Visual walkthrough — FEEP, MEMS thrusters — micro-propulsion

3.3.43 · D2 · Physics › Rocket Propulsion › FEEP, MEMS thrusters — micro-propulsion

Hum is order mein build up karenge: ek charged liquid tip par → woh cone jo woh banata hai → ek ion voltage se girta hai → uski final speed → ions ki poori stream → woh force jo stream deliver karti hai → edge cases. Kuch bhi use nahi hoga pehle draw kiye bina.

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Step 1 — Ek liquid metal sharp needle par baitha hai

KYA. Socho ek chhoti si liquid metal ki drop (jaise indium) spacecraft ke andar ek needle ki bilkul tip par rakh hai. Abhi kuch nahi ho raha — koi field on nahi hai. Drop rounded hai kyunki surface tension hai.

YAH SE KYUN SHURU KAREIN. Baad mein aane wala har symbol (field $E$, voltage $V$, ion) is tip par hi paida hota hai. Agar hum pehle surface ko picture nahi karte, toh Step 2 ka cone kahin se bhi aa jaata hai.

PICTURE. Figure s01 dekho. Blue blob liquid hai; grey wedge needle hai. Do words abhi define karte hain:

Is stage par koi electricity nahi hai. Drop ek shant rounded cap hai. Iski shape yaad rakho — agla step ise deform kar dega.

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Step 2 — Field on karo: Taylor cone

KYA. Ab hum ek doori par ek doosri metal plate rakhte hain aur gap ke across ek bada voltage lagate hain. Isse ek electric field $E$ banta hai jo positive surface charges ko bahar ki taraf kheenchta hai.

YEH STEP KYUN. Ek hi surface par do opposite pulls ek aisi shape par settle ho jaate hain jahan woh balance karte hain. Woh balance shape ek round cap nahi hai — woh ek cone hai. Humein ise dekhna hai kyunki sharp cone tip hi woh jagah hai jahan se ions escape karte hain.

PICTURE. Figure s02 mein andar ki taraf mint arrows surface tension hain; bahar ki taraf lavender arrows electric pull hain. Jahan yeh balance karte hain wahan surface Taylor cone mein pinch ho jaati hai — ek cone jiska half-angle lagbhag $49.3^{\circ }$ hota hai (dekho Taylor Cone aur Electrospray & Colloid Thrusters).

Bahar ki taraf ka electric pull surface par ek pressure ki tarah kaam karta hai: $P_{\text{elec}}=\underset{\text{geometry}}{\underbrace{\frac{1}{2}}}\underset{\begin{array}{c}\text{vacuum}\\ \text{constant}\end{array}}{\underbrace{{\epsilon }_0}}\underset{\begin{array}{c}\text{field}\\ \text{squared}\end{array}}{\underbrace{E^2}}$

Har piece: $\frac{1}{2}$ field pressure ka ek fixed geometric factor hai; ${\epsilon }_0$ free space ka constant hai ($8.85\times 10^{-12}$); $E^2$ kehta hai ki pull field ke square ke saath badhti hai, toh voltage double karne se outward push chaar guna ho jaati hai. Jab yeh pressure $\gamma$ ki inward pull se match karta hai, cone stable ho jaata hai.

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Step 3 — Ek ion tip se tear off hota hai

KYA. Needle-sharp apex par field itni zyada strong hai ki woh liquid surface se ek ion pull off kar leti hai. Iske charge ko $q$ aur mass ko $m$ bolo. Jis moment yeh chhod ta hai, woh essentially rest par hota hai.

YEH STEP KYUN. Hum "beam" ki speed ek saath compute nahi kar sakte — woh bahut zyada hai. Toh hum ek ion mein zoom in karte hain aur poochte hain: yeh single ion kitni tez ho jaati hai? Ek sahi karo, aur stream sirf kai copies hain.

PICTURE. Figure s03 mein akela coral dot apex chhod kar jaata hai, aur field lines (lavender) woh rasta dikhati hain jis taraf woh accelerate karega.

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Step 4 — Ion voltage $V$ se girta hai aur speed pakad leta hai

KYA. Ion, ek baar free hone ke baad, voltage gap $V$ ke across drag hota hai. Woh gap cross karna stored electric potential energy ko motion energy (kinetic energy) mein convert karta hai.

