Tsiolkovsky rocket equation — full first-principles derivation from momentum: Visual walkthrough

3.3.1 · D2 · Physics › Rocket Propulsion › Tsiolkovsky rocket equation — full first-principles derivati

Hum poore time ek hi dimension mein kaam karte hain: sab kuch ek horizontal line par move karta hai, "right" positive hai, "left" negative hai. Bas itna hi — koi angles nahi, koi triangles nahi, sirf ek number line jisme arrows hain.

Step 0 — Characters se milte hain (koi bhi physics shuru hone se pehle)

KYA. Teen quantities hain jo poori story chalati hain. Mujhe inhe picture mein point karke naam dene do, formula se nahi.

$m$ — rocket ki mass abhi is waqt (rocket plus us mein bacha hua har drop of fuel). Kilograms mein measure hoti hai.
$v$ — rocket ki velocity abhi is waqt, ground se measure ki gayi. Hamare line par ek number: positive = right ki taraf ja raha hai.
$v_{e}$ — effective exhaust velocity: gas kitni tez bahar nikalti hai jaise ek astronaut rocket par baith ke dekhta hai. Hamesha ek fixed positive number, engine nozzle se set hoti hai. Zaroori baat — yeh rocket ke relative measure hoti hai, ground se nahi.

SIRF YEH TEEN KYUN, AUR KOI NAHI. Momentum "mass times velocity" hai. Momentum bookkeeping karne ke liye hume sirf kitna stuff aur kitni tez ja raha hai — yahi chahiye hota hai. Yeh hain $m$ aur $v$ . Engine apna ek fact add karti hai — throw-speed $v_{e}$ . Teen numbers, poora problem.

PICTURE. Rocket number line par right ki taraf $v$ speed se ja raha hai; peeche ek chota sa nozzle hai jahan se gas $v_{e}$ relative speed par ship ke relative bahar aayegi.

Figure — Tsiolkovsky rocket equation — full first-principles derivation from momentum

Step 1 — System ko ABHI freeze karo aur momentum count karo

KYA. Time $t$ par ek snapshot lo. Sab kuch — rocket body aur fuel ka woh chhota puff jo abhi fire hone wala hai — abhi bhi ek lump hai, mass $m$ , saath mein speed $v$ par chal rahe hain.

$p (t) = poori mass, kg m \cdot uski speed, m/s v$

KYUN. Hume poore system ko count karna hai kuch bhi phenkne se pehle, taaki baad mein compare karne ke liye kuch ho. Jo fuel jane wala hai woh abhi andar hi hai, isliye woh $v$ par chal raha hai aur $m v$ ka hissa count hota hai. Ise bhoolna classic galti hai — tum mass ko "kahin se appear" nahi hone de sakte ek moment baad.

PICTURE. Ek black arrow of momentum $p (t) = m v$ right ki taraf point karta hua. Yeh hamara "before" total hai; ise pakad ke rakho.

Step 2 — Thoda sa time $d t$ guzarne do: split

KYA. Ek tiny sliver of time $d t$ baad, rocket ne ek chota sa gas puff peeche fire kar diya. Ek lump ab do objects ban gaya hai:

Rocket, ab thoda halka aur thoda teez.
Gas puff, ab peeche ki taraf ud raha hai.

$d m$ ko rocket ki mass mein change maan lo. Rocket halka hua, isliye $d m$ negative hai. Jo mass gayi woh isliye $- d m$ hai, ek positive amount.

$rocket mass = m + d m (chhoti, kyunki d m < 0)$ $gas mass = - d m (positive)$

NEGATIVE SIGN KYUN. $m$ ka matlab rocket ki mass hai, aur rocket ka weight kam ho raha hai. Jab koi quantity ghatti hai, uska change negative hota hai — yeh bas "change" ka matlab hai. Isliye $d m < 0$ , aur jo chunk nikala gaya woh uska mirror hai, $- d m > 0$ . Yeh sign sahi karna hi final answer ko positive instead of upside-down banata hai.

PICTURE. Upar "before" lump; neeche "after" split into rocket (abhi bhi right ja raha hai) aur ek chota gas blob dusri taraf ja raha hai.

Step 3 — Gas GROUND frame mein kitni tez move karti hai?

KYA. Engine gas ko $v_{e}$ rocket ke relative, peeche ki taraf phenkti hai. Lekin hamara momentum ledger ground se likha gaya hai. Isliye hume gas ki ground speed chahiye.

$v_{gas, ground} = rocket ki ground speed v - backward throw speed v_{e} = v - v_{e}$

SUBTRACT KYUN. Socho tum rocket par khade ho: tum gas ko $v_{e}$ left ki taraf jaate hue dekhte ho. Ab ground par aa jao, jahan rocket khud right ki taraf $v$ par drift kar raha hai. "Rocket se dekhi gayi speed" ko "ground se dekhi gayi speed" mein convert karne ke liye, rocket ki apni velocity add karo. Gas ki rocket-frame velocity $- v_{e}$ hai (leftward), isliye uski ground velocity $v + (- v_{e}) = v - v_{e}$ hai. Isliye humne Step 0 mein yeh insist kiya tha ki $v_{e}$ ek relative speed hai — yahi do frames ko seedha rakhta hai.

