Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX - LH2 (~450s), ion engines (~3000s): Worked examples

3.3.5 · D3 · Physics › Rocket Propulsion › Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX -

Yeh page parent topic ki practice ground hai. Kuch bhi compute karne se pehle, hum har tarah ke questions lay out karte hain jo specific-impulse ka idea tumhare saamne rakh sakta hai. Phir hum ek example har box mein solve karte hain, taaki koi bhi scenario exam mein tumhe surprise na kare.

Jo kuch bhi hum use karte hain woh parent note mein build kiya gaya tha. Ek-line reminder ke roop mein, teen tools jo hum baar baar use karte hain:

Definition Do alag "g" symbols — inhe confuse mat karo

Is poori page ka hinge in do dikh-alike symbols ko alag rakhne par hai:

$g_{0} = 9.81 m/s^{2}$ ek fixed unit-conversion constant hai. Yeh $I_{s p}$ ki definition mein baked in hai taaki answer seconds mein aaye. Yeh kabhi nahi badalta — Earth par, Moon par, ya deep space mein.
$g_{local}$ woh actual gravitational field hai jahan rocket physically hai (Earth surface $\approx 9.81$ , Moon $\approx 1.62 m/s^{2}$ , deep space $\approx 0$ ). Hum $g_{local}$ sirf tab use karte hain jab real weight $W = m g_{local}$ compute karna ho lift-off comparison ke liye.

Dono ka number $9.81$ Earth's surface par same hota hai, aur yehi wajah hai ki dono confuse ho jaate hain (Example 9 dekho). Jab bhi tum " $g$ " dekho, pucho: kya main units convert kar raha hoon ( $g_{0}$ ) ya kuch weigh kar raha hoon ( $g_{local}$ )?

The scenario matrix

$I_{s p}$ ke baare mein har sawaal in case classes mein se kisi ek mein aata hai. Last column us example ka naam deta hai jo ise cover karta hai.

#	Case class	Isme kya special hai	Covered by
A	Forward: $I_{s p} \to v_{e}$	Seedha multiply	Ex 1
B	Backward: $v_{e} \to I_{s p}$	Divide (inverse direction)	Ex 2
C	Thrust from flow rate	Do relations combine karo	Ex 3
D	Same- $Δ v$ comparison across engines	Ratio / exponential, do engines	Ex 4
E	Degenerate input: $\overset{m}{˙} \to 0$ (ion limit)	Huge $I_{s p}$ ke bawajood tiny thrust	Ex 5
F	Degenerate input: no fuel burned ( $m_{0} = m_{f}$ )	$Δ v \to 0$ , "zero" case	Ex 6
G	Limiting/ceiling case — chemical cap near 450 s	Kyun $M$ (molar mass) wall set karta hai	Ex 7
H	Real-world word problem — staging choice	Mission ke liye engine choose karo	Ex 8
I	Exam twist — the $g_{0}$ trap (Moon)	Constant vs local gravity	Ex 9

Neeche figure ko kaise padhein. Yeh woh map hai jise hum baar baar point karte hain. Horizontal axis specific impulse $I_{s p}$ hai (efficiency knob) aur vertical axis ek representative thrust $F$ hai (force knob, log scale par taaki bada range fit ho). Charo coloured dots mein se har ek parent table se ek engine type hai. Dashed arrow ek fundamental trade-off trace karta hai: jab tum right move karte ho (zyada efficient, higher $I_{s p}$ ) toh neeche bhi move karte ho (less thrust). Examples 3, 5 aur 8 sab actually is baare mein hain ki sawaal kis axis ki parwah karta hai — toh inme se har ek se pehle is map par ek nazar daalo.

Figure — Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX - LH2 (~450s), ion engines (~3000s)

(Alt text: char rocket engines ka log–log scatter. Magenta "Solid ~260 s" top-left mein baitha hai (high thrust, low efficiency); orange "LOX/RP-1 ~311 s" thoda neeche aur right; violet "LOX/LH2 ~450 s" aur neeche aur further right; navy "Ion ~3000 s" far right aur bottom par (huge efficiency, near-zero thrust). Ek dashed violet arrow top-left se bottom-right tak jaata hai, trade-off ke label ke saath.)

