Exercises — FEEP, MEMS thrusters — micro-propulsion

What we do: compare $80\mu \text{N}$ against the defining range. Micro-propulsion means thrust in the band $1\mu \text{N}$ to $\sim 1\text{mN}$ (i.e. $10^{-6}$ N to $10^{-3}$ N). Since $80\mu \text{N}=8\times 10^{-5}\text{N}$ sits inside that band, yes, it is micro-propulsion.

Speed comes from $v_e=\sqrt{2qV/m}$. The current $I$ does not appear in it. So to raise $v_e$ you raise $V$. (Current controls how many ions leave per second, i.e. thrust — not how fast each one goes.)

What we do: apply $I_{sp}=v_e/g_0$ (definition from Specific Impulse). $I_{sp}=\frac{1.0\times 10^5}{9.81}\approx 1.02\times 10^4\text{s}\approx 10,200\text{s}.$ A number in the thousands of seconds is the signature of electric propulsion — that is why these thrusters barely sip propellant.

What we do: energy conservation — the ion's electrical potential energy $qV$ becomes kinetic energy $\frac{1}{2}m{v}_e^2$, so $v_e=\sqrt{2qV/m}$. $v_e=\sqrt{\frac{2(1.6\times 10^{-19})(8000)}{2.2\times 10^{-25}}}=\sqrt{\frac{2.56\times 10^{-15}}{2.2\times 10^{-25}}}=\sqrt{1.164\times 10^{10}}\approx 1.08\times 10^5\text{m/s}.$ About $108\text{km/s}$.

The valid calculation (part a). Work the fraction inside the root to a clean power of ten first, then take the square root: $\frac{2mV}{q}=\frac{2(2.2\times 10^{-25})(8000)}{1.6\times 10^{-19}}=\frac{3.52\times 10^{-21}}{1.6\times 10^{-19}}=2.2\times 10^{-2}.$ $\sqrt{2mV/q}=\sqrt{2.2\times 10^{-2}}=0.148,F=I\cdot 0.148=(15\times 10^{-6})(0.148)=2.2\times 10^{-6}\text{N}\approx 2.2\mu \text{N}.$ The valid calculation (part b), independent route. $\frac{m}{q}=\frac{2.2\times 10^{-25}}{1.6\times 10^{-19}}=1.375\times 10^{-6}\text{kg/C}$. $\dot{m}=\frac{m}{q}I=(1.375\times 10^{-6})(15\times 10^{-6})=2.06\times 10^{-11}\text{kg/s}.$ $F=\dot{m}{v}_e=(2.06\times 10^{-11})(1.08\times 10^5)=2.23\times 10^{-6}\text{N}\approx 2.2\mu \text{N}.$ Both routes give $\approx 2.2\mu \text{N}$. The direct formula and the mass-flow picture are the same physics, so this agreement is the check we wanted. Lesson baked in: always sanity-check a µN device against the $\dot{m}{v}_e$ picture, and always collapse the exponent inside the root before square-rooting (see the L2 trap below for the classic slip).

$v_e=\sqrt{2qV/m}\propto 1/\sqrt{m}$: lighter is faster. So light ion $>$ indium $>$ caesium in $v_e$. (This is exactly the trade-off explored more in Ion Thrusters.)

What we do: compare via the formulas, holding $I$ and $V$ fixed. Thrust: $F=I\sqrt{2mV/q}\propto \sqrt{m}$. So $\frac{F_B}{F_A}=\sqrt{\frac{m_B}{m_A}}=\sqrt{1.16}\approx 1.077$. Caesium gives ~7.7% more thrust per amp. $I_{sp}$: $v_e\propto 1/\sqrt{m}$, so $\frac{I_{sp,B}}{I_{sp,A}}=\sqrt{\frac{m_A}{m_B}}=\frac{1}{\sqrt{1.16}}\approx 0.928$. Caesium gives ~7% less $I_{sp}$. What it means: heavy ions buy thrust (momentum per charge) at the cost of fuel economy. Neither is "better" — it depends on whether the mission is thrust-limited or propellant-limited.

Read the figure first. The green inward arrows on the cone flanks are the pull of surface tension — it wants the smallest possible surface area, so it tugs the liquid inward. The red outward arrows near the sharp apex are the electrostatic pressure produced by the electric field $E$: its size is $\frac{1}{2}{\epsilon }_0{E}^2$ (both $E$ and ${\epsilon }_0$ are defined in the symbol reminder). The yellow arc marks the half-angle, and the yellow arrows at the top are the ions being flung off the apex. The dashed circle marks a test point a distance $r$ from the apex where we weigh the two pulls.

