= Rocket equation
= Tsiolkovsky rocket equation
{c}
{synonym}
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> $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,
$$
M\,dv=-u\,dM,\qquad
\Delta v=v_{\mathrm{final}}-v_{\mathrm{initial}}
=u\ln\frac{M_{\mathrm{initial}}}{M_{\mathrm{final}}}.
$$
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>. https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/ideal-rocket-equation/[NASA Glenn Research Center: Ideal Rocket Equation]{ref}
For a toy rocket with $u=2\,\mathrm{km/s}$, $M_{\mathrm{initial}}=1000\,\mathrm{kg}$ and <rocket dry mass> $M_{\mathrm{final}}=250\,\mathrm{kg}$ after all <rocket propellant> is spent, the <rocket mass ratio> is $4$ and
$$
\Delta v=2\ln4\,\mathrm{km/s}\simeq2.7725887\,\mathrm{km/s}.
$$
Burning from $1000$ to $500\,\mathrm{kg}$ gains $2\ln2\,\mathrm{km/s}\simeq1.3863\,\mathrm{km/s}$. 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.
\Image[/rocket-mass-ratio-delta-v.png]
{title=Equal mass halvings give equal ideal velocity gains for an exhaust speed of 2 kilometres per second}
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 $u\ln2$ 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> gives
$$
Mv=(M+dM)(v+dv)+(-dM)(v-u),\qquad
0=M\,dv+u\,dM+dM\,dv.
$$
Discarding the second-order product $dM\,dv$ gives $M\,dv=-u\,dM$; 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.
\Image[/rocket-momentum-ejection.png]
{title=Momentum balance for a tiny expelled parcel: rearward relative exhaust can still move forward in the inertial frame}
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 $\Delta v=-\int u\,dM/M$, 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.
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