PICTURE. Ek chota velocity-addition diagram: rocket par gas $- v_{e}$ read karti hai; rocket ka apna $+ v$ add karo; ground meter $v - v_{e}$ read karta hai.

Step 4 — Split ke BAAD momentum count karo

KYA. Total "after" momentum = rocket ka naya momentum + gas ka momentum.

$p (t + d t) = rocket ki nayi mass (m + d m) rocket ki nayi speed (v + d v) + gas mass (- d m) gas ki ground speed (v - v_{e})$

Yahan $d v$ woh tiny speed hai jo rocket ne time $d t$ mein gain ki — wahi cheez jo hum dhundh rahe hain.

KYUN. Momentum additive hai: jab ek system split hota hai, total bas pieces ka sum hota hai. Humne har piece Steps 2 aur 3 mein banaya — ab hum simply "rocket part plus gas part" likhte hain.

PICTURE. Do momentum arrows side by side: ek bada black wala rocket ke liye (rightward), ek chota gas ke liye (leftward), unka tail-to-tip sum Step 2 ke single "before" arrow ke barabar.

Step 5 — Expand karo, aur dekho terms cancel hote hain

KYA. Sab kuch multiply karo:

$p (t + d t) = m v + m d v + v d m + tiny \times tiny d m d v - v d m + v_{e} d m$

Do cheezein hoti hain:

$+ v d m$ aur $- v d m$ bilkul cancel ho jaate hain.
$d m d v$ ek tiny number times ek tiny number hai — negligibly small, isliye hum ise drop karte hain.

$\Rightarrow p (t + d t) = m v + m d v + v_{e} d m$

$d m d v$ kyun drop karein. Dono $d m$ aur $d v$ "jitna chahein utna chhota" hain. Unka product doubly chhota hai — jaise $0.001 \times 0.001 = 0.000001$ . Jaise-jaise hum $d t$ ko zero ki taraf shrink karte hain, yeh term doosron se kahin zyada tez mar jaati hai, isliye limit mein kuch contribute nahi karti. Yeh calculus ka standard move hai: first-order rakho, second-order drop karo.

PICTURE. Chhe expanded terms laid out; do $v d m$ terms crossed through (woh annihilate ho jaate hain) aur $d m d v$ term greyed out aur crossed (matter karne ke liye bahut chhota). Jo bacha hai woh highlighted hai.

Step 6 — Ek law apply karo: momentum conserved hai

KYA. Koi gravity nahi, koi drag nahi → koi bahari push nahi → total momentum badal nahi sakta:

$p (t + d t) = p (t)$ $m v + m d v + v_{e} d m = m v$

Dono sides ka $m v$ cancel ho jaata hai, aur sab kuch ka dil reh jaata hai:

$0 = m d v + v_{e} d m$

KYUN. Yeh Newton's law hai momentum form mein: $F_{e x t} = d p / d t$ . Jab $F_{e x t} = 0$ to momentum simply move nahi karta — pehle equals baad. Yahan se aage sab pure calculus hai; poori physics isi ek line mein hai, jo actually Conservation of Momentum hai disguise mein.

PICTURE. "Before" arrow aur "after" arrow same length draw kiye gaye, beech mein equals sign — momentum unchanged — aur leftover balance $m d v = - v_{e} d m$ ek see-saw ki tarah dikhaya gaya.

Step 7 — Saare tiny puffs add karo (integrate karo)

KYA. Balance ko rearrange karo taaki ek speed-gain isolate ho:

$d v = - v_{e} \frac{d m}{m}$

Har puff ek sliver of speed $d v$ deta hai. Total $Δ v$ paane ke liye, launch ( $m = m_{0}$ , $v = v_{i}$ ) se burnout ( $m = m_{f}$ , $v = v_{f}$ ) tak har sliver add karo. "Infinitely many slivers add karna" exactly wahi hai jo integral $\int$ karta hai:

$\int_{v_{i}}^{v_{f}} d v = - v_{e} \int_{m_{0}}^{m_{f}} \frac{d m}{m}$

LOGARITHM KYUN AATA HAI. Hume $\frac{d m}{m}$ ka running-total chahiye — ek fractional change. Woh function jiske chhote changes "change divided by current value" jaisi dikhti hain woh natural logarithm hai: $\int \frac{d m}{m} = ln m$ . Yeh koi coincidence nahi hai jo hum impose karte hain; yeh Step 6 ke balance ne force kiya hai. Log us sawaal ka jawab hai — "kaunsa function fractional changes add karta hai?"

