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Loss of matter from a star or other astronomical system, for example through a stellar wind or explosive ejection. Mass loss rate conventions must distinguish the positive ejected rate $|\dot{M}|$ from the negative change of the retained mass, $\dot{M}$. In a binary star, the fate of an orbit also depends on the specific angular momentum carried away.

A stellar wind carries matter away from a star, altering its mass, angular momentum and surroundings. Hot coronal gas can escape through pressure-driven expansion; radiation transfers momentum to gas through continuum or spectral line absorption, or to dust that then drags gas. A magnetic field can transfer rotational energy and angular momentum to an outflow. These mechanisms need not give the same terminal speed or mass loss rate.

For a steady spherical pressure-driven model, conservation of mass gives $\dot{M}=4\pi r^2\rho u$. Combining the radial Euler equation with an isothermal equation of state $p=\rho a^2$ gives the Parker wind equationA smooth transonic branch passes through $u=a$ at $r_s=GM/(2a^2)$, connecting a subsonic inner atmosphere to an escaping supersonic wind. The isothermal Parker wind requires continuing heat input as the gas expands; a polytropic stellar wind instead tracks a changing temperature and must satisfy its own terminal-energy condition. A subsonic isothermal breeze generally tends to a nonzero external pressure, so outward velocity alone does not select the escaping solution.

The solar wind supplies a nearby example of an outward plasma flow carrying a magnetic field. Profiles of spectral lines, especially absorption shifted by the Doppler effect together with emission, diagnose more distant stellar winds. The critical-point condition and boundary heating constrain $\dot{M}$; it is not determined by an escape-speed estimate alone.

The outward mass loss rate is $\dot{M}=4\pi r^2\rho u>0$. Put $\beta =4/(\gamma +1)$, $\alpha =4(\gamma -1)/(\gamma +1)$ and $y=\rho r^{\beta }$. Both terms in the Bernoulli equation then have the same radial scaling: $\beta (\gamma -1)=4-2\beta =\alpha$. Eliminate $u$ and multiply by $r^{\alpha }$ to obtainFor positive terminal speed, $C>0$. Differentiation gives $F^{\prime }=r^{\alpha -2}[\alpha Cr-(1-\alpha )GM]$. It has an interior minimum precisely when $0<\alpha <1$, or $1<\gamma <5/3$, atAt a sonic point, the Bernoulli equation and $GM/r_c=2c_{s,c}^2$ give $C=c_{s,c}^2(5-3\gamma )/[2(\gamma -1)]$, recovering this radius. The limiting $C=0$, $\gamma =5/3$ case has no isolated interior minimum selected by these formulas.

The function $G$ diverges at both ends of $y>0$ and has a unique minimum whereThis condition is $u_c^2=c_{s,c}^2$. Thus a smooth crossing requires minimum matching at a polytropic sonic point: $F(r_c)=G(y_c)$. If the minimum of $F$ is lower, an interval has no real positive solution; if it is higher, the two $y$ branches stay separate and cannot cross the sonic value. When the minima agree, their positive second derivatives give $G^{\prime \prime }(y_c)(y-y_c{)}^2\sim F^{\prime \prime }(r_c)(r-r_c{)}^2$, allowing a finite-slope branch to pass between them. The wind with finite positive terminal speed takes the lower-$y$ branch outside $r_c$, because $\rho \sim \dot{M}/(4\pi \sqrt{2C}{r}^2)$ and $\beta <2$ make $y\to 0$.

At the matching point,Insert these into $\dot{M}=4\pi r_c^2{\rho }_c{c}_{s,c}$:The positive-$C$, $1<\gamma <5/3$ range is part of this finite-radius transonic wind result, rather than an unrestricted formula for every $\gamma >1$.

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.

A steady spherical polytropic flow escaping a point-mass potential conserves the outward mass loss rate and the Bernoulli function $C=u^2/2+c_s^2/(\gamma -1)-GM/r$. For positive terminal speed, a finite-radius smooth sonic point requires $C>0$ and $1<\gamma <5/3$. Its sound speed obeys $c_{s,c}^2=2C(\gamma -1)/(5-3\gamma )$, and evaluating the conserved flux there selects the transonic mass loss rate. Other parameter limits need separate treatment.

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 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]