Electric propulsion — thrust, power, Isp trade-off: Worked examples

Definition The symbols used on this page (defined once, up front)

Every worked example below reuses these. Nail them now so no example ever surprises you:

$P_{in}$ ::: input power — the electrical watts the solar array or reactor feeds the thruster.
$η$ (eta) ::: efficiency $= \frac{P _{j e t}}{P _{in}}$ — the fraction of input power that actually ends up in the exhaust beam. It is a pure number between 0 and 1. The leftover $(1 - η) P_{in}$ becomes heat.
$v_{e}$ ::: exhaust speed (m/s) — how fast propellant leaves the nozzle.
$I_{s p}$ ::: specific impulse (seconds) — the fuel-economy score, related by $v_{e} = g_{0} I_{s p}$ with $g_{0} = 9.81 m/s^{2}$ .
$\overset{m}{˙}$ ::: mass flow rate (kg/s) — kilograms of propellant thrown per second.
$Δ v$ (delta-vee) ::: the change in the spacecraft's velocity a maneuver must produce (m/s) — the "budget" a mission must pay for.
$m_{0}$ ::: initial mass — spacecraft plus all propellant, before the burn.
$m_{f}$ ::: final mass — spacecraft after the burn, i.e. $m_{0}$ minus the propellant used.

#	Cell (case class)	What makes it tricky	Covered by
A	Ordinary forward calc	Given $P_{in}, η, I_{s p}$ → find $F, \overset{m}{˙}$	Ex 1
B	Inverse calc	Given the thrust you want → find required power / $I_{s p}$	Ex 2
C	Fixed power, vary $I_{s p}$	The seesaw: does thrust move the way you predict?	Ex 3
D	Degenerate: $I_{s p} \to \infty$	Photon-rocket limit — thrust $\to 0$	Ex 4
E	Degenerate: $v_{e} \to 0$ / $\overset{m}{˙}$ huge	Chemical-like limit — huge thrust, awful economy	Ex 4
F	Efficiency & waste heat	Split $P_{in}$ into beam vs heat; ties to Thermal Control (radiators)	Ex 5
G	Real-world word problem	Mission: how long to reach a target $Δ v$ ? Uses Tsiolkovsky Rocket Equation	Ex 6
H	Exam twist / trap	Units trap: forgetting $g_{0}$ , or "input vs jet power"	Ex 7
I	Comparison across regimes	EP vs chemical at same $Δ v$ — propellant saved	Ex 8
J	Boundary: $P_{in} \to 0$ or $η \to 0$	Zero power / zero efficiency — physical impossibility	Ex 9

2. Worked examples

Cell A — the plain forward calculation

Worked example Example 1 — Full state of a Hall thruster (Cell A)

A Hall thruster runs at $P_{in} = 4.5 kW$ , efficiency $η = 0.55$ , and $I_{s p} = 1800 s$ . Find the exhaust speed $v_{e}$ , thrust $F$ , mass flow $\overset{m}{˙}$ , and jet power $P_{j e t}$ .

Forecast: $I_{s p}$ here (1800 s) is lower than the 3000 s ion engine in the parent, so expect more thrust but a larger mass flow. Guess: a couple hundred mN.

Step 1 — economy score to real speed. $v_{e} = g_{0} I_{s p} = 9.81 \times 1800 = 1.766 \times 1 0^{4} m/s$ . Why this step? Every equation below wants a speed in m/s, not a score in seconds. $I_{s p}$ is just $v_{e}$ rescaled by $g_{0} = 9.81 m/s^{2}$ .

Step 2 — thrust from the trade-off. $F = \frac{2 η P _{in}}{v _{e}} = \frac{2 ( 0.55 ) ( 4500 )}{1.766 \times 1 0 ^{4}} = 0.280 N = 280 mN$ . Why this step? The trade-off equation gives thrust directly from wall power and speed; no need to know $\overset{m}{˙}$ yet.

Step 3 — mass flow. $\overset{m}{˙} = \frac{F}{v _{e}} = \frac{0.280}{1.766 \times 1 0 ^{4}} = 1.585 \times 1 0^{- 5} kg/s = 15.9 mg/s$ . Why this step? $F = \overset{m}{˙} v_{e}$ rearranged — thrust is momentum per second, so mass per second is thrust divided by speed.

Step 4 — jet power. $P_{j e t} = η P_{in} = 0.55 \times 4500 = 2475 W$ . Why this step? Efficiency $η = P_{j e t} / P_{in}$ by definition, so the beam carries $η$ of the wall power.

