Thrust equation F = ṁv_e + (P_e − P_a)A_e — derivation: Worked examples

This page hammers the thrust equation $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ through every case it can hand you. Before we compute anything, let's lay out the full map so no scenario surprises you later.

Recall What each symbol means (from the

parent)

$\overset{m}{˙}$ = mass flow rate of exhaust, in kg/s (how many kilograms leave the nozzle each second).
$v_{e}$ = exhaust speed relative to the rocket, in m/s.
$P_{e}$ = static pressure of gas at the exit plane, in Pa.
$P_{a}$ = ambient (outside air) pressure, in Pa.
$A_{e}$ = area of the nozzle exit hole, in m².
$F$ = net forward thrust, in newtons (N).

The scenario matrix

Every problem this topic throws lives in one of these cells. Read the "sign of pressure term" column as the story: is the exit gas over-pressured, under-pressured, or balanced against the outside?

#	Cell class	Condition	Sign of $(P_{e} - P_{a}) A_{e}$	Covered by
1	Overexpanded	$P_{e} < P_{a}$	negative (loss)	Ex 1
2	Underexpanded	$P_{e} > P_{a}$	positive (bonus)	Ex 2
3	Perfectly expanded	$P_{e} = P_{a}$	zero	Ex 3
4	Vacuum (degenerate $P_{a} = 0$ )	$P_{a} = 0$	max positive	Ex 4
5	Zero pressure thrust from zero area (degenerate)	$A_{e} \to 0$	zero regardless of $P$	Ex 5
6	Limiting behaviour: thrust vs altitude	$P_{a}$ decreasing	rises as you climb	Ex 6
7	Reverse-solve (find $v_{e}$ or $\overset{m}{˙}$ from $F$ )	given $F$	either	Ex 7
8	Real-world word problem	sea-level launch	negative-ish	Ex 8
9	Exam twist: effective exhaust velocity	uses $v_{e f f}$	folded into one number	Ex 9

Each example below is tagged with the cell(s) it fills.

Worked examples

Example 1 — Overexpanded nozzle (Cell 1)

Worked example Statement

A first-stage engine at sea level: $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $P_{e} = 80$ kPa, $P_{a} = 101$ kPa, $A_{e} = 1.2 m^{2}$ . Find $F$ .

Forecast: The exit pressure (80 kPa) is below the outside air (101 kPa). Guess: pressure thrust is negative, so $F$ comes out a little below the pure momentum thrust. Jot a number before reading on.

Step 1 — Momentum thrust. $\overset{m}{˙} v_{e} = 300 \times 2800 = 840, 000 N$ Why this step? This is the Newton's-3rd-law core: mass thrown per second times how fast it leaves. It's always the biggest term.

Step 2 — Pressure thrust. $(P_{e} - P_{a}) A_{e} = (80, 000 - 101, 000) (1.2) = (- 21, 000) (1.2) = - 25, 200 N$ Why this step? Since $P_{e} < P_{a}$ , the outside atmosphere presses harder on the exit region than the exhaust does — a net backward push. That's the overexpanded penalty.

Step 3 — Add them. $F = 840, 000 - 25, 200 = 814, 800 N \approx 814.8 kN$

Verify: Units of momentum thrust: $(kg/s) (m/s) = kg\cdotm/s^{2} = N$ ✓. Units of pressure thrust: $(Pa) (m^{2}) = (N/m^{2}) (m^{2}) = N$ ✓. The result sits just below 840 kN as forecast — the atmosphere stole 25.2 kN.

Example 2 — Underexpanded nozzle (Cell 2)

Worked example Statement

Same $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $A_{e} = 1.2 m^{2}$ , but at high altitude where $P_{a} = 30$ kPa and the exit gas is $P_{e} = 80$ kPa. Find $F$ .

Forecast: Now $P_{e} > P_{a}$ . Guess: pressure thrust flips to positive — a bonus — so $F$ exceeds 840 kN.

Step 1 — Momentum thrust (unchanged, $\overset{m}{˙}$ and $v_{e}$ same): $\overset{m}{˙} v_{e} = 840, 000 N$ Why this step? The engine hasn't changed its throwing rate; only the surrounding air changed.