ENERGY CONSERVATION KYUN, force × time nahi? Hum $F=ma$ use kar sakte the aur ion ko ek messy, non-uniform field mein step by step track kar sakte the — bahut painful. Energy conservation yah sab skip kar deta hai: use sirf start (rest par, energy $qV$ available) aur end (chal raha hai, energy $\frac{1}{2}m{v}_e^2$) ki parwah hai. Koi path details nahi chahiye. Yahi sahi tool hai is kaam ke liye.

PICTURE. Figure s04 "energy ramp" hai: ion upar se shuru hota hai (saari potential energy $qV$) aur neeche pahuncha hai (saari kinetic energy $\frac{1}{2}m{v}_e^2$), jaise ek marble slope se roll karti ho.

$\underset{\begin{array}{c}\text{energy given}\\ \text{by the voltage}\end{array}}{\underbrace{qV}}=\underset{\text{½}}{\underbrace{\frac{1}{2}}}\underset{\text{ion mass}}{\underbrace{m}}\underset{\begin{array}{c}\text{exit speed}\\ \text{squared}\end{array}}{\underbrace{v_e^2}}$

Exit speed $v_e$ (exhaust velocity) ke liye solve karo:

$\boxed{v_e=\sqrt{\frac{2qV}{m}}}$

Term by term: $2$ aata hai $\frac{1}{2}$ cancel karne se; upar $q$ aur $V$ kehte hain zyada charge ya zyada voltage → zyada fast; neeche $m$ kehta hai bhaare ions → slower (inhe speed up karna mushkil hai). Square root isliye hai kyunki energy mein $v^2$ tha, $v$ nahi.

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Step 5 — Ek ion se poori stream tak: current $I$

KYA. Ek thruster sirf ek ion nahi fire karta balki ek steady stream karta hai. Stream count karne ke do tarike:

• mass per second se, mass flow $\dot{m}$ ($\text{kg/s}$);
• charge per second se, beam current $I$ ($\text{amps}=\text{C/s}$).

UNHE LINK KYUN KAREIN. Hum current $I$ ko electronically measure aur dial kar sakte hain (yahi FEEP ko precise banata hai), lekin thrust momentum ke baare mein hai, jiske liye mass chahiye. Toh humein current ko mass flow mein translate karna hoga.

PICTURE. Figure s05: ions ki ek conveyor. Har ion charge count mein $q$ aur mass count mein $m$ contribute karta hai. Ek chhote time $\Delta t$ mein, $N$ ions nikalte hain, current $I=Nq/\Delta t$ aur mass flow $\dot{m}=Nm/\Delta t$ dete hain. Divide karo:

$\frac{\dot{m}}{I}=\frac{Nm/\Delta t}{Nq/\Delta t}=\frac{m}{q}⟹\dot{m}=\frac{m}{q}I$

Ise padhte hain: $\frac{m}{q}$ "mass carried per unit charge" hai. Current ko isse multiply karo aur tum charge-per-second ko mass-per-second mein convert kar lete ho.

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Step 6 — Thrust: har second nikalta momentum

KYA. Thrust Newton's third law in action hai: ions ek taraf shoot karte hain, spacecraft doosri taraf push hota hai. Push equal hoti hai momentum carry away per second ke.

AB COMBINE KYUN KAREIN. Hamare paas har piece hai: $v_e$ (Step 4) aur $\dot{m}=\frac{m}{q}I$ (Step 5). Dono ko $F=\dot{m}{v}_e$ mein substitute karo aur simplify karo.

PICTURE. Figure s06: beam satellite ko push karti hai, substitution chain arrows se dikhaayi gayi hai. Algebra dekho:

$F=\dot{m}{v}_e=\underset{\dot{m}}{\underbrace{\frac{m}{q}I}}\cdot \underset{v_e}{\underbrace{\sqrt{\frac{2qV}{m}}}}=I\sqrt{\frac{m^2}{q^2}\cdot \frac{2qV}{m}}=I\sqrt{\frac{2mV}{q}}$

$\boxed{F=I\sqrt{\frac{2mV}{q}}}$

Term by term: $I$ front mein — har second zyada ions, zyada push (yahi woh knob hai jo hum ghumate hain). $V$ root ke andar — zyada voltage, faster ions, zyada push. $m$ root ke andar — bhaare ions per charge zyada momentum carry karte hain, toh per amp zyada thrust (lekin lower $v_e$, toh lower $I_{sp}$: parent ne jo trade-off flag kiya tha). $q$ neeche — per ion zyada charge ka matlab hai per amp kam kg, kam thrust.

