Mass ratio m₀ - m_f — why it's so critical: Worked examples

3.3.3 · D3 · Physics › Rocket Propulsion › Mass ratio m₀ - m_f — why it's so critical

Yahan sab kuch parent par based hai: the mass-ratio note, aur Tsiolkovsky Rocket Equation par. Hum sirf iske teen symbols reuse karte hain:

Recall Teen symbols jo hum yahan kabhi re-derive nahi karenge

$R = m_{0} / m_{f}$ (mass ratio, wet over dry). $u$ = exhaust speed rocket ke relative ( $u = I_{s p} g_{0}$ , dekho Specific Impulse Isp). $Δ v = u ln R$ (the Tsiolkovsky Rocket Equation).

Scenario matrix

Kisi bhi number se pehle, chaliye har distinct tarah ki problem list karte hain jo ye ek equation pose kar sakti hai. Har row ek "cell" hai — ek scenario class apne trap ke saath. Neeche ke worked examples har ek ke saath (Cell A) jaisi tag lagate hain taaki tum dekh sako ki poora grid cover hua hai.

Cell	Kya diya hai	Kya poochha hai	Trap / edge jise dhyaan rakhna hai
A	$u$ , $R$	$Δ v$ nikalo	seedha direction, aasaan — "warm-up"
B	$u$ , $Δ v$	$R$ nikalo	$e^{(\cdot)}$ se invert karna hoga
C	$u$ , $Δ v$ , $m_{f}$	fuel $m_{p}$ nikalo	$m_{f}$ fixed hai, toh $m_{0} = R m_{f}$ , phir subtract karo
D	$u$ , $m_{0}$ , $m_{f}$	$Δ v$ nikalo	raw masses ko $ln (m_{0} / m_{f})$ mein daalein
E	degenerate: $R = 1$ (koi fuel nahi) ya $R \to \infty$	$Δ v$ ?	limiting behaviour — $ln$ ends par kya karta hai
F	do stages	total $Δ v$	$Δ v$ add hota hai, $R$ multiply hota hai — Multistage Rockets
G	word problem (real mission)	required $R$ + feasibility	mission Delta-v Budget ko fuel mein convert karo, phir judge karo
H	exam twist: $u$ ko ek factor se badlao	$R$ kaise respond karta hai	$R$ exponential hai $1/ u$ mein — non-linear

Matrix ek baar padho. Notice karo cells E, F, H woh hain jo log skip karte hain — aur jinhe exams pasand karte hain. Hum har ek ko apna example dete hain.

Cell A — forward: diya $R$ , nikalo $Δ v$

Worked example Ek rocket tab tak jalta hai jab tak sirf

$\frac{1}{4}$ mass bachta hai, $u = 3 km/s$ ke saath. $Δ v$ nikalo. Forecast: padhne se pehle andaaza lagao — kya $Δ v$ , $u$ se bada hoga ya chhota? (Sirf $1/4$ bachi hai matlab bahut fuel jala, toh $u$ se zyada…)

Mass ratio likho. "Sirf $1/4$ bachi hai" ka matlab hai $m_{f} = \frac{1}{4} m_{0}$ , toh $R = \frac{m _{0}}{m _{f}} = \frac{m _{0}}{\frac{1}{4} m _{0}} = 4.$ Ye step kyun? $R$ define hota hai wet-over-dry ke roop mein; payload/structure cancel ho jaata hai, hume sirf fraction chahiye jo bachi hai.
Tsiolkovsky apply karo. $Δ v = u ln R = 3 ln 4 = 3 \times 1.3863 = 4.159 km/s .$ Ye step kyun? Ye seedha direction hai — kuch bhi invert nahi karna.

Verify: $ln 4 = 2 ln 2 \approx 1.386$ , aur $1.386 > 1$ , toh $Δ v > u$ . Forecast se match karta hai: quarter tak jalana ek exhaust-speed se zyada deta hai. Units: $(km/s) \times (dimensionless) = km/s$ . ✓

Cell B — inverse: diya $Δ v$ , nikalo $R$

Worked example Tumhe

$Δ v = 6 km/s$ chahiye aur tumhara engine $u = 3 km/s$ deta hai. Tumhe kaun sa mass ratio build karna hoga? Forecast: exponent hai $6/3 = 2$ . Guess karo: kya $R$ lagbhag 4, 7, ya 20 ke aas-paas hai?