$v_{f} - v_{i} = - v_{e} [ln m]_{m_{0}}^{m_{f}} = - v_{e} (ln m_{f} - ln m_{0})$

$ln A - ln B = ln (A / B)$ use karke aur fraction ko flip karke minus sign khatam karo (yeh legal hai kyunki $m_{0} > m_{f}$ ):

PICTURE. Curve $1/ m$ ke neeche ka area $m_{f}$ se $m_{0}$ tak shaded — woh shaded area hi $ln (m_{0} / m_{f})$ hai, aur use $v_{e}$ se multiply karne par $Δ v$ milta hai.

Yahi balance, total karne ki jagah per second padha jaye, to thrust ban jaata hai $F = v_{e} \overset{m}{˙}$ .

Step 8 — Edge cases (taaki tum kabhi wall se na takrao)

KYA & KYUN — formula ke chaar corners:

Koi fuel mat jalao: $m_{f} = m_{0} \Rightarrow R = 1 \Rightarrow ln 1 = 0 \Rightarrow Δ v = 0$ . Koi throw nahi, koi push nahi. ✔ Sanity sahi hai.
Sab kuch jalao (impossible ideal): $m_{f} \to 0 \Rightarrow R \to \infty \Rightarrow ln R \to \infty$ . Formula vaada karta hai unlimited speed — lekin sirf agar rocket end mein kuch bhi na weighe. Yeh ho nahi sakta; hamesha structure bachta hai. Isliye $m_{f}$ kabhi truly zero nahi pahuncha.
Kya $Δ v$ , $v_{e}$ se zyada ho sakta hai? Haan! Jaise hi $R > e \approx 2.718$ , $ln R > 1$ , toh $Δ v > v_{e}$ . Rocket apne exhaust se aage nikal jaata hai kyunki use baar baar push milti hai.
Fuel share double karna: $R \to 2 R$ jaana har baar ek fixed $v_{e} ln 2$ add karta hai, $Δ v$ kabhi double nahi hota. Brutal diminishing returns — yahi reason hai Multistage Rockets ka.

PICTURE. Curve $Δ v = v_{e} ln R$ : $R = 1$ par flat-zero, $R = e$ par $v_{e}$ line cross karta hai, aur dhire-dhire rise karta rehta hai — equal-height "staircase" ke saath jo dikhata hai ki $R$ har baar double hone par same step add hoti hai.

Ek-picture summary

PICTURE. Poori derivation ek single canvas par: (1) intact rocket $v$ par, (2) lighter rocket $(v + d v)$ aur backward gas $(v - v_{e})$ mein split, (3) balance $m d v = - v_{e} d m$ , (4) integral sign slivers gather karta hua, (5) boxed result $Δ v = v_{e} ln (m_{0} / m_{f})$ — ek arrow sab paanch se left-to-right flow karta hua.

Recall Poore walkthrough ki Feynman retelling

Tum space mein float kar rahe ho aur tumhare haath mein baseballs ka ek bag hai — total weight tumhari mass $m$ hai, aur tum speed $v$ par glide kar rahe ho. Snapshot ek: tum aur har ball saath move karte ho; yahi tumhara momentum hai, mass times speed. Ab ek ball peeche phenko. Tum do mein split ho jaate ho: ek thoda halka tum aage nudge hua, aur ek ball peeche ud gayi. Ground se ball ki speed tumhari speed minus jitna tumne phenka, $v - v_{e}$ hai. Throw ke baad momentum count karo aur pehle ke barabar set karo — kyunki space mein push karne ko kuch nahi hai, total badal nahi sakta. Sab multiply karo aur beech ke messy bits ya toh cancel ho jaate hain ya bahut chhote hain rakhne ke liye; jo bachta hai woh ek clean deal hai: jo speed gain karte ho woh throw-speed times tumne abhi kitna fraction mass phenka hai ke barabar hai. Yeh baar baar karo, har tiny fraction add karte jao — aur "apni ghatti hua mass ke fractions add karna" exactly wahi hai jo ek logarithm count karta hai. Isliye tumhari total speed-up throw-speed times (start weight over end weight) ka log hai. Kuch mat phenko, kuch gain nahi. Almost sab phenko, aur tum apni khud ki baseballs ki speed se bhi aage nikal sakte ho — kyunki har throw ek lighter-and-lighter tumhe thoda aur zyada push karta hai.

Recall

Ek sentence mein, logarithm kahan se aata hai? ::: $d m / m$ — fractional mass change — ko integrate karne se, aur $\int d m / m = ln m$ . Gas ki ground velocity $v - v_{e}$ kyun hai aur $- v_{e}$ kyun nahi? ::: Kyunki $v_{e}$ moving rocket ke relative measure hoti hai; rocket ki apni ground velocity $v$ add karne par $v - v_{e}$ milta hai. Kis mass ratio par $Δ v$ pehli baar $v_{e}$ se zyada hota hai? ::: $R = e \approx 2.718$ par, jahan $ln R = 1$ . Poori derivation kaun sa ek physical law chalata hai? ::: Conservation of momentum (free space mein zero external force).