Systems with variable mass — rocket equation derivation preview: Exercises

1.4.12 · D4 · Physics › Momentum & Collisions › Systems with variable mass — rocket equation derivation prev

Yeh page ek self-testing ladder hai. Har problem ek difficulty level par hai (L1 → L5). Pehle solution collapse karke try karo, phir [!recall]- block kholo aur har step check karo. Is sab ke liye jo bhi physics chahiye woh parent note mein hai: the rocket-equation preview.

Symbols ke baare mein ek reminder taaki kuch bhi pehle se use na ho:

Level 1 — Recognition

Recall Solution

$u = 3000$ m/s (exhaust velocity rocket ke relative measure ki jaati hai).
$∣ \overset{m}{˙} ∣ = 40$ kg/s (burn rate).
$m_{0} = 5000$ kg ("wet mass" = fuelled = start).
$m_{f} = 2000$ kg ("dry mass" = empty = end).
Mass ratio $= \frac{m _{0}}{m _{f}} = \frac{5000}{2000} = 2.5$ .

Humne kya kiya: rozmarra ke engineering words ko apne symbols se match kiya. Abhi kuch compute nahi kiya — sirf recognition.

Recall Solution

$F_{thrust} = u ∣ \overset{m}{˙} ∣ = 3000 \times 40 = 1.2 \times 1 0^{5} N = 120 kN .$ Yeh formula kyun: thrust woh reaction force hai jo mass ko backward throw karne se milti hai. Har second, $40$ kg, $3000$ m/s rocket ke relative se jaata hai, toh momentum $40 \times 3000$ kg·m/s per second ki rate se jaata hai — aur force hai hi momentum per second (Impulse–Momentum Theorem).

Level 2 — Application

Recall Solution

Koi external force nahi, toh Tsiolkovsky directly apply hota hai: $Δ v = u ln \frac{m _{0}}{m _{f}} = 2000 ln \frac{4000}{1000} = 2000 ln 4.$ $ln$ kyun: har kg burn hone se bacha hua, chhota mass aage dhakelta hai, toh payoff compound hota hai — $d m / m$ integrate karne se $ln$ milta hai. $= 2000 \times 1.3863 \approx 2773 m/s .$ Rest se shuru kiya, toh final speed $= 2773$ m/s. $3/4$ mass burn kiya phir bhi sirf $\approx 1.39 u$ gain hua — log kanji hai.

Recall Solution

$F_{thrust} = u ∣ \overset{m}{˙} ∣ = 2000 \times 25 = 5.0 \times 1 0^{4} N .$ $a = \frac{F _{thrust}}{m} = \frac{5.0 \times 1 0 ^{4}}{2000} = 25 m/s^{2} .$ Current mass kyun: acceleration hamesha abhi ke mass ka use karta hai. Jaise $m$ girta hai, $a$ badhta hai chahe thrust constant ho.

Level 3 — Analysis

Recall Solution

Tsiolkovsky se shuru karo aur use invert karo — hum $Δ v$ jaante hain aur ratio chahiye: $Δ v = u ln \frac{m _{0}}{m _{f}} \Rightarrow \frac{Δ v}{u} = ln \frac{m _{0}}{m _{f}} \Rightarrow \frac{m _{0}}{m _{f}} = e^{Δ v / u} .$ Exponentiate kyun: $ln$ aur $e^{(\cdot)}$ ek doosre ko undo karte hain — ratio ko log se bahar nikalne ke liye, dono sides ko $e$ ki power mein raise karo (Logarithms and Exponential Growth). $\frac{m _{0}}{m _{f}} = e^{6000/3000} = e^{2} \approx 7.389.$ Fuel fraction $= \frac{m _{0} - m _{f}}{m _{0}} = 1 - \frac{m _{f}}{m _{0}} = 1 - e^{- 2} \approx 1 - 0.1353 = 0.8647$ . Toh rocket ka lagbhag 86.5% fuel hona chahiye — yahi "tyranny of the rocket equation" action mein hai.

Recall Solution

1-D near-Earth equation (parent se) yeh hai: $m \frac{d v}{d t} = u ∣ \overset{m}{˙} ∣ - m g .$ Hover ka matlab $\frac{d v}{d t} = 0$ hai, toh thrust weight balance karta hai: $u ∣ \overset{m}{˙} ∣ = m g \Rightarrow ∣ \overset{m}{˙} ∣ = \frac{m g}{u} = \frac{8000 \times 9.8}{2500} = \frac{78400}{2500} = 31.36 kg/s .$ Thrust-to-weight $= 1.5$ ke liye, thrust $1.5 m g$ hona chahiye: $u ∣ \overset{m}{˙} ∣ = 1.5 m g \Rightarrow ∣ \overset{m}{˙} ∣ = 1.5 \times 31.36 = 47.04 kg/s .$ Kyun: thrust-to-weight ratio burn rate ko linearly scale karta hai, kyunki thrust $= u ∣ \overset{m}{˙} ∣$ mein $∣ \overset{m}{˙} ∣$ linear hai.