Why surface tension scales as $1/r$. Surface tension produces an inward pressure equal to $\gamma$ (the tension) times the surface's curvature. On a cone, the surface at a slant distance $r$ from the apex curves around a circular cross-section whose radius is proportional to $r$ (the cone widens linearly as you move away from the tip). Curvature is one over that radius, so the curvature $\propto 1/r$. Hence the inward surface-tension pressure $\propto \gamma /r$. Closer to the tip (small $r$) the surface is more sharply curved, so tension squeezes harder.

Why the electrostatic pull also scales as $1/r$. A charged conductor concentrates its field wherever the surface is sharpest. Near a conical tip the electric field grows as you approach the apex like $E\propto r^{-1/2}$ (a standard result for a conductor with a conical point). The outward electrostatic pressure is $\frac{1}{2}{\epsilon }_0{E}^2$, and squaring $r^{-1/2}$ gives $E^2\propto r^{-1}$. So the outward pressure $\propto \frac{1}{2}{\epsilon }_0/r$ — it also blows up as $1/r$ toward the tip.

Why a cone specifically? Both the inward pull ($\propto \gamma /r$) and the outward pull ($\propto \frac{1}{2}{\epsilon }_0{E}^2\propto 1/r$) carry the same $1/r$ dependence. That is special: it means if they balance at one radius, they balance at every radius simultaneously. A cone is the unique self-similar shape where the red arrows and green arrows cancel all the way down to the tip. That balanced shape is the Taylor cone, half-angle $\approx 49.3^{\circ }$ (the yellow arc; more in Taylor Cone). A sphere gives a curvature that does not match the field's $1/r$ growth, so the field keeps winning at the tip and pulls it out further (unstable); a flat film has no sharp point, so the field never concentrates and nothing is emitted. Only the cone balances everywhere — and its apex is where the ions (yellow) are ripped off.

What $Re$ measures: $Re=\rho vL/\mu$ is (inertia of the flow)/(viscous friction). A small $Re$ (here 40) means friction dominates. (Contrast: an airliner wing has $Re\sim 10^7$.) Consequence: in a tiny channel the sticky boundary layer against the walls is a large fraction of the whole flow cross-section. That friction turns directed kinetic energy into heat and slows the gas. So the real exhaust speed is lower than 700 m/s, and efficiency drops below the ideal-expansion prediction. This is the fundamental MEMS penalty — see Reynolds Number.

(a) Propellant. Because $\Delta v\ll v_e$, use the small-$x$ approximation of the Rocket Equation, $e^{-x}\approx 1-x$: $m_p=m_0\left(1-e^{-\Delta v/v_e}\right)\approx m_0\frac{\Delta v}{v_e}=4\cdot \frac{12}{10^5}=4.8\times 10^{-4}\text{kg}=0.48\text{g}.$ Under half a gram — the payoff of huge $v_e$. (b) Firing time. Total impulse needed is $J=m_0\Delta v=4\times 12=48\text{kg⋅m/s}$ (using $m_0$ since propellant is negligible). Firing at constant $F$: $t=J/F$. $t=\frac{48}{2.0\times 10^{-6}}=2.4\times 10^7\text{s}=\frac{2.4\times 10^7}{86400}\approx 278\text{days}.$ What it means: micro-thrust means you fire almost constantly, gently, for most of the mission — precisely the "eyedropper, not firehose" picture from the parent note.

What we do: invert $F=\dot{m}{v}_e=(m/q)I{v}_e$ to solve for $I$. $I=\frac{Fq}{m{v}_e}=\frac{(2.0\times 10^{-6})(1.6\times 10^{-19})}{(1.9\times 10^{-25})(1.0\times 10^5)}.$ Numerator: $3.2\times 10^{-25}$. Denominator: $1.9\times 10^{-20}$. $I=\frac{3.2\times 10^{-25}}{1.9\times 10^{-20}}=1.68\times 10^{-5}\text{A}\approx 17\mu \text{A}.$ A ~17-microamp beam gives ~2 µN — and we can dial that current electronically down to nanoamps, which is why FEEP hits µN thrust resolution.

