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

3.3.40 · D3 · Physics › Rocket Propulsion › Electric propulsion — thrust, power, Isp trade-off

Definition Is page par use hone wale symbols (ek baar define, aage se yaad)

Neeche har worked example inhi ko reuse karta hai. Inhe abhi pakad lo taaki koi example kabhi surprise na kare:

$P_{in}$ ::: input power — electrical watts jo solar array ya reactor thruster ko deta hai.
$η$ (eta) ::: efficiency $= \frac{P _{j e t}}{P _{in}}$ — input power ka woh fraction jo actually exhaust beam mein jaata hai. Yeh 0 aur 1 ke beech ek pure number hai. Bacha hua $(1 - η) P_{in}$ heat ban jaata hai.
$v_{e}$ ::: exhaust speed (m/s) — propellant nozzle se kitni tez nikalti hai.
$I_{s p}$ ::: specific impulse (seconds) — fuel-economy score, $v_{e} = g_{0} I_{s p}$ se juda hua jahan $g_{0} = 9.81 m/s^{2}$ .
$\overset{m}{˙}$ ::: mass flow rate (kg/s) — har second kitne kilogram propellant pheka jaata hai.
$Δ v$ (delta-vee) ::: spacecraft ki velocity mein badlaav jo maneuver produce karna chahta hai (m/s) — woh "budget" jo mission ko pay karna hoga.
$m_{0}$ ::: initial mass — spacecraft plus saara propellant, burn se pehle.
$m_{f}$ ::: final mass — burn ke baad spacecraft, yaani $m_{0}$ minus use kiya gaya propellant.

1. Scenario matrix

Kuch bhi calculate karne se pehle, har tarah ka case list karte hain jo is topic mein aa sakta hai. Har row ek "cell" hai; neeche har worked example tagged hai us cell(s) ke saath jo woh cover karta hai.

#	Cell (case class)	Tricky kyun hai	Covered by
A	Ordinary forward calc	Diya $P_{in}, η, I_{s p}$ → nikalo $F, \overset{m}{˙}$	Ex 1
B	Inverse calc	Diya chaahiye thrust → nikalo required power / $I_{s p}$	Ex 2
C	Fixed power, vary $I_{s p}$	Seesaw: kya thrust waise move karta hai jaisa tumne predict kiya?	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 — bahut zyada thrust, buri economy	Ex 4
F	Efficiency & waste heat	$P_{in}$ ko beam vs heat mein baanto; Thermal Control (radiators) se juda	Ex 5
G	Real-world word problem	Mission: target $Δ v$ tak pahunchne mein kitna time lagega? Tsiolkovsky Rocket Equation use hoga	Ex 6
H	Exam twist / trap	Units trap: $g_{0}$ bhool jaana, ya "input vs jet power"	Ex 7
I	Comparison across regimes	Same $Δ v$ par EP vs chemical — propellant ki bachat	Ex 8
J	Boundary: $P_{in} \to 0$ ya $η \to 0$	Zero power / zero efficiency — physical impossibility	Ex 9

Ab hum har cell clear karte hain.

2. Worked examples

Cell A — seedha forward calculation

Worked example Example 1 — Hall thruster ki poori state (Cell A)

Ek Hall thruster $P_{in} = 4.5 kW$ , efficiency $η = 0.55$ , aur $I_{s p} = 1800 s$ par run karta hai. Exhaust speed $v_{e}$ , thrust $F$ , mass flow $\overset{m}{˙}$ , aur jet power $P_{j e t}$ nikalo.

Forecast: Yahan $I_{s p}$ (1800 s) parent ke 3000 s ion engine se kam hai, to expect karo zyada thrust lekin bada mass flow. Guess: kuch sau mN.

Step 1 — economy score ko real speed mein badlo. $v_{e} = g_{0} I_{s p} = 9.81 \times 1800 = 1.766 \times 1 0^{4} m/s$ . Yeh step kyun? Neeche har equation speed m/s mein chahti hai, seconds mein score nahi. $I_{s p}$ bas $v_{e}$ hai jo $g_{0} = 9.81 m/s^{2}$ se rescale kiya gaya hai.

