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Choose downward velocity $v$ positive. The remaining rocket mass is $m(t)=m_0-\alpha t$, and the upward rocket thrust has magnitude $\alpha u$, where $u>0$ is the relative rocket exhaust speed. The rocket equation including gravity isThis follows by applying momentum conservation to the rocket and its just-expelled rocket propellant, with gravity supplying the external impulse. Applying $d(mv)/dt=mg$ to the rocket alone would omit the exhaust momentum flux.

Integrating the rocket equation with $v(0)=v_0$ givesThis formula holds during the constant-flow burn, while $m_0-\alpha t>0$ and fuel remains. The negative logarithmic term represents the slowing due to upward rocket thrust.

Height decreases at rate $\dot{h}=-v$. Set $m=m_0-\alpha t$; an antiderivative needed for the rocket equation isConsequentlyDifferentiating this expression recovers $\dot{h}=-v$ and substituting $t=0$ gives $h(0)=h_0$, which check both the sign and the integration constant.

During $dt$, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the expelled positive mass $-dM$ has velocity $v-u$ to first order. Subtract the initial momentum $Mv$ from their final momentum to obtain $Mdv+udM$. Equating this to the external impulse gives the rocket equationWith $F=-Mg-kMv$ and $\dot{M}/M=-b$, this reduces to $\dot{v}+kv=ub-g$. Therefore, for initial velocity $v_0$ and $k>0$,For a launch from rest,The inequality makes the initial upward acceleration positive. Equality gives no upward launch from rest. In the zero-drag limit, $v=(ub-g)t$. These expressions apply while the prescribed burn law and constant-$g$ approximation remain valid.

Powered ascent. Use upward signed velocity, constant gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, constant relative rocket exhaust speed $u=2500\mathrm{m}{\mathrm{s}}^{-1}$, vertical flight from rest, no drag force, and no impulse at stage separation. The constant-$g$ approximation is an explicit modelling assumption; a varying Newtonian gravitational field would require integrating a different ascent equation. Put $m_0=540000\mathrm{kg}$ and $q=2500\mathrm{kg}{\mathrm{s}}^{-1}$. During ascent $m(t)=m_0-qt$, and momentum balance gives the rocket equation with gravity:The launch rocket thrust $uq=6.25\times 10^6\mathrm{N}$ exceeds $m_{0}g$, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has mass $m_b=140000\mathrm{kg}$, not $40000\mathrm{kg}$; the latter is the separated first stage. The burn duration and endpoint speed are

Coasting height. Integrating the ascent velocity from zero height givesFor example this follows by substituting $m=m_0-qt$ in ${\int }_0^t\ln [m_0/(m_0-qs)]ds$. At separation,The separated stage then undergoes free fall in a constant gravitational field with initial upward speed $v_b$. Conservation of mechanical energy during this coast givesIts smaller mass after separation does not change that coasting height.

Reserve burn. On descending to $z_i=100000\mathrm{m}$, conservation of mechanical energy gives downward speedReset time to $s=0$ at reserve ignition. Now $m_i=40000\mathrm{kg}$, $m_f=14000\mathrm{kg}$ and $q_r=26000/92\mathrm{kg}{\mathrm{s}}^{-1}$. Downward ejection supplies upward rocket thrust. The upward signed velocity isThus the formal free-flight burnout value iscorresponding to a formal downward speed of $2.26\mathrm{m}{\mathrm{s}}^{-1}$. Keeping $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ throughout instead gives $3.82\mathrm{m}{\mathrm{s}}^{-1}$ downward; premature rounding is especially damaging because three large velocity terms almost cancel.

The free-flight endpoint is below ground. Integrating the same reserve-burn velocity givesAt the stated endpoint this gives $z_r(92)=-167.27\mathrm{m}$ for $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The surface is crossed at approximately $s=89.15\mathrm{s}$, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout speed cannot be inferred without a ground-contact model. The boxed velocity is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The constant-rate retro-rocket landing must satisfy both its velocity and height conditions.

Take upward velocity as positive, and let $w>0$ be the rearward rocket exhaust speed. During $dt$, the ejected parcel has mass $-dm>0$ and laboratory velocity $v-w$ to first order. The momentum change of the rocket and this parcel equals the gravitational impulse:Discarding the second-order product $dmdv$ gives the vertical rocket equationWith $\dot{m}=-\alpha$ and $m=m_0-\alpha t$, integration for free flight starting from rest at $t=0$ givesThe logarithmic term is the ideal Tsiolkovsky rocket equation contribution; $gt$ is the gravity loss. Immediate upward departure requires positive initial acceleration, orAt exact equality the initial acceleration is zero, but further burning reduces the weight and permits departure immediately thereafter. If departure is allowed to be delayed while fuel burns on the pad, the condition is instead $\alpha w>m(t)g$ at some time before burnout. With final rocket dry mass $m_f$, any powered departure before burnout requires $w>g{m}_f/\alpha$; the free-flight formula then starts at the actual departure time $t_{\ell }$ with $m(t_{\ell })$, giving $v=w\ln [m(t_{\ell })/m(t)]-g(t-t_{\ell })$.

For the specified data, the initial mass includes the payload. The rocket fuel fraction givesThe $100\mathrm{kg}$ payload is already part of the $2500\mathrm{kg}$ final mass; it is not the entire rocket dry mass and must not be added a second time. Here $m_{0}g/\alpha =196m/s$, well below the stated rocket exhaust speed, so departure is immediate. The burnout speed isThis is the speed at burnout. Subsequent motion under the assumed uniform gravity has $v(t)=v(T)-g(t-T)$.