Case A — Forward direction ( $I_{s p} \to v_{e}$ )

Worked example Example 1 — Hydrolox engine ki exhaust speed

Ek LOX/LH2 engine $I_{s p} = 450$ s par rated hai. Uski effective exhaust velocity $v_{e}$ nikalo. (Map par: chemical cluster ke top-left mein violet dot.)

Forecast: Table kehti hai hydrolox exhaust roughly $4, 400$ m/s par nikalti hai. Compute karne se pehle guess karo — kya $450$ ko $\sim 10$ se multiply karne par $4, 500$ ke paas pahunchoge? (Haan — yeh ek achhi aadat hai: $g_{0} \approx 10$ .)

Master relation likho: $v_{e} = I_{s p} g_{0}$ . Yeh step kyun? $I_{s p}$ ko define kiya gaya tha $v_{e} / g_{0}$ ke roop mein, isliye $g_{0}$ se multiply karna simply definition ko undo karta hai aur physical speed wapas deta hai.
Numbers substitute karo: $v_{e} = 450 \times 9.81$ . Yeh step kyun? Hum given $I_{s p}$ aur fixed constant $g_{0}$ plug in karte hain taaki sirf ek number compute karna bache.
Compute karo: $v_{e} = 4414.5 m/s \approx 4.41 km/s$ . Yeh step kyun? Multiply out karne se symbolic relation physical answer mein badal jaati hai jiske baare mein pucha gaya tha.

Verify: Units check — $[s] \times [m/s^{2}] = [m/s]$ , ek velocity. ✓ Aur $4414.5$ parent table ke $\sim 4410$ m/s se match karta hai. ✓

Case B — Backward direction ( $v_{e} \to I_{s p}$ )

Worked example Example 2 — Test stand par measure kiya gaya ek naya engine

Engineers ek effective exhaust velocity $v_{e} = 3050$ m/s measure karte hain. Yeh kaisa $I_{s p}$ hai, aur yeh kis propellant se match karta hai? (Map par: yeh orange dot par land karta hai.)

Forecast: $3050$ m/s kerolox ( $\sim 3050$ ) aur solid ( $\sim 2550$ ) ke beech hai. Predict karo answer around $311$ s hoga.

Master relation ko rearrange karo $I_{s p}$ isolate karne ke liye: $I_{s p} = \frac{v _{e}}{g _{0}}$ . Yeh step kyun? Ab hum $v_{e}$ jaante hain aur $I_{s p}$ chahiye — Example 1 ka inverse. Divide karna multiply ko undo karta hai.
Substitute karo: $I_{s p} = \frac{3050}{9.81}$ . Yeh step kyun? Hum measured $v_{e}$ aur fixed $g_{0}$ drop in karte hain taaki division purely numeric ho.
Compute karo: $I_{s p} = 310.9 s \approx 311 s$ . Yeh step kyun? Division carry out karne se seconds figure milta hai jise hum benchmark table se match kar sakte hain.

Verify: Wapas multiply karo — $311 \times 9.81 = 3050.9$ m/s, hamara input recover hota hai. ✓ $311$ s exactly LOX/RP-1 (kerolox) benchmark hai. ✓

Case C — Flow rate se Thrust

Worked example Example 3 — Ek bade kerolox engine se force

Ek LOX/RP-1 engine $\overset{m}{˙} = 500$ kg/s par propellant burn karta hai with $I_{s p} = 311$ s. Yeh kitna thrust produce karta hai? (Map par: orange dot, lekin ab hum vertical (thrust) axis padhte hain, horizontal nahi.)