Step 2 — trade-off se thrust. $F = \frac{2 η P _{in}}{v _{e}} = \frac{2 ( 0.55 ) ( 4500 )}{1.766 \times 1 0 ^{4}} = 0.280 N = 280 mN$ . Yeh step kyun? Trade-off equation wall power aur speed se seedha thrust deta hai; abhi $\overset{m}{˙}$ jaanna zaroori nahi.

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

Step 4 — jet power. $P_{j e t} = η P_{in} = 0.55 \times 4500 = 2475 W$ . Yeh step kyun? Efficiency $η = P_{j e t} / P_{in}$ definition se, to beam wall power ka $η$ fraction carry karti hai.

Verify: Cross-check karo $P_{j e t} = \frac{1}{2} F v_{e} = \frac{1}{2} (0.280) (1.766 \times 1 0^{4}) = 2472 W$ ✓ (2475 W se rounding ke andar match karta hai). Forecast sahi nikla: kam $I_{s p}$ → 280 mN vs parent mein 102 mN. ✓

Cell B — equations ko ulta chalana

Worked example Example 2 — "Mujhe itna thrust chahiye" (Cell B)

Ek mission ko $F = 150 mN$ chahiye $I_{s p} = 2500 s$ aur $η = 0.6$ ke saath. Input power kitna hona chahiye jo solar array deliver kare?

Forecast: Ex 1 se zyada $I_{s p}$ lekin similar thrust — thrust ko power aur speed dono chahiye, to expect karo kuch kW.

Step 1 — speed. $v_{e} = 9.81 \times 2500 = 2.4525 \times 1 0^{4} m/s$ . Yeh step kyun? Trade-off equation ulta karne ke liye $v_{e}$ chahiye.

Step 2 — power ke liye invert karo. $F = \frac{2 η P _{in}}{v _{e}}$ se $P_{in} = \frac{F v _{e}}{2 η}$ nikalo. Yeh step kyun? Hume thrust diya gaya hai aur power chahiye, to $P_{in}$ ko subject banao.

Step 3 — plug in karo. $P_{in} = \frac{( 0.150 ) ( 2.4525 \times 1 0 ^{4} )}{2 ( 0.6 )} = 3066 W \approx 3.07 kW$ . Yeh step kyun? Numbers seedha daal do.

Verify: Forward run karo: $F = \frac{2 ( 0.6 ) ( 3066 )}{2.4525 \times 1 0 ^{4}} = 0.150 N$ ✓. Units: $\frac{N \cdot ( m/s )}{1} = W$ (kyunki $N\cdotm/s = W$ ) ✓.

Cell C — fixed-power seesaw

Neeche ka figure thrust $F$ (mN, vertical axis) ko specific impulse $I_{s p}$ (seconds, horizontal axis) ke against plot karta hai fixed power $P_{in} = 4.5 kW$ aur $η = 0.55$ par. Lavender curve hai $F = 2 η P_{in} / (g_{0} I_{s p})$ — ek girta hua hyperbola. Kaise padhein: bottom axis par koi bhi $I_{s p}$ chuno, curve tak upar jao, left axis par thrust padho. Coral dot Ex 1 mark karta hai (1800 s, 280 mN); mint dot Ex 3 mark karta hai (5400 s, 93 mN). Right slide karna (zyada economy) hamesha neeche lejata hai (kam thrust).

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

Worked example Example 3 — Economy teen guna karo, thrust ka kya hoga? (Cell C)

$P_{in} = 4.5 kW$ , $η = 0.55$ fixed rakho (Ex 1 jaisa hardware). $I_{s p}$ 1800 s se 5400 s (teen guna) kar do. Nayi thrust predict karo, phir compute karo.

Forecast: Fixed power par $F \propto 1/ I_{s p}$ . Teen guna $I_{s p}$ → thrust ek tihai ho jaani chahiye 280 mN ki ≈ 93 mN. Figure mein curve dekho — hum coral dot se right aur neeche same hyperbola par mint dot par slip karte hain.

Step 1 — nayi speed. $v_{e} = 9.81 \times 5400 = 5.297 \times 1 0^{4} m/s$ . Yeh step kyun? Nayi economy score → nayi speed; power aur $η$ untouched.

Step 2 — nayi thrust. $F = \frac{2 ( 0.55 ) ( 4500 )}{5.297 \times 1 0 ^{4}} = 0.0934 N = 93.4 mN$ . Yeh step kyun? Same trade-off equation; sirf $v_{e}$ badla.

