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

3.3.1 · D3 · Physics › Rocket Propulsion › Tsiolkovsky rocket equation — full first-principles derivati

Yeh page ek drill floor hai. Parent derivation ne formula ek baar prove kar diya. Ab hum isme har tarah ke sawaal daalte hain — forward, backward, degenerate, real-world, aur exam-trap — taaki koi bhi scenario aisa na ho jo tumne pehle na dekha ho.

Sab kuch ek hi equation par tika hai, jise tumhe teen tareekon se padhna aana chahiye:

Koi bhi symbol use karne se pehle, plain-word meaning yaad karo: $v_{e}$ = speed of exhaust gas relative to the rocket (metres per second); $m_{0}$ = mass at the start of the burn; $m_{f}$ = mass at the end; $Δ v$ = the change in the rocket's own speed. Natural log $ln$ yeh sawaal hai "e ki kaunsi power yeh number degi?" — yeh exactly $e^{x}$ ka inverse hai, isliye Roop 1 aur Roop 2 ek doosre ko undo karte hain.

Scenario matrix

Is topic ka har sawaal inhi cells mein se kisi ek mein aata hai. Table ke neeche, Examples 1–8 apne-apne cell(s) ko mark karte hain.

Cell	Kya alag hai isme	Covered by
A. Forward, basic	given $m_{0}, m_{f}, v_{e}$ → find $Δ v$	Ex 1
B. Backward (mass ratio)	given $Δ v, v_{e}$ → find $R = m_{0} / m_{f}$ or fuel fraction	Ex 2
C. Backward (find $v_{e}$ )	given $Δ v$ and masses → find $v_{e}$	Ex 3
D. Doubling / log-scaling	show a fixed boost per multiplicative fuel step	Ex 4
E. Degenerate: no fuel burned	$m_{f} = m_{0} \Rightarrow R = 1$	Ex 5
F. Limiting: burn everything	$m_{f} \to$ payload only, and $m_{f} \to 0$	Ex 5
G. $Δ v > v_{e}$ "impossible" case	rocket apne exhaust se bhi aage nikal jaata hai	Ex 6
H. Real-world word problem	staging, add-the- $Δ v$ logic	Ex 7
I. Exam twist: gravity loss	non-ideal $Δ v$ , subtract $g t$	Ex 8

Related tools jab zaroorat padegi tab aayenge: Specific Impulse (Ex 3), Thrust and Mass Flow Rate (Ex 8), Multistage Rockets (Ex 7), Gravity Loss and Drag Loss (Ex 8).

Example 1 — Cell A: forward, basic

Worked example

Ek rocket $v_{e} = 2500 m/s$ par gas eject karta hai. Uska mass $m_{0} = 12, 000 kg$ se $m_{f} = 3000 kg$ tak girta hai. $Δ v$ nikalo.

Forecast: Ratio hai $12000/3000 = 4$ . Kya $Δ v$ lagbhag $v_{e}$ ke barabar hoga, bahut kam, ya bahut zyada? Pehle guess karo.

Step 1 — Mass ratio banao. $R = \frac{m _{0}}{m _{f}} = \frac{12000}{3000} = 4$ . Yeh step kyun? Formula sirf ratio use karta hai, raw kilograms kabhi nahi — isliye hum do numbers ko ek mein compress karte hain.

Step 2 — Roop 1 lagao. $Δ v = v_{e} ln R = 2500 \times ln 4$ . Yeh step kyun? Dono masses aur $v_{e}$ pata hain, toh Roop 1 (speed nikalo) direct raasta hai.

Step 3 — Calculate karo. $ln 4 = 1.3863$ , toh $Δ v = 2500 \times 1.3863 = 3466 m/s$ . Yeh step kyun? Symbolic answer ko ek aisi number mein badlo jo mission planner use kar sake.

