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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 $m_0/m_f$.

For a circular orbit, gravity supplies centripetal acceleration, giving $v_c^2/r=\mu /r^2$. ThusA body in a circular orbit is bound despite its continuous free fall. At the same $r$, escape needs twice its kinetic energy per unit mass. An instantaneous forward tangential velocity increment from the circular orbit requires $\Delta v=(\sqrt{2}-1)v_c$, rather than the full escape speed from rest. In the synthetic example this is exactly $2(\sqrt{2}-1)m/s\approx 0.828427m/s$.^[ref]

Escape speed is a coasting threshold, not a required continuously maintained speed or a measure of rocket propellant. A powered vehicle can start at rest and acquire energy gradually. The rocket equation relates an ideal velocity increment to rocket exhaust speed and rocket mass ratio; rocket thrust, drag, and the timing and direction of propulsion affect a real ascent. Rotation also changes the relation between surface-relative velocity and the inertial frame used in the calculation.

Reaching another body is a finite-destination spacecraft trajectory problem. In the synthetic radial example, a launch at $2m/s$ reaches $r=4\mathrm{m}$ while remaining bound. A spacecraft can similarly encounter the Moon on a bound geocentric transfer; actual planning must include the Moon's gravity, target motion, arrival velocity, and any propulsion needed for capture. Escaping Earth alone does not imply escaping the Sun, because the relevant reference frame and attracting mass have changed. The isolated formula describes one gravitational well; a full mission requires more than its scalar threshold.

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.

Let $s=m_0/T$ be the constant fuel mass rate, and let $q(t)=M+m_0-st$ be the remaining rocket mass. Take positive motion opposite to the exhaust direction. Over a short interval $dt$, momentum conservation including the mass $sdt$ expelled at velocity $v-u$ givesConsequently, while the rocket slides forward,This is the constant-burn rocket with ground friction. Assuming forward sliding begins immediately and persists, integration from rest givesand henceA necessary physical assumption is missing from the printed request. With a common static friction and kinetic friction coefficient $\mu >0$, immediate motion requires $su\ge \mu g(M+m_0)$; equality allows motion as soon as fuel loss lowers the normal force. Then the acceleration is nonnegative initially and increases thereafter, so forward sliding is consistent. Without this assumption the displayed expression need not be the final speed and can even be negative.

For completeness, the physical rest/sliding solution with these equal coefficients is as follows. If $su\le \mu gM$, the rocket never starts before burnout and its final speed is zero. If $\mu gM<su<\mu g(M+m_0)$, static friction initially balances the thrust and release occurs atIntegrating only the sliding interval yieldsThis speed is positive because, with $y=q_s/M>1$, it equals $u[\log y-1+1/y]$, whose derivative with respect to $y$ is $(y-1)/y^2>0$. The $\mu =0$ case recovers the frictionless rocket equation. If static and kinetic coefficients differ, the release condition uses the static coefficient and the subsequent integral uses the kinetic coefficient; the prompt supplies no second coefficient.

Take upward velocity as positive, and consider the closed material system consisting of the rocket and the fuel it ejects during a short interval $dt$. Initially its mass and momentum are $m$ and $mv$. Let $\delta m>0$ be the ejected mass, so the rocket's mass change is $dm=-\delta m$. At the end its velocity is $v+dv$, while the ejected gas has inertial velocity $v-u$ to first order; its downward relative velocity need not be downward in the inertial frame.

The final total momentum, to first order, isNeglect air resistance and take the gravitational acceleration $g$ constant. The external gravitational impulse on this material system is $-mgdt$ to first order. Newton's second law therefore gives $mdv-u\delta m=-mgdt$, orThe positive thrust is $-u\dot{m}$ because fuel loss makes $\dot{m}<0$. Applying $d(mv)/dt$ to the rocket alone without accounting for the escaping momentum would give the wrong equation.

