Systems with variable mass — rocket equation derivation preview: Visual walkthrough

1.4.12 · D2 · Physics › Momentum & Collisions › Systems with variable mass — rocket equation derivation prev

Step 1 — "Momentum" kya hai aur hum ise kyun track karte hain

KYA. Hum sab kuch ek fixed ground frame mein label karte hain (ek camera jo launchpad se bolted hai) aur picture mein sab cheez ka total momentum padhte hain.

MOMENTUM KYUN, FORCE KYUN NAHI? Force ko simple rehne ke liye fixed mass chahiye. Momentum $p = m v$ tab bhi kaam karta hai jab mass change ho raha ho, kyunki hum har piece ka $m v$ alag-alag add kar sakte hain. Isliye hum $p$ track karte hain.

PICTURE. Ek arrow, rocket, right ki taraf $v$ speed se drift kar raha hai. Uska momentum neele bar ke roop mein neeche hai — length $= m v$ .

Figure — Systems with variable mass — rocket equation derivation preview

Step 2 — Ek fixed lump of stuff ke around box banao

KYA. Hum "rocket" ko track karne se mana karte hain (uska mass shrink hota hai — cheating hai). Iske bajaye hum rocket aur us tiny puff of fuel ke around ek dashed box banate hain jo abhi abhi bahaar nikalne wala hai. Yeh box poore instant $d t$ (time ka ek tiny slice) ke liye same total matter hold karta hai.

YEH BOX KYUN? Newton ka honest law $F_{ext} = \frac{d p}{d t}$ sirf un systems pe allowed hai jinka membership change nahi hota. Humara dashed box qualify karta hai: koi bhi matter box mein enter ya exit nahi karta; fuel sirf box ke ek hisse se doosre mein move karta hai.

PICTURE. Dashed box mein do cheezein hain jo alag hone wali hain: main rocket (mass $m + d m$ , jahan $d m < 0$ kyunki mass lose ho raha hai) aur ek red puff (mass $- d m > 0$ ).

Step 3 — Puff actually kitni tezi se move karta hai? (relative velocity)

KYA. Hum puff ki ground-frame velocity pin down karte hain: $v - u$ .

RELATIVE KYUN, GROUND SPEED KYUN NAHI? Kyunki $u$ engine ki ek fixed property hai (nozzle gas ko kitni tezi aur garam se throw karta hai, ~2000–4500 m/s), koi aisi cheez nahi jo rocket accelerate hone ke saath change ho. Derivation ko $u$ par build karna ise har speed ke liye honest rakhta hai jo rocket reach karta hai.

PICTURE. Split point se do arrows: rocket ab $v + d v$ pe (thoda faster), red puff $v - u$ pe (ship ke relative backward, toh ground frame mein slower ya negative bhi ho sakta hai).

Step 4 — Ek heartbeat baad momentum likho

KYA. Time $t + d t$ pe box mein do moving pieces hain. Unke momenta add karo.

ADD KYUN KAREIN? Momentum additive hai: box ka total $p$ sirf har piece ke $m v$ ka sum hai. Kyunki box closed hai, yeh sum $F_{ext} d t = d p$ mein feed karna bilkul legal hai.

PICTURE. Do stacked momentum bars — rocket bar $(m + d m) (v + d v)$ aur puff bar $(- d m) (v - u)$ — jinka total hum Step 1 ke single bar se compare karenge.

Step 5 — Expand karo, aur dekho do terms ek doosre ko kaise cancel karte hain

KYA. Sab kuch multiply out karo:

$p (t + d t) = m v + m d v + v d m + d m d v - v d m + u d m$

EXPAND KYUN? Cancellation dhundhne ke liye. Do circled terms dekho: $+ v d m$ (rocket se) aur $- v d m$ (puff se). Yeh equal aur opposite hain — jaane wale mass ki ground-frame speed $v$ khud ko cancel kar leti hai.

YEH KYUN MAYNE RAKHTA HAI. Cancellation ke baad exhaust ko describe karne wali sirf $u$ speed bachti hai, woh relative wali. Picture prove karti hai ki push karne wali cheez relative velocity hai — ground speed nahi.

PICTURE. Chha terms ek row mein; do $v d m$ terms green mein strike-through ke saath drawn, aur tiny $d m d v$ term "second-order, negligible" ke roop mein shaded out.

$v d m$ cancel karne aur $d m d v$ drop karne ke baad:

$d p = p (t + d t) - p (t) = m d v + u d m$

Step 6 — Outside force laao → master equation

KYA. Closed box pe Newton ka honest law: $F_{ext} d t = d p$ . $d p = m d v + u d m$ substitute karo aur $d t$ se divide karo.

$F = \overset{p}{˙}$ KYUN, $F = ma$ KYUN NAHI? $F = ma$ fixed mass assume karta hai. Humara box construction ke hisaab se fixed-mass hai, toh $F_{ext} = \frac{d p}{d t}$ yahan bilkul legal hai — yeh impulse–momentum theorem hai jo instant $d t$ pe apply ki gayi hai.

PICTURE. Ek free-body sketch: box pe ek external arrow $F_{ext}$ , internal thrust arrow $- u \overset{m}{˙}$ ke roop mein emerging, aur rocket pe resulting acceleration $a$ .