Verify: Units: $m/s \times (dimensionless log) = m/s$ ✓. Sanity: $Δ v = 3466 > v_{e} = 2500$ , jo bilkul theek hai kyunki $R = 4 > e \approx 2.718$ (yeh Ex 6 mein detail mein samjhayenge). $Δ v \approx 3.47 km/s$ .

Example 2 — Cell B: fuel fraction nikalte hue ulta kaam karo

Worked example

Tumhe $v_{e} = 3.0 km/s$ ke engine se $Δ v = 6.0 km/s$ hasil karni hai. Launch mass ka kitna hissa propellant hona chahiye?

Forecast: $Δ v$ , $v_{e}$ se do guna hai. Kya fuel fraction 50%, 75%, ya 85% se upar hoga? Guess karo.

Step 1 — Roop 2 se ratio nikalo. $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{6.0/3.0} = e^{2} = 7.389$ . Yeh step kyun? Humein speed chahiye, masses nahi — toh hum equation ko $e^{x}$ se ulta chalate hain, jo $ln$ ka inverse hai.

Step 2 — Bacha hua fraction nikalo. $\frac{m _{f}}{m _{0}} = \frac{1}{7.389} = 0.1353$ . Yeh step kyun? "Bacha hua mass fraction" mass ratio ka reciprocal hai — yeh dry+payload ka hissa hai.

Step 3 — Propellant fraction baaki sab kuch hai. $f_{p r o p} = 1 - \frac{m _{f}}{m _{0}} = 1 - 0.1353 = 0.8647$ . Yeh step kyun? Total mass = propellant + jo bacha; jo nahi bacha woh jal gaya.

Verify: Wapas plug karo: $ln (1/0.1353) = ln (7.39) = 2.0$ , times $v_{e} = 3.0 = 6.0 km/s$ ✓. ≈ 86.5% propellant hona chahiye. $Δ v$ ka $v_{e}$ se do guna hona pehle hi rocket ko almost all-fuel banana impose kar deta hai — log ki warning.

Example 3 — Cell C: exhaust velocity nikalo (specific impulse ke saath)

Worked example

Ek stage $m_{0} = 500 kg$ se start karti hai, $m_{f} = 200 kg$ par khatam hoti hai, aur is burn mein measured speed gain $Δ v = 3200 m/s$ rahi. $v_{e}$ kya tha? Phir ise Specific Impulse $I_{s p} = v_{e} / g_{0}$ ke roop mein express karo jahan $g_{0} = 9.81 m/s^{2}$ .

Forecast: $R = 2.5$ , toh $ln R \approx 0.916$ . Kya $v_{e}$ , $Δ v$ se bada hoga ya chhota? Guess karo.

Step 1 — Roop 1 ko $v_{e}$ ke liye rearrange karo. $v_{e} = \frac{Δ v}{ln ( m _{0} / m _{f} )}$ . Yeh step kyun? Ek hi unknown hai $v_{e}$ ; use divide karke free karo.

Step 2 — Numbers daalo. $R = 500/200 = 2.5$ , $ln 2.5 = 0.9163$ , toh $v_{e} = \frac{3200}{0.9163} = 3492 m/s$ . Yeh step kyun? Ab $v_{e}$ ke alaawa sab kuch pata hai.

Step 3 — Specific impulse mein convert karo. $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{3492}{9.81} = 356 s$ . Yeh step kyun? Engineers engine quality ko seconds mein quote karte hain; $I_{s p}$ bas $v_{e}$ hai jo gravity se scale ho jaata hai taaki units time ban jaayein.

Verify: Reverse-check: $v_{e} ln R = 3492 \times 0.9163 = 3200 m/s$ ✓. Units of $I_{s p}$ : $(m/s) / (m/s^{2}) = s$ ✓. $v_{e} \approx 3490 m/s$ , $I_{s p} \approx 356 s$ — ek accha chemical rocket.