Verify: Cross-check $P_{j e t} = \frac{1}{2} F v_{e} = \frac{1}{2} (0.280) (1.766 \times 1 0^{4}) = 2472 W$ ✓ (matches 2475 W within rounding). Forecast held: lower $I_{s p}$ → 280 mN vs 102 mN in the parent. ✓

Cell B — running the equations backward

Cell C — the fixed-power seesaw

The figure below plots thrust $F$ (mN, vertical axis) against specific impulse $I_{s p}$ (seconds, horizontal axis) at fixed power $P_{in} = 4.5 kW$ and $η = 0.55$ . The lavender curve is $F = 2 η P_{in} / (g_{0} I_{s p})$ — a falling hyperbola. How to read it: pick any $I_{s p}$ on the bottom axis, go up to the curve, read the thrust on the left axis. The coral dot marks Ex 1 (1800 s, 280 mN); the mint dot marks Ex 3 (5400 s, 93 mN). Sliding right (more economy) always drops you down (less thrust).

Figure — Electric propulsion — thrust, power, Isp trade-off

Worked example Example 3 — Triple the economy, what happens to thrust? (Cell C)

Hold $P_{in} = 4.5 kW$ , $η = 0.55$ fixed (same hardware as Ex 1). Change $I_{s p}$ from 1800 s to 5400 s (tripled). Predict, then compute, the new thrust.

Forecast: At fixed power $F \propto 1/ I_{s p}$ . Triple $I_{s p}$ → thrust should fall to one third of 280 mN ≈ 93 mN. Look at the curve in the figure — we slide from the coral dot right and down to the mint dot on the same hyperbola.

Step 1 — new speed. $v_{e} = 9.81 \times 5400 = 5.297 \times 1 0^{4} m/s$ . Why this step? New economy score → new speed; power and $η$ untouched.

Step 2 — new thrust. $F = \frac{2 ( 0.55 ) ( 4500 )}{5.297 \times 1 0 ^{4}} = 0.0934 N = 93.4 mN$ . Why this step? Same trade-off equation; only $v_{e}$ moved.

Verify: Ratio $\frac{93.4}{280} = 0.334 \approx \frac{1}{3}$ ✓ — exactly the inverse of the ×3 in $I_{s p}$ . The mass flow, meanwhile, drops even harder: $\overset{m}{˙} = F / v_{e}$ falls because $F$ shrank and $v_{e}$ grew. This is the seesaw pinned at fixed power, visualized as the coral point sliding down the lavender curve to the mint point.

Cells D & E — the two degenerate extremes

The figure below plots two curves against exhaust speed $v_{e}$ (km/s, horizontal axis) at fixed $P_{in} = 4.5 kW$ , $η = 0.55$ . The lavender solid curve (left axis) is thrust $F$ in mN; the coral dashed curve (right axis) is mass flow $\overset{m}{˙}$ in mg/s. How to read it: at the far left ( $v_{e} \to 0$ ) both curves shoot up — the chemical corner. At the far right ( $v_{e} \to \infty$ ) both sink toward zero — the photon corner. The dashed coral curve falls faster than the solid lavender one, showing $\overset{m}{˙}$ collapses harder than $F$ as you speed up the exhaust.

Worked example Example 4 — The two ends of the seesaw (Cells D & E)

Same fixed $P_{in} = 4.5 kW$ , $η = 0.55$ . Examine the limits $I_{s p} \to \infty$ (photon-like) and $v_{e} \to 0$ (chemical-like). What do $F$ and $\overset{m}{˙}$ do?

Forecast: Push exhaust infinitely fast and thrust should vanish (you throw almost no mass). Push exhaust very slowly and you need a torrent of mass — huge $\overset{m}{˙}$ , huge $F$ but terrible economy.

Step 1 — limit $v_{e} \to \infty$ ( $I_{s p} \to \infty$ ). In $F = \frac{2 η P _{in}}{v _{e}}$ , as $v_{e} \to \infty$ , $F \to 0$ . Why this step? Fixed power spread over infinitely fast particles means almost no mass leaves per second, so momentum-per-second → 0. This is the maximum-economy, zero-push corner (the right edge of the figure).

Step 2 — check $\overset{m}{˙}$ there. $\overset{m}{˙} = F / v_{e} \to 0/\infty = 0$ : essentially no propellant used. That's why high $I_{s p}$ = fuel lasts "forever" — the price is no thrust.