Step 2 — Pressure thrust. $(P_{e} - P_{a}) A_{e} = (80, 000 - 30, 000) (1.2) = (50, 000) (1.2) = 60, 000 N$ Why this step? Exit gas now pushes forward harder than the thin outside air pushes back → net forward force.

Step 3 — Add. $F = 840, 000 + 60, 000 = 900, 000 N = 900 kN$

Verify: Same engine, thinner air → bigger thrust, matching the forecast. Compare to Example 1 (814.8 kN at sea level): climbing gained $900 - 814.8 = 85.2$ kN. That's the whole reason rockets thrust harder as they rise.

Example 3 — Perfect expansion (Cell 3)

Example 4 — Vacuum, the degenerate $P_{a} = 0$ (Cell 4)

Worked example Statement

Take the engine to deep space: $P_{a} = 0$ . Keep $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $P_{e} = 80$ kPa, $A_{e} = 1.2 m^{2}$ . Find $F$ .

Forecast: With no outside air to push back, the full exit pressure now contributes forward. Guess: this is the largest thrust of all our cases.

Step 1 — Pressure thrust with $P_{a} = 0$ . $(P_{e} - 0) A_{e} = 80, 000 \times 1.2 = 96, 000 N$ Why this step? Setting $P_{a} = 0$ removes the subtracted back-push entirely — this is the physical meaning of "vacuum." Nothing outside opposes the exit gas.

Step 2 — Total. $F = 840, 000 + 96, 000 = 936, 000 N = 936 kN$

Verify: This exceeds every earlier case (814.8, 900, 840 kN) ✓. Going from sea level (Ex 1) to vacuum, the same engine gains $936 - 814.8 = 121.2$ kN — that's the whole $P_{a} A_{e} = 101, 000 \times 1.2 = 121, 200$ N the atmosphere used to steal. This is why the linked Specific Impulse of an engine is quoted higher "in vacuum."

Example 5 — Zero exit area, the degenerate $A_{e} \to 0$ (Cell 5)

Worked example Statement

A thought experiment: shrink the nozzle exit to a pinhole, $A_{e} \to 0$ , while somehow keeping $\overset{m}{˙} = 300$ kg/s and $v_{e} = 2800$ m/s. What happens to the pressure term? Take $P_{e} = 80$ kPa, $P_{a} = 101$ kPa.

Forecast: The pressure term has $A_{e}$ as a factor. Guess: no matter what the pressures are, if $A_{e} = 0$ the whole pressure thrust is zero.

Step 1 — Pressure thrust. $(P_{e} - P_{a}) A_{e} = (80, 000 - 101, 000) (0) = 0 N$ Why this step? Pressure is force per area; with zero area there's no area for it to act on. The degenerate case collapses the whole second term regardless of the pressure mismatch.

Step 2 — Total. $F = 840, 000 + 0 = 840, 000 N = 840 kN$

Verify: A physically real nozzle can't hold this flow through zero area, but the equation behaves sensibly: $A_{e} = 0 \Rightarrow$ only momentum thrust survives. This confirms the pressure thrust is entirely an exit-area phenomenon, exactly as the parent derivation argued (net pressure survives only over the open exit).

Example 6 — Limiting behaviour: thrust vs altitude (Cell 6) 📈

Worked example Statement

Fix the engine ( $\overset{m}{˙} = 300$ kg/s, $v_{e} = 2800$ m/s, $P_{e} = 80$ kPa, $A_{e} = 1.2 m^{2}$ ) and let $P_{a}$ fall as the rocket climbs: $P_{a} = 101, 70, 30, 0$ kPa. Tabulate $F$ and describe the trend.

Forecast: Lower $P_{a}$ → smaller subtraction → larger $F$ . Guess a smooth, straight-line rise.

Step 1 — Write $F$ as a function of $P_{a}$ . $F (P_{a}) = \overset{m}{˙} v_{e} 840, 000 + (80, 000 - P_{a}) (1.2)$ Why this step? Only $P_{a}$ changes, so isolate it. Everything else is a constant. This makes $F$ a linear function of $P_{a}$ with slope $- A_{e} = - 1.2$ N per Pa.

