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

3.3.1 · D4 · Physics › Rocket Propulsion › Tsiolkovsky rocket equation — full first-principles derivati

Every problem uses the master equation from the parent note: $Δ v = v_{e} ln (\frac{m _{0}}{m _{f}}), R = \frac{m _{0}}{m _{f}}$ jahaan $v_{e}$ = exhaust speed relative to the rocket, $m_{0}$ = initial mass, $m_{f}$ = final mass, $R$ = mass ratio. Agar in mein se koi bhi word shaky lage, toh aage badhne se pehle parent note dobara padho. Hum $ln$ = natural logarithm (jo $e^{x}$ ka inverse hai: "kis power par $e$ yeh number deta hai?") aur iska inverse $e^{x}$ bhi use karte hain (physics ke neeche dekho Conservation of Momentum aur Newton's Second Law (momentum form)).

Ek quick numerical anchor jo tum har jagah reuse karoge:

$ln 2 = 0.6931$
$ln 4 = 2 ln 2 = 1.3863$
$ln 10 = 2.3026$
$e^{1} = 2.7183$

Level 1 — Recognition

Kya tum equation padh ke uske pieces identify kar sakte ho?

Recall Solution L1·1

Galat. $m_{f}$ final total mass hai (dry rocket plus koi bhi bacha hua fuel), jalaya gaya fuel nahi.

$Δ v$ = rocket ki speed mein change ( $m/s$ mein).
$v_{e}$ = exhaust speed rocket ke relative (constant).
$m_{0}$ = shuru mein total mass (structure + payload + saara propellant).
$m_{f}$ = end mein total mass (structure + payload + unburned propellant). Jalaya gaya fuel $m_{0} - m_{f}$ hai, jo ek alag quantity hai.

Recall Solution L1·2

$R = \frac{m _{0}}{m _{f}} = \frac{12 , 000}{3000} = 4.$ Yeh dimensionless kyun hai: dono masses kg mein hain, toh units cancel ho jaate hain — $R$ ek pure number hai.

Recall Solution L1·3

$Δ v = v_{e} ln R = 2500 \times ln 4 = 2500 \times 1.3863 = 3466 m/s \approx 3.47 km/s .$

Level 2 — Application

Kya tum equation ko aage aur peechhe chala sakte ho?

Recall Solution L2·1

$R = \frac{50 , 000}{10 , 000} = 5, ln 5 = 1.6094.$ $Δ v = 3200 \times 1.6094 = 5150 m/s \approx 5.15 km/s .$

Recall Solution L2·2

Log ko $e^{x}$ se undo karke rearrange karo (isliye hum yahan exponential use karte hain — woh $ln$ ka exact inverse hai): $\frac{Δ v}{v _{e}} = ln R \Rightarrow R = e^{Δ v / v_{e}} = e^{6000/4000} = e^{1.5} = 4.482.$ Toh rocket ko launch par burnout se lagbhag $4.48 \times$ bhaari hona chahiye.

Recall Solution L2·3

Propellant fraction hai $f = \frac{m _{0} - m _{f}}{m _{0}} = 1 - \frac{m _{f}}{m _{0}} = 1 - \frac{1}{R} = 1 - \frac{1}{4.482} = 0.7769.$ Launch mass ka lagbhag 77.7% fuel hona zaroori hai.

Recall Solution L2·4

Parent note se, $F_{t h r u s t} = v_{e} \overset{m}{˙}$ : $F = 2800 \times 250 = 700, 000 N = 700 kN .$

Level 3 — Analysis

Kya tum reason kar sakte ho ki pieces change ke saath kaise respond karte hain?

Recall Solution L3·1

Mass ratio $R$ aur $2 R$ ko compare karo: $Δ v (2 R) - Δ v (R) = v_{e} ln (2 R) - v_{e} ln R = v_{e} [ln 2 + ln R - ln R] = v_{e} ln 2.$ $ln R$ terms cancel ho jaate hain, ek constant $v_{e} ln 2$ reh jaata hai jo $R$ par depend nahi karta. Yeh log ka fingerprint hai: input ko multiply karna output mein fixed amount add karta hai. $v_{e} ln 2 = 3000 \times 0.6931 = 2079 m/s \approx 2.08 km/s .$ Figure s01 dekho: orange curve flatten hoti hai — wahi horizontal doubling increasingly chhote-dikhne wale (par numerically equal) vertical steps deta hai.