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Step 7 — Edge cases: kya hoga agar koi knob zero (ya bahut bada) ho jaye?

KYA. Ek formula tabhi trustworthy hai jab woh apne extremes par sahi behave kare. Har knob ko push karte hain.

KYUN. Reader ko kabhi aisa case nahi milna chahiye jo humne check na kiya ho. Micro-thrusters sach mein sabse halki nudges ke liye near-zero current par dial ki jaati hain — toh limits physical hain, academic nahi.

PICTURE. Figure s07 $F=I\sqrt{2mV/q}$ ko $I$ ke against (ek straight line) aur $V$ ke against (ek $\sqrt{V}$ curve) plot karta hai, special points marked hain.

• $I\to 0$ (koi ion fire nahi): $F\to 0$. Koi stream nahi, koi push nahi. Sahi hai — thruster idle hai. Exactly isi tarah FEEP sub-µN resolution reach karta hai: $I$ ko nanoamps tak down karo.
• $V\to 0$ (koi accelerating voltage nahi): $v_e=\sqrt{2qV/m}\to 0$ aur $F\to 0$. Ions barely chalte hain; koi useful push nahi. Tumhe kilovolts chahiye hi chahiye.
• $V$ bahut bada: $F$ $\sqrt{V}$ ki tarah badhta hai — slowly. $V$ double karne se thrust sirf $\sqrt{2}\approx 1.41$ se multiply hoti hai. Toh voltage se akele bada thrust nahi nikal sakte; zyada thrust ka sahi tarika hai zyada current $I$ (ek straight line, koi diminishing returns nahi).
• Bhaari ion, $m$ bada: $v_e\downarrow$ (worse $I_{sp}$) lekin per amp $F$ $\uparrow$. Na "hamesha acha" na "hamesha bura" — ek genuine trade-off. Yahi wajah hai ki parent bhaare ions ko useless kehne ke against warn karta hai.

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Worked numbers, picture par re-derived

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Ek-picture summary

Figure s08 poori chain ko ek single strip par compress karta hai: surface → Taylor cone → ek ion → $V$ se girta hai, $v_e$ gain karta hai → current $I$ ki stream, mass flow $\frac{m}{q}I$ → thrust $F=I\sqrt{2mV/q}$. Arrows ko left se right follow karo aur tumne FEEP re-derive kar liya.

Recall Feynman retelling — plain words mein bolo

Ek needle par liquid metal ka blob baitha hai. Strong electric field on karo: field surface ko bahar ki taraf kheenchti hai, surface tension andar ki taraf kheenchti hai, aur jahan woh balance karte hain wahan blob ek sharp cone mein pinch ho jaata hai. Us sharp tip se field individual charged atoms — ions — ko yank off kar leti hai. Har ion, ek baar free hone ke baad, voltage $V$ se "downhill" slide karta hai; energy conservation kehti hai ki use di gayi energy, $qV$, motion energy $\frac{1}{2}m{v}_e^2$ ban jaati hai, toh woh $v_e=\sqrt{2qV/m}$ par exit karta hai — tens of km/s. Ab poori stream fire karo: current $I$ charges per second count karta hai, aur kyunki har ion charge $q$ per mass $m$ carry karta hai, per second nikalna wala mass $\frac{m}{q}I$ hai. Thrust sirf momentum per second hai — mass-flow times speed — jo tidy ho ke $F=I\sqrt{2mV/q}$ ban jaata hai. Current ko nanoamps tak down karo aur push sub-micro-newtons tak girta hai — itna precise ki ek floating leaf ko nudge kar sako.

Recall Self-test

Ion ke liye $F=ma$ ki jagah energy conservation kyun? ::: Humein sirf start aur end energies chahiye, path ke saath non-uniform field ki messiness nahi. Thrust kaun set karta hai, $I_{sp}$ kaun set karta hai? ::: Current $I$ thrust set karta hai; voltage $V$ $v_e$ set karta hai aur isliye $I_{sp}$. $V$ double karne se thrust double kyun nahi hoti? ::: Thrust $\propto \sqrt{V}$ hai, toh yeh sirf $\sqrt{2}\approx 1.41$ se badhti hai. $I\to 0$ par $F$ ka kya hota hai? ::: $F\to 0$ — isi tarah FEEP sub-µN resolution reach karta hai.