Example 4 — Cell D: doubling law ko concrete banao

Worked example

$v_{e} = 3000 m/s$ ke saath, mass ratios $R = 2, 4, 8, 16$ ke liye $Δ v$ ka table banao. Dikhao ki $R$ ki har doubling same $Δ v$ add karti hai.

Forecast: $R : 2 \to 4$ se extra $Δ v$ , $R : 8 \to 16$ ke extra $Δ v$ se zyada, kam, ya barabar hoga? Guess karo.

Step 1 — Har ek calculate karo. $Δ v (2) = 3000 ln 2 = 2079 m/s$ $Δ v (4) = 3000 ln 4 = 4159 m/s$ $Δ v (8) = 3000 ln 8 = 6238 m/s$ $Δ v (16) = 3000 ln 16 = 8318 m/s$ Yeh step kyun? Chaar ratios pe seedha Roop 1 lagao.

Step 2 — Gaps dekho. Har step pe: $4159 - 2079 = 2080$ , $6238 - 4159 = 2079$ , $8318 - 6238 = 2080$ . Yeh step kyun? Differences woh pattern expose karte hain jo raw numbers chhupa deti hain.

Step 3 — Explain karo. $ln (2 R) - ln (R) = ln 2$ hamesha, toh har doubling $v_{e} ln 2 = 3000 \times 0.6931 = 2079 m/s$ add karti hai, chahe shuru kahan se karo. Yeh step kyun? Constant gap hi diminishing-returns ka message hai: speed add karne ke liye fuel multiply karna padega.

Verify: $v_{e} ln 2 = 2079 m/s$ teeno gaps se match karta hai ✓. Figure mein curve concave hai — logarithm ki pehchaan.

Example 5 — Cells E & F: degenerate aur limiting inputs

Worked example

Same engine $v_{e} = 3000 m/s$ . Teen edge cases: (a) Koi fuel nahi jaala: $m_{f} = m_{0}$ . (b) Sirf payload bachta hai: $m_{0} = 10, 000 kg$ , $m_{f} = 1000 kg$ (90% fuel). (c) Theoretical limit $m_{f} \to 0$ .

Forecast: Inme se kis mein $Δ v = 0$ hoga, kis mein bada finite number, aur kis mein "blow up"?

Step 1 — Case (a). $R = m_{0} / m_{0} = 1$ , aur $ln 1 = 0$ , toh $Δ v = 0$ . Yeh step kyun? Agar kuch nahi feinka, toh kuch nahi mila — formula zero return karna chahiye, aur karta hai. Yeh poori equation ka sanity anchor hai.

Step 2 — Case (b). $R = 10000/1000 = 10$ , $Δ v = 3000 ln 10 = 3000 \times 2.3026 = 6908 m/s$ . Yeh step kyun? Ek realistic "90% fuel pe kharch karo" case — note karo ki phir bhi sirf ≈2.3 exhaust-speeds milti hain.

Step 3 — Case (c). Jaise $m_{f} \to 0$ , $R \to \infty$ aur $ln R \to \infty$ , toh $Δ v \to \infty$ . Yeh step kyun? Mathematically unbounded — lekin physically impossible: kabhi bhi zero mass nahi ho sakta (engine, tank, payload hamesha hoga). Yeh limit warn karti hai ki equation ka promise sirf tab fulfill hota jab tum pure fuel ship kar sako, jo tum kar nahi sakte.

Verify: (a) $ln 1 = 0$ exactly ✓. (b) $ln 10 = 2.3026$ , product $= 6908 m/s$ ✓. (c) ek limit hai, number nahi — divergence hi baat hai.

Example 6 — Cell G: apne exhaust ko beat karna ( $Δ v > v_{e}$ )

Worked example

"Rocket apne faanke gaye gas se tez nahi ja sakta." Galat sabit karo. $v_{e} = 2000 m/s$ ke saath, kaunsa mass ratio $Δ v = v_{e}$ exactly deta hai? Kaunsa ratio $Δ v = 2 v_{e}$ deta hai?

