FEEP, MEMS thrusters — micro-propulsion: Worked examples

Our three tools (all earned in the parent):

The scenario matrix

Every micro-propulsion problem lives in one of these cells. The examples below are labelled with the cell they cover.

Cell	What varies / degenerates	Example
A. Baseline	ordinary $q, m, V, I$ → find $v_{e}$ , $F$	Ex 1
B. Low- $V$ limit	small voltage → slow, weak thrust	Ex 2
C. High- $V$ limit	kilovolt → very fast ions	Ex 3
D. Zero / degenerate input	$I = 0$ , or $Δ v = 0$ , or $V = 0$	Ex 4
E. Heavy vs light ion	swap Cs ↔ In at fixed $I, V$	Ex 5
F. Real-world word problem	mission $Δ v$ , propellant mass, burn time	Ex 6
G. MEMS / low- $R e$ twist	gas expansion + Reynolds-number sanity	Ex 7
H. Exam twist (inverse)	given $F$ and $I_{s p}$ , back out $V$ and $I$	Ex 8

We use two ions throughout:

Indium In $^{+}$ : $m = 1.9 \times 1 0^{- 25} kg$ , $q = 1.6 \times 1 0^{- 19} C$ .
Caesium Cs $^{+}$ : $m = 2.2 \times 1 0^{- 25} kg$ , same $q$ .

Cell A — Baseline

Worked example Ex 1 — Indium ion at 6 kV, 10 µA beam

$m = 1.9 \times 1 0^{- 25}$ , $q = 1.6 \times 1 0^{- 19}$ , $V = 6000$ V, $I = 10 μ A = 1 0^{- 5}$ A. Find $v_{e}$ , $I_{s p}$ , $F$ .

Forecast: $v_{e} \sim 1 0^{5}$ m/s? $F$ a few µN?

$v_{e} = \frac{2 q V}{m} = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 6000 )}{1.9 \times 1 0 ^{- 25}} \approx 1.005 \times 1 0^{5} m/s$ . Why this step? Energy conservation: all the electric potential energy $q V$ becomes kinetic $\frac{1}{2} m v_{e}^{2}$ .
$I_{s p} = v_{e} / g_{0} = 1.005 \times 1 0^{5} /9.81 \approx 1.02 \times 1 0^{4} s$ . Why this step? $I_{s p}$ is $v_{e}$ measured in "seconds of gravity" — the standard fuel-economy yardstick (Specific Impulse).
$F = I \frac{2 mV}{q} = I \frac{m}{q} v_{e} = (1 0^{- 5}) (1.1875 \times 1 0^{- 6}) (1.005 \times 1 0^{5}) \approx 1.19 \times 1 0^{- 6} N$ . Why this step? $F = \overset{m}{˙} v_{e}$ and $\overset{m}{˙} = (m / q) I$ (charge/sec → mass/sec).

Verify: units of $F$ : $[A] [kg/C] [m/s] = [C/s] [kg/C] [m/s] = kg\cdotm/s^{2} = N$ ✓. A microamp giving a micro-newton is exactly the "eyedropper" regime.

Cell D — Zero / degenerate inputs

Worked example Ex 4 — The three "what if it's zero?" traps

(a) Beam current $I = 0$ . (b) Accelerating voltage $V = 0$ . (c) Required $Δ v = 0$ .

Forecast: which give zero thrust? Which give zero propellant?

(a) $I = 0$ : $F = I 2 mV / q = 0$ . No ions per second → no momentum flow → no thrust, but $v_{e}$ (an ion's speed) is still $1 0^{5}$ m/s if $V$ is on. Why this step? Thrust needs stuff leaving; a switched-off beam still has an accelerating field, so $v_{e} \neq = 0$ but $\overset{m}{˙} = 0$ . Never confuse "each ion is fast" with "thrust exists".
(b) $V = 0$ : $v_{e} = 2 q (0) / m = 0$ , hence $F = \overset{m}{˙} \cdot 0 = 0$ . Even if ions leak out, they carry no directed momentum. Why this step? No potential drop → no energy given → ions dribble at ~0 speed. This is a genuine degenerate case, not a math error.
(c) $Δ v = 0$ : $m_{p} = m_{0} (1 - e^{0}) = m_{0} (1 - 1) = 0$ . Why this step? No velocity change requested → burn nothing. The rocket equation correctly returns zero fuel.

Verify: $1 - e^{0} = 0$ ✓; $2 q \cdot 0/ m = 0$ ✓; $I \cdot (\dots) = 0$ when $I = 0$ ✓. All three degeneracies behave sensibly.