Set-up. At fixed $I,V,q$: thrust $F=I\sqrt{2mV/q}\propto \sqrt{m}$ (heavy wins thrust), and $v_e=\sqrt{2qV/m}\propto 1/\sqrt{m}$ (light wins $I_{sp}$).

(a) Thrust. Compute $\sqrt{2mV/q}$ (collapse the exponent inside the root first), then multiply by $I=20\times 10^{-6}$.

• Indium: $\frac{2(1.9\times 10^{-25})(7000)}{1.6\times 10^{-19}}=\frac{2.66\times 10^{-21}}{1.6\times 10^{-19}}=1.6625\times 10^{-2}$; $\sqrt{}=0.1289$; $F=20\times 10^{-6}\times 0.1289=2.58\times 10^{-6}\text{N}\approx 2.6\mu \text{N}$.
• Caesium: $\frac{2(2.2\times 10^{-25})(7000)}{1.6\times 10^{-19}}=\frac{3.08\times 10^{-21}}{1.6\times 10^{-19}}=1.925\times 10^{-2}$; $\sqrt{}=0.1387$; $F=20\times 10^{-6}\times 0.1387=2.77\times 10^{-6}\text{N}\approx 2.8\mu \text{N}$. Caesium gives more thrust (~2.8 vs ~2.6 µN), as expected from $F\propto \sqrt{m}$.

(b) $I_{sp}$. $v_e=\sqrt{2qV/m}$, then $I_{sp}=v_e/g_0$ with $g_0=9.81$.

• Indium: $v_e=\sqrt{\frac{2(1.6\times 10^{-19})(7000)}{1.9\times 10^{-25}}}=\sqrt{1.179\times 10^{10}}=1.086\times 10^5\text{m/s}$; $I_{sp}=1.086\times 10^5/9.81=1.107\times 10^4\text{s}$.
• Caesium: $v_e=\sqrt{\frac{2(1.6\times 10^{-19})(7000)}{2.2\times 10^{-25}}}=\sqrt{1.018\times 10^{10}}=1.009\times 10^5\text{m/s}$; $I_{sp}=1.009\times 10^5/9.81=1.029\times 10^4\text{s}$. Indium gives more $I_{sp}$ (~11,100 vs ~10,300 s), as expected from $v_e\propto 1/\sqrt{m}$.

(c) Propellant for $\Delta v=15\text{m/s}$, $m_0=5\text{kg}$, using the small-$x$ form $m_p\approx m_0\Delta v/v_e$:

• Indium: $m_p=5\cdot \frac{15}{1.086\times 10^5}=6.91\times 10^{-4}\text{kg}\approx 0.69\text{g}$.
• Caesium: $m_p=5\cdot \frac{15}{1.009\times 10^5}=7.43\times 10^{-4}\text{kg}\approx 0.74\text{g}$.

Decision. The propellant difference is ~0.05 g — utterly negligible on a 5 kg satellite. The thrust difference is also tiny (~0.2 µN). So the choice is not decided by these numbers; it is decided by engineering: indium melts at ~157 °C and does not chemically contaminate optics the way caesium (highly reactive, low work-function, plates onto surfaces) does. Fly indium — near-identical performance with far less contamination risk. The lesson: when two designs tie on the equations, the tie-breaker is the physics the equations left out.

What we do: minimum thrust $F_{\min }=(m/q)I_{\min }{v}_e$, then impulse bit $J_{\min }=F_{\min }\Delta t$. $F_{\min }=(1.19\times 10^{-6})(100\times 10^{-9})(1.0\times 10^5)=1.19\times 10^{-8}\text{N}\approx 12\text{nN}.$ $J_{\min }=F_{\min }\Delta t=(1.19\times 10^{-8})(1\times 10^{-3})=1.19\times 10^{-11}\text{N⋅s}\approx 12\text{pN⋅s}.$ Why this is the headline number: a ~$10^{-11}\text{N⋅s}$ impulse bit is what lets a CubeSat point to arc-second precision — the electronically-tunable beam current is what makes such absurdly small, repeatable "taps" possible. That precision, not raw thrust, is the entire reason FEEP exists.

The single sentence that ties every level together ::: Voltage sets exhaust speed and $I_{sp}$; current sets thrust; heavy ions trade $I_{sp}$ for thrust-per-amp; and once the numbers tie, real-world material properties decide.