Δv = v_e · ln(m₀ - m_f) — understanding each term: Exercises

3.3.2 · D4 · Physics › Rocket Propulsion › Δv = v_e · ln(m₀ - m_f) — understanding each term

Yeh page ek self-test ladder hai parent note Δv = v_e · ln(m₀/m_f) ke liye. Har level difficulty mein upar jaata hai. Har problem ko solution collapsed rakh ke try karo, phir usse kholo.

Shuru karne se pehle, yeh symbols yaad rakh lo (parent se har ek symbol, simple words mein):

Definition Woh symbols jo tum baar baar use karoge

$Δ v$ = change in speed jo rocket fuel jalaa ke haasil kar sakta hai (ek budget, m/s).
$v_{e}$ = exhaust velocity, gas rocket ke relative nozzle se kitni tez nikalti hai (m/s).
$m_{0}$ = wet mass (shuru mein sab kuch, saara fuel shamil).
$m_{f}$ = dry mass (burn ke baad jo bacha — structure + payload).
$R = m_{0} / m_{f}$ = mass ratio, ek pure number jisme koi units nahi.
$I_{s p}$ = specific impulse — engine ki efficiency rate karne ka ek tarika, seconds mein measure hota hai. Zyada $I_{s p}$ matlab zyada push per kilogram fuel. Yeh exhaust velocity se is tarah juda hai: $v_{e} = g_{0} I_{s p}$ — dekho Specific Impulse (Isp).
$g_{0}$ = standard gravity, ek fixed constant $9.81$ m/s² jo sirf specific impulse ko speed mein convert karne ke liye use hota hai. Yeh kisi bhi planet ki local gravity nahi hai — bas ek bookkeeping number hai.
Core relation: $Δ v = v_{e} ln R$ , aur uska inverse $R = e^{Δ v / v_{e}}$ .

Do tools hain jinhe hum baar baar use karte hain, toh chalo inhe yahan dobara samjhte hain:

Intuition Woh do moves jo tum baar baar karoge

$ln$ ("log") ek shrinking machine hai: yeh ratio $R$ ko ek bahut chote number mein badal deta hai. $ln 1 = 0$ , $ln e = 1$ , $ln R$ bahut dheere badhta hai jab $R$ barhta hai. Ise kyun use karte hain? Kyunki velocity gain $d v = - v_{e} \frac{d m}{m}$ ke roop mein build hoti hai. Yahan $m$ rocket's instantaneous mass hai us burn ke us pal mein, aur $d m$ ek instant mein chota mass change hai. Kyunki rocket mass khho raha hai, $d m < 0$ (yeh negative hai), aur aage minus sign ise wapas palatda hai taaki $d v$ positive aaye — rocket speed up hota hai. $1/ m$ factor ka matlab hai ki ek ek kilogram phenka jaane se halke rocket ko zyada kick milta hai, aur $1/ m$ ko add karna (integrate karna) bilkul wahi hai jo $ln$ produce karta hai — dekho Natural Logarithm and Integration of 1/x.
$e^{x}$ ("exponential") $ln$ ko undo karta hai. Agar $ln R = x$ , toh $R = e^{x}$ . Hum ise tab use karte hain jab humein ek zaroori $Δ v$ diya ho aur mass ratio dhundhna ho — hum equation ko ulta chalate hain.

Neeche ke do chhote plots par gaur se dekho — yeh is equation ki poori personality hain.

Figure — Δv = v_e · ln(m₀ - m_f) — understanding each term

Level 1 — Recognition

L1.1

$Δ v = v_{e} ln (m_{0} / m_{f})$ mein kaun sa symbol dimensionless hai (koi units nahi), aur kyun?

Recall Solution

Ratio $R = m_{0} / m_{f}$ dimensionless hai. Kyun: yeh mass divided by mass hai (kg ÷ kg), toh units cancel ho jaati hain. Yeh zaroori hai — tum kisi aisi cheez ka $ln$ nahi le sakte jisme units hon. Toh $ln R$ bhi ek pure number hai, aur $v_{e} ln R$ sahi se $v_{e}$ ki units carry karta hai: metres per second. ✔

L1.2

Ek rocket ka $m_{0} = 400, 000$ kg wet hai aur $250, 000$ kg propellant jalata hai. $m_{f}$ kya hai?