Cell E — Heavy vs light ion (Cs vs In)

Figure — FEEP, MEMS thrusters — micro-propulsion

Worked example Ex 5 — Same

$I$ , same $V$ : which gives more thrust, which more $I_{s p}$ ? Fix $V = 6$ kV, $I = 10 μ A$ . Compare In ( $m = 1.9 \times 1 0^{- 25}$ ) with Cs ( $m = 2.2 \times 1 0^{- 25}$ ).

Forecast: heavier caesium — more thrust or less? More $I_{s p}$ or less?

$v_{e} \propto q / m = 1/ m$ , so heavier Cs is slower: $v_{e, Cs} = v_{e, In} 1.9/2.2 \approx 1.005 \times 1 0^{5} \times 0.929 = 9.34 \times 1 0^{4}$ m/s. Why this step? More mass per charge → each charge drags more inertia → lower speed → lower $I_{s p}$ . Look at the shorter teal arrow in the figure.
$F = I 2 mV / q \propto m$ , so heavier Cs gives more thrust per amp: $F_{Cs} = F_{In} 2.2/1.9 = 1.19 \times 1 0^{- 6} \times 1.076 \approx 1.28 \times 1 0^{- 6}$ N. Why this step? At fixed current the ion count is fixed; heavier ions carry more momentum each → more push. Longer plum arrow in the figure.

Verify: product check $v_{e} \cdot F$ should be independent of $m$ (since $m \cdot 1/ m = 1$ ): In gives $1.005 \times 1 0^{5} \times 1.19 \times 1 0^{- 6} = 0.1196$ ; Cs gives $9.34 \times 1 0^{4} \times 1.28 \times 1 0^{- 6} = 0.1196$ ✓. Lesson: heavy ≠ worse — it's a thrust-vs- $I_{s p}$ trade, echoing the parent's Ion Thrusters logic.

Cell F — Real-world word problem

Worked example Ex 6 — Full mission budget for a CubeSat

A 3 kg CubeSat (CubeSats & Attitude Control) must perform $Δ v = 5$ m/s of drag make-up over its life, using the Ex 1 indium FEEP ( $v_{e} = 1.005 \times 1 0^{5}$ m/s, $F = 1.19 μ N$ ). Find (a) propellant mass, (b) total impulse, (c) burn time if run continuously.

Forecast: grams of fuel? Days or years of firing?

(a) $m_{p} = m_{0} (1 - e^{- Δ v / v_{e}})$ . Since $Δ v / v_{e} = 5/1 0^{5} = 5 \times 1 0^{- 5} ≪ 1$ , use $e^{- x} \approx 1 - x$ : $m_{p} \approx m_{0} \frac{Δ v}{v _{e}} = 3 \times 5 \times 1 0^{- 5} = 1.5 \times 1 0^{- 4}$ kg $= 0.15$ g. Why this step? Rocket Equation; the linear approximation is justified because $x$ is tiny.
(b) Total impulse $J = m_{p} v_{e} = 1.5 \times 1 0^{- 4} \times 1.005 \times 1 0^{5} \approx 15.1$ N·s. (Or $J = m_{0} Δ v = 3 \times 5 = 15$ N·s.) Why this step? Impulse = momentum delivered = fuel mass × exhaust speed = spacecraft mass × $Δ v$ . Two routes must agree.
(c) $t = J / F = 15/ (1.19 \times 1 0^{- 6}) \approx 1.26 \times 1 0^{7}$ s $\approx 146$ days. Why this step? Impulse = force × time (constant thrust), so $t = J / F$ .

Verify: two impulse routes agree: 15.1 vs 15.0 N·s (0.7% from the linear approx) ✓. Units of $t$ : $N\cdots / N = s$ ✓. Lesson: 0.15 g of metal buys the whole mission — the payoff of huge $v_{e}$ .

Cell G — MEMS / low-Reynolds twist

Worked example Ex 7 — Cold-gas MEMS exhaust speed and the

$R e$ reality check Nitrogen through a micro-nozzle: $γ = 1.4$ , $R = 8.314$ J/mol·K, molar mass $M = 0.028$ kg/mol, $T_{0} = 300$ K, pressure ratio $p_{e} / p_{0} = 0.01$ . Also estimate $R e$ with $ρ = 0.05 kg/m^{3}$ , $v = 300$ m/s, channel $L = 100 μ m$ , $μ = 1.8 \times 1 0^{- 5}$ Pa·s.

Forecast: ideal $v_{e}$ near 700–800 m/s? Is $R e$ big (inviscid) or small (viscous)?