For a fixed set of bodies in an inertial frame, total linear momentum is constant if the resultant external force vanishes. Summing Newton's second law cancels internal force pairs by Newton's third law. For an open subsystem, momentum carried across its boundary must also be included.

For an equal-intake-fuel mass-loading rocket, enclose the original craft and the stationary material collected during $dt$. Its initial momentum is $mu$. At the end, the craft has mass $m-\mu dt$ and the exhaust has mass $2\mu dt$. To first order,The first term is rocket thrust; the second is the momentum cost of collecting stationary matter. It vanishes in the conventional rocket equation, which therefore has no analogous fixed exhaust-speed ceiling in Newtonian mechanics. Here the acceleration vanishes at $u_{\max }=2u_0$; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.

For constant $\mu$, mass conservation gives $m=m_i-\mu t$. Separating variables,ThereforeThe formal limiting speed would require $t=m_i/\mu$ and zero remaining craft mass. A positive rocket dry mass and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.

For particles in an inertial frame, total momentum and angular momentum about a fixed origin areThe momentum derivative is the total external force when internal forces cancel pairwise. Angular momentum is conserved when the total external torque vanishes and the internal torques sum to zero, as for pairwise equal-and-opposite central forces. Equal-and-opposite internal forces alone need not cancel internal torque if their lines of action are different. An isolated system with these internal-force properties conserves both quantities; if fields carry momentum, the complete system must include them.

For a straight rocket, let its instantaneous remaining mass be $M$ and let $dM<0$ be ejected during a velocity increase $dv$. The exhaust's inertial velocity is $v-v_e$ to first order. Conservation of momentum givesIntegrating from initial mass $m+m_f$ to dry mass $m$ yields the rocket equationIt is the rocket-plus-exhaust system that is isolated; the rocket alone is a variable-mass open system.

For the firework, the original diagram places one tangentially firing point rocket at the end of each radial rod. The parallel axis theorem changes each rod's central inertia to $m{a}^2/12+m(a/2{)}^2=m{a}^2/3$. Adding all rods and fueled rockets givesWhen fuel is exhausted, $I_f=4m{a}^2$.

Let $M_f$ now denote the total remaining fuel in all three rockets, so $I=4m{a}^2+a^2{M}_f$. Fuel at the rim co-rotates before ejection. Its backward tangential exhaust velocity is $a\omega -v_e$, and its angular momentum per unit mass about the axle is $a(a\omega -v_e)$. For an infinitesimal ejection,Since $dI=a^{2}d{M}_f$, the co-rotation terms cancel and leave $Id\omega =-a{v}_{e}d{M}_f$. The tangential-exhaust rotor with a fixed dry inertia therefore obeysNeglecting the exhaust's initial tangential motion would give an incorrect angular-momentum balance. The rod mass remains in the denominator throughout the burn.

After the burn, the opposing constant axle torque $G>0$ acts on the fixed inertia $I_f$. Its work over the remaining turn equals the initial rotational kinetic energy:

Powered vertical flight. Take upward velocity $v$ as positive, and let $v_e>0$ be the downward exhaust speed relative to the rocket. During $dt$ the rocket's mass changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame velocity $v-v_e$. Compare the momentum of the rocket plus this newly expelled exhaust before and after the interval:Discard the second-order product $dmdv$. The result is $mdv+v_{e}dm=-mgdt$, orThe positive rocket thrust is $-v_e\dot{m}$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the rocket equation with gravitational loss:Using $d(mv)/dt$ for the rocket alone without the exhaust momentum flux would give the wrong equation for this open system.

Orbital speed under the stated gravity assumption. A circular orbit of radius $r_o=R_E+h_o=7.90\times 10^6\mathrm{m}$ requires centripetal acceleration $v_o^2/r_o$. If the magnitude of gravitational acceleration is the stipulated constant $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ even at the orbit,This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.

Burnout height. During the unpowered projectile motion, horizontal velocity is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital speed $v_o$. At burnout, therefore,These follow either by integrating $\ddot{y}=-g$ or by the vertical work-energy theorem. Substituting the preceding constant-$g$ orbital result givesThe vertical component is initially positive, as required to reach the orbit after burnout.

Alternative inverse-square interpretation. A real Earth's exterior Newtonian gravitational field is not constant over a $1500\mathrm{km}$ altitude change. If $9.8\mathrm{m}{\mathrm{s}}^{-2}$ instead means the surface value $g_0$, then Newton's law of universal gravitation gives $GM=g_0{R}_E^2$ andUsing this orbital speed but the paper's constant-$g$ flat projectile motion approximation for the coast would giveThis alternative changes the global constant-gravity assumption; it is not the same model as the $183\mathrm{km}$ result. An exact curved Kepler orbit coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.

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 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.

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, $v=(u\beta /\alpha )[1-(M/M_0{)}^{\alpha /(\beta -\alpha )}]$.

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.

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 $m_0$.

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 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]

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]

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.

A body exchanging mass with material of inertial velocity $\mathbf{u}$ satisfies the displayed momentum conservation law for one stream, with signed $\dot{m}$ positive for entry and negative for exit. The force $\mathbf{F}$ excludes the incoming or outgoing momentum flux. A short time interval gives $d(m\mathbf{v})/dt=\mathbf{F}+\mathbf{u}\dot{m}$, proving the formula. For stationary incoming matter, $d(m\mathbf{v})/dt=\mathbf{F}$; for matter leaving at the body's own velocity, $m\dot{\mathbf{v}}=\mathbf{F}$. The rocket equation and accreting-drop motion are applications.

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.