Step 3 — limit $v_{e} \to 0$ . $F = \frac{2 η P _{in}}{v _{e}} \to \infty$ and $\overset{m}{˙} = \frac{2 η P _{in}}{v _{e}^{2}} \to \infty$ even faster. Why this step? Slow, heavy exhaust converts fixed power into big momentum flux — but you dump propellant at a ruinous rate. This is the maximum-thrust, worst-economy corner (the left edge of the figure, the chemical end of the spectrum).

Step 4 — a concrete number between the extremes. At a moderate $v_{e} = 1000 m/s$ ( $I_{s p} \approx 102 s$ ): $F = \frac{2 ( 0.55 ) ( 4500 )}{1000} = 4.95 N$ — about 17× the Ex 1 thrust — but $\overset{m}{˙} = F / v_{e} = 4.95/1000 = 4.95 \times 1 0^{- 3} kg/s = 4.95 g/s$ , roughly 310× the Ex 1 mass flow. Great push, disastrous mileage.

Verify: The figure plots $F$ vs $v_{e}$ : a rectangular hyperbola pinned to both axes — $F \to \infty$ as $v_{e} \to 0$ (left) and $F \to 0$ as $v_{e} \to \infty$ (right). Both asymptotes confirmed, and the dashed $\overset{m}{˙}$ curve sinks faster on the right, matching $\overset{m}{˙} \propto 1/ v_{e}^{2}$ vs $F \propto 1/ v_{e}$ . ✓

Cell F — where the wasted power goes

Cell G — a mission word problem

Worked example Example 6 — How long to raise the orbit? (Cell G)

A 1200 kg spacecraft carries 300 kg of xenon propellant. Its thruster gives $F = 280 mN$ (the Ex 1 value) with $\overset{m}{˙} = 15.9 mg/s$ . The mission needs a velocity budget $Δ v = 4.0 km/s$ (recall from the top of the page: $Δ v$ is the change in the ship's velocity the maneuver must produce). (a) Does it have enough propellant? (b) Roughly how long does the burn take?

Forecast: Tiny thrust → this will take months, not minutes. Xenon is precious, so check the rocket equation first.

Step 1 — propellant needed via the rocket equation. With $v_{e} = 1.766 \times 1 0^{4} m/s$ (Ex 1), Tsiolkovsky gives $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{4000/17660} = e^{0.2265} = 1.2542$ , where $m_{0}$ is the initial (loaded) mass and $m_{f}$ the final (burned-out) mass. Why this step? The rocket equation tells us the mass ratio for a target $Δ v$ — the fundamental fuel-economy check.

Step 2 — final vs initial mass. Here $m_{0} = 1200 kg$ (ship + all propellant), so $m_{f} = 1200/1.2542 = 956.8 kg$ . Propellant burned $= m_{0} - m_{f} = 1200 - 956.8 = 243.2 kg$ . Why this step? $m_{0} - m_{f}$ is the propellant the maneuver actually consumes.

Step 3 — enough? Needed 243.2 kg < 300 kg carried. Yes, with ~57 kg margin. ✓

Step 4 — burn duration. Time $= \frac{propellant}{m ˙} = \frac{243.2}{1.585 \times 1 0 ^{- 5}} = 1.534 \times 1 0^{7} s$ . Why this step? Mass burned divided by mass-burn rate = seconds of firing.

Step 5 — convert. $1.534 \times 1 0^{7} s \div (86400 s/day) = 177.6 days \approx 5.9 months$ . Why this step? Seconds are unreadable; days show the deep-space cruise character of EP.

Verify: Sanity: impulse $= F \times t = 0.280 \times 1.534 \times 1 0^{7} = 4.30 \times 1 0^{6} N\cdots$ ; and $Δ v \approx$ impulse $/ \overset{m}{ˉ}$ with mean mass $\overset{m}{ˉ} \approx (1200 + 956.8) /2 = 1078 kg$ gives $4.30 \times 1 0^{6} /1078 \approx 3.99 \times 1 0^{3} m/s \approx 4.0 km/s$ ✓. The ~6-month burn is exactly why EP is for cruising, not launch.

Cell H — the exam trap

Worked example Example 7 — Spot the units trap (Cell H)

Exam statement: "A thruster has $I_{s p} = 2000 s$ , $η = 0.7$ , $P_{in} = 3 kW$ . A student writes $F = \frac{2 η P _{in}}{I _{s p}} = \frac{2 ( 0.7 ) ( 3000 )}{2000} = 2.1 N$ . Find the error and the correct $F$ ."