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

Recall Solution L3·2

Humein $Δ v > v_{e}$ chahiye, yaani $v_{e} ln R > v_{e} \Rightarrow ln R > 1 \Rightarrow R > e \approx 2.718.$ Toh $\approx 2.72$ se upar koi bhi mass ratio rocket ko uski apni exhaust speed se tez bana deta hai. Yeh isliye kaam karta hai kyunki rocket ko baar baar push milte hain, har baar ek aur halke body par — pushes $v_{e}$ se aage accumulate ho jaate hain.

Recall Solution L3·3

Baseline: $Δ v = 3000 ln 4 = 3000 (1.3863) = 4159 m/s$ . (a) $v_{e} \to 3300$ : $Δ v = 3300 \times 1.3863 = 4575 m/s$ . Gain $= 416 m/s$ (flat 10% kyunki $v_{e}$ multiply karta hai). (b) $R \to 4.4$ : $ln 4.4 = 1.4816$ , toh $Δ v = 3000 \times 1.4816 = 4445 m/s$ . Gain $= 286 m/s$ . $v_{e}$ badhana jeet jaata hai. Kyun: $v_{e}$ , $Δ v$ ko linearly scale karta hai (woh log ke bahar rehta hai), lekin $R$ sirf dheere-dheere badhne wale log ke through enter karta hai. Isliye engineers higher Specific Impulse (better $v_{e}$ ) ke peechhe itni aggressively bhaagte hain.

Level 4 — Synthesis

Kya tum equation ko staging, thrust, aur losses ke saath combine kar sakte ho?

Recall Solution L4·1

(a) $Δ v = 3000 ln 16 = 3000 (2.7726) = 8318 m/s$ . (b) Stage burns add up hote hain: $Δ v = 3000 ln 4 + 3000 ln 4 = 3000 (ln 4 + ln 4) = 3000 ln (4 \times 4) = 3000 ln 16 = 8318 m/s$ . Dono tie karte hain — agar mass ratios sach mein same total tak multiply hon. Staging ka asli faayda yeh hai ki khaali tanks drop karne se har stage ek high ratio achieve kar sakta hai jo woh dead structural mass kheechtey hue kabhi nahi pa sakta, isliye real multistage rockets ek single stage se zyada effective total $R$ paate hain. Yahan humne $R_{tot}$ construction se fixed rakha, toh log ki additivity tie dikhati hai.

Recall Solution L4·2

Real velocity gain ideal minus gravity loss term hai: $Δ v_{actual} = Δ v_{ideal} - g t = 6000 - (9.8) (120) = 6000 - 1176 = 4824 m/s .$ Lagbhag 4.82 km/s. Gravity akele ne kareeb 1.18 km/s kha liya — isliye launches seedhe upar ladne mein kam waqt bitaane ke liye jaldi pitch over karte hain.

Recall Solution L4·3

(i) Burn time = propellant divided by flow rate: $t_{b} = \frac{M _{p}}{m ˙} = \frac{18 , 000}{300} = 60 s .$ (ii) Thrust $= v_{e} \overset{m}{˙} = 3000 \times 300 = 900, 000 N = 900 kN .$ (iii) Initial mass $m_{0} = m_{f} + M_{p} = 6000 + 18, 000 = 24, 000 kg$ , toh $R = 24, 000/6000 = 4$ : $Δ v = 3000 ln 4 = 3000 (1.3863) = 4159 m/s \approx 4.16 km/s .$

Level 5 — Mastery

Kya tum derive, invert, aur edge cases par reason kar sakte ho?