Forecast: Ek first-stage engine. Guess karo: kya yeh $1$ N, $1, 000$ N, ya $1, 000, 000$ N ke paas hai? (Launch engine ko kai tonnes lift karna hota hai — toh millions of newtons.)

Do relations chain karo: $F = \overset{m}{˙} v_{e} = \overset{m}{˙} (I_{s p} g_{0})$ . Yeh step kyun? Thrust momentum per second thrown hai, $\overset{m}{˙} v_{e}$ — exactly woh relation jo humne upar intuition callout mein build kiya. Hum $v_{e}$ directly nahi jaante, lekin $v_{e} = I_{s p} g_{0}$ se milta hai.
Substitute karo: $F = 500 \times 311 \times 9.81$ . Yeh step kyun? Teeno quantities known hain, isliye product ek single number hai.
Compute karo: $F = 1, 525, 455 N \approx 1.53 \times 1 0^{6} N = 1.53$ MN. Yeh step kyun? Multiply out karne se formula us newton figure mein convert hoti hai jiske baare mein pucha gaya tha.

Verify: Units — $[kg/s] \times [s] \times [m/s^{2}] = [kg \cdot m/s^{2}] = [N]$ . ✓ $1.5$ MN ek realistic large first-stage thrust hai. ✓

Case D — Same- $Δ v$ comparison across engines

Worked example Example 4 — Same maneuver ke liye Solid vs ion

Ek maneuver ko $Δ v = 3000$ m/s chahiye. Ek solid ( $I_{s p} = 260$ s) aur ek ion engine ( $I_{s p} = 3000$ s) ke liye mass ratio $m_{0} / m_{f}$ nikalo. Kise kam propellant chahiye? (Map par: sabse left ka magenta dot versus far-right navy dot.)

Forecast: Ion engine $\sim 11 \times$ zyada efficient hai per kg. Predict karo uska mass ratio $1$ ke just upar baithega (barely koi fuel), jabki solid ko $3$ se upar ka fat ratio chahiye.

Tsiolkovsky ko ratio ke liye rearrange karo: $\frac{m _{0}}{m _{f}} = e^{Δ v / (I_{s p} g_{0})}$ . Yeh step kyun? Hum $Δ v$ given hone par mass ratio chahte hain, isliye $Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})$ invert karte hain. $ln$ undo karne ke liye exponential $e^{x}$ chahiye — yehi wajah hai ki $e$ yahan aata hai. Dekho Tsiolkovsky Rocket Equation aur Staging and Mass Ratio.
Solid: exponent $= \frac{3000}{260 \times 9.81} = 1.176$ , toh $\frac{m _{0}}{m _{f}} = e^{1.176} = 3.243$ . Propellant fraction $ζ = 1 - 1/3.243 = 0.692 \Rightarrow 69%$ . Yeh step kyun? Hum solid ka low $I_{s p}$ exponent mein plug karte hain, phir $e$ ko us par raise karte hain; $ζ$ (upar define kiya) mass ratio ko fuel share mein turn karta hai.
Ion: exponent $= \frac{3000}{3000 \times 9.81} = 0.1019$ , toh $\frac{m _{0}}{m _{f}} = e^{0.1019} = 1.107$ . Propellant fraction $ζ = 1 - 1/1.107 = 0.097 \Rightarrow 9.7%$ . Yeh step kyun? Same formula, lekin ion ka huge $I_{s p}$ exponent ko tiny banata hai, isliye ratio barely 1 se upar jaata hai aur $ζ$ chhota hai.

Verify: $ln$ wapas lo: $260 \times 9.81 \times ln (3.243) = 3000$ m/s ✓ aur $3000 \times 9.81 \times ln (1.107) = 3003$ m/s ✓ (ratio ko 3 d.p. tak round karne se $\sim 3$ m/s residue rehta hai) — dono required $Δ v$ recover karte hain. Ion engine $\sim 7 \times$ kam propellant use karta hai. ✓

Case E — Degenerate input: $\overset{m}{˙} \to 0$ (the ion trap)

Worked example Example 5 — Huge

$I_{s p}$ , vanishing thrust Ek ion engine ka $I_{s p} = 3000$ s hai lekin sirf $\overset{m}{˙} = 5$ mg/s $= 5 \times 1 0^{- 6}$ kg/s flow karta hai. Uska thrust nikalo. Kya yeh apna $1$ kg mass Earth par ground se lift kar sakta hai? (Map par: navy dot — $I_{s p}$ ke liye far right, phir bhi thrust ke liye way down at the bottom.)