Step 2 — Evaluate at each altitude.

$P_{a}$ (kPa)	pressure thrust (N)	$F$ (kN)
101	$(80 - 101) (1.2) \times 1 0^{3} = - 25, 200$	814.8
70	$(80 - 70) (1.2) \times 1 0^{3} = + 12, 000$	852.0
30	$(80 - 30) (1.2) \times 1 0^{3} = + 60, 000$	900.0
0	$(80 - 0) (1.2) \times 1 0^{3} = + 96, 000$	936.0

Why this step? Sweeping $P_{a}$ from sea level to vacuum shows every sign of the pressure term in one shot: negative at 101, zero somewhere near 80, positive above.

Look at the figure: the red line is straight, crossing from below the dashed "momentum-only" line to above it exactly at $P_{a} = P_{e} = 80$ kPa.

Verify: The crossing point is where pressure thrust $= 0$ , i.e. $P_{a} = P_{e} = 80$ kPa, giving $F = 840$ kN — matching Example 3's logic. Slope check: from $P_{a} = 101$ to $P_{a} = 0$ , $F$ rose by $936 - 814.8 = 121.2$ kN over a $101$ kPa drop, i.e. $121, 200/101, 000 = 1.2$ N/Pa $= A_{e}$ ✓.

Example 7 — Reverse-solve for exhaust velocity (Cell 7)

Worked example Statement

An engine must deliver $F = 750$ kN at sea level ( $P_{a} = 101$ kPa). Its nozzle gives $P_{e} = 95$ kPa, $A_{e} = 0.9 m^{2}$ , and $\overset{m}{˙} = 260$ kg/s. What exhaust velocity $v_{e}$ is required?

Forecast: We're running the equation backward. Guess $v_{e}$ near 3000 m/s.

Step 1 — Rearrange the thrust equation for $v_{e}$ . $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e} \Rightarrow v_{e} = \frac{F - ( P _{e} - P _{a} ) A _{e}}{m ˙}$ Why this step? $v_{e}$ sits inside the momentum term multiplied by $\overset{m}{˙}$ ; isolate it by subtracting the (known) pressure term and dividing.

Step 2 — Compute the pressure term. $(P_{e} - P_{a}) A_{e} = (95, 000 - 101, 000) (0.9) = (- 6, 000) (0.9) = - 5, 400 N$ Why this step? Slightly overexpanded, so it's a small negative — the momentum term must make up for it.

Step 3 — Solve. $v_{e} = \frac{750 , 000 - ( - 5 , 400 )}{260} = \frac{755 , 400}{260} \approx 2905.4 m/s$ Why this step? Subtracting a negative adds; the engine must throw slightly faster to cover the pressure loss.

Verify: Plug back: $260 \times 2905.4 = 755, 404$ ; add $- 5, 400 \Rightarrow 750, 004 \approx 750, 000$ N ✓ (rounding). Forecast of ~3000 m/s was in the right ballpark.

Example 8 — Real-world word problem (Cell 8)

Worked example Statement

A launch pad engineer reads off a test-stand firing at sea level: measured thrust $F = 1, 860$ kN, exit pressure $P_{e} = 70$ kPa, ambient $P_{a} = 101$ kPa, exit area $A_{e} = 2.5 m^{2}$ , and mass flow $\overset{m}{˙} = 620$ kg/s. She wants the effective exhaust velocity she'd advertise and how much thrust the atmosphere is costing.

Forecast: The engine is overexpanded ( $P_{e} < P_{a}$ ), so atmosphere costs thrust. Guess a loss of tens of kN.

Step 1 — Pressure thrust (the atmospheric cost). $(P_{e} - P_{a}) A_{e} = (70, 000 - 101, 000) (2.5) = (- 31, 000) (2.5) = - 77, 500 N$ Why this step? This negative number is the atmosphere's toll: the pad loses 77.5 kN of thrust here.

