Visual walkthrough — Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ

3.3.8 · D2 · Physics › Rocket Propulsion › Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ

---

Step 1 — Rocket aur uski exhaust se milte hain

WHAT. Hum ek rocket ko space mein freeze karte hain aur do arrows mark karte hain: gas left (backward) ja rahi hai, rocket right (forward) ja raha hai.

WHY. Thrust ki poori kahani momentum ka gas se rocket tak transfer hone ki kahani hai. Uss hand-off ko measure karne se pehle hum directions aur names par agree karna zaroori hai. Hum forward (jis taraf rocket move karta hai) ko positive direction kehte hain.

PICTURE. Figure s01 dekho. Amber shape rocket hai. Mouth se nikalta cyan arrow escape ho raha gas hai speed $v_e$ par. Note karo ki ye opposite directions mein point karte hain — yahi rockets ka poora raaz hai.

---

Step 2 — "Rate" ka matlab kya hai: $\dot{m}$ padhna

WHAT. Time ke ek chhote slice $dt$ mein (padho: "bahut chhota moment"), mass $dm=\dot{m}dt$ ka ek chhota gas chunk bahar nikalта hai.

WHY yeh tool — kyun sirf "ek mass" nahi? Ek rocket ek lump fire karke nahi rukta; woh gas continuously dalta rehta hai. Isliye natural sawaal yeh hai "har second kitna?", aur jawaab ek rate hai. Ek rate ($\dot{m}$, kg per second) ko ek duration ($dt$, seconds) se multiply karne par ek actual mass milti hai ($dm$, kg) — seconds cancel ho jaate hain.

$\underset{\text{small mass, kg}}{\underbrace{dm}}=\underset{\text{kg per second}}{\underbrace{\dot{m}}}\cdot \underset{\text{small time, s}}{\underbrace{dt}}$

PICTURE. Figure s02 mein ek stopwatch aur ek chhota cyan blob $dm$ dikhta hai jo interval $dt$ ke dauran nikalta hai. Jaise clock chalta hai, blob ke baad blob depart hota hai — woh steady drip hai $\dot{m}$.

---

Step 3 — Momentum thrust: moving gas se dhakka

WHAT. Har departing blob $dm$ backward momentum $dm\cdot v_e$ carry karta hai. Newton ka law kehta hai force bas momentum per second hand over hona hai.

WHY yeh tool — kyun "bas ek push" nahi? Kyunki jo push tum feel kar sakte ho woh tab tak exist karti hai jab tak kuch apni motion change kar raha ho. Newton ka second law kehta hai force momentum ke rate of change ke barabar hai, aur Newton ka third law kehta hai gas ka backward momentum rocket ke forward momentum se match hota hai. Isliye forward shove ko count karne ka sabse clean tarika backward momentum ko per second count karna hai.

$F_{\text{mom}}=\frac{d{p}_{\text{gas}}}{dt}=\frac{v_{e}dm}{dt}=\underset{\text{kg/s}}{\underbrace{\dot{m}}}\underset{\text{m/s}}{\underbrace{v_e}}\Rightarrow \text{units }\frac{\text{kg}}{\text{s}}\cdot \frac{\text{m}}{\text{s}}=\text{N}$

Yahan $p_{\text{gas}}$ gas ka momentum hai; $\frac{dp}{dt}$ padha jaata hai "woh momentum har second kitni fast pile up hota hai."

PICTURE. Figure s03: cyan gas left fire karta hai ek long backward-momentum arrow ke saath; ek equal-length amber arrow dikhata hai reaction rocket ko right push karta hai. Same length = equal and opposite = Newton's third law visible ho gaya.

Recall Units khud check karo

$\dot{m}{v}_e$ newtons mein kyun measure hota hai? ::: $\frac{\text{kg}}{\text{s}}\times \frac{\text{m}}{\text{s}}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}$, aur $1\text{N}=1\frac{\text{kg}\cdot \text{m}}{{\text{s}}^2}$. Toh yeh hai ek force.

---

Step 4 — Aakhir doosra push kyun hota hai: pressure

WHAT. Exit mouth par exhaust outward press karta hai pressure $P_e$ ke saath. Surrounding air inward press karta hai pressure $P_a$ ke saath ("$a$" for ambient, rocket ke around ki hawa).