For constant $u$ and positive mass, divide by $m$ and integrate from release at $t=0$:Thus the gravity-corrected rocket equation is equivalent to

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

Take upward as positive and let $dm<0$ be the change in rocket mass during $dt$. The exhaust emitted has mass $-dm$ and, to first order, upward velocity $v-U$. The linear momentum of the rocket and just-emitted gas changes byThe external gravitational impulse is $-mgdt$ to first order. Applying the momentum balance to this material system, rather than applying a fixed-mass formula to the rocket alone, gives the rocket equationThe term $-U\dot{m}$ is the upward thrust.

For the specified mass loss and exhaust law, $\dot{m}=-\alpha$ andThus the condition for strictly positive upward acceleration at launch isAt equality the initial thrust just balances the weight; under this particular exhaust law the acceleration becomes positive immediately afterwards. If the initial thrust is smaller than the weight and the rocket initially rests on a launch surface, a nonzero supporting force persists until a later lift-off, so the free-flight equation does not describe its earlier supported motion.

For immediate free flight, integrate the rocket equation with $v(0)=0$:This applies while $0\le t<m_0/\alpha$ and the nonrelativistic approximation remains valid. The formal divergence as the model's remaining mass approaches zero is not a prediction within that approximation.

For dimensional analysis, write the mass, length and time dimensions as $\mathsf{M},\mathsf{L},\mathsf{T}$. Then $[m_0]=[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[t]=\mathsf{T}$, $[U_0]=[U]=[v]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Both $m_0$ and $\alpha t$ have mass dimension, so $\alpha t/(m_0-\alpha t)$ is dimensionless; both terms in the boxed velocity have dimension $\mathsf{L}{\mathsf{T}}^{-1}$. The launch inequality compares two forces, each of dimension $\mathsf{ML}{\mathsf{T}}^{-2}$.

Without dust collection, $dM=-dm$. The rocket equation reduces toIntegrating from $v=0$, $M=M_0$ givesThis is the usual rocket equation with constant relative exhaust speed, valid while the mass remains positive and the prescribed ejection continues.

Use zero-relative-velocity mass shedding: the released sand initially shares the balloon's vertical velocity, rather than being actively ejected. Let $q=M+m(t)$ and let a positive mass $\delta m$ be shed in a short interval $dt$. The momentum of the balloon and that newly released sand changes byThe leading external impulse is $(T-qg)dt$. Newton's second law applied to this combined material system therefore givesWriting $\frac{d}{dt}(qv)=T-qg$ for the remaining balloon alone would omit the outgoing momentum flux. A nonzero relative ejection velocity would instead produce a rocket equation thrust term.

Initial mechanical equilibrium gives $T=(M+m_0)g$. Write $\lambda =m_0/t_0$, so $m=m_0-\lambda t$ up to depletion. ThenIntegrating from rest,Since $A/t_0=1+M/m_0$ and $A/(A-t_0)=1+m_0/M$, the final upward speed isThe acceleration is nonnegative throughout release, consistent with the upward motion.

First apply momentum conservation to the rocket and the infinitesimal amount of expelled gas. During $dt$, the exhaust mass is $\alpha dt$ and its inertial velocity, to first order, is $v-u$. With no dust, the momentum balance isIt gives the rocket equation $\boxed{m\dot{v}=\alpha u}$, while $\dot{m}=-\alpha$.

For a dust-collecting rocket, the incoming dust mass is $\beta vdt$ with zero initial momentum. The same balance has rocket mass $m+(\beta v-\alpha )dt$ after the interval, soConsequentlyThe second term in the acceleration equation accounts for the momentum required to bring collected dust up to rocket speed.

Set $\lambda =\sqrt{\alpha /(\beta u)}$ and $V=\lambda u$, so $\alpha u=\beta V^2$. As long as the engines operate and $0\le v<V$, the velocity increases and time can be eliminated:Integrating from $v=0$, $m=m_0$, givesThus the required mass–velocity relation is

If $\lambda >1$, this formula would give $m\to 0$ as $v\to V^{-}$. In this interval, $\dot{m}=\beta (v-{\lambda }^{2}u)<0$. A real rocket retains positive body mass even after its fuel is gone, and collected dust only increases that residual mass. It therefore cannot follow this branch all the way to zero mass: its fuel must run out while $v<\lambda u$. Indeed the hypothetical time integral $dt/dv=m/[\beta (V^2-v^2)]$ is integrable at $V$ when $\lambda >1$, so even the formal zero-mass endpoint would occur at finite time.

