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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.

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.

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 $u_e>0$ 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.

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.

The specified rocket fuel fraction gives $m(T)=2m_0/3$, so $\alpha T=m_0/3$. Imposing $v(T)=0$ yieldsThen $h(T)=0$ fixesNumerically,The associated constant mass loss rate is $\alpha =m_0/(3T)\simeq 88.78\mathrm{kg}{\mathrm{s}}^{-1}$. The initial upward thrust acceleration already exceeds $g$ and increases as mass falls, so the downward speed decreases monotonically to zero. This calculation assumes constant lunar gravitational acceleration, constant relative rocket exhaust speed, and negligible drag.

Take the rocket exhaust speed $u$ relative to the hovering ship, with exhaust directed vertically downward. If $\alpha =-\dot{M}>0$ is the mass flow rate, the upward rocket thrust is $u\alpha$. For example, over a short interval $dt$, the expelled mass $\alpha dt$ acquires downward momentum $-u\alpha dt$, giving the opposite recoil to the ship. Balancing its weight requiresSolving this differential equation with $M(0)=M_0$ yields the hovering rocket mass flowThe required mass flow rate decreases because the mass to support decreases. These expressions apply while fuel is available and $g,u$ remain constant. If the rocket dry mass is $M_d>0$, hovering ends at $t=(u/g)\log (M_0/M_d)$; the mathematical limit $M\to 0$ is not a model of an indefinitely fuelled physical ship.

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 $v(t)=v(T)-g(t-T)$.

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.

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.

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]