Step 2 — Momentum thrust (back it out). $\overset{m}{˙} v_{e} = F - (P_{e} - P_{a}) A_{e} = 1, 860, 000 - (- 77, 500) = 1, 937, 500 N$ Why this step? Total minus pressure gives the pure Newton's-3rd-law part.

Step 3 — True relative exhaust velocity. $v_{e} = \frac{1 , 937 , 500}{620} \approx 3125.0 m/s$

Step 4 — Effective exhaust velocity (the advertised number). $v_{e f f} = \frac{F}{m ˙} = \frac{1 , 860 , 000}{620} = 3000.0 m/s$ Why this step? $v_{e f f}$ folds the pressure term into one figure via $F = \overset{m}{˙} v_{e f f}$ ; it's what feeds directly into Specific Impulse and the Tsiolkovsky Rocket Equation.

Verify: $v_{e f f} < v_{e}$ (3000 < 3125), which must be true when pressure thrust is negative ✓. Reconstruct: $\overset{m}{˙} v_{e f f} = 620 \times 3000 = 1, 860, 000 = F$ ✓.

Example 9 — Exam twist: effective exhaust velocity in vacuum (Cell 9)

Worked example Statement

An exam gives you $v_{e f f} = 3000$ m/s and $\overset{m}{˙} = 620$ kg/s at sea level (from Example 8's engine), and asks: what is $v_{e f f}$ in vacuum, given $P_{e} = 70$ kPa, $A_{e} = 2.5 m^{2}$ ?

Forecast: In vacuum the pressure term jumps up (no $P_{a}$ ), so $v_{e f f}$ rises. Guess a few hundred m/s higher.

Step 1 — Recall $v_{e f f}$ 's definition. $v_{e f f} = v_{e} + \frac{( P _{e} - P _{a} ) A _{e}}{m ˙}$ Why this step? Only the pressure part of $v_{e f f}$ changes with altitude; $v_{e}$ (relative speed) stays fixed at 3125 m/s from Example 8.

Step 2 — Vacuum pressure term per unit mass flow. $\frac{( P _{e} - 0 ) A _{e}}{m ˙} = \frac{70 , 000 \times 2.5}{620} = \frac{175 , 000}{620} \approx 282.3 m/s$ Why this step? Setting $P_{a} = 0$ gives the maximum pressure contribution, converted into velocity units by dividing by $\overset{m}{˙}$ .

Step 3 — Add to $v_{e}$ . $v_{e f f, vac} = 3125.0 + 282.3 = 3407.3 m/s$

Verify: Vacuum $v_{e f f}$ (3407.3) $>$ sea-level $v_{e f f}$ (3000) ✓ — the engine is more efficient in space, consistent with Example 4's story. Cross-check the vacuum thrust: $\overset{m}{˙} v_{e f f, vac} = 620 \times 3407.3 \approx 2, 112, 500$ N; direct: $\overset{m}{˙} v_{e} + P_{e} A_{e} = 1, 937, 500 + 175, 000 = 2, 112, 500$ N ✓.

Recall Quick self-test

Which cell is overexpanded and what's the sign of its pressure thrust? ::: $P_{e} < P_{a}$ , pressure thrust negative Same engine: which gives more thrust, sea level or vacuum? ::: Vacuum, because $P_{a} = 0$ removes the atmospheric back-push If $A_{e} = 0$ , what is the pressure thrust? ::: Zero, regardless of the pressure mismatch How do you get $v_{e}$ from a measured $F$ ? ::: $v_{e} = (F - (P_{e} - P_{a}) A_{e}) / \overset{m}{˙}$ Relationship of $v_{e f f}$ to $v_{e}$ when overexpanded? ::: $v_{e f f} < v_{e}$ (pressure term is negative)

Connections

Parent topic — the derivation these examples exercise
Specific Impulse — built from $v_{e f f}$ (Examples 8, 9)
Tsiolkovsky Rocket Equation — consumes $v_{e f f}$
De Laval Nozzle — sets $P_{e}$ and $A_{e}$ (over/under/perfect expansion)
Newton's Third Law — origin of the momentum thrust
Conservation of Momentum — foundation of the derivation