits the outgoing momentum flux. Nonzero relative ejection velocity instead produces the thrust term of the rocket equation.

A classical rocket ejecting fuel at rate $\alpha >0$ and relative exhaust speed $u>0$, while collecting stationary dust of mass $\beta dx$ per traveled distance, obeysThese are mass conservation and momentum conservation for the rocket, incoming dust and outgoing gas. The dust-loading term supplies the momentum needed to accelerate incoming stationary material. Put $\lambda =\sqrt{\alpha /(\beta u)}$ and $V=\lambda u$. While the engines operate and $0\le v<V$, an initially stationary rocket satisfiesFor $\lambda >1$, continuing to $V$ would drive total mass to zero. For $0<\lambda <1$, total mass instead grows without bound and $V$ is approached only after infinite time under indefinitely maintained thrust. In both cases a finite fuel supply runs out before that limiting speed. Total stored mass must not be confused with remaining fuel.

For a rocket equation with exhaust mass rate $\dot{m}$ and incoming stationary mass rate $\dot{D}$, mass balance is $\dot{M}=\dot{D}-\dot{m}$. A short-interval momentum conservation calculation gives $M\dot{v}=u\dot{m}-v\dot{D}$. Constant time rates $\dot{D}=\alpha$, $\dot{m}=\beta$ give $M\dot{v}+\alpha v=u\beta$, unlike a constant accretion rate per distance, whose incoming mass rate is proportional to speed. For unequal rates and initial rest, $v=(u\beta /\alpha )[1-(M/M_0{)}^{\alpha /(\beta -\alpha )}]$.

A Newtonian rocket collecting stationary matter at rate $\mu$ and combining it with the same rate of carried fuel ejects mass at rate $2\mu$. If relative exhaust speed is $u_0$, momentum conservation gives the title equation and mass conservation gives $\dot{m}=-\mu$. With constant $\mu$ and initial rest, $m=m_i-\mu t$ and $u=2u_0\mu t/m_i$. The limiting speed $2u_0$ would require zero remaining mass in this ideal solution; positive dry mass terminates it sooner. A constant collection rate per time is distinct from a constant collection cross-section through a uniform medium.

If stationary mass arrives at the same constant time rate as fuel is ejected, total rocket mass stays fixed. The rocket equation with stationary mass accretion is then $M_0\dot{v}+\alpha v=\alpha u$. For initial rest, speed approaches the relative exhaust speed exponentially with relaxation time $M_0/\alpha$. This is the regular equal-rate limit of the unequal-rate mass-power formula.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.

A photon rocket converts stored mass-energy into a collimated beam of photons. In its instantaneous rest frame, energy conservation gives $c^{2}dM=-d{E}_{\gamma }$, and momentum conservation gives $Mcd\eta =d{E}_{\gamma }/c$. Therefore a monotone change in rapidity costs $M_f/M_i=e^{-|\Delta \eta |}$. The rocket's rest mass includes its fuel and stored internal energy; inefficiency or an angular spread of the exhaust increases the fuel requirement.

Accelerating from rest to rapidity $\eta$ and braking to rest requires two separate photon-rocket burns. Their mass ratios multiply, giving $M_f/M_i=e^{-2\eta }$. For ideal equal-mass matter–antimatter annihilation, the initial antimatter mass fraction is $(1-e^{-2\eta })/2$. Braking fuel must be carried and accelerated during the first burn.

For zero initial velocity, body masses $kM,(1-k)M$ and fuel masses $k{m}_0,(1-k)m_0$, the rocket equation gives the displayed final second-stage speed after burning the first-stage fuel, discarding its body without impulse, and burning the second-stage fuel. It assumes $M>0$ and $0\le k<1$. The speed exceeds the single-stage value when $0<k<1$ and fuel is positive. Its limit as $k\to 1^{-}$ concerns a surviving second stage of vanishing mass; at $k=1$ no such stage exists.

If a vehicle moving at speed $v$ ejects mass forward at relative speed $u>0$, thenAs $dM<0$, the vehicle slows down.