Verify: Ratio $\frac{93.4}{280} = 0.334 \approx \frac{1}{3}$ ✓ — bilkul $I_{s p}$ ke ×3 ka inverse. Mass flow, meanwhile, aur tezi se girta hai: $\overset{m}{˙} = F / v_{e}$ isliye girti hai kyunki $F$ chota hua aur $v_{e}$ bada hua. Yahi hai fixed power par seesaw, lavender curve par coral point ko mint point tak slide hote dekhna.

Cells D & E — do degenerate extremes

Neeche ka figure do curves ko exhaust speed $v_{e}$ (km/s, horizontal axis) ke against plot karta hai fixed $P_{in} = 4.5 kW$ , $η = 0.55$ par. Lavender solid curve (left axis) thrust $F$ in mN hai; coral dashed curve (right axis) mass flow $\overset{m}{˙}$ in mg/s hai. Kaise padhein: bahut left par ( $v_{e} \to 0$ ) dono curves upar jaati hain — chemical corner. Bahut right par ( $v_{e} \to \infty$ ) dono zero ki taraf sinkti hain — photon corner. Dashed coral curve solid lavender se tezi se girti hai, dikhata hai ki exhaust speed badhane par $\overset{m}{˙}$ $F$ se zyada tezi se collapse karta hai.

Worked example Example 4 — Seesaw ke do ends (Cells D & E)

Same fixed $P_{in} = 4.5 kW$ , $η = 0.55$ . Limits $I_{s p} \to \infty$ (photon-like) aur $v_{e} \to 0$ (chemical-like) examine karo. $F$ aur $\overset{m}{˙}$ ka kya hota hai?

Forecast: Exhaust ko infinitely fast push karo aur thrust vanish honi chahiye (almost koi mass nahi phekta). Exhaust ko bahut dhire push karo aur mass ka torent chahiye — bada $\overset{m}{˙}$ , bada $F$ lekin buri economy.

Step 1 — limit $v_{e} \to \infty$ ( $I_{s p} \to \infty$ ). $F = \frac{2 η P _{in}}{v _{e}}$ mein, jab $v_{e} \to \infty$ , $F \to 0$ . Yeh step kyun? Fixed power jo infinitely fast particles par spread ho matlab almost koi mass har second nahi jaata, to momentum-per-second → 0. Yeh maximum-economy, zero-push corner hai (figure ka right edge).

Step 2 — wahan $\overset{m}{˙}$ check karo. $\overset{m}{˙} = F / v_{e} \to 0/\infty = 0$ : essentially koi propellant use nahi hua. Isliye high $I_{s p}$ = fuel "forever" chalta hai — price hai koi thrust nahi.

Step 3 — limit $v_{e} \to 0$ . $F = \frac{2 η P _{in}}{v _{e}} \to \infty$ aur $\overset{m}{˙} = \frac{2 η P _{in}}{v _{e}^{2}} \to \infty$ aur bhi tezi se. Yeh step kyun? Dheema, bhaari exhaust fixed power ko bade momentum flux mein convert karta hai — lekin tum propellant barbad rate par dump karte ho. Yeh maximum-thrust, worst-economy corner hai (figure ka left edge, chemical end of the spectrum).

Step 4 — extremes ke beech ek concrete number. Moderate $v_{e} = 1000 m/s$ ( $I_{s p} \approx 102 s$ ) par: $F = \frac{2 ( 0.55 ) ( 4500 )}{1000} = 4.95 N$ — Ex 1 thrust ka 17× lagbhag — lekin $\overset{m}{˙} = F / v_{e} = 4.95/1000 = 4.95 \times 1 0^{- 3} kg/s = 4.95 g/s$ , Ex 1 mass flow ka roughly 310×. Zabardast push, katastrofic mileage.

Verify: Figure $F$ vs $v_{e}$ plot karta hai: ek rectangular hyperbola dono axes se pinned — $F \to \infty$ jab $v_{e} \to 0$ (left) aur $F \to 0$ jab $v_{e} \to \infty$ (right). Dono asymptotes confirm hue, aur dashed $\overset{m}{˙}$ curve right par tezi se sinkti hai, $\overset{m}{˙} \propto 1/ v_{e}^{2}$ vs $F \propto 1/ v_{e}$ se match karta hai. ✓

Cell F — waste power kahan jaata hai

Worked example Example 5 — Waste heat budget (Cell F)

Ex 1 Hall thruster ( $P_{in} = 4.5 kW$ , $η = 0.55$ ). Kitna power heat ke roop mein jaata hai jo radiators ko reject karna padega? Agar radiators $200 W/m^{2}$ shed karte hain, to sirf is thruster ke waste ke liye kitna radiator area chahiye?

