Systems with variable mass — rocket equation derivation preview: Worked examples

1.4.12 · D3 · Physics › Momentum & Collisions › Systems with variable mass — rocket equation derivation prev

The scenario matrix

Har variable-mass problem is grid ka ek cell hai. Columns poochte hain mass aur baahri duniya ke saath kya ho raha hai; rows poochte hain tum kaunsi quantity chahte ho.

Cell	Sign of $\overset{m}{˙}$	External force $F_{ext}$	What you compute	Example
A Deep-space $Δ v$	$\overset{m}{˙} < 0$ (loses)	$0$	Burnout speed via $ln$	Ex 1
B Thrust + accel now	$\overset{m}{˙} < 0$	$0$	$F_{thrust}$ , instantaneous $a$	Ex 2
C Launch vs gravity	$\overset{m}{˙} < 0$	$- m g$	Lift-off / hover condition	Ex 3
D Mass gained	$\overset{m}{˙} > 0$ (accretes)	given	Force to keep it moving	Ex 4
E Degenerate $u = 0$	$\overset{m}{˙} < 0$	$0$	Shows no thrust	Ex 5
F Degenerate $m_{f} \to m_{0}$	$\overset{m}{˙} < 0$	$0$	$Δ v \to 0$ limit	Ex 6
G Mass-ratio design (inverse)	$\overset{m}{˙} < 0$	$0$	Solve for required $m_{0} / m_{f}$	Ex 7
H Word problem / exam twist	mixed	mixed	Multi-step reasoning	Ex 8

"Sign of $\overset{m}{˙}$ " ke do cases (loses vs gains) asli deep split hain. Baaki sab kaunsa force add karo ya kaunsa unknown solve karo hai.

Figure — Systems with variable mass — rocket equation derivation preview

Cell A — Deep-space burnout speed

Worked example Ex 1 (Cell A) — the pure Tsiolkovsky case

Deep space mein rest par ek probe: $m_{0} = 5000$ kg, $m_{f} = 2000$ kg tak burn karta hai, exhaust speed $u = 3000$ m/s. Uski final speed nikalo.

Forecast: compute karne se pehle guess karo — kya yeh $u = 3000$ m/s se zyada hoga ya kam? (Log usually tumhe punish karta hai; ek number note karo.)

Koi external force nahi, isliye $Δ v = u ln \frac{m _{0}}{m _{f}}$ use karo. Yeh step kyun? $F_{ext} = 0$ master equation ko seedha Tsiolkovsky tak collapse kar deta hai — koi gravity term saath nahi uthana padta.
Mass ratio: $\frac{m _{0}}{m _{f}} = \frac{5000}{2000} = 2.5$ . Yeh step kyun? Equation ko jo bhi chahiye woh sab is ek dimensionless number mein squeeze ho jaata hai — poori fuel ki kahani ek ratio mein.
$Δ v = 3000 ln 2.5 = 3000 \times 0.9163 = 2748.9$ m/s. Yeh step kyun? Rest se start kiya, isliye $v_{f} = Δ v$ . $ln$ "60% mass jalaya" ko "sirf $0.92 u$ pahuncha" mein badal deta hai.

Verify: $ln 2.5 < 1$ , isliye $Δ v < u$ — "the log is unkind" se match karta hai. Units: $(m/s) \times (dimensionless) = m/s$ . ✓

Cell B — Thrust aur instantaneous acceleration

Worked example Ex 2 (Cell B) — force aur

$a$ ek instant par Ex 1 wala probe $∣ \overset{m}{˙} ∣ = 40$ kg/s par burn karta hai. Thrust nikalo, aur jab uski mass $3000$ kg ho tab uska acceleration nikalo.

Forecast: burn ke dauran thrust constant hai, lekin kya $a$ constant hai? Haan/nahi guess karo.

$F_{thrust} = u ∣ \overset{m}{˙} ∣ = 3000 \times 40 = 120000$ N $= 1.2 \times 1 0^{5}$ N. Yeh step kyun? Thrust sirf is par depend karta hai ki tum kitni tez aur kitna fek rahe ho — $u$ aur $∣ \overset{m}{˙} ∣$ — na ki kitna rocket bacha hai.
Deep space $\Rightarrow F_{ext} = 0$ , isliye $a = \frac{F _{thrust}}{m}$ current mass ke saath. Yeh step kyun? Newton's law abhi jo mass hai usi ko use karta hai, aur woh shrink ho rahi hai.
$m = 3000$ kg par: $a = \frac{1.2 \times 1 0 ^{5}}{3000} = 40$ m/s $^{2}$ . Yeh step kyun? Same thrust, chhoti mass ⇒ bada $a$ . Isliye $a$ constant nahi hai — jaise fuel jalta hai, $a$ badhta hai.