Recall Solution

$m_{f}$ woh hai jo burn ke baad bachta hai, na ki jo jalaya gaya. $m_{f} = m_{0} - (propellant burned) = 400, 000 - 250, 000 = 150, 000 kg .$

L1.3

Agar koi fuel nahi jalaya ( $m_{0} = m_{f}$ ), toh $Δ v$ kya hoga? Ek line mein reason dikhao.

Recall Solution

$R = m_{0} / m_{f} = 1$ , aur $ln 1 = 0$ , toh $Δ v = v_{e} \cdot 0 = 0 m/s .$ Kyun sense banta hai: koi mass peeche nahi pheka ⇒ koi forward kick nahi (Conservation of Momentum se).

Level 2 — Application

L2.1

$m_{0} = 600, 000$ kg, $m_{f} = 120, 000$ kg, $v_{e} = 3200$ m/s. $Δ v$ nikalo.

Recall Solution

Pehle ratio (yahi toh log khaata hai): $R = \frac{600 , 000}{120 , 000} = 5.$ $R$ pehle kyun nikaalte hain? Equation kehti hai $Δ v$ sirf wet to dry mass ke ratio par depend karta hai, kabhi bhi raw kilograms par alag se nahi. $(600, 000, 120, 000)$ aur $(5, 1)$ masses wale do rockets ko same $Δ v$ milega. Toh hum donon masses ko log ko dene se pehle single number $R = 5$ mein collapse karte hain — sirf yahi quantity log ko matter karti hai. $Δ v = 3200 \cdot ln 5 = 3200 \times 1.60944 \approx 5150 m/s .$

L2.2

Tumhe $Δ v = 8000$ m/s chahiye aur $v_{e} = 3000$ m/s hai. Kaun sa mass ratio $R$ chahiye?

Recall Solution

Equation ko exponential ke saath ulta chalao (woh tool jo $ln$ undo karta hai): $ln R = \frac{Δ v}{v _{e}} = \frac{8000}{3000} = 2.6667.$ Divide karne ke baad exponentiate kyun karte hain? $v_{e}$ se divide karna multiplier hatata hai aur $ln R$ akela chhod deta hai — lekin woh ratio ka log hai, ratio khud nahi. Ratio $R$ abhi bhi log ke andar band hai. Isse azaad karne ke liye hum dono sides par $e^{(\cdot)}$ lagate hain, kyunki $e^{l n R} = R$ exactly — exponential ek aisi operation hai jo natural log cancel karta hai. Ise skip karne par tumhare paas $ln R \approx 2.67$ bachega, jo ek logarithm hai, mass ratio nahi. $R = e^{2.6667} \approx 14.39.$ Matlab: target tak pahunchne ke liye lagbhag $14.4$ kg wet mass har $1$ kg ke liye — fuel fraction bahut bada hai.

L2.3

Ek engine ka specific impulse $I_{s p} = 380$ s hai. Uska exhaust velocity $v_{e}$ nikalo ( $g_{0} = 9.81$ m/s² use karo).

Recall Solution

Specific Impulse (Isp) se, $v_{e} = g_{0} I_{s p}$ , jahan $I_{s p}$ specific impulse hai aur $g_{0} = 9.81$ m/s² standard-gravity conversion constant hai, dono symbol list mein define hain: $v_{e} = 9.81 \times 380 \approx 3728 m/s .$ Yeh link kyun hai: $I_{s p}$ bas $v_{e}$ hai jo "fuel ke weight per thrust ke seconds" mein measure ki gayi hai; $g_{0}$ se multiply karne par yeh wapas speed ban jaati hai.

Level 3 — Analysis

L3.1

Ek rocket mass ratio $R$ se $Δ v_{1}$ haasil karta hai. Agar mass ratio square karke $R^{2}$ kar do ( $v_{e}$ same rakho), toh naya $Δ v_{2}$ , $Δ v_{1}$ se kaise compare hoga?

Recall Solution

$Δ v_{2} = v_{e} ln (R^{2}) = v_{e} \cdot 2 ln R = 2 (v_{e} ln R) = 2 Δ v_{1} .$ $ln (R^{2}) = 2 ln R$ kyun hota hai? Yeh power rule of logarithms hai: $ln (a^{b}) = b ln a$ . Yeh isliye hota hai kyunki log exponents ko multipliers mein badal deta hai (kisi number ko square karna uski log khud mein add karta hai, yaani double karta hai). Dekho Natural Logarithm and Integration of 1/x. Interpretation: ratio square karne se $Δ v$ sirf double hoti hai. Speed budget double karne ke liye tumhe wet rocket ka dry se ratio square karna padega — "tyranny of the rocket equation." Isliye parent note log ko villain kehta hai.