Forecast: $I_{s p}$ enormous hai — lekin $\overset{m}{˙}$ nearly zero hai. Predict karo thrust ek tiny fraction of a newton hoga, $1$ kg lift karne ke liye zaruri $\sim 9.81$ N se kaafi neeche.

Phir se $F = \overset{m}{˙} I_{s p} g_{0}$ use karo. Yeh step kyun? Same thrust formula jaise Ex 3 — lekin ab $\overset{m}{˙}$ degenerate (near-zero) input hai, jo dikhata hai ki thrust $\overset{m}{˙}$ follow karta hai, $I_{s p}$ nahi. Dhyan raho ki yahan $g_{0}$ woh unit-conversion constant hai jo $I_{s p}$ ke andar hai, koi gravity field nahi.
Substitute karo: $F = 5 \times 1 0^{- 6} \times 3000 \times 9.81$ . Yeh step kyun? Tiny flow rate drop in karte hain dekhne ke liye ki yeh product par kaise dominate karta hai.
Compute karo: $F = 0.1472 N$ . Yeh step kyun? Multiply out karne se actual force reveal hoti hai — ek fraction of a newton.
$1$ kg Earth par lift karne ka weight: $W = m g_{local} = 1 \times 9.81 = 9.81$ N, Earth ki local gravity $g_{local} = 9.81 m/s^{2}$ use karke. Yeh step kyun? Yahan hum genuine local gravitational field $g_{local}$ use karte hain, kyunki lift off karna ek real weight comparison hai — yeh $I_{s p}$ ke andar constant $g_{0}$ se ek alag role hai, chahe dono Earth's surface par $9.81$ ke equal hon (upar "Two different g" definition aur Example 9 dekho). Kyunki $0.147 ≪ 9.81$ , yeh lift off nahi kar sakta.

Verify: $F / W = 0.147/9.81 = 0.015$ — thrust weight ka $1.5%$ hai. Jaise $\overset{m}{˙} \to 0$ , $F \to 0$ chahe $I_{s p}$ kitna bhi bada ho. ✓ Yeh quantitative wajah hai ki ion engines sirf cruise ke liye hain (Ion and Electric Propulsion).

Case F — Degenerate input: koi propellant nahi jala

Worked example Example 6 — Agar

$m_{0} = m_{f}$ ho toh? Ek stage full tanks carry karta hai lekin engine kabhi fire nahi karta, isliye uska start aur end mass equal hai: $m_{0} = m_{f} = 20, 000$ kg. Ise kitna $Δ v$ milta hai? (Yeh Tsiolkovsky ka boundary "zero" case hai.)

Forecast: Koi fuel nahi nikla → koi momentum change nahi → predict karo $Δ v = 0$ , kisi bhi $I_{s p}$ ke liye.

$Δ v = I_{s p} g_{0} ln \frac{m _{0}}{m _{f}}$ apply karo. Yeh step kyun? Hum formula ko uski degenerate edge par test karte hain confirm karne ke liye ki yeh sensibly behave karta hai.
Ratio substitute karo: $\frac{m _{0}}{m _{f}} = \frac{20000}{20000} = 1$ , toh $Δ v = I_{s p} g_{0} ln (1)$ . Yeh step kyun? Equal masses ratio ko exactly 1 par collapse karte hain, jo woh key hai jo poora result unlock karti hai.
Kyunki $ln (1) = 0$ hai, hum paate hain $Δ v = 0$ chahe $I_{s p}$ kuch bhi ho. Yeh step kyun? $ln (1) = 0$ kyunki $e^{0} = 1$ — "1 par rehne ke liye koi growth ki zarurat nahi." Koi mass change nahi matlab koi velocity change nahi. Propellant fraction $ζ = 1 - 1/1 = 0$ confirm karta hai ki kuch nahi jala.

