Visual walkthrough — Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ

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Step 1 — Meet the rocket and its exhaust

WHAT. We freeze a rocket in space and mark two arrows: gas going left (backward), rocket going right (forward).

WHY. Everything about thrust is a story of momentum being handed from the gas to the rocket. Before we can measure that hand-off we must agree on directions and names. We call forward (the way the rocket moves) the positive direction.

PICTURE. Look at figure s01. The amber shape is the rocket. The cyan arrow leaving the mouth is the escaping gas at speed $v_e$. Notice they point opposite ways — that is the whole secret of rockets.

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Step 2 — What a "rate" means: reading $\dot{m}$

WHAT. In a tiny slice of time $dt$ (read: "a very short moment"), a little chunk of gas of mass $dm=\dot{m}dt$ squirts out.

WHY this tool — why a rate and not just "a mass"? A rocket does not fire one lump and stop; it pours gas out continuously. The natural question is therefore "how much per second?", and the answer is a rate. Multiplying a rate ($\dot{m}$, kg per second) by a duration ($dt$, seconds) gives an actual mass ($dm$, kg) — the seconds cancel.

$\underset{\text{small mass, kg}}{\underbrace{dm}}=\underset{\text{kg per second}}{\underbrace{\dot{m}}}\cdot \underset{\text{small time, s}}{\underbrace{dt}}$

PICTURE. Figure s02 shows a stopwatch and a small cyan blob $dm$ leaving during the interval $dt$. As the clock runs, blob after blob departs — that steady drip is $\dot{m}$.

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Step 3 — Momentum thrust: the shove from moving gas

WHAT. Each departing blob $dm$ carries backward momentum $dm\cdot v_e$. Newton's law says a force is just momentum handed over per second.

WHY this tool — why momentum and not "just a push"? Because a push you can feel only exists as long as something is changing its motion. Newton's second law states the force equals the rate of change of momentum, and Newton's third law says the gas's backward momentum is matched by the rocket's forward momentum. So the cleanest way to count the forward shove is to count the backward momentum leaving per second.

$F_{\text{mom}}=\frac{d{p}_{\text{gas}}}{dt}=\frac{v_{e}dm}{dt}=\underset{\text{kg/s}}{\underbrace{\dot{m}}}\underset{\text{m/s}}{\underbrace{v_e}}\Rightarrow \text{units }\frac{\text{kg}}{\text{s}}\cdot \frac{\text{m}}{\text{s}}=\text{N}$

Here $p_{\text{gas}}$ is the gas's momentum; $\frac{dp}{dt}$ reads "how fast that momentum piles up each second."

PICTURE. Figure s03: the cyan gas fires left with a long backward-momentum arrow; an equal-length amber arrow shows the reaction shoving the rocket right. Same length = equal and opposite = Newton's third law made visible.

Recall Check the units yourself

Why is $\dot{m}{v}_e$ measured in newtons? ::: $\frac{\text{kg}}{\text{s}}\times \frac{\text{m}}{\text{s}}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}$, and $1\text{N}=1\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}$. So it is a force.

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Step 4 — Why there is a second push at all: pressure

WHAT. At the exit mouth the exhaust presses outward with pressure $P_e$. The surrounding air presses inward with pressure $P_a$ ("$a$" for ambient, the air around the rocket).

WHY this matters — why isn't the gas pressure already zero? A perfect nozzle would keep expanding the gas until its pressure exactly equalled the outside air. But real nozzles have a finite length and must stop somewhere. Often the gas is still slightly squeezed when it reaches the lip: $P_e\ne P_a$. That leftover pressure difference pushes on the exit ring — and there is no nozzle wall there to cancel it, so it leaks out as extra thrust.

PICTURE. Figure s04: at the mouth, cyan arrows push outward (labelled $P_e$) and white arrows push inward (labelled $P_a$). Along the solid nozzle walls the inward and outward pushes cancel — but at the open mouth they do not.

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Step 5 — Pressure thrust: turning a squeeze into a force

WHAT. Force from pressure = pressure $\times$ area. The net outward pressure at the mouth is $(P_e-P_a)$, and it acts over the exit area $A_e$.

WHY subtract? Because the inward air push $P_a$ genuinely opposes the outward gas push $P_e$. Only the leftover, $P_e-P_a$, survives to shove the rocket. Multiplying that leftover pressure by the area it acts on converts "force per square metre" into an actual force.

$F_{\text{pres}}=\underset{\text{net Pa}}{\underbrace{(P_e-P_a)}}\underset{{\text{m}}^2}{\underbrace{A_e}}\Rightarrow \frac{\text{N}}{{\text{m}}^2}\cdot {\text{m}}^2=\text{N}$

• $P_e$ — exit gas pressure (how squeezed the exhaust still is).
• $P_a$ — ambient air pressure pushing back.
• $A_e$ — the ring the leftover push acts on.

PICTURE. Figure s05 fades away the cancelling walls and keeps only the surviving amber arrow $(P_e-P_a)A_e$ acting forward on the ring — pressure thrust, alone.

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Step 6 — Add the two shoves

WHAT. Total thrust is the momentum shove plus the pressure shove.

WHY add? They are two independent forward forces on the same body during the same second, so they simply sum — like two people pushing a stalled car from behind.