Verify: start par ( $m = 5000$ ) $a = 1.2 \times 1 0^{5} /5000 = 24$ m/s $^{2}$ $< 40$ m/s $^{2}$ . Mass girne ke saath acceleration badha — consistent. Units: $N/kg = m/s^{2}$ . ✓

Cell C — Gravity ke against Launch

Worked example Ex 3 (Cell C) — kya yeh lift off bhi karega?

Pad par ek rocket: $m_{0} = 4000$ kg, $u = 2000$ m/s, burn rate $∣ \overset{m}{˙} ∣ = 25$ kg/s. $g = 10$ m/s $^{2}$ lo. Kya yeh lift off karta hai? Agar nahi, toh kaunsa burn rate karega?

Forecast: pehle apne dimag mein weight compute karo — kya thrust bada hai?

Gravity ke saath master equation: $m \frac{d v}{d t} = u ∣ \overset{m}{˙} ∣ - m g$ . Yeh step kyun? Ab $F_{ext} = - m g$ (weight neeche ki taraf hai), isliye hum us term ko drop karne ki bajaye rakhte hain.
Thrust: $u ∣ \overset{m}{˙} ∣ = 2000 \times 25 = 50000$ N. Weight: $m_{0} g = 4000 \times 10 = 40000$ N. Yeh step kyun? Lift-off ke liye upar ki push ko sabse bhaari moment — start — par neeche ke weight ko beat karna hoga.
$50000 > 40000$ , isliye lift hota hai. Initial acceleration $a = \frac{50000 - 40000}{4000} = 2.5$ m/s $^{2}$ . Yeh step kyun? Net upward force ko current mass se divide karo toh rocket ka launch acceleration milta hai.
Hover burn rate (thrust = weight): $u ∣ \overset{m}{˙} ∣_{hover} = m_{0} g \Rightarrow ∣ \overset{m}{˙} ∣_{hover} = \frac{40000}{2000} = 20$ kg/s. Yeh step kyun? $a = 0$ set karne se woh break-even burn rate milta hai jiske neeche woh kabhi pad nahi chhodta.

Verify: thrust-to-weight $= 50000/40000 = 1.25 > 1$ ⇒ lift-off, parent ki condition $u ∣ \overset{m}{˙} ∣ > m_{0} g$ se match karta hai. $∣ \overset{m}{˙} ∣ = 20$ kg/s par thrust $= 40000 =$ weight ⇒ $a = 0$ hover. ✓

Cell D — Mass GAIN ho raha hai (accretion / loading)

Yeh mirror image hai: ab $\overset{m}{˙} > 0$ .

Worked example Ex 4 (Cell D) — conveyor belt loading

Reeti (sand) $\frac{d m}{d t} = + 10$ kg/s rate se ek horizontal belt par vertically girta hai. Belt $v = 3$ m/s constant speed par chalti hai. Isse usi speed par rakhne ke liye kaunsa horizontal force chahiye?

Forecast: belt constant speed par chalti hai — kya required force zero hai (kyunki $d v = 0$ )? Guess karo.

Master equation: $m \frac{d v}{d t} = F_{ext} - u \frac{d m}{d t}$ . Yahan aane wale sand ki zero horizontal velocity hai, aur woh $v$ par chal rahi cheez mein join hota hai, isliye belt ke relative uski speed $u = v = 3$ m/s hai (belt ko use $0$ se $3$ tak kheenchna padta hai). Yeh step kyun? $u$ hamesha transfer ho rahi mass aur body ke beech ki relative speed hoti hai. Girta hua sand koi forward speed se start nahi karta, isliye belt ke relative woh $v$ se peeche hai.
Constant speed ⇒ $\frac{d v}{d t} = 0$ , isliye $0 = F_{ext} - u \frac{d m}{d t}$ , giving $F_{ext} = u \frac{d m}{d t}$ . Yeh step kyun? Koi acceleration nahi, toh applied force sirf ek kaam karta hai: nayi add hui mass ko belt speed tak accelerate karna.
$F_{ext} = v \frac{d m}{d t} = 3 \times 10 = 30$ N. Yeh step kyun? Har second tumhe $10$ kg ko extra $3$ m/s dena padta hai; force woh momentum flow deliver karta hai.