If $0<\lambda <1$, the same relation instead gives $m\to \infty$ as $v\to V^{-}$. The mass first decreases, reaches its minimum at $v={\lambda }^{2}u$, and then increases because dust arrives faster than fuel is expelled. The time integral now diverges at $V$: under indefinitely maintained thrust, $v$ approaches $\lambda u$ asymptotically while stored dust grows without bound. At that limiting speed, thrust $\alpha u$ balances the dust-loading term $\beta v^2$, and $\dot{m}\to \beta u\lambda (1-\lambda )>0$. This is the physical reason the total rocket mass no longer signals fuel exhaustion. A rocket with a finite initial fuel supply still runs out at finite time $t=M_{\mathrm{fuel}}/\alpha$ and before reaching $V$; increasing total mass does not create new fuel.

Work in one fixed inertial frame, and take positive velocity in the rocket's direction. During $d\tau$, the rocket ejects the positive amount of mass $-m^{\prime }(\tau )d\tau$. Its exhaust has inertial velocity $v(\tau )-u$, by the Galilean transformation of velocities. Once expelled, that material retains its velocity because it experiences no external force. The system's momentum is therefore the rocket momentum plus the sum of all these exhaust momenta:The minus sign comes from the positive expelled mass being $-m^{\prime }(\tau )d\tau$.

Differentiate, applying the product rule to the rocket momentum and the fundamental theorem of calculus to the integral:The momentum conservation law thus gives the nonrelativistic rocket equation

The same bookkeeping for kinetic energy givesIts derivative iswhere the rocket equation cancels the first two terms. Thus kinetic energy increases during mass ejection, with $\boxed{K^{\prime }=-u^2{m}^{\prime }/2>0}$ when $u>0$ and $m^{\prime }<0$. This does not violate conservation of energy: the increase comes from stored internal energy.

For fixed finite velocities, the nonrelativistic limit $c\to \infty$ gives ${\gamma }^2=(1-v^2/c^2{)}^{-1}\to 1$. Hence the relativistic rocket equation reduces towhich is the nonrelativistic rocket equation.

The conserved quantities called energy in the two calculations include different contributions. The nonrelativistic calculation considered only kinetic energy, which increases as fuel releases internal energy; total energy including that internal reservoir is still conserved. Relativistic energy includes rest energy, and therefore includes the fuel's stored internal energy through the rocket's rest mass. That rest mass decreases as energy is transferred into the exhaust. Thus conservation of relativistic energy and the nonconservation of kinetic energy express the same physical conversion.

Write the total remaining rocket mass as $\mathcal{M}=M+m(t)$, with $\dot{m}=-\alpha$. Over an interval $dt$, an exhaust mass $d{m}_e=\alpha dt$ is expelled backwards with inertial velocity $v-u$ to first order. Applying momentum conservation to the rocket and this exhaust givesCancellation and division by $dt$ give the rocket equationWith $M>0$, $m(t)=m_0-\alpha t$, and $0\le t\le m_0/\alpha$, integration givesAt burnout,The logarithm depends on the mass ratio, not on the burn rate. This follows from momentum conservation of the complete rocket–exhaust system; applying a constant-mass equation directly to the rocket alone would omit the momentum carried away.

For $0\le k<1$, the first burn reduces the total mass from $M+m_0$ to $M+(1-k)m_0$. The rocket equation therefore givesDetachment exerts negligible impulse, so the surviving second stage keeps velocity $v_1$. Its initial and final masses are $(1-k)(M+m_0)$ and $(1-k)M$, respectively. A second application of the rocket equation givesFor the single-stage comparison from rest, $v_{\mathrm{single}}=u\log (1+m_0/M)$. HenceEquality holds at $k=0$; for positive fuel and exhaust speed it is strict when $0<k<1$. This two-stage rocket with proportional fuel and body masses gains speed by discarding inert body mass before the last burn.