WHY yeh matter karta hai — gas pressure already zero kyun nahi hoti? Ek perfect nozzle gas ko tab tak expand karta rehta jab tak uska pressure bahar ki hawa se exactly equal na ho jaaye. Lekin real nozzles ki finite length hoti hai aur kahin rukna padta hai. Aksar gas thodi si squeezed rehti hai jab woh lip par pahunchti hai: $P_e\ne P_a$. Woh leftover pressure difference exit ring par push karta hai — aur wahan koi nozzle wall nahi hai use cancel karne ke liye, isliye woh extra thrust ke roop mein leak out ho jaata hai.

PICTURE. Figure s04: mouth par, cyan arrows outward push karte hain (labelled $P_e$) aur white arrows inward push karte hain (labelled $P_a$). Solid nozzle walls ke along inward aur outward pushes cancel ho jaate hain — lekin open mouth par nahi hote.

---

Step 5 — Pressure thrust: ek squeeze ko force mein badalna

WHAT. Pressure se force = pressure $\times$ area. Mouth par net outward pressure $(P_e-P_a)$ hai, aur yeh exit area $A_e$ par act karta hai.

WHY subtract? Kyunki inward air push $P_a$ genuinely outward gas push $P_e$ ko oppose karta hai. Sirf leftover, $P_e-P_a$, rocket ko shove karne ke liye bachta hai. Us leftover pressure ko us area se multiply karna jis par woh act karta hai "force per square metre" ko actual force mein convert karta hai.

$F_{\text{pres}}=\underset{\text{net Pa}}{\underbrace{(P_e-P_a)}}\underset{{\text{m}}^2}{\underbrace{A_e}}\Rightarrow \frac{\text{N}}{{\text{m}}^2}\cdot {\text{m}}^2=\text{N}$

• $P_e$ — exit gas pressure (exhaust abhi bhi kitna squeezed hai).
• $P_a$ — ambient air pressure jo push back kar raha hai.
• $A_e$ — woh ring jis par leftover push act karta hai.

PICTURE. Figure s05 cancelling walls ko fade away karta hai aur sirf surviving amber arrow $(P_e-P_a)A_e$ ko ring par forward act karte hue rakhta hai — pressure thrust, akela.

---

Step 6 — Dono shoves ko add karo

WHAT. Total thrust momentum shove plus pressure shove hai.

WHY add? Yeh ek hi body par same second mein do independent forward forces hain, isliye yeh simply sum ho jaate hain — jaise do log ek stalled car ko peechhe se push kar rahe hain.

$F=\underset{\text{momentum thrust}}{\underbrace{\dot{m}{v}_e}}+\underset{\text{pressure thrust}}{\underbrace{(P_e-P_a)A_e}}$

PICTURE. Figure s06 do amber arrows ko tip-to-tail stack karta hai; unki combined length full thrust $F$ hai.

---

Step 7 — Sab kuch ek velocity $c$ mein fold karo

WHAT. $F=\dot{m}c$ set karo aur Step 6 equation ko $\dot{m}$ se divide karke $c$ solve karo.

WHY $\dot{m}$ se divide karein? Kyunki hume ek speed chahiye. Thrust divided by mass-flow ($\frac{\text{N}}{\text{kg/s}}$) deta hai $\frac{\text{m}}{\text{s}}$ — exactly velocity ke units. Division cleanly ek force plus pressure term ko ek speed mein repackage karta hai.

$c=\frac{F}{\dot{m}}=\underset{\text{real gas speed}}{\underbrace{v_e}}+\underset{\text{pressure extra speed ke roop mein likha gaya}}{\underbrace{\frac{(P_e-P_a)A_e}{\dot{m}}}}$

$\boxed{c=v_e+\frac{(P_e-P_a)A_e}{\dot{m}}}$

Bother kyun karein? Kyunki ab $I_{sp} = c/g_0$ aur Tsiolkovsky equation $\Delta v = c\ln(m_0/m_f)$ dono char ki jagah EK number use karte hain.

PICTURE. Figure s07: Step 6 ke two-arrow stack se ek single long cyan arrow $c$ labelled mein morph hota hai — same total length, ek clean vector.

---

Step 8 — Saare cases: pressure term ka sign

Pressure term $\frac{(P_e-P_a)A_e}{\dot{m}}$ positive, zero, ya negative ho sakta hai. Kyunki $A_e$ aur $\dot{m}$ hamesha positive hain, iska sign poori tarah $(P_e-P_a)$ se decide hota hai — jo Nozzle Expansion (Under/Over/Optimal) se set hota hai.

