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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 isThe pressure term motivates an 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 $I_{\mathrm{sp}}=v_e/g_0$, where $g_0=9.80665\mathrm{m}{\mathrm{s}}^{-2}$ is conventional standard gravity; using this unit does not imply that the engine must operate near Earth.^[ref]

For an illustrative pressure-matched engine, $\dot{m}=2\mathrm{kg}{\mathrm{s}}^{-1}$ and $u_e=3000\mathrm{m}{\mathrm{s}}^{-1}$ 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 $2079\mathrm{m}{\mathrm{s}}^{-1}$. Thus higher effective exhaust velocity increases the velocity change available from a specified propellant fraction; it does not by itself specify a large thrust.

Specific impulse measures the impulse delivered by a rocket per unit propellant consumed. In its conventional seconds form, divide total axial thrust impulse by propellant mass $m_p$ and standard gravity $g_0$:For constant conditions this becomes $I_{\mathrm{sp}}=F/(\dot{m}{g}_0)$, where $\dot{m}$ is the positive expelled mass flow rate. Multiplying by $g_0$ gives the effective exhaust velocity. The conventional value $g_0=9.80665\mathrm{m}{\mathrm{s}}^{-2}$ is a unit convention, not a requirement that the engine operate near Earth.^{[ref] [ref]}

For a worked example, a steady $6000\mathrm{N}$ engine consuming $2\mathrm{kg}{\mathrm{s}}^{-1}$ has $I_{\mathrm{sp}}\approx 305.9\mathrm{s}$ and effective exhaust velocity $3000\mathrm{m}{\mathrm{s}}^{-1}$. A ten-second firing supplies $60000\mathrm{N}\mathrm{s}$ using $20\mathrm{kg}$ of propellant. The unit of seconds does not mean that this engine can burn for 305.9 seconds: duration also depends on the available propellant and its flow rate.

In a laboratory illustration, integrate the force recorded on a calibrated engine stand and measure the mass consumed during the same firing. Report the ambient pressure, since nozzle pressure forces contribute to thrust. At fixed specific impulse, doubling the mass flow rate doubles thrust; a large specific impulse alone does not specify a large force. For varying performance, the burn-integrated value is weighted by propellant flow, rather than generally equaling a simple time average.