As $k\to 1^{-}$ the expression tends to $2u\log (1+m_0/M)$, but the surviving second-stage mass also tends to zero. At exactly $k=1$ there is no second stage, so its final speed is undefined and that limiting formula cannot be assigned to it. The first stage alone reaches the single-stage burnout speed before detachment. If $m_0=0$, neither design gains speed.

Let $M(t)=m_0-\alpha t$ be the trolley's remaining mass. During $dt$, the expelled mass $-dM$ has ground-frame velocity $v+u$. Conservation of momentum gives, to first order,and hence $Mdv=udM$. Integration yields the rocket equationAt the stopping time, $M=m_0{e}^{-v_0/u}$. Therefore the time spent ejecting gas is

Choose positive x along the rocket's motion. If its instantaneous total mass is $\mu$ and it ejects a positive amount $d\eta =-d\mu$, the exhaust velocity in the original inertial frame is $v-u_0$ to first order. Conservation of momentum over that increment givesIntegrating the rocket equation from $\mu =m+M$ to $\mu =m$ givesThe constant exhaust speed is relative to the rocket, not the initial frame. Kinetic energy of rocket plus exhaust is not conserved because stored fuel energy is released. The calculation is Newtonian and requires resulting speeds small compared with the speed of light.

Put $m_A=M=m$ and $m_B=2m$. At reversal the transferred mass is $q_r=0.45m$, leaving A with $1.55m$ and, after capture of that exhaust, B with $2.45m$. The first burn givesAfter reversal A emits its remaining exhaust away from B, so B receives no further fuel and eventually coasts at this velocity. Applying the rocket equation with reversed thrust to A givesBoth now move toward negative x, but A moves faster in that direction. The asymptotic closing speed isThis conclusion does not require a burn rate or an initial separation: after a finite burn, the positive eventual closing speed closes any finite remaining gap. Those additional inputs would be required to calculate the meeting time. Any pre-reversal exhaust still in transit affects the timing, not B's eventual momentum after collecting that fixed amount.

Let $u_e>0$ be the constant rearward rocket exhaust speed, and let $\mathcal{M}$ denote the rocket's instantaneous mass. Apply momentum conservation over a short ejection interval: the remaining rocket gains $\mathcal{M}dv$, while the expelled positive mass $-d\mathcal{M}$ carries backward relative momentum $u_{e}d\mathcal{M}$. To first order,Integrating gives the rocket equationThis is a velocity increment, measured along the thrust direction, for motion in free space without external forces.

For the two-stage vehicle, take $M_1$ and $M_2$ to be the initial masses of the respective stages, each including its fuel $m$. The first burn changes the total mass from $M_1+M_2$ to $M_1+M_2-m$. Discarding the empty first-stage shell of mass $M_1-m$ leaves mass $M_2$; ideal jettison without a relative impulse does not change the velocity. The second burn then changes $M_2$ to $M_2-m$. Adding the two rocket equation increments givesHere $M_1>m$ and $M_2>m$ leave positive dry masses. Different stage exhaust speeds would multiply their respective logarithms. A multistage rocket gains its advantage by avoiding acceleration of the spent first-stage shell during the second burn.

Let $\mu$ be the instantaneous total mass of the rocket, including its remaining rocket propellant. During a short ejection it changes by $d\mu <0$. The expelled mass $-d\mu$ has velocity $v-v_0$ to first order in a fixed inertial frame. With no external force, momentum conservation readsDiscarding the second-order product gives $\mu dv=-v_{0}d\mu$. Integrate from initial total mass $m_0+M$ to rocket dry mass $m_0$:For constant mass flow rate $\alpha =-\dot{\mu }>0$, the same rocket equation gives $\mu (t)=m_0+M-\alpha t$ and $v(t)=v_0\ln [(m_0+M)/\mu (t)]$ until $t=M/\alpha$. The final result depends on the rocket mass ratio, not on how rapidly the fuel is ejected. These are Newtonian velocities; a very large rocket mass ratio would require a relativistic treatment even if the exhaust is slow in the instantaneous rocket frame.