$F=\underset{\text{momentum thrust}}{\underbrace{\dot{m}{v}_e}}+\underset{\text{pressure thrust}}{\underbrace{(P_e-P_a)A_e}}$

PICTURE. Figure s06 stacks the two amber arrows tip-to-tail; their combined length is the full thrust $F$.

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Step 7 — Fold everything into one velocity $c$

WHAT. Set $F=\dot{m}c$ and solve for $c$ by dividing the Step 6 equation by $\dot{m}$.

WHY divide by $\dot{m}$? Because we want a speed. Thrust divided by mass-flow ($\frac{\text{N}}{\text{kg/s}}$) gives $\frac{\text{m}}{\text{s}}$ — exactly the units of velocity. The division cleanly repackages a force plus a pressure term into one speed.

$c=\frac{F}{\dot{m}}=\underset{\text{real gas speed}}{\underbrace{v_e}}+\underset{\text{pressure written as extra speed}}{\underbrace{\frac{(P_e-P_a)A_e}{\dot{m}}}}$

$\boxed{c=v_e+\frac{(P_e-P_a)A_e}{\dot{m}}}$

Why bother? Because now $I_{sp} = c/g_0$ and the Tsiolkovsky equation $\Delta v = c\ln(m_0/m_f)$ each use ONE number instead of four.

PICTURE. Figure s07: the two-arrow stack of Step 6 morphs into a single long cyan arrow labelled $c$ — same total length, one clean vector.

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Step 8 — All the cases: sign of the pressure term

The pressure term $\frac{(P_e-P_a)A_e}{\dot{m}}$ can be positive, zero, or negative. Because $A_e$ and $\dot{m}$ are always positive, its sign is decided entirely by $(P_e-P_a)$ — set by Nozzle Expansion (Under/Over/Optimal).

WHAT / WHY / PICTURE for each case (figure s08 shows all three side by side):

• Optimally expanded — $P_e=P_a$: the leftover push is exactly zero, so $c=v_e$. The make-believe speed equals the real speed. (Middle panel: arrows cancel, $c$ arrow = $v_e$ arrow.)
• Underexpanded — $P_e>P_a$: gas still squeezed, leftover push forward, $c>v_e$. Extra thrust. Under → Up. (Left panel: amber arrow adds on.)
• Overexpanded — $P_e<P_a$: outside air wins, leftover push backward, $c<v_e$. Thrust penalty and possible flow separation. Over → Off. (Right panel: amber arrow points back, shortening $c$.)

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Step 9 — Degenerate and limiting checks

WHAT / WHY. A good formula must survive extreme inputs. Figure s09 plots $c$ versus altitude for one fixed engine to show these limits at a glance.

• Vacuum limit ($P_a\to 0$): the back-push disappears, so $c$ reaches its maximum, $c=v_e+\frac{P_e{A}_e}{\dot{m}}$. This is the top of the s09 curve.
• Ground limit (dense air, large $P_a$): the term is most negative, so $c$ is smallest. Bottom of the curve.
• $\dot{m}\to 0$ (engine off): the formula $c=F/\dot{m}$ divides by zero — meaningless, because with no gas flowing there is no "exhaust velocity" to speak of. The equation only applies while the engine is firing.
• $A_e=0$ (no opening): pressure term vanishes; but then $\dot{m}$ would also be zero — degenerate, no rocket.

${\lim }_{P_a\to 0}c=v_e+\frac{P_e{A}_e}{\dot{m}}\text{(highest }c\text{)}$

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The one-picture summary

Figure s10 compresses the whole story: a rocket with its exit mouth; the cyan momentum arrow $\dot{m}{v}_e$; the amber pressure arrow $(P_e-P_a)A_e$; their tip-to-tail sum $F$; and the single merged $c$ arrow, all labelled — the derivation in one glance.

Recall Feynman retelling of the walkthrough

A rocket shoots gas out the back. The rushing gas carries "amount of motion" (momentum) away every second, and by reaction it shoves the rocket forward — that is the first push, $\dot{m}{v}_e$. But the gas leaving the mouth is often still a little squeezed, harder than the outside air right at the opening. That extra squeeze presses on the ring of the mouth with nothing to cancel it, giving a second push, $(P_e-P_a)A_e$. Add the two shoves and you have the total thrust. To keep the maths tidy, we pretend the gas actually came out at one make-believe speed $c$ that gives the same total push. Up high where the air is thin, nothing pushes back, so $c$ grows and the rocket pushes harder. Down low, the thick air fights back and $c$ shrinks. Turn the engine off and the whole idea of "$c$" evaporates — no gas, no exhaust speed.

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Connections

• Thrust Equation — the parent; $F=\dot{m}{v}_e+(P_e-P_a)A_e$ derived here in pictures.
• Conservation of Momentum — the engine behind Step 3's momentum thrust.
• Nozzle Expansion (Under/Over/Optimal) — fixes the sign in Step 8.
• Atmospheric Pressure vs Altitude — why Step 9's curve slopes upward.
• Specific Impulse — $I_{sp}=c/g_0$, uses the $c$ we just built.
• Tsiolkovsky Rocket Equation — $\Delta v=c\ln (m_0/m_f)$.
• Yeh page Hinglish mein →

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