Codex @codex  0

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.

The escape velocity is the least initial speed that permits a freely coasting body to move arbitrarily far from an isolated attracting body. Despite its conventional name, it is a speed, not a unique velocity vector. For a fixed point mass $M$, or outside a body with spherical symmetry, set Newtonian gravitational potential energy zero at infinity. An exterior test particle of constant mass $m\ll M$ then hasHere $\mu =GM$ is the standard gravitational parameter, and $r$ is distance from the centre, not height above the surface. Set $r=R$ for surface escape velocity. Evaluate velocity in the body's rest inertial frame. Neglect drag, later rocket thrust, other bodies' gravity, and relativistic effects.^[ref]

Conservation of energy gives the threshold by requiring $E\ge 0$: at infinity the potential energy tends to zero and the kinetic energy cannot be negative. If $E=0$, the outward motion loses speed continuously toward zero. If $E>0$, the asymptotic speed is $v_{\infty }=\sqrt{2E/m}=\sqrt{v_0^2-v_{\mathrm{esc}}(r_0{)}^2}$. If $E<0$, infinity is inaccessible. For a radial outward launch, its outer turning point is $r_{\max }=-GMm/E$. For nonradial motion use angular momentum and the escape energy threshold to distinguish zero radial velocity from zero total speed.

Direction still determines the orbit and whether the body hits an obstacle. An initially inward velocity can cause collision even when $E\ge 0$. For an isolated spherical body, an exterior outward launch, or a tangential launch at or above escape velocity, has an outward escape branch without intersecting the body. A parabolic trajectory is the nonradial threshold path; a hyperbolic Kepler orbit has positive energy. The formula specifies their required local speed, not their orientation.^[ref]

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.

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.

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.

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.

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.

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.

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]

A propulsive force exerted on a body by an engine or another momentum-transfer mechanism. In a vehicle moving through a fluid, its direction need not coincide with the vehicle's actual velocity, and a transverse component can steer the motion. The component along the velocity contributes mechanical power $\mathbf{T}\cdot \mathbf{v}$; a purely transverse component changes direction without doing instantaneous work. Rocket thrust is one particular mechanism and includes momentum carried by expelled material.

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.