Exercises — Ion engine — ionization, acceleration grid, neutralizer

Stage 1 — Ionization → done by the discharge chamber / hollow cathode (fires electrons that knock electrons off xenon atoms). Stage 2 — Acceleration → done by the grids (screen grid at high $+V$, accelerator grid at strong $-V$). Stage 3 — Neutralization → done by the neutralizer cathode (sprays electrons into the exiting beam). Mnemonic from the parent note: Ionize, Accelerate, Neutralize → "I Am Nimble".

False. An electric field only pushes charged objects. A neutral Xe atom feels no electric force — that is the entire reason Stage 1 (ionization) must happen first. Once it is Xe⁺ (charge $+q$), the field can grab it.

WHAT: all the electrical energy $qV$ becomes kinetic energy $\frac{1}{2}m{v}_e^2$ (work–energy theorem, see Work-Energy Theorem). WHY this formula: no burning, no heat loss in the ideal case, so energy in = kinetic energy out. $v_e=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.602\times 10^{-19})(1500)}{2.18\times 10^{-25}}}.$ Numerator: $2\times 1.602\times 10^{-19}\times 1500=4.806\times 10^{-16}$. Divide by mass: $4.806\times 10^{-16}/2.18\times 10^{-25}=2.204\times 10^9$. $v_e=\sqrt{2.204\times 10^9}\approx 4.70\times 10^4\text{ m/s}=46.9\text{ km/s}.$

WHAT: thrust = momentum leaving per second. Using $F=I\sqrt{2mV/q}$ from the parent note. WHY this form: beam current tells us how many ions leave per second ($N=I/q$), and each carries momentum $m{v}_e$. $F=I\sqrt{\frac{2mV}{q}}=2.0\sqrt{\frac{2(2.18\times 10^{-25})(1500)}{1.602\times 10^{-19}}}.$ Inside: $2\times 2.18\times 10^{-25}\times 1500=6.54\times 10^{-22}$; divide by $1.602\times 10^{-19}=4.082\times 10^{-3}$; square root $=0.06389$. $F=2.0\times 0.06389\approx 0.128\text{ N}.$ About the weight of a couple of paperclips.

$E_i=12.13\text{ eV}=12.13\times 1.602\times 10^{-19}\text{ J}\approx 1.94\times 10^{-18}\text{ J}.$ (See Plasma Physics — Ionization Energy for why this specific value.)

Exhaust speed scales as $v_e\propto \sqrt{V}$. Doubling $V$ multiplies $v_e$ by $\sqrt{2}\approx 1.41$. Energy per ion is $qV$, which scales linearly: doubling $V$ doubles the energy each ion costs. Analysis: you pay twice the energy to get only 1.41× the speed. Speed gets expensive fast — the square-root is a diminishing return. This is exactly why raising $V$ mainly buys efficiency, not free performance.

The curve is $v_e=\sqrt{2q/m}\sqrt{V}$ — a sideways square-root (a parabola on its side). Its steepness flattens as $V$ grows: near $V=0$ a small voltage boost adds a lot of speed, but far to the right the curve is nearly flat, so equal voltage steps add ever-smaller speed slices. Mathematically the slope is $\frac{d{v}_e}{dV}=\sqrt{\frac{q}{2mV}}$, which shrinks like $1/\sqrt{V}$. That shrinking slope is the diminishing return.

At equal speed $v$, momentum ratio = mass ratio (per particle, and there are equal numbers): $\frac{p_e}{p_{ion}}=\frac{m_e}{m}=\frac{9.11\times 10^{-31}}{2.18\times 10^{-25}}\approx 4.18\times 10^{-6}.$ So electrons carry about 4 millionths of the momentum — utterly negligible. And in reality electrons are added downstream at low speed, so their contribution is even smaller. Thrust is momentum, not charge, so the neutralizer fixes charge without stealing push.