Verify: Physically, agar kuch peeche nahi pheka, toh conservation of momentum koi bhi speed gain forbid karta hai. ✓ Math aur physics boundary par agree karte hain. ✓ (Sanity check: $I_{s p} = 3000$ wala ion engine bhi $3000 \times 9.81 \times 0 = 0$ deta hai.)

Case G — Limiting case: chemical ceiling

Intuition

$v_{e} \propto T_{c} / M$ kahan se aata hai? Yeh magic nahi hai — yeh ek hot gas ke nozzle se guzarne ka energy bookkeeping hai (Exhaust Velocity and Nozzle Design). Exhaust ka har mole hot chamber se roughly $T_{c}$ ke proportional thermal energy ke saath nikalta hai (temperature hai hi stored molecular motion). Nozzle us thermal energy ko directed kinetic energy $\frac{1}{2} M v_{e}^{2}$ mein convert karta hai (yahan $M$ exhaust ke ek mole ki mass hai). "Thermal energy in" $\propto T_{c}$ ko "kinetic energy out" $= \frac{1}{2} M v_{e}^{2}$ ke barabar set karne par milta hai $v_{e}^{2} \propto T_{c} / M$ , yaani $v_{e} \propto T_{c} / M$ . Hidden proportionality constant gas constant, specific-heat ratio, aur nozzle expansion bundle karta hai — ek given gas ke liye ek fixed number, aur yehi wajah hai ki hum ise neeche ratio lete waqt cancel kar sakte hain. Poora constant Exhaust Velocity and Nozzle Design mein derive kiya gaya hai; do exhausts ko equal $T_{c}$ par compare karne ke liye hum uski value ki zarurat nahi.

Worked example Example 7 — Chemistry

$\sim 450$ s ke paas kyun cap karti hai $v_{e} \propto T_{c} / M$ use karte hue (jahan $T_{c}$ = chamber temperature kelvin, K mein, aur $M$ = exhaust ki mean molar mass kg/mol mein), maano ek engine fixed $T_{c} = 3600$ K par chalta hai. Heavy exhaust ( $M = 0.030$ kg/mol, CO $_{2}$ -rich) versus light exhaust ( $M = 0.010$ kg/mol, hydrogen-rich) ke saath exhaust speed compare karo. $\sim 450$ s reach karne ke liye $M$ ko kis taraf push karna hai?

Forecast: Kyunki $M$ denominator mein square root ke andar hai, lighter exhaust (smaller $M$ ) matlab faster $v_{e}$ . Predict karo light case $3$ ke factor se faster hoga.

Hidden proportionality constant cancel karne ke liye ratio form karo: $\frac{v _{e, light}}{v _{e, heavy}} = \frac{M _{heavy}}{M _{light}}$ . Yeh step kyun? Same $T_{c}$ aur same gas-type constant, isliye dono cancel ho jaate hain; sirf molar-mass ratio bachta hai. Ratio lena hume constant ki value jaane bina compare karne deta hai — yehi wajah hai ki hum yahan absolute numbers ki jagah ratio use karte hain.
Substitute karo: $0.030/0.010 = 3 = 1.732$ . Yeh step kyun? Do molar masses plug in karne se symbolic ratio ek plain number mein badal jaata hai jise hum interpret kar sakte hain.
Interpret karo: light-exhaust engine ko same temperature se $73%$ zyada $v_{e}$ milta hai. Hydrogen-rich exhaust (lowest possible $M$ ) wahi hai jo chemical $I_{s p}$ ko $\sim 450$ s wall tak push karta hai. Yeh step kyun? Tum $M$ ko hydrogen se neeche nahi le ja sakte, aur tum $T_{c}$ chamber melt kiye bina nahi badhaa sakte (Combustion Chamber Temperature) — dono limits milkar chemical $I_{s p}$ cap karte hain.

