= Exhaust velocity
{title2=$v_e$}
<Exhaust velocity> is the <velocity> of expelled propellant measured relative to the engine. In a simple straight jet it points backward, while the engine's <thrust> points forward. Rocket calculations often use its positive magnitude, as in <rocket exhaust speed>. The relative frame matters: the propellant can be moving forward relative to a distant observer while still leaving the engine backward.
For a steady rocket nozzle with outward <mass flow rate> $\dot m>0$, exit speed $u_e$, exit <pressure> $p_e$, ambient <pressure> $p_a$ and exit <area> $A_e$, the axial <thrust> is
$$
F=\dot m u_e+(p_e-p_a)A_e.
$$
The pressure term motivates an \i[effective] <exhaust velocity> $v_e=F/\dot m$, which includes both momentum flux and nozzle pressure effects. When exit and ambient <pressures> match, $v_e=u_e$. <Specific impulse> in seconds is $I_{\rm sp}=v_e/g_0$, where $g_0=9.80665\,\mathrm{m\,s^{-2}}$ is conventional standard gravity; using this unit does not imply that the engine must operate near Earth.<Specific Impulse by NASA Glenn Research Center>[Thrust Equation and Specific Impulse sections; use the thrust equation before the intermediate division]{ref}
For an illustrative pressure-matched engine, $\dot m=2\,\mathrm{kg\,s^{-1}}$ and $u_e=3000\,\mathrm{m\,s^{-1}}$ give $F=6000\,\mathrm N$. At the same effective <exhaust velocity>, halving the <mass flow rate> halves <thrust>. For an ideal burn without external forces and with constant $v_e$, the <rocket equation> gives $\Delta v=v_e\ln(m_0/m_f)$: a mass ratio of $2$ yields approximately $2079\,\mathrm{m\,s^{-1}}$. Thus higher effective <exhaust velocity> increases the velocity change available from a specified propellant fraction; it does not by itself specify a large <thrust>.
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