Forecast: The student divided by $I_{s p}$ (seconds) instead of $v_{e}$ (m/s). Since $v_{e} = g_{0} I_{s p}$ is ~9.81× bigger, the true thrust is ~9.81× smaller — around 0.2 N, not 2.1 N.

Step 1 — name the error. The trade-off equation is $F = \frac{2 η P _{in}}{v _{e}}$ , not $\frac{2 η P _{in}}{I _{s p}}$ . $I_{s p}$ is in seconds; you must first convert to a speed. Why this step? Dimensional check: $\frac{W}{s} = \frac{J/s}{s} = J/s^{2}$ — not newtons. The student's answer is dimensionally wrong.

Step 2 — convert. $v_{e} = g_{0} I_{s p} = 9.81 \times 2000 = 1.962 \times 1 0^{4} m/s$ . Why this step? Restore the missing $g_{0}$ .

Step 3 — correct thrust. $F = \frac{2 ( 0.7 ) ( 3000 )}{1.962 \times 1 0 ^{4}} = 0.2141 N \approx 214 mN$ . Why this step? Now the units work: $\frac{W}{m/s} = \frac{J/s}{m/s} = J/m = N$ ✓.

Verify: Ratio of wrong to right $= 2.1/0.2141 = 9.81 = g_{0}$ ✓ — exactly the factor the student dropped.

Cell I — comparison across propulsion regimes

Worked example Example 8 — EP vs chemical, propellant saved (Cell I)

Two stages must both deliver $Δ v = 6.0 km/s$ to a 1000 kg dry spacecraft (so the final mass $m_{f} = 1000 kg$ in each case). Stage A is chemical ( $I_{s p} = 350 s$ ); Stage B is electric ( $I_{s p} = 3000 s$ ). Compare the propellant each needs.

Forecast: EP's ~8.6× higher $I_{s p}$ should slash propellant dramatically — chemical might need tonnes, EP a couple hundred kg.

Step 1 — exhaust speeds. Chemical: $v_{e} = 9.81 \times 350 = 3434 m/s$ . Electric: $v_{e} = 9.81 \times 3000 = 2.943 \times 1 0^{4} m/s$ . Why this step? The rocket equation uses $v_{e}$ , so convert both economy scores.

Step 2 — mass ratios. Chemical: $\frac{m _{0}}{m _{f}} = e^{6000/3434} = e^{1.7472} = 5.739$ . Electric: $\frac{m _{0}}{m _{f}} = e^{6000/29430} = e^{0.2039} = 1.2262$ . Why this step? $m_{0} / m_{f} = e^{Δ v / v_{e}}$ — the core of Tsiolkovsky.

Step 3 — initial masses and propellant burned. With $m_{f} = 1000 kg$ : chemical $m_{0} = 5.739 \times 1000 = 5739 kg$ , so propellant $= m_{0} - m_{f} = 5739 - 1000 = 4739 kg$ . Electric $m_{0} = 1.2262 \times 1000 = 1226.2 kg$ , so propellant $= 1226.2 - 1000 = 226.2 kg$ . Why this step? Propellant $= m_{0} - m_{f}$ for each — this is the number the mission actually pays.

Step 4 — the payoff. Ratio $= \frac{4739}{226.2} = 21.0$ : the chemical stage burns ~21× more propellant for the same job. Why this step? This single number is the reason EP exists — high $I_{sp}$ saves enormous mass.

Verify: Forward-check EP: $Δ v = v_{e} ln (m_{0} / m_{f}) = 29430 \times ln (1.2262) = 29430 \times 0.2039 = 6001 m/s \approx 6.0 km/s$ ✓. The trade-off's cost — recall Ex 6 — is that EP takes months, while chemical burns in minutes.

Cell J — the boundary cases: zero power, zero efficiency

Worked example Example 9 — What happens at

$P_{in} \to 0$ and $η \to 0$ ? (Cell J) Keep $I_{s p} = 1800 s$ ( $v_{e} = 1.766 \times 1 0^{4} m/s$ ). Push the input power $P_{in}$ toward zero; separately push efficiency $η$ toward zero. What thrust do you get, and what does it mean physically?

Forecast: Both are the "engine off" corner — no power in, or none of it reaching the beam, means no thrust at all. The formula should give exactly zero, and physically the ship just coasts.