Forecast: $η = 0.55$ matlab 45% lost → lagbhag 2 kW heat → kai square metre.

Step 1 — beam vs waste. $P_{j e t} = η P_{in} = 0.55 \times 4500 = 2475 W$ ; waste $= P_{in} - P_{j e t} = (1 - η) P_{in} = 0.45 \times 4500 = 2025 W$ . Yeh step kyun? Energy conservation: jo beam mein nahi jaata woh heat banta hai.

Step 2 — radiator area. $A = \frac{2025}{200} = 10.1 m^{2}$ . Yeh step kyun? Area × flux = power rejected, to area = power ÷ flux.

Verify: $P_{j e t} + P_{w a s t e} = 2475 + 2025 = 4500 W = P_{in}$ ✓ (kuch lost ya create nahi hua). Units: $W \div (W/m^{2}) = m^{2}$ ✓.

Cell G — ek mission word problem

Worked example Example 6 — Orbit raise karne mein kitna time lagega? (Cell G)

Ek 1200 kg spacecraft 300 kg xenon propellant carry karta hai. Uska thruster $F = 280 mN$ deta hai (Ex 1 ki value) $\overset{m}{˙} = 15.9 mg/s$ ke saath. Mission ko velocity budget $Δ v = 4.0 km/s$ chahiye (page ke top se recall karo: $Δ v$ woh badlaav hai jo maneuver spacecraft ki velocity mein produce karna chahta hai). (a) Kya iske paas kaafi propellant hai? (b) Roughly burn kitna time lega?

Forecast: Tiny thrust → yeh months lega, minutes nahi. Xenon precious hai, to pehle rocket equation check karo.

Step 1 — rocket equation se propellant needed. $v_{e} = 1.766 \times 1 0^{4} m/s$ (Ex 1) ke saath, Tsiolkovsky deta hai $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{4000/17660} = e^{0.2265} = 1.2542$ , jahan $m_{0}$ initial (loaded) mass hai aur $m_{f}$ final (burned-out) mass. Yeh step kyun? Rocket equation target $Δ v$ ke liye mass ratio batata hai — fundamental fuel-economy check.

Step 2 — final vs initial mass. Yahan $m_{0} = 1200 kg$ (ship + saara propellant), to $m_{f} = 1200/1.2542 = 956.8 kg$ . Propellant burned $= m_{0} - m_{f} = 1200 - 956.8 = 243.2 kg$ . Yeh step kyun? $m_{0} - m_{f}$ woh propellant hai jo maneuver actually consume karta hai.

Step 3 — kaafi hai? Needed 243.2 kg < 300 kg carried. Haan, ~57 kg margin ke saath. ✓

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

Step 5 — convert karo. $1.534 \times 1 0^{7} s \div (86400 s/day) = 177.6 days \approx 5.9 months$ . Yeh step kyun? Seconds unreadable hain; days EP ka deep-space cruise character dikhate hain.

Verify: Sanity: impulse $= F \times t = 0.280 \times 1.534 \times 1 0^{7} = 4.30 \times 1 0^{6} N\cdots$ ; aur $Δ v \approx$ impulse $/ \overset{m}{ˉ}$ mean mass $\overset{m}{ˉ} \approx (1200 + 956.8) /2 = 1078 kg$ ke saath deta hai $4.30 \times 1 0^{6} /1078 \approx 3.99 \times 1 0^{3} m/s \approx 4.0 km/s$ ✓. ~6-month burn exactly isliye hai kyunki EP cruising ke liye hai, launch ke liye nahi.

Cell H — exam trap

Worked example Example 7 — Units trap pakdo (Cell H)

Exam statement: "Ek thruster ka $I_{s p} = 2000 s$ , $η = 0.7$ , $P_{in} = 3 kW$ hai. Ek student likhta hai $F = \frac{2 η P _{in}}{I _{s p}} = \frac{2 ( 0.7 ) ( 3000 )}{2000} = 2.1 N$ . Error dhundo aur sahi $F$ nikalo."