Bracket term: $1 - (p_{e} / p_{0})^{(γ - 1) / γ} = 1 - (0.01)^{0.4/1.4} = 1 - (0.01)^{0.2857} = 1 - 0.2683 = 0.7317$ . Why this step? This is the fraction of enthalpy released across the pressure drop — the engine of the expansion.
$v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{0}}{M} \times 0.7317 = 7.0 \times \frac{8.314 \times 300}{0.028} \times 0.7317$ . Inner: $R T_{0} / M = 8.314 \times 300/0.028 = 8.908 \times 1 0^{4}$ . So $v_{e} = 7.0 \times 8.908 \times 1 0^{4} \times 0.7317 = 4.563 \times 1 0^{5} \approx 676$ m/s. Why this step? Enthalpy → directed kinetic energy through the nozzle; $\frac{2 γ}{γ - 1} = 7$ for $γ = 1.4$ .
$R e = \frac{ρ v L}{μ} = \frac{0.05 \times 300 \times 1 0 ^{- 4}}{1.8 \times 1 0 ^{- 5}} = \frac{1.5 \times 1 0 ^{- 3}}{1.8 \times 1 0 ^{- 5}} \approx 83$ . Why this step? Reynolds Number compares inertia to viscosity; $R e \sim 80$ is low → thick boundary layer, so real $v_{e}$ falls below the ideal 676 m/s.

Verify: $(0.01)^{0.2857}$ : take logs, $0.2857 \times ln 0.01 = 0.2857 \times (- 4.605) = - 1.316$ , $e^{- 1.316} = 0.268$ ✓. $R e \approx 83 ≪ 2000$ (laminar, viscous-dominated) ✓. Lesson: the ideal formula ignores size, but $R e$ tells you micro-nozzles bleed efficiency.

Cell H — Exam twist (inverse problem)

Worked example Ex 8 — Given performance, back out the settings

Design spec: indium FEEP must deliver $F = 2.0 μ N$ at $I_{s p} = 8000$ s. Find the required $V$ and $I$ . ( $m = 1.9 \times 1 0^{- 25}$ , $q = 1.6 \times 1 0^{- 19}$ , $g_{0} = 9.81$ .)

Forecast: kilovolts of $V$ ? tens of µA of $I$ ?

From $I_{s p}$ : $v_{e} = I_{s p} g_{0} = 8000 \times 9.81 = 7.848 \times 1 0^{4}$ m/s. Why this step? $I_{s p}$ fixes $v_{e}$ directly — voltage's job.
Invert $v_{e} = 2 q V / m$ → $V = \frac{m v _{e}^{2}}{2 q} = \frac{1.9 \times 1 0 ^{- 25} \times ( 7.848 \times 1 0 ^{4} ) ^{2}}{2 \times 1.6 \times 1 0 ^{- 19}} = \frac{1.9 \times 1 0 ^{- 25} \times 6.159 \times 1 0 ^{9}}{3.2 \times 1 0 ^{- 19}} \approx 3.66 \times 1 0^{3}$ V. Why this step? Solve the energy equation for the one unknown $V$ . That's ≈3.7 kV — a realistic FEEP voltage.
From $F = \overset{m}{˙} v_{e} = (m / q) I v_{e}$ → $I = \frac{F q}{m v _{e}} = \frac{2.0 \times 1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 19}}{1.9 \times 1 0 ^{- 25} \times 7.848 \times 1 0 ^{4}} = \frac{3.2 \times 1 0 ^{- 25}}{1.491 \times 1 0 ^{- 20}} \approx 2.15 \times 1 0^{- 5}$ A. Why this step? With $v_{e}$ known, thrust pins down the current — the two knobs are independent (the parent's key point).

Verify: plug back: $v_{e} = 2 (1.6 \times 1 0^{- 19}) (3660) /1.9 \times 1 0^{- 25} = 6.16 \times 1 0^{9} = 7.85 \times 1 0^{4}$ ✓; $F = (m / q) I v_{e} = 1.1875 \times 1 0^{- 6} \times 2.15 \times 1 0^{- 5} \times 7.848 \times 1 0^{4} = 2.00 \times 1 0^{- 6}$ N ✓. Both targets recovered.

Recall Which knob does what?

Voltage sets ::: exhaust speed $v_{e}$ and hence $I_{s p}$ . Beam current sets ::: thrust $F$ (ions per second). Doubling $V$ multiplies $v_{e}$ by ::: $2$ , not 2. Heavier ion at fixed $I, V$ gives ::: more thrust, less $I_{s p}$ (a trade, not a defect). $R e ≪ 2000$ in a micro-nozzle means ::: viscous losses dominate → real $v_{e}$ below ideal.

Worked examples — FEEP, MEMS thrusters — micro-propulsion