Forecast: Kya crossover ratio lagbhag 2 hai, lagbhag $e \approx 2.7$ , ya lagbhag 10? Guess karo.

Step 1 — $Δ v = v_{e}$ set karo. Toh $v_{e} = v_{e} ln R \Rightarrow ln R = 1 \Rightarrow R = e = 2.718$ . Yeh step kyun? Hum exact ratio dhundhte hain jahan rocket apni exhaust speed match karta hai — $v_{e}$ se divide karne par $ln R = 1$ isolate hota hai.

Step 2 — Interpret karo. Kisi bhi $R > e \approx 2.718$ ke liye, $Δ v > v_{e}$ milta hai. Example 1 mein already $R = 4 > e$ tha, toh uska $3466 m/s$ genuinely $v_{e} = 2500$ se zyada tha. Yeh step kyun? Yeh exactly batata hai kab "impossible" hota hai — ek modest 2.72:1 mass ratio par.

Step 3 — $Δ v = 2 v_{e}$ set karo. $ln R = 2 \Rightarrow R = e^{2} = 7.389$ . Yeh step kyun? Dikhata hai ki tum exhaust speed double bhi kar sakte ho — $R \approx 7.4$ ki keemat par (Ex 2 ke ratio se milta hai!).

Yeh jaadu nahi hai: gas us waqt rocket ke relative $v_{e}$ par jaati hai. Lekin rocket ko laakhon choti-choti pushes milti hain; woh accumulate hoti hain. Ek frictionless cart pe baithe insaan ki tarah jo 5 m/s ke dhar par balls fainke — kaafi balls fainkne ke baad, woh khud 5 m/s se tez hoga.

Verify: $e^{1} = 2.718$ , $e^{2} = 7.389$ ✓; $R = 4$ par, $ln 4 = 1.386 > 1$ toh $Δ v > v_{e}$ ✓.

Example 7 — Cell H: real-world staging word problem

Worked example

Ek two-stage rocket (Multistage Rockets). Stage 1: $v_{e} = 2800 m/s$ , $m_{0} = 100, 000 kg$ se $m_{1} = 40, 000 kg$ tak jalti hai (phir khali pehla stage drop ho jaata hai). Stage 2: $v_{e} = 3400 m/s$ , $25, 000 kg$ se $8, 000 kg$ tak jalti hai. Total $Δ v$ nikalo.

Forecast: Total $Δ v$ = do logs ka sum. Kya yeh Ex 5 ke ≈6.9 km/s single-stage-90% case ko beat karega? Guess karo.

Step 1 — Stage 1 ka $Δ v$ . $R_{1} = 100000/40000 = 2.5$ , $Δ v_{1} = 2800 ln 2.5 = 2800 \times 0.9163 = 2566 m/s$ . Yeh step kyun? Har stage apna alag rocket-equation problem hai, apne $m_{0}, m_{f}, v_{e}$ ke saath.

Step 2 — Stage 2 ka $Δ v$ . $R_{2} = 25000/8000 = 3.125$ , $Δ v_{2} = 3400 ln 3.125 = 3400 \times 1.1394 = 3874 m/s$ . Yeh step kyun? Fresh masses kyunki dead pehla stage jettison ho gaya — staging ka yahi poora fayeda hai.

Step 3 — Add karo. $Δ v_{t o t} = 2566 + 3874 = 6440 m/s$ . Yeh step kyun? Sequential burns se speed gains simply add hote hain — $Δ v$ additive hai, chahe $R$ multiplicative ho.

Verify: $ln 2.5 = 0.9163$ , $ln 3.125 = 1.1394$ ; products $2566$ aur $3874 m/s$ ; sum $6440 m/s$ ✓. Staging ne hume mid-flight dead weight discard karne diya, isliye do modest stages ek extreme single stage se compete karte hain.