Forecast: Student ne $v_{e}$ (m/s) ki jagah $I_{s p}$ (seconds) se divide kiya. Kyunki $v_{e} = g_{0} I_{s p}$ lagbhag 9.81× bada hai, sach mein thrust ~9.81× chota hoga — around 0.2 N, 2.1 N nahi.

Step 1 — error name karo. Trade-off equation hai $F = \frac{2 η P _{in}}{v _{e}}$ , nahi $\frac{2 η P _{in}}{I _{s p}}$ . $I_{s p}$ seconds mein hai; pehle speed mein convert karna hoga. Yeh step kyun? Dimensional check: $\frac{W}{s} = \frac{J/s}{s} = J/s^{2}$ — newtons nahi. Student ka answer dimensionally galat hai.

Step 2 — convert karo. $v_{e} = g_{0} I_{s p} = 9.81 \times 2000 = 1.962 \times 1 0^{4} m/s$ . Yeh step kyun? Missing $g_{0}$ restore karo.

Step 3 — sahi thrust. $F = \frac{2 ( 0.7 ) ( 3000 )}{1.962 \times 1 0 ^{4}} = 0.2141 N \approx 214 mN$ . Yeh step kyun? Ab units kaam karte hain: $\frac{W}{m/s} = \frac{J/s}{m/s} = J/m = N$ ✓.

Verify: Galat aur sahi ka ratio $= 2.1/0.2141 = 9.81 = g_{0}$ ✓ — exactly woh factor jo student ne drop kiya.

Cell I — propulsion regimes across comparison

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

Do stages ko dono ko $Δ v = 6.0 km/s$ deliver karna hai ek 1000 kg dry spacecraft ko (to dono cases mein final mass $m_{f} = 1000 kg$ ). Stage A chemical hai ( $I_{s p} = 350 s$ ); Stage B electric hai ( $I_{s p} = 3000 s$ ). Dono ko kitna propellant chahiye compare karo.

Forecast: EP ka ~8.6× zyada $I_{s p}$ propellant dramatically kam karega — chemical ko tonnes chahiye honge, EP ko kuch sau kg.

Step 1 — exhaust speeds. Chemical: $v_{e} = 9.81 \times 350 = 3434 m/s$ . Electric: $v_{e} = 9.81 \times 3000 = 2.943 \times 1 0^{4} m/s$ . Yeh step kyun? Rocket equation $v_{e}$ use karta hai, to dono economy scores convert karo.

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

Step 3 — initial masses aur propellant burned. $m_{f} = 1000 kg$ ke saath: chemical $m_{0} = 5.739 \times 1000 = 5739 kg$ , to propellant $= m_{0} - m_{f} = 5739 - 1000 = 4739 kg$ . Electric $m_{0} = 1.2262 \times 1000 = 1226.2 kg$ , to propellant $= 1226.2 - 1000 = 226.2 kg$ . Yeh step kyun? Propellant $= m_{0} - m_{f}$ dono ke liye — yeh number hai jo mission actually pay karta hai.

Step 4 — payoff. Ratio $= \frac{4739}{226.2} = 21.0$ : chemical stage same kaam ke liye ~21× zyada propellant jaalta hai. Yeh step kyun? Yeh single number hi reason hai kyun EP exist karta hai — high $I_{sp}$ enormous mass bachata hai.

Verify: EP forward-check: $Δ v = v_{e} ln (m_{0} / m_{f}) = 29430 \times ln (1.2262) = 29430 \times 0.2039 = 6001 m/s \approx 6.0 km/s$ ✓. Trade-off ka cost — Ex 6 yaad karo — hai ki EP months leta hai, jabki chemical minutes mein burn karta hai.

Cell J — boundary cases: zero power, zero efficiency

Worked example Example 9 —

$P_{in} \to 0$ aur $η \to 0$ par kya hota hai? (Cell J) $I_{s p} = 1800 s$ ( $v_{e} = 1.766 \times 1 0^{4} m/s$ ) rakho. Input power $P_{in}$ ko zero ki taraf push karo; alag se efficiency $η$ ko zero ki taraf push karo. Kitna thrust milta hai, aur physically kya matlab hai?

Forecast: Dono "engine off" corner hain — zero power in, ya uska kuch bhi beam tak nahi pahuncha, matlab zero thrust. Formula exactly zero dena chahiye, aur physically ship sirf coast karti hai.