Log isolate karo. Equation ko $u$ se divide karo: $\frac{Δ v}{u} = ln R \Rightarrow \frac{6}{3} = 2 = ln R .$ Ye step kyun? $ln R$ unknown hai; isko undo karne se pehle akele kar lo.
Log ko $e$ se undo karo. Exponential $e^{(\cdot)}$ exactly $ln$ ka inverse hai — ye jawaab deta hai "kis number ka natural log ye hai?" $R = e^{2} = 7.389.$ Ye step kyun? Isliye hum $e$ use karte hain aur, say, base-10 nahi: rocket equation ne ek natural log produce kiya, toh iska inverse $e^{x}$ hi hona chahiye.

Verify: isko aage push karo — $u ln R = 3 \times ln (7.389) = 3 \times 2.0 = 6 km/s$ . ✓ Aur $R \approx 7.4$ "4" aur "20" guesses ke beech hai, jaise exponent 2 ke liye expected tha.

Cell C — fuel mass ke liye solve karo

Worked example Ek satellite bus ka dry mass

$m_{f} = 200 kg$ hai. Uske thruster ka $u = 2.5 km/s$ hai aur mission ko $Δ v = 1.5 km/s$ chahiye. Kitna propellant $m_{p}$ carry karna hoga? Forecast: chhota $Δ v$ ( $u$ se kaafi kam), toh guess karo: kya fuel dry mass se kam hai ya zyada?

Mass ratio nikalo (Cell B move). $R = e^{Δ v / u} = e^{1.5/2.5} = e^{0.6} = 1.8221.$ Ye step kyun? Fuel nahi nikal sakte jab tak pata na ho ki wet, dry se kitna bada hai — wahi $R$ hai.
Wet mass recover karo. $m_{f}$ fixed hai (hardware nahi badal sakte), toh $m_{0} = R m_{f} = 1.8221 \times 200 = 364.42 kg .$ Ye step kyun? $R = m_{0} / m_{f}$ rearrange hota hai $m_{0} = R m_{f}$ mein; dry mass yahan anchor hai.
Fuel difference hai. $m_{p} = m_{0} - m_{f} = 364.42 - 200 = 164.42 kg .$ Ye step kyun? Propellant = jo shuru mein tha minus jo bacha.

Verify: propellant fraction $ζ = m_{p} / m_{0} = 164.42/364.42 = 0.451$ , aur formula $ζ = 1 - 1/ R = 1 - 1/1.8221 = 0.451$ . ✓ Fuel ( $164$ kg) dry mass ( $200$ kg) se kam hai — chhote $Δ v$ ke forecast se match karta hai.

Cell D — raw masses seedha daalein

Cell E — degenerate aur limiting cases

Ye woh cell hai jo log bhool jaate hain. $R$ ki range ke ends par kya hota hai? Yaad raho $R \geq 1$ hamesha (negative fuel nahi ho sakta).

Figure — Mass ratio m₀ - m_f — why it's so critical

Cell F — do stages (Δv add hota hai, R multiply hota hai)

Worked example Ek two-stage rocket.

Stage 1: $R_{1} = 4$ , $u_{1} = 3 km/s$ . Stage 2: $R_{2} = 5$ , $u_{2} = 4 km/s$ . Total $Δ v$ nikalo. Forecast: kya tum ratios add karoge ( $4 + 5$ ) ya multiply ( $4 \times 5$ )? Guess karo konsi quantity add hoti hai.

Har stage ka Δv alag alag. $Δ v_{1} = u_{1} ln R_{1} = 3 ln 4 = 4.1589 km/s,$ $Δ v_{2} = u_{2} ln R_{2} = 4 ln 5 = 6.4378 km/s .$ Ye step kyun? Har stage apna khud ka rocket equation hai; stage 1 drop hone ke baad, stage 2 apne $u$ aur $R$ ke saath fresh start karta hai.
Total Δv add hota hai. Velocity changes motion ki same line pe simple additions hain: $Δ v_{tot} = Δ v_{1} + Δ v_{2} = 4.1589 + 6.4378 = 10.597 km/s .$ Ye step kyun? Tumhari jo speed pehle se hai woh rakhte ho aur upar aur daalte ho — tumhari velocity kahi reset nahi hoti.

Verify: log ratios ke product ko sum mein badalta hai: agar $u$ barabar hota, $u ln (R_{1} R_{2}) = u (ln R_{1} + ln R_{2})$ . Toh $R$ multiply hota hai ( $4 \times 5 = 20$ effective) jabki $Δ v$ add hota hai — exactly forecast test. ✓ Total $10.6 km/s$ LEO budget clear karta hai jahaan ek stage (Cell G) nahi kar sakta tha.