Step 7 — Gravity band karo aur integrate karo → Tsiolkovsky

KYA. Deep space mein $F_{ext} = 0$ , toh $m d v = - u d m$ , matlab $d v = - u \frac{d m}{m}$ . Launch $(v_{0}, m_{0})$ se burnout $(v_{f}, m_{f})$ tak har tiny gain sum karo.

LOGARITHM KYUN AATA HAI. Burn kiya hua har kilogram bachi hui, hamesha-chhoti-hoti mass ko push karta hai. Gain per kg hai $- u \frac{d m}{m}$ — ek change divided by current amount. "Change over current amount" ko add up karna exactly wahi hai jo ek logarithm measure karta hai. Isliye $\int \frac{d m}{m} = ln m$ .

PICTURE. $\frac{1}{m}$ vs $m$ ka curve; $m_{f}$ se $m_{0}$ tak shaded area $ln \frac{m _{0}}{m _{f}}$ ke barabar hai, aur har thin strip ek kilogram ka ever-growing contribution hai jab $m$ leftward shrink karta hai.

Step 8 — Edge cases (reader ko kabhi stranded mat chhodho)

KYA & KYUN. Har degenerate input, ek baar draw kiya — taaki koi bhi scenario tumhe surprise na kare.

Koi fuel burn nahi hua ( $m_{f} = m_{0}$ ): $ln \frac{m _{0}}{m _{0}} = ln 1 = 0 \Rightarrow Δ v = 0$ . Kuch throw nahi kiya, koi speed gain nahi. ✔
Sab kuch ek point tak burn kar do ( $m_{f} \to 0$ ): $ln \frac{m _{0}}{m _{f}} \to + \infty$ . Infinite $Δ v$ — lekin impossible hai, kyunki real rockets engines aur structure carry karte hain jo burn nahi ho sakta. Math tumhe warn karta hai ki payoff sirf fantasy mein diverge karta hai.
Gravity ke khilaaf launch ( $F_{ext} = - m g$ ): master equation ban jaata hai $m \overset{v}{˙} = u ∣ \overset{m}{˙} ∣ - m g$ . Agar $u ∣ \overset{m}{˙} ∣ = m g$ , toh $a = 0$ — rocket hover karta hai. Lift-off ke liye thrust > weight chahiye: $u ∣ \overset{m}{˙} ∣ > m_{0} g$ .
Slow burn / bada mass ratio contrast: $u$ double karne se $Δ v$ linearly double hota hai, lekin sirf fuel se $Δ v$ double karne ke liye mass ratio squared chahiye — yahi rocket equation ki tyranny hai.

PICTURE. Chaar mini-panels: $m_{f} = m_{0}$ (flat), $m_{f} \to 0$ (blow-up), hover balance, aur lift-off jahan thrust arrow weight arrow ko beat karta hai.

Ek-picture summary

Poori derivation ek canvas pe: closed lump ko box karo → rocket + puff mein split karo → $v d m$ terms cancel ho jaate hain → sirf relative $u$ bachta hai → $d m / m$ integrate karo → logarithm.

Recall Poore walkthrough ki Feynman retelling

Socho tum ek frictionless skateboard pe ho aur bricks pakde ho. Pehle hum agree karte hain ki momentum — mass times speed — watch karna hai, kyunki yeh tab bhi nicely add up hota hai jab tum halke ho rahe ho (Step 1). Hum tum aur agla brick dono ke around ek dashed box draw karte hain, taaki ek throw ke dauran kuch andar ya bahar na jaaye (Step 2). Tum brick ko apne relative ek fixed speed se backward throw karte ho — woh $u$ hai, tumhari arm ki property, yeh nahi ki tum already kitni tezi se roll kar rahe ho (Step 3). Hum ek heartbeat baad momentum tally karte hain: tum thode faster, brick peeche fly kar raha hai (Step 4). Jab hum sab kuch multiply out karte hain, woh part jo tumhari ground speed pe depend karta hai khud cancel ho jaata hai — proof ki sirf relative throw speed matter karti hai (Step 5). Woh closed box ke liye Newton's law mein feed karo aur master equation nikalta hai: tumhara acceleration equals outside forces plus ek forward push $u ∣ \overset{m}{˙} ∣$ (Step 6). Full se empty tak har throw add karo; kyunki har throw ek halke tum ko push karta hai, gains interest ki tarah compound karte hain — wahi logarithm hai, $Δ v = u ln (m_{0} / m_{f})$ (Step 7). Finally hum corners check karte hain: koi bricks throw nahi = koi speed nahi; apna saara mass throw karna = fantasy infinite speed; aur gravity ko beat karne ke liye tumhara throwing tumhare weight se zyada push karna chahiye (Step 8).

Recall Quick self-test

Step 5 mein kaun se do terms cancel hote hain aur kya bachta hai? ::: $+ v d m$ aur $- v d m$ cancel hote hain; sirf relative-speed term $u d m$ bachta hai. $d m$ ki jagah $d m / m$ integrate kyun karte hain? ::: Har kg ever-smaller remaining mass ko push karta hai, toh gain per kg (change ÷ current amount) hota hai — woh add karne se logarithm milta hai. Deep space mein master equation kya ban jaata hai? ::: $m d v = - u d m$ , jo integrate hoke $Δ v = u ln (m_{0} / m_{f})$ deta hai. Launch karte waqt hover condition kya hai? ::: $u ∣ \overset{m}{˙} ∣ = m g$ se $a = 0$ milta hai; lift-off ke liye $u ∣ \overset{m}{˙} ∣ > m_{0} g$ chahiye.