Verify: momentum flux check: $\frac{d p}{d t} = v \frac{d m}{d t} = 3 \times 10 = 30$ N. Kyunki $\overset{m}{˙} > 0$ hai, mass term ek drag hai — tum steady speed par bhi aage dhakelte ho. Units: $(m/s) (kg/s) = kg\cdotm/s^{2} = N$ . ✓

Cell E — Degenerate: $u = 0$

Worked example Ex 5 (Cell E) — zero relative speed ke saath mass girana

Ek cart jiska mass $m_{0} = 200$ kg hai deep space mein $v = 6$ m/s par glide kar raha hai. Woh gently mass release karta hai (use cart ke relative zero speed se jaane deta hai, yani $u = 0$ ), $80$ kg khota hai. Cart ki final speed?

Forecast: apni ek-tihai mass dump karne se kya woh tez ho jaata hai? Padhne se pehle guess karo.

Thrust $= - u \frac{d m}{d t}$ with $u = 0$ ⇒ thrust $= 0$ . Yeh step kyun? Yeh exactly parent ki pehli galti hai jo deliberately ki gayi hai: mass ko cart ki khud ki velocity par drop karna koi push nahi deta.
Koi external force nahi aur koi thrust nahi, toh $\frac{d v}{d t} = 0$ : speed unchanged. Yeh step kyun? Closed system ka momentum pehle hi $v$ par carry ho raha tha; use do pieces mein split karna jo dono $v$ par chal rahe hain, kisi ki bhi velocity nahi badlata.
$v_{f} = 6$ m/s. (Ya Tsiolkovsky se: $Δ v = 0 \cdot ln (200/120) = 0$ .) Yeh step kyun? $u = 0$ set karne se poora Tsiolkovsky expression zero ho jaata hai — equation khud confirm karti hai "koi relative speed nahi, koi gain nahi."

Verify: $Δ v = 0 \times ln (200/120) = 0$ , isliye $v_{f} = 6$ m/s. Yeh woh degenerate case hai jo expose karta hai ki kyun $u$ (na ki $v$ ) sab kuch drive karta hai. ✓

Cell F — Degenerate: $m_{f} \to m_{0}$ (almost no fuel)

Worked example Ex 6 (Cell F) — tiny-burn limit

Ek satellite $m_{0} = 1000$ kg ka $u = 3000$ m/s ke saath ek thruster fire karta hai lekin sirf $m_{ej} = 5$ kg eject karta hai, isliye $m_{f} = 995$ kg. $Δ v$ nikalo, aur crude estimate " $u \times (fraction of mass ejected)$ " se compare karo.

Forecast: ek chhote burn ke liye, kya $ln (m_{0} / m_{f})$ simple fraction $m_{ej} / m_{0}$ ke close hoga? Guess karo.

Exact: $Δ v = u ln \frac{m _{0}}{m _{f}} = 3000 ln \frac{1000}{995} = 3000 ln (1.005025)$ . Yeh step kyun? Ek tiny burn bhi same log use karta hai — koi alag "small" formula nahi chahiye.
$ln (1.005025) = 0.0050126$ , isliye $Δ v = 15.04$ m/s. Yeh step kyun? Near-1 ratio ko log ke through plug karna honest answer deta hai.
Crude estimate: $u \cdot \frac{m _{ej}}{m _{0}} = 3000 \times \frac{5}{1000} = 15.0$ m/s. Yeh step kyun? Chhote $x$ ke liye, $ln (1 + x) \approx x$ (dekho Logarithms and Exponential Growth); yahan $x = m_{ej} / m_{f}$ tiny hai, isliye log ≈ fraction.

Verify: exact $15.04$ vs crude $15.0$ — $< 0.3%$ se agree karte hain. Jaise $m_{f} \to m_{0}$ , $ln \to 0$ aur $Δ v \to 0$ : limiting behaviour smooth hai, koi blow-up nahi. ✓

Cell G — Inverse design: mass ratio ke liye solve karo

Worked example Ex 7 (Cell G) — kitna fuel chahiye?

Ek mission $Δ v = 6000$ m/s demand karta hai $u = 2500$ m/s ke engine ke saath deep space mein. Kaunsa mass ratio $m_{0} / m_{f}$ required hai? Agar dry (empty) mass $m_{f} = 1500$ kg hai, toh fuelled mass kya hai?

Forecast: $Δ v$ $u$ se zyada se zyada double hai — kya mass ratio around 2 hoga, ya bahut bada? Guess karo.