Cell G — real-world mission + feasibility

Worked example Earth se escape karne ke liye mission ko

$Δ v \approx 11.2 km/s$ chahiye (Delta-v Budget se). Tumhara best single engine $u = 3.5 km/s$ deta hai. Real tanks force karte hain $m_{f} \geq 0.08 m_{0}$ (structure $\geq 8%$ hai). Kya single-stage escape possible hai? Forecast: required $R$ bada hoga — kya ye structural ceiling $R = 1/0.08 = 12.5$ se exceed karega?

Required mass ratio (Cell B). $R_{req} = e^{Δ v / u} = e^{11.2/3.5} = e^{3.2} = 24.533.$ Ye step kyun? Mission ki speed demand ko fuel-fraction demand mein badlo.
Structural ceiling. Agar $m_{f} \geq 0.08 m_{0}$ , toh $R = m_{0} / m_{f} \leq 1/0.08 = 12.5$ . Ye step kyun? Real hardware ek maximum achievable $R$ set karta hai; tanks ko lighter nahi wish kar sakte.
Compare karo. $R_{req} = 24.5 > R_{max} = 12.5$ . Ek stage se Impossible hai. Ye step kyun? Demand jo bhi koi bhi single structure hold kar sakta hai usse baahar nikal jaati hai — ye exponential wall in action hai.

Verify: best ek stage kar sakta hai $Δ v_{m a x} = u ln R_{m a x} = 3.5 ln (12.5) = 3.5 \times 2.5257 = 8.84 km/s < 11.2$ . $2.36 km/s$ short ⇒ tumhe zaroor stage karna hoga (dekho Cell F) ya $u$ badhana hoga (Cell H). ✓

Cell H — exam twist: $u$ scale karo, $R$ dekho

Worked example Ek mission

$Δ v = 8 km/s$ fix karta hai. Engine X ka $u = 2.0 km/s$ hai; engine Y ka $u = 4.0 km/s$ (double). $u$ double karne par required $R$ kis factor se girta hai? Forecast: $u$ double karo — kya required $R$ aadha ho jaata hai, ya bahut zyada girta hai?

Engine X ke liye $R$ . $R_{X} = e^{8/2} = e^{4} = 54.598.$
Engine Y ke liye $R$ . $R_{Y} = e^{8/4} = e^{2} = 7.389.$ Ye steps kyun? Har ek Cell-B inversion hai; sirf exponent $Δ v / u$ badalta hai.
Factor. $\frac{R _{X}}{R _{Y}} = \frac{e ^{4}}{e ^{2}} = e^{4 - 2} = e^{2} = 7.389.$ Ye step kyun? Exponentials divide karne par exponents subtract hote hain — isliye effect sirf halving nahi hai.

Verify: $u$ ko 2 se 4 tak double karne par required $R$ 7.4× factor se gira, 2× nahi. Kyunki $R = e^{Δ v / u}$ $1/ u$ par ek exponential ke andar depend karta hai, $u$ mein 2× improvement exponentially payoff karta hai. ✓ Ye deep reason hai ki Specific Impulse Isp kyun chase karne laayak hai.

Active recall

Recall Cell E mein

$Δ v = 0$ kyun hota hai jab $R = 1$ ? $R = 1$ ka matlab wet mass = dry mass = zero propellant; $ln 1 = 0$ , toh $Δ v = u \cdot 0 = 0$ . Koi fuel nahi pheka, koi momentum swap nahi hua.

Recall Stages ke across, konsi quantity add hoti hai aur konsi multiply?

$Δ v$ add hota hai ( $Δ v_{1} + Δ v_{2}$ ); mass ratios multiply hote hain ( $R_{1} R_{2}$ ). Log unhe link karta hai: $ln (R_{1} R_{2}) = ln R_{1} + ln R_{2}$ .

Recall Fixed

$Δ v$ par $u$ double karne se required $R$ kis factor se badalta hai (Cell H, $Δ v = 8$ , $u : 2 \to 4$ )? $e^{2} \approx 7.39$ factor se — ek exponential drop, halving nahi.

Connections

Tsiolkovsky Rocket Equation — har cell iska ek use hai.
Specific Impulse Isp — Cell H isliye hai ki bada $u$ exponentially kyun matter karta hai.
Multistage Rockets — Cells F aur G dikhate hain ki staging kyun forced hai.
Delta-v Budget — Cell G ek mission budget ko required $R$ mein convert karta hai.
Conservation of Momentum — woh equation ka origin jise cells apply karti hain.
Exhaust Velocity and Thrust — jahaan $u$ physically aata hai.