(a) Mass flow. Number of ions per second $N=I/q$. Each has mass $m$, so $\dot{m}=\frac{I}{q}m=\frac{1.76}{1.602\times 10^{-19}}\times 2.18\times 10^{-25}.$ $I/q=1.099\times 10^{19}\text{ ions/s}$; times $m$: $\dot{m}=2.395\times 10^{-6}\text{ kg/s}$ ($\approx 2.4\text{ mg/s}$). (b) Exhaust speed. $v_e=\sqrt{\frac{2(1.602\times 10^{-19})(1200)}{2.18\times 10^{-25}}}=\sqrt{1.764\times 10^9}\approx 4.20\times 10^4\text{ m/s}.$ (c) Thrust, way 1: $F=\dot{m}{v}_e=2.395\times 10^{-6}\times 4.20\times 10^4\approx 0.1006\text{ N}$. Way 2: $F=I\sqrt{2mV/q}=1.76\times 0.05716\approx 0.1006\text{ N}$. ✅ They match — the two formulas are the same physics written differently.

WHAT $I_{sp}$ means: specific impulse (in seconds) $=v_e/g_0$ — how many seconds 1 kg of propellant weight could hover its own weight; a fuel-economy score. $I_{sp}=\frac{v_e}{g_0}=\frac{4.20\times 10^4}{9.81}\approx 4281\text{ s}.$ Chemical: $I_{sp}^{chem}=4500/9.81\approx 459\text{ s}$. Ratio: $4281/459\approx 9.3$. The ion engine is about 9× more fuel-efficient — same conclusion the parent note reached from raw speeds.

Burn time = propellant / mass flow: $t=\frac{M_p}{\dot{m}}=\frac{80}{2.395\times 10^{-6}}\approx 3.34\times 10^7\text{ s}.$ Convert to days: $3.34\times 10^7/86400\approx 387\text{ days}$ — over a year of thrusting. Tiny 0.1 N force, but running for a year is what makes ion engines mighty.

(a) Tsiolkovsky: $\Delta v=v_e\ln \frac{m_{initial}}{m_{final}}$, where $m_{initial}=580\text{ kg}$, $m_{final}=500\text{ kg}$. $\Delta v=4.20\times 10^4\times \ln \frac{580}{500}=4.20\times 10^4\times \ln (1.16).$ $\ln (1.16)=0.14842$, so $\Delta v=4.20\times 10^4\times 0.14842\approx 6234\text{ m/s}\approx 6.2\text{ km/s}$. (b) Starting acceleration: $a=F/m_{initial}=0.1006/580\approx 1.73\times 10^{-4}{\text{ m/s}}^2$. Insight: the acceleration is minuscule (~$1.8\times 10^{-5}g$), yet over a year of firing it accumulates to a $\Delta v$ that would eat an enormous chemical propellant load. Patience beats power in deep space.

First form: $F=\dot{m}{v}_e=\dot{m}\sqrt{2qV/m}$. Since $\dot{m}=(I/q)m$, substituting recovers $F=I\sqrt{2mV/q}$ — consistent. For fixed $\dot{m}$, indeed $F\propto \sqrt{V}$. Power form: beam power $P=IV$. From energy, each ion gets $qV$; power = (ions/s)$\times qV=(I/q)(qV)=IV$. Also $\frac{1}{2}\dot{m}{v}_e^2=P$ (kinetic power of the beam), so $F=\dot{m}{v}_e=\frac{2P}{v_e^2}\cdot v_e=\frac{2P}{v_e}.$ So at fixed power, $F=2P/v_e$: thrust is inversely proportional to exhaust speed. Faster exhaust ⇒ less thrust for the same power. This is the fundamental thrust–efficiency trade of all electric propulsion.

$F=\frac{2P}{v_e}=\frac{2\times 2100}{4.20\times 10^4}=\frac{4200}{42000}=0.100\text{ N}.$ ✅ Matches the momentum-based answer. Three independent routes ($\dot{m}{v}_e$, $I\sqrt{2mV/q}$, $2P/v_e$) all land on 0.10 N — a strong self-consistency check.