Verify: $3 = 1.732$ , aur $1.73 2^{2} = 3$ molar-mass ratio recover karta hai. ✓ Direction sahi hai: smaller $M$ → larger $v_{e}$ , jo match karta hai ki kyun LOX/LH2 har doosre chemical propellant ko beat karta hai. ✓

Case H — Real-world word problem: engine choose karna

Worked example Example 8 — Mission designer ka dilemma

Ek probe ko pehle Earth se launch hona hai (kuch minutes ke liye $\sim 9$ MN thrust chahiye) aur baad mein ek slow $Δ v = 6000$ m/s deep-space cruise perform karna hai jahan fuel mass scarce hai. Tumhare paas solids ( $260$ s, huge thrust) aur ion engines ( $3000$ s, tiny thrust) hain. Kaun sa engine kis phase ke liye, aur ion cruise ko kaun sa mass ratio chahiye? (Map par: launch tumhe high-thrust magenta dot ki taraf kheenchta hai, cruise tumhe high- $I_{s p}$ navy dot ki taraf — tum literally chart ke across travel karte ho.)

Forecast: Launch ko brute force chahiye → solids. Cruise ko efficiency chahiye, time hai → ion. Predict karo ion mass ratio chhota hoga ( $\sim 1.2$ ke paas).

Launch phase: thrust dominate karta hai. $F = \overset{m}{˙} I_{s p} g_{0}$ — MN-scale $F$ paane ke liye huge $\overset{m}{˙}$ chahiye, jo solids provide karte hain. Solid boosters choose karo. Yeh step kyun? Ex 5 se humne dekha ki thrust $\overset{m}{˙}$ track karta hai, $I_{s p}$ nahi; liftoff ek thrust problem hai.
Cruise phase: propellant precious hai, isliye high- $I_{s p}$ ion engine choose karo. Mass ratio: $\frac{m _{0}}{m _{f}} = e^{6000/ (3000 \times 9.81)} = e^{0.2039}$ . Yeh step kyun? Cruise ek $Δ v$ /efficiency problem hai — exactly jahan $I_{s p}$ pay off karta hai; hum Ex 4 se inverted Tsiolkovsky reuse karte hain.
Compute karo: $e^{0.2039} = 1.226$ . Propellant fraction $ζ = 1 - 1/1.226 = 0.184 \Rightarrow 18.4%$ . Yeh step kyun? $e$ ko exponent par raise karne se mass ratio milta hai, aur $ζ$ batata hai ki probe ka kitna hissa fuel hai.
Compare karo: same cruise ke liye ek solid ko $e^{6000/ (260 \times 9.81)} = e^{2.352} = 10.5$ chahiye — $90%$ se zyada propellant. Ek chhote probe ke liye impossible. Yeh step kyun? Same numbers solid ke low $I_{s p}$ ke saath run karne se pata chalta hai ki mission ion choice kyun force karta hai cruise ke liye.

Verify: Ion back-check: $3000 \times 9.81 \times ln (1.226) = 6001$ m/s ✓. Solid back-check: $260 \times 9.81 \times ln (10.5) = 6001$ m/s ✓. Staged choice (solid boost + ion cruise) standard real design pattern hai. ✓

Case I — Exam twist: $g_{0}$ Moon trap

Worked example Example 9 — Kya

$I_{s p}$ Moon par change hota hai? Example 1 ka same hydrolox engine ( $I_{s p} = 450$ s on Earth) Moon ke paas fly kiya jaata hai, jahan local gravity $g_{local} = 1.62 m/s^{2}$ hai. Ek student Moon ki gravity se rescale karke $I_{s p}$ "correct" karta hai: $450 \times (9.81/1.62) = 2725$ s. Kya student sahi hai? Moon ke paas true $I_{s p}$ aur $v_{e}$ kya hai? (Map par: student violet dot ko ion region ki taraf rightward slide karne ki koshish kar raha hai — lekin dot move nahi kar sakta, kyunki engine hardware unchanged hai.)