Step 1 — limit $P_{in} \to 0$ . In $F = \frac{2 η P _{in}}{v _{e}}$ , with $η, v_{e}$ fixed and finite, $F \to \frac{2 η \cdot 0}{v _{e}} = 0$ . Why this step? Thrust is powered momentum flux. With zero electrical input there is no energy to accelerate propellant, so nothing is thrown — $F = 0$ . This is the physical statement "an unpowered EP thruster produces no thrust," a natural floor of the matrix.

Step 2 — limit $η \to 0$ . In $F = \frac{2 η P _{in}}{v _{e}}$ , with $P_{in}, v_{e}$ fixed, $F \to \frac{2 \cdot 0 \cdot P _{in}}{v _{e}} = 0$ , while the heat $= (1 - η) P_{in} \to P_{in}$ . Why this step? $η = 0$ means none of the input power reaches the beam — all of it becomes waste heat (Ex 5's leftover taken to its extreme). A perfectly inefficient thruster is just an electric heater: zero thrust, full power dumped as heat. Physically impossible as a rocket, but it bounds the efficiency axis.

Step 3 — sanity on a small-but-nonzero value. At $P_{in} = 1 W$ , $η = 0.55$ : $F = \frac{2 ( 0.55 ) ( 1 )}{1.766 \times 1 0 ^{4}} = 6.23 \times 1 0^{- 5} N = 62 μ N$ — vanishingly small, confirming thrust smoothly approaches 0 as power drops. Why this step? Shows the limit isn't a jump — thrust glides continuously to zero, so there's no hidden surprise near the boundary.

Verify: Both limits give $F = 0$ : substituting $P_{in} = 0$ or $η = 0$ into $F = 2 η P_{in} / v_{e}$ yields exactly $0$ ✓. Heat check at $η \to 0$ : $(1 - 0) P_{in} = P_{in}$ — all input becomes heat ✓. Physically: you cannot get thrust from zero power or zero efficiency, closing the bottom-left corner of the matrix.

3. Matrix coverage check

Recall Which example cleared which cell?

A (forward) ::: Example 1 B (inverse) ::: Example 2 C (fixed power, vary $I_{s p}$ ) ::: Example 3 D ( $I_{s p} \to \infty$ ) & E ( $v_{e} \to 0$ ) ::: Example 4 F (waste heat) ::: Example 5 G (mission word problem) ::: Example 6 H (exam units trap) ::: Example 7 I (EP vs chemical comparison) ::: Example 8 J ( $P_{in} \to 0$ or $η \to 0$ boundary) ::: Example 9

Recall One-line lessons

Fixed power thrust law ::: $F = 2 η P_{in} / v_{e}$ , so $F \propto 1/ I_{s p}$ (Ex 3). The $v_{e} \to 0$ corner ::: $F \to \infty$ but $\overset{m}{˙} \to \infty$ faster — no free lunch (Ex 4). The units trap ::: divide by $v_{e} = g_{0} I_{s p}$ , never by $I_{s p}$ (Ex 7). Why EP saves mass ::: high $I_{s p}$ shrinks the mass ratio $e^{Δ v / v_{e}}$ (Ex 8). The zero-power / zero-efficiency corner ::: $F \to 0$ ; no push without power reaching the beam (Ex 9).

4. Connections

Parent topic (Hinglish) — the core derivations these examples exercise.
Tsiolkovsky Rocket Equation — used in Ex 6 and Ex 8 to turn $Δ v$ into propellant.
Specific Impulse — the $I_{s p} = v_{e} / g_{0}$ conversion at the heart of every trap and forecast.
Chemical vs Electric Propulsion — the two ends of Ex 4 and the comparison in Ex 8.
Ion and Hall Thrusters — the hardware in Ex 1 and Ex 6.
Spacecraft Power Systems — supplies the $P_{in}$ we invert for in Ex 2 and drive to zero in Ex 9.
Thermal Control (radiators) — absorbs the waste heat computed in Ex 5 and Ex 9.
Newton's Third Law — the origin of $F = \overset{m}{˙} v_{e}$ used throughout.

Worked examples — Electric propulsion — thrust, power, Isp trade-off

1. The scenario matrix

2. Worked examples

Cell A — the plain forward calculation

Cell B — running the equations backward

Cell C — the fixed-power seesaw

Cells D & E — the two degenerate extremes

Cell F — where the wasted power goes

Cell G — a mission word problem

Cell H — the exam trap

Cell I — comparison across propulsion regimes

Cell J — the boundary cases: zero power, zero efficiency

3. Matrix coverage check

4. Connections