Step 1 — limit $P_{in} \to 0$ . $F = \frac{2 η P _{in}}{v _{e}}$ mein, $η, v_{e}$ fixed aur finite ke saath, $F \to \frac{2 η \cdot 0}{v _{e}} = 0$ . Yeh step kyun? Thrust powered momentum flux hai. Zero electrical input ke saath propellant accelerate karne ki koi energy nahi, to kuch nahi pheka jaata — $F = 0$ . Yeh physical statement hai "unpowered EP thruster koi thrust produce nahi karta," matrix ka natural floor.

Step 2 — limit $η \to 0$ . $F = \frac{2 η P _{in}}{v _{e}}$ mein, $P_{in}, v_{e}$ fixed ke saath, $F \to \frac{2 \cdot 0 \cdot P _{in}}{v _{e}} = 0$ , jabki heat $= (1 - η) P_{in} \to P_{in}$ . Yeh step kyun? $η = 0$ matlab input power ka kuch bhi beam tak nahi pahuncha — sab waste heat ban jaata hai (Ex 5 ka leftover extreme par le jaana). Perfectly inefficient thruster bas ek electric heater hai: zero thrust, full power heat ke roop mein dump. Physically rocket ke roop mein impossible, lekin yeh efficiency axis ko bound karta hai.

Step 3 — chhoti-lekin-nonzero value par sanity. $P_{in} = 1 W$ , $η = 0.55$ par: $F = \frac{2 ( 0.55 ) ( 1 )}{1.766 \times 1 0 ^{4}} = 6.23 \times 1 0^{- 5} N = 62 μ N$ — vanishingly small, confirm karta hai ki thrust smoothly 0 ki taraf approach karta hai jab power girti hai. Yeh step kyun? Dikhata hai limit ek jump nahi hai — thrust continuously zero ki taraf glide karta hai, to boundary ke paas koi hidden surprise nahi.

Verify: Dono limits $F = 0$ dete hain: $P_{in} = 0$ ya $η = 0$ substitute karo $F = 2 η P_{in} / v_{e}$ mein exactly $0$ milta hai ✓. Heat check $η \to 0$ par: $(1 - 0) P_{in} = P_{in}$ — saara input heat banta hai ✓. Physically: zero power ya zero efficiency se thrust nahi milti, matrix ka bottom-left corner close karta hai.

3. Matrix coverage check

Recall Kaun se example ne kaun sa cell clear kiya?

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 ( $P_{in} \to 0$ ya $η \to 0$ boundary) ::: Example 9

Recall Ek-line lessons

Fixed power thrust law ::: $F = 2 η P_{in} / v_{e}$ , to $F \propto 1/ I_{s p}$ (Ex 3). $v_{e} \to 0$ corner ::: $F \to \infty$ lekin $\overset{m}{˙} \to \infty$ aur tezi se — koi free lunch nahi (Ex 4). Units trap ::: $v_{e} = g_{0} I_{s p}$ se divide karo, kabhi bhi sirf $I_{s p}$ se nahi (Ex 7). EP mass kyun bachata hai ::: high $I_{s p}$ mass ratio $e^{Δ v / v_{e}}$ ko chota karta hai (Ex 8). Zero-power / zero-efficiency corner ::: $F \to 0$ ; beam tak power pahunche bina push nahi milti (Ex 9).

4. Connections

Parent topic (Hinglish) — core derivations jinhe yeh examples exercise karte hain.
Tsiolkovsky Rocket Equation — Ex 6 aur Ex 8 mein use hua $Δ v$ ko propellant mein badalne ke liye.
Specific Impulse — $I_{s p} = v_{e} / g_{0}$ conversion jo har trap aur forecast ke core mein hai.
Chemical vs Electric Propulsion — Ex 4 ke do ends aur Ex 8 mein comparison.
Ion and Hall Thrusters — Ex 1 aur Ex 6 mein hardware.
Spacecraft Power Systems — woh $P_{in}$ supply karta hai jise hum Ex 2 mein invert karte hain aur Ex 9 mein zero tak drive karte hain.
Thermal Control (radiators) — Ex 5 aur Ex 9 mein compute kiya gaya waste heat absorb karta hai.
Newton's Third Law — $F = \overset{m}{˙} v_{e}$ ka origin jo poore mein use hota hai.