Exercises — Thrust equation F = ṁv_e + (P_e − P_a)A_e — derivation

Momentum thrust is $\dot{m}{v}_e$ — mass thrown per second times how fast it's thrown. Pressure thrust is $(P_e-P_a)A_e$ — the mismatch between exit pressure and outside pressure acting on the exit area. Perfectly expanded means $P_e=P_a$, so $(P_e-P_a)=0$ and the pressure term vanishes. Only $\dot{m}{v}_e$ survives.

Perfect expansion → pressure term is $0$, so $F=\dot{m}{v}_e$. $F=300\times 3000=900,000\text{N}=900\text{kN}.$

(a) $P_e>P_a$ → underexpanded → pressure thrust positive. (b) $P_e<P_a$ → overexpanded → pressure thrust negative. (c) $P_e=P_a$ → perfectly expanded → pressure thrust zero.

Momentum thrust: $\dot{m}{v}_e=250\times 2800=700,000\text{N}$. Pressure thrust: $(P_e-P_a)A_e=(95,000-101,000)(1.2)=(-6,000)(1.2)=-7,200\text{N}$ (overexpanded, so negative). Total: $F=700,000-7,200=692,800\text{N}=692.8\text{kN}$.

Momentum thrust unchanged: $700,000\text{N}$. Pressure thrust: $(95,000-0)(1.2)=114,000\text{N}$ (now fully positive — nothing pushes back). Total: $F=700,000+114,000=814,000\text{N}=814\text{kN}$. Gain over sea level: $814,000-692,800=121,200\text{N}=121.2\text{kN}$. That gain equals $P_a{A}_e=101,000\times 1.2=121,200\text{N}$ — exactly the missing atmospheric back-push.

Start from $F=\dot{m}{v}_e+P_e{A}_e$ (since $P_a=0$). Momentum thrust: $180\times 3100=558,000\text{N}$. So the pressure term must supply $600,000-558,000=42,000\text{N}$. $A_e=\frac{42,000}{P_e}=\frac{42,000}{40,000}=1.05{\text{m}}^2.$

Why collect the pressure term into a velocity first? The thrust has two visually different pieces — one is $(\text{mass/s})\times (\text{speed})$, the other is $(\text{pressure})\times (\text{area})$. That's awkward to carry around in later formulas (like Tsiolkovsky) that only know about a single "exhaust velocity." So we ask: what single speed, if the whole thrust came purely from throwing mass, would give the same force? Set $F=\dot{m}{v}_{eff}$ and solve for that pretend speed. Dividing the pressure force $(P_e-P_a)A_e$ by $\dot{m}$ converts "extra newtons" into "extra m/s," so it can sit next to $v_e$. That is exactly the formula given.

The pressure term was $-7,200\text{N}$. $v_{eff}=2800+\frac{-7,200}{250}=2800-28.8=2771.2\text{m/s}.$ Check: $\dot{m}{v}_{eff}=250\times 2771.2=692,800\text{N}$ ✓. Overexpansion effectively slowed the exhaust by $28.8\text{m/s}$ worth of push.

Only the term $-P_a{A}_e$ depends on $P_a$. Since it carries a minus sign, decreasing $P_a$ increases $F$. So thrust rises monotonically all the way to vacuum, where it is maximal. Total change from sea level to vacuum: $\Delta F=-A_e(0-101,000)=1.2\times 101,000=121,200\text{N}=121.2\text{kN}.$ This is why a booster's thrust climbs as it ascends, even with the engine running unchanged.

Pressure thrust vanishes when $P_e-P_a=0$, i.e. $P_a=P_e=60\text{kPa}$ (perfect expansion). There: $F=\dot{m}{v}_e=220\times 2600=572,000\text{N}=572\text{kN}.$ At sea level $P_a=101\text{kPa}$ (now overexpanded, $P_e<P_a$): $F=572,000+(60,000-101,000)(1.5)=572,000-61,500=510,500\text{N}=510.5\text{kN}.$ Overexpansion at sea level costs $61.5$ kN.

$I_{sp}=\frac{814,000}{250\times 9.81}=\frac{814,000}{2452.5}=331.9\text{s}.$ Because thrust is higher in vacuum, so is $I_{sp}$ — the same reason engines are quoted with a separate (larger) vacuum $I_{sp}$.

Vacuum thrust (from L2·Q2) $F=814,000\text{N}$, constant here. Impulse $J=Ft=814,000\times 2=1,628,000\text{N⋅s}$. Exhaust mass ejected $=\dot{m}t=250\times 2=500\text{kg}$. By Newton's third law and momentum conservation, this total impulse $J$ equals the forward momentum the rocket gains. Note that $J$ already bundles both thrust pieces together: the momentum-thrust part is the exhaust's backward momentum, and the pressure-thrust part is the exit-pressure push — the single number $J=Ft$ counts them both, since $F$ itself is the full thrust.

First $v_{eff}=F/\dot{m}=814,000/250=3256\text{m/s}$. Mass ratio $m_0/m_f=20,000/8,000=2.5$, so $\ln 2.5=0.91629\dots$ $\Delta v=3256\times 0.91629=2983.4\text{m/s}.$ The pressure thrust quietly raised $v_{eff}$ above the raw $v_e=2800$, buying extra $\Delta v$.

Momentum thrust (fixed): $\dot{m}{v}_e=400\times 3200=1,280,000\text{N}$. (i) Design altitude, $P_e=P_a$ → pressure term $0$: $F=1,280,000\text{N}=1280\text{kN}.$ (ii) Sea level, $P_e=26\text{kPa}<P_a=101\text{kPa}$ (overexpanded): $F=1,280,000+(26,000-101,000)(2.0)=1,280,000-150,000=1,130,000\text{N}=1130\text{kN}.$ (iii) Vacuum, $P_a=0$ (underexpanded): $F=1,280,000+(26,000)(2.0)=1,280,000+52,000=1,332,000\text{N}=1332\text{kN}.$ Same engine, three altitudes: $1130$ → $1280$ → $1332$ kN as air thins. Notice thrust keeps climbing past the "perfect" point — perfection was a design target for that altitude, not a peak of raw force (echo of L3).

Momentum thrust: $50\times 200=10,000\text{N}$. Pressure thrust: $(500,000-0)(0.8)=400,000\text{N}$. Total: $F=10,000+400,000=410,000\text{N}=410\text{kN}$. Pressure fraction: $400,000/410,000=0.9756=97.56\%$. This is the degenerate regime where the pressure term dominates — physically a cold, barely-moving, highly pressurised jet. Real rocket nozzles avoid this by expanding the gas so momentum thrust dominates instead.

Rearrange $F=\dot{m}{v}_e+(P_e-P_a)A_e$ for $v_e$: $v_e=\frac{F-(P_e-P_a)A_e}{\dot{m}}.$ Pressure term: $(80,000-101,000)(1.0)=-21,000\text{N}$. $v_e=\frac{445,000-(-21,000)}{200}=\frac{445,000+21,000}{200}=\frac{466,000}{200}=2330\text{m/s}.$ Note the double-negative: subtracting a negative pressure term adds back the $21$ kN the atmosphere stole, so the true $v_e$ is larger than a naive $445,000/200=2225$ would suggest.

For a fixed engine, thrust rises as ambient pressure falls, is highest in vacuum, and equals pure $\dot{m}{v}_e$ exactly at the altitude where $P_e=P_a$ ::: True — this single sentence contains L1–L5.