Recall Solution L5·1

Fractional mass change isolate karne ke liye rearrange karo: $d v = - v_{e} \frac{d m}{m} .$ $(v_{i}, m_{0})$ se $(v_{f}, m_{f})$ tak integrate karo: $\int_{v_{i}}^{v_{f}} d v = - v_{e} \int_{m_{0}}^{m_{f}} \frac{d m}{m} .$ Right integrand $1/ m$ wahi ek function hai jiska antiderivative $ln m$ hai — isliye log humpar force hota hai; kuch aur choose nahi kiya gaya. Toh $v_{f} - v_{i} = - v_{e} (ln m_{f} - ln m_{0}) = v_{e} ln \frac{m _{0}}{m _{f}},$ jahaan minus sign ne ratio ko seedha kar diya (kyunki $m_{0} > m_{f}$ hai, log positive rehta hai). General framework ke liye dekho Variable Mass Systems.

Recall Solution L5·2

Required ratio: $R = e^{9400/3000} = e^{3.1333} = 22.95$ . Launch mass $m_{0} = 1$ lo (normalise karo). Toh $m_{f} = 1/ R = 0.04357$ . Propellant $M_{p}$ aur structure $M_{s} = ε M_{p}$ split karo. Burned mass $= M_{p} = m_{0} - m_{f} - (payload)$ ... saaf karo: propellant burned $= m_{0} - m_{f} = 1 - 0.04357 = 0.95643$ sirf tab agar payload $\to 0$ aur structure $m_{f}$ ke andar count ho. Structure required $= ε M_{p} = 0.08 \times 0.95643 = 0.07651$ . Lekin $m_{f} = 0.04357$ ko saari bachi hui mass (structure + payload) hold karni hai. Kyunki required structure $0.07651 > m_{f} = 0.04357$ hai, structure akela allowed final mass ke andar fit nahi hoga. Conclusion: koi payload possible nahi — yeh design ek single stage ke roop mein impossible hai. Yahi classic "tyranny of the rocket equation" hai, aur yahi reason hai ki Multistage Rockets exist karte hain.

Recall Solution L5·3

(a) $m_{f} = m_{0} \Rightarrow R = 1 \Rightarrow ln 1 = 0 \Rightarrow Δ v = 0.$ Sensible: koi fuel nahi jalaya, koi speed nahi mili. (b) $m_{f} \to 0 \Rightarrow R \to \infty \Rightarrow ln R \to \infty \Rightarrow Δ v \to \infty.$ Mathematically unbounded — lekin physically impossible kyunki real rockets mein irreducible structural mass hoti hai, toh $m_{f}$ kabhi 0 nahi pahunchta. Log ki slow growth ka matlab hai ki bahut bada $R$ bhi sirf thoda sa bada $Δ v$ deta hai (Figure s01). (c) $v_{e} = 0 \Rightarrow Δ v = 0$ kisi bhi $R$ ke liye. Agar gas zero relative speed par nikalti hai toh woh koi momentum nahi le jaati, toh Conservation of Momentum se rocket ko kuch nahi milta. Dono levers ( $v_{e}$ aur $R$ ) non-trivial hone chahiye.

Recall Solution L5·4

$Δ v_{tot} = v_{e} ln R_{1} + v_{e} ln R_{2} = v_{e} ln (R_{1} R_{2}) = v_{e} ln 9.$ $= 3500 \times 2.1972 = 7690 m/s \approx 7.69 km/s,$ kisi bhi split ke liye same jab $R_{1} R_{2} = 9$ ho. Reality mein isko kya tod deta hai: har stage apna structural mass carry karta hai aur (lower stages ke liye) upar wale stages ka dead weight bhi, isliye achievable $R_{i}$ per stage free nahi hai — optimal staging woh splits choose karta hai jo in structural penalties ko balance kare. Yahan clean additivity idealised skeleton hai.

Recall Ek-line self-check

Stage ke $Δ v$ add kyun hote hain jab stage mass ratios multiply hote hain? ::: Kyunki $ln (R_{1} R_{2}) = ln R_{1} + ln R_{2}$ — log ratios ke product ko velocity contributions ki sum mein convert kar deta hai.