Level 4 — Synthesis

Recall Solution

(a) Single stage: $Δ v_{a} = 3000 ln \frac{10000}{2000} = 3000 ln 5 \approx 3000 \times 1.6094 = 4828 m/s .$ (b) Stage 1 ( $10000 \to 4000$ burns karta hai, toh jettison se pehle uska final mass $4000$ kg hai): $Δ v_{1} = 3000 ln \frac{10000}{4000} = 3000 ln 2.5 \approx 3000 \times 0.9163 = 2749 m/s .$ Phir $2000$ kg empty casing drop karo (koi momentum change nahi — yeh rocket ki speed par jaata hai). Stage 2 ab $2000$ kg se shuru hota hai: $Δ v_{2} = 3000 ln \frac{2000}{500} = 3000 ln 4 \approx 3000 \times 1.3863 = 4159 m/s .$ Total: $Δ v_{b} = 2749 + 4159 = 6908 m/s .$ Staging kyun jeetta hai: single stage mein tum bhaari empty tank poore raaste kheeenchte ho. Mid-flight mein dead mass drop karne ka matlab hai baad ka fuel ek halke vehicle ko dhakelta hai — log tumhe reward karta hai. Same total structure, $\approx 2080$ m/s zyada $Δ v$ .

Recall Solution

(i) Constant burn, toh mass linearly girta hai: $m (t) = m_{0} - ∣ \overset{m}{˙} ∣ t$ . Burnout jab $m = m_{f}$ ho: $t_{b} = \frac{m _{0} - m _{f}}{∣ m ˙ ∣} = \frac{6000 - 2000}{20} = 200 s .$ (ii) Deep space mein, $v (t) = u ln \frac{m _{0}}{m ( t )}$ . $m (t) = m_{0} - ∣ \overset{m}{˙} ∣ t$ substitute karo: $v (t) = 2500 ln \frac{6000}{6000 - 20 t} .$ Burnout $t = 200$ s par: $m = 2000$ kg, $v_{f} = 2500 ln \frac{6000}{2000} = 2500 ln 3 \approx 2500 \times 1.0986 = 2747 m/s .$ Shape (figure dekho): $v$ pehle slowly badhta hai (bhaari rocket) aur end ki taraf steep hota jaata hai (halka rocket, same thrust ⇒ bada acceleration). Yeh straight line nahi hai — acceleration badhta hai.

Figure — Systems with variable mass — rocket equation derivation preview

Level 5 — Mastery

Recall Solution

Constant gravity ke saath 1-D equation yeh hai: $m \frac{d v}{d t} = u ∣ \overset{m}{˙} ∣ - m g .$ Burn par integrate karo. Thrust term Tsiolkovsky log tak integrate hota hai; gravity term $- g t_{b}$ tak integrate hota hai (burn time par ek steady drain). Result: $v_{f} = ideal u ln \frac{m _{0}}{m _{f}} - gravity loss g t_{b} .$ Gravity term sirf $g t_{b}$ kyun hai: $m d v = u ∣ \overset{m}{˙} ∣ d t - m g d t$ mein, $m$ se divide karne par gravity piece $- g d t$ reh jaata hai, jiska burn par integral simply $- g t_{b}$ hai — yeh is baat se independent hai ki mass kaise change karta hai. Burn time: $t_{b} = \frac{m _{0} - m _{f}}{∣ m ˙ ∣} = \frac{500000 - 150000}{2500} = \frac{350000}{2500} = 140 s .$ Ideal $Δ v$ : $u ln \frac{m _{0}}{m _{f}} = 3000 ln \frac{500000}{150000} = 3000 ln \frac{10}{3} \approx 3000 \times 1.20397 = 3611.9 m/s .$ Gravity loss: $g t_{b} = 9.8 \times 140 = 1372 m/s .$ Sach mein burnout speed: $v_{f} \approx 3611.9 - 1372 = 2239.9 m/s .$ Gravity ne $\approx 1372$ m/s chura liya — ideal $Δ v$ ka lagbhag 38%. Isliye launches hard aur fast burn karte hain (chhota $t_{b}$ ) aur jaldi horizontal ho jaate hain: $g t_{b}$ ko chhota karne ke liye.

Recall Solution

$m_{0}$ ke liye Tsiolkovsky invert karo: $\frac{m _{0}}{m _{f}} = e^{Δ v / u} = e^{9000/4000} = e^{2.25} \approx 9.4877.$ $m_{0} = m_{f} e^{2.25} = 3000 \times 9.4877 \approx 28463 kg .$ Fuel chahiye $= m_{0} - m_{f} \approx 28463 - 3000 = 25463 kg .$ Iska matlab: $3000$ kg payload+structure ko $9000$ m/s se move karne ke liye $\approx 25.5$ tonnes fuel carry karna padega — dry mass ka $8 \times$ se zyada. Kyunki ratio $e^{Δ v / u}$ hai, $Δ v$ upar push karne se (ya $u$ neeche) fuel budget exponentially badh jaata hai. Yahi tyranny hai, quantified.

Recall Har level ke liye one-line self-check

L1 — $u$ rocket ke relative hai ::: ek engine property, ~2000–4500 m/s, ground speed nahi. L2 — acceleration kaun sa mass use karta hai ::: current mass, toh $a$ badhta hai jaise fuel burns hota hai. L3 — Tsiolkovsky ko kaise invert karo ::: $m_{0} / m_{f} = e^{Δ v / u}$ (log undo karne ke liye exponentiate karo). L4 — staging single stage ko kyun beat karta hai ::: mid-flight mein dead mass drop karne se baad ka fuel ek halke vehicle ko dhakelta hai. L5 — gravity loss barabar hai ::: $g t_{b}$ , ideal $Δ v$ se ek baar subtract karo.