Forecast: $I_{s p}$ ek engine/propellant property hai. Predict karo yeh $450$ s rehta hai aur $v_{e}$ $4414.5$ m/s rehta hai — student galat hai.

Do-g distinction yaad karo: $I_{s p} = v_{e} / g_{0}$ ke andar ka $g_{0}$ woh fixed unit-conversion constant hai ( $9.81$ m/s²). Yeh $g_{local}$ nahi hai, woh real field jahan tum fly karte ho. Yeh step kyun? Poora "trap" constant $g_{0}$ ko local field $g_{local}$ se swap karna hai; us swap ko naam dena hi confusion dissolve karta hai (upar "Two different g" definition dekho).
Moon ke paas true exhaust velocity: $v_{e} = I_{s p} g_{0} = 450 \times 9.81 = 4414.5$ m/s — unchanged. Yeh step kyun? Engine physics (chamber temperature, molar mass, nozzle) jahan bhi tum fly karo identical hai, isliye $v_{e} = I_{s p} g_{0}$ mein kuch bhi location ke saath nahi badalta. $g_{0}$ definition se $9.81$ rehta hai.
Student ka $2725$ s meaningless hai: $g_{0}$ constant ki jagah $g_{local}$ substitute karna $I_{s p}$ ki definition tod deta hai. Yeh step kyun? Agar $I_{s p}$ sach mein location ke saath badalta, toh ek fixed engine ke infinitely many "efficiencies" hote — ek single piece of hardware ke liye impossible.

Verify: $450 \times (9.81/1.62) = 2725$ s woh hai jo student compute karta hai, lekin yeh ek valid $I_{s p}$ nahi hai; sahi value $450$ s rehti hai aur $v_{e}$ $4414.5$ m/s. ✓ Conclusion: $I_{s p}$ engine par depend karta hai, us sky par nahi jo uske upar hai — isliye exam answer hai "nahi, yeh Moon par nahi badalta." ✓

Recall checkpoint

Recall Har cue kaun se case se belong karta hai?

"Given $v_{e}$ , find $I_{s p}$ " — kaun sa direction aur operation? ::: Backward (Case B): $g_{0}$ se divide karo. " $m_{0} = m_{f}$ , find $Δ v$ " — answer kya hai aur kyun? ::: $Δ v = 0$ kyunki $ln (1) = 0$ ; koi fuel nahi nikla (Case F). "Ion engine lift off nahi kar sakta" — kaun si quantity degenerate hai? ::: $\overset{m}{˙} \to 0$ , isliye thrust $F = \overset{m}{˙} I_{s p} g_{0}$ tiny hai (Case E). "Same $Δ v$ , do engines, kam fuel?" — kaun sa tool? ::: Mass ratio $m_{0} / m_{f} = e^{Δ v / (I_{s p} g_{0})}$ ; higher $I_{s p}$ → smaller ratio (Case D). "Kya $I_{s p}$ Moon par change hota hai?" — trap aur fix? ::: Nahi; $g_{0}$ ek fixed constant hai (na ki $g_{local}$ ), $I_{s p}$ ek engine property hai (Case I). "Chemical $I_{s p}$ ~450 s se kyun exceed nahi kar sakti?" ::: $v_{e} \propto T_{c} / M$ ; $T_{c}$ melting se limited hai aur $M$ hydrogen exhaust se neeche nahi ja sakta (Case G). Propellant fraction $ζ$ ka matlab kya hai? ::: Starting mass ka woh hissa jo fuel tha, $ζ = 1 - m_{f} / m_{0} = 1 - 1/ (m_{0} / m_{f})$ .