Tsiolkovsky invert karo: $Δ v = u ln \frac{m _{0}}{m _{f}} \Rightarrow \frac{m _{0}}{m _{f}} = e^{Δ v / u}$ . Yeh step kyun? $ln$ ko undo karne ke liye tum uska inverse apply karte ho, exponential $e^{(\cdot)}$ — dekho Logarithms and Exponential Growth. Yahan "tyranny" visible hoti hai.
$\frac{Δ v}{u} = \frac{6000}{2500} = 2.4$ , isliye $\frac{m _{0}}{m _{f}} = e^{2.4} = 11.02$ . Yeh step kyun? Exponential ek modest ratio $Δ v / u = 2.4$ ko 11-fold mass ratio mein blow up kar deta hai — exponentially mehenga fuel.
Fuelled mass: $m_{0} = 11.02 \times 1500 = 16530$ kg. Yeh step kyun? Dry mass ko required ratio se multiply karo toh launch mass milta hai; uska $15000$ kg se zyada fuel hai.

Verify: back-substitute: $2500 ln (16530/1500) = 2500 ln 11.02 = 2500 \times 2.3997 = 5999$ m/s $\approx 6000$ . ✓ Ratio $> 1$ , mass positive. Units consistent. ✓

Cell H — Exam-style word twist

Worked example Ex 8 (Cell H) — two-stage trap

Ek student claim karta hai: "Ek single-stage rocket $u = 2500$ m/s ke saath kabhi $Δ v = 9000$ m/s nahi pahunch sakta kyunki us ke liye mass ratio $e^{3.6} \approx 36.6$ chahiye, jo banaya nahi ja sakta." Phir woh ise two stages mein split karta hai, har ek $Δ v_{1} = Δ v_{2} = 4500$ m/s deta hai, aur claim karta hai ki required total mass ratio chhota hai. Dono claims check karo. Kya stage $Δ v$ 's add hote hain?

Forecast: kya do $4500$ 's add hokar $9000$ bante hain? Aur kya split sach mein sasta hai? Dono guess karo.

Single stage: $m_{0} / m_{f} = e^{9000/2500} = e^{3.6} = 36.6$ . Yeh step kyun? Same inverse-Tsiolkovsky jaise Ex 7 mein — "unbuildable" ratio confirm karta hai.
$Δ v$ 's add hote hain kyunki velocity sequential burns mein additive hoti hai: $Δ v_{total} = Δ v_{1} + Δ v_{2} = 4500 + 4500 = 9000$ m/s. ✓ Yeh step kyun? Har stage wahan se start hoti hai jahan pichli chhodi thi; speeds linearly pile hoti hain (mass ki tarah nahi, jo multiply karta hai).
Har stage ratio: $e^{4500/2500} = e^{1.8} = 6.05$ . Do identical stages ⇒ ratios ka product $= 6.05 \times 6.05 = 36.6$ . Yeh step kyun? Jab stages chain hoti hain toh mass ratios multiply hote hain, aur $6.05 \times 6.05 = e^{1.8} \cdot e^{1.8} = e^{3.6} = 36.6$ — wahi total.
Isliye staging ideal mass ratio reduce nahi karti; iska asli faayda burns ke beech dead structural mass fenkna hai (is idealised sum mein capture nahi hua). Yeh step kyun? Exam trap yeh sochna hai ki log split karne se "help" karta hai; algebraically $e^{a} e^{b} = e^{a + b}$ , isliye ideal cost identical hai. Asli saving khaali tanks drop karna hai.

Verify: $6.05 \times 6.05 = 36.6 = e^{3.6}$ , aur $ln (36.6) \times 2500 = 3.6 \times 2500 = 9000$ m/s. Claims: " $Δ v$ add hote hain" ✓ sach; "split ideally sasta hai" ✗ galat. ✓

Recall Quick self-test on the matrix

Which cell has zero thrust despite mass leaving? ::: Cell E ( $u = 0$ ) — no relative speed, no push. When does the mass term act as a drag instead of thrust? ::: Cell D — when $\overset{m}{˙} > 0$ (mass arriving). How do you go from a required $Δ v$ to a mass ratio? ::: Take $e^{Δ v / u}$ — the inverse of the $ln$ . Do chained stages reduce the ideal total mass ratio? ::: No — ratios multiply, $e^{a} e^{b} = e^{a + b}$ ; only shedding dead mass helps.

Back to the parent topic · related: Conservation of Linear Momentum, Impulse–Momentum Theorem, Relative Velocity, Logarithms and Exponential Growth.