WHAT / WHY / PICTURE har case ke liye (figure s08 teeno ko side by side dikhata hai):

• Optimally expanded — $P_e=P_a$: leftover push exactly zero hai, isliye $c=v_e$. Make-believe speed real speed ke barabar hai. (Middle panel: arrows cancel ho jaate hain, $c$ arrow = $v_e$ arrow.)
• Underexpanded — $P_e>P_a$: gas abhi bhi squeezed hai, leftover push forward hai, $c>v_e$. Extra thrust. Under → Up. (Left panel: amber arrow add ho jaata hai.)
• Overexpanded — $P_e<P_a$: outside air jeet jaata hai, leftover push backward hai, $c<v_e$. Thrust penalty aur possible flow separation. Over → Off. (Right panel: amber arrow back point karta hai, $c$ ko short kar raha hai.)

---

Step 9 — Degenerate aur limiting checks

WHAT / WHY. Ek achha formula extreme inputs mein survive karna chahiye. Figure s09 ek fixed engine ke liye $c$ vs altitude plot karta hai taaki yeh limits ek nazar mein dikhein.

• Vacuum limit ($P_a\to 0$): back-push gayab ho jaata hai, isliye $c$ apni maximum par pahunchta hai, $c=v_e+\frac{P_e{A}_e}{\dot{m}}$. Yeh s09 curve ka top hai.
• Ground limit (dense air, large $P_a$): term sabse zyada negative hai, isliye $c$ sabse chhota hai. Curve ka bottom.
• $\dot{m}\to 0$ (engine off): formula $c=F/\dot{m}$ zero se divide karta hai — meaningless, kyunki koi gas flow nahi toh koi "exhaust velocity" ki baat nahi. Equation tab hi apply hoti hai jab engine firing ho.
• $A_e=0$ (koi opening nahi): pressure term vanish ho jaata hai; lekin phir $\dot{m}$ bhi zero hoga — degenerate, koi rocket nahi.

${\lim }_{P_a\to 0}c=v_e+\frac{P_e{A}_e}{\dot{m}}\text{(highest }c\text{)}$

---

Ek-picture summary

Figure s10 poori kahani compress karta hai: ek rocket uske exit mouth ke saath; cyan momentum arrow $\dot{m}{v}_e$; amber pressure arrow $(P_e-P_a)A_e$; unka tip-to-tail sum $F$; aur single merged $c$ arrow, sab labelled — ek nazar mein derivation.

Recall Walkthrough ki Feynman retelling

Ek rocket gas ko peechhe shoot karta hai. Rushing gas har second "amount of motion" (momentum) carry karke jaati hai, aur reaction se woh rocket ko forward shove karti hai — yeh pehla push hai, $\dot{m}{v}_e$. Lekin mouth chhod rahi gas aksar thodi si squeezed rehti hai, opening par bahar ki hawa se zyada. Woh extra squeeze mouth ki ring par press karta hai bina kuch cancel karne ke — ek doosra push milta hai, $(P_e-P_a)A_e$. Dono shoves add karo aur tumhare paas total thrust hai. Maths ko tidy rakhne ke liye, hum pretend karte hain ki gas actually ek make-believe speed $c$ par aayi jo same total push deti hai. Upar jahan hawa patli hai, kuch push back nahi karta, isliye $c$ badhta hai aur rocket harder push karta hai. Neeche, thick air fight back karti hai aur $c$ shrink ho jaata hai. Engine band karo aur "$c$" ka poora idea evaporate ho jaata hai — koi gas nahi, koi exhaust speed nahi.

---

Connections

• Thrust Equation — parent; $F=\dot{m}{v}_e+(P_e-P_a)A_e$ yahan pictures mein derived.
• Conservation of Momentum — Step 3 ke momentum thrust ke peeche ka engine.
• Nozzle Expansion (Under/Over/Optimal) — Step 8 mein sign fix karta hai.
• Atmospheric Pressure vs Altitude — kyun Step 9 ka curve upar slope karta hai.
• Specific Impulse — $I_{sp}=c/g_0$, wahi $c$ use karta hai jo humne abhi banaya.
• Tsiolkovsky Rocket Equation — $\Delta v=c\ln (m_0/m_f)$.
• Yeh page Hinglish mein →

Concept Map