Example 8 — Cell I: gravity loss ke saath exam twist

Worked example

Ek rocket $t = 90 s$ seedha upar jalti hai. Ideal $Δ v_{i d e a l}$ aata hai $v_{e} = 3000 m/s$ , $R = m_{0} / m_{f} = 4$ se. Lekin us time mein gravity $Δ v_{g r a v} = g_{0} t$ le jaati hai jahan $g_{0} = 9.81 m/s^{2}$ (Gravity Loss and Drag Loss). Actual speed gain nikalo.

Forecast: 90 s mein gravity loss — kya yeh ideal $Δ v$ ka chhota sa nibble hai ya bada bhaag?

Step 1 — Ideal $Δ v$ . $Δ v_{i d e a l} = 3000 ln 4 = 3000 \times 1.3863 = 4159 m/s$ . Yeh step kyun? Tsiolkovsky equation frictionless, gravity-free speed deta hai — yeh hamara baseline hai.

Step 2 — Gravity loss. $Δ v_{g r a v} = g_{0} t = 9.81 \times 90 = 883 m/s$ . Yeh step kyun? Upar thrust karte waqt, gravity poore time neeche kheenchti rehti hai; constant $g_{0}$ ko burn pe integrate karne par $g_{0} t$ milta hai. (Yeh wahi $- \int g d t$ term hai jiske baare mein parent ke fine print ne warn kiya tha.)

Step 3 — Actual gain. $Δ v_{a c t u a l} = 4159 - 883 = 3276 m/s$ . Yeh step kyun? Real speed = ideal minus jo gravity ne chheena.

Verify: $9.81 \times 90 = 882.9 m/s$ ; $4159 - 883 = 3276 m/s$ ✓. Units sab m/s ✓. Gravity ne ideal $Δ v$ ka ≈21% nigal liya — isliye rockets jaldi pitch over karte hain taaki seedha upar gravity se ladne mein kam waqt lagao.

Recall Har example kaun si cell hit karta hai?

Ex 1 → A (forward basic) ::: masses diye, $Δ v$ nikalo Ex 2 → B (backward to fuel fraction) ::: $Δ v$ diya, $R$ nikalo Ex 3 → C ( $v_{e}$ nikalo) + $I_{s p}$ ::: exhaust speed ke liye rearrange karo Ex 4 → D (doubling law) ::: har fuel doubling $v_{e} ln 2$ add karti hai Ex 5 → E aur F (degenerate + limits) ::: $R = 1$ se 0 milta hai, $m_{f} \to 0$ diverge karta hai Ex 6 → G ( $Δ v > v_{e}$ ) ::: crossover $R = e$ par Ex 7 → H (staging word problem) ::: do rocket equations, add kiye Ex 8 → I (gravity loss twist) ::: $g_{0} t$ ghataao

Active Recall Flashcards

Backward solve karne ke liye (jaante ho

Δ v

, mass ratio chahiye), kaunsa function lagate ho?

Exponential:

m_{0} / m_{f} = e^{Δ v / v_{e}}

—

ln

ka inverse.

Kitne mass ratio ke upar

Δ v

v_{e}

se zyada hota hai?

R > e \approx 2.718

, kyunki wahan

ln R > 1

hota hai.

Jab koi fuel nahi jaalta (

m_{f} = m_{0}

), toh

Δ v

kya hoga?

Exactly

0

, kyunki

ln 1 = 0

Successive rocket stages ke

Δ v

values kaise combine hote hain?

Woh add hote hain:

Δ v_{t o t} = Δ v_{1} + Δ v_{2} + \dots

Seedhi chadhaai mein

t

duration ke burn ka gravity loss kaise estimate karte hain?

Δ v_{g r a v} = g_{0} t

, ideal

Δ v

mein se ghataate hain.

Mass ratio ki har doubling kitna

Δ v

add karti hai?

Ek fixed

v_{e} ln 2

, starting ratio se independent.