L3.2

Dono designs ko $Δ v = 6000$ m/s deliver karna hai. Design A: $v_{e} = 3000$ m/s. Design B: $v_{e} = 4500$ m/s. Har ek ka zaroori mass ratio nikalo aur comment karo ki kaun sa per kg payload kam fuel maangta hai.

Recall Solution

Design A: $ln R_{A} = 6000/3000 = 2.000$ , toh $R_{A} = e^{2} \approx 7.389$ . Design B: $ln R_{B} = 6000/4500 = 1.3333$ , toh $R_{B} = e^{1.3333} \approx 3.793$ . Comment: zyada $v_{e}$ wala engine (B) sirf $\sim 3.8$ mass ratio maangta hai jabki same kaam ke liye A ko $\sim 7.4$ chahiye — same kaam ke liye roughly aadha wet-to-dry ratio. Zyada exhaust speed $Δ v$ bahut saste mein deta hai. Yahi ion engines ki poori appeal hai.

L3.3

Ek single stage mein $m_{f} = 30, 000$ kg hai aur $270, 000$ kg propellant carry karta hai, $v_{e} = 3400$ m/s. (a) $Δ v$ nikalo. (b) Ek mission ko $10, 000$ m/s chahiye. Yeh single stage kitna short hai?

Recall Solution

(a) $m_{0} = 270, 000 + 30, 000 = 300, 000$ kg, toh $R = 300, 000/30, 000 = 10$ . $Δ v = 3400 \cdot ln 10 = 3400 \times 2.302585 \approx 7829 m/s .$ (b) Shortfall $= 10, 000 - 7829 = 2171$ m/s. Single stage budget tak nahi pahunch sakta — yeh Multistage Rockets ki taraf ek hint hai.

Level 4 — Synthesis

L4.1 — Two-stage rocket

Ek rocket do stages mein udta hai, har ek depletion tak fire karta hai. Total $Δ v$ har stage ke $Δ v$ ka sum hai (velocities add hoti hain).

Stage 1: $v_{e, 1} = 3000$ m/s, wet $= 500, 000$ kg, dry-at-separation $= 200, 000$ kg.
Stage 2: $v_{e, 2} = 4000$ m/s, wet $= 90, 000$ kg, dry $= 30, 000$ kg.

Total $Δ v$ nikalo.

Neeche ka figure dono burns mein velocity badhte track karta hai. Left se right padho: lavender line Stage 1 ki climb hai, mint arrow wahan mark karta hai jahan spent Stage-1 shell drop hoti hai, aur coral line Stage 2 ki steeper climb hai (halka rocket, tez exhaust). Dots har milestone par cumulative speed hain — final dot woh total hai jo hum neeche compute karte hain.

Recall Solution

Add kyun karte hain? Har stage apna khud ka velocity change produce karta hai; same direction mein velocity changes simply add ho jaate hain ( $d v$ ko har burn par integrate karne se). Stage 1: $R_{1} = 500, 000/200, 000 = 2.5$ , $Δ v_{1} = 3000 \cdot ln 2.5 = 3000 \times 0.916291 \approx 2749 m/s .$ Stage 2: $R_{2} = 90, 000/30, 000 = 3$ , $Δ v_{2} = 4000 \cdot ln 3 = 4000 \times 1.098612 \approx 4394 m/s .$ Total: $Δ v = Δ v_{1} + Δ v_{2} \approx 2749 + 4394 = 7144 m/s .$

L4.2 — Staging ek bade tank se kyun behtar hai

L4.1 ka staged rocket lo aur poochho: agar hum poora kaam ek single stage mein karne ki koshish karte — beeche mein spent hardware drop na karke? Ek sach muchi single stage ko apna saara structure poori taraf carry karna padega, toh uska dry mass donon stages ke hardware aur payload ka combined dead weight hai. Maano woh honest single-stage dry mass $m_{f} = 90, 000$ kg hai (donon stage shells + payload), total wet $= 500, 000$ kg, aur $v_{e} = 3500$ m/s use karo (donon engines ka simple average $\frac{3000 + 4000}{2}$ , kyunki single rocket mein ek hi engine type hoti hai). $Δ v$ nikalo aur L4.1 ke staged total $7144$ m/s se compare karo.

Recall Solution

$R = 500, 000/90, 000 = 5.5 \overline{5}$ . $Δ v = 3500 \cdot ln (5.5556) = 3500 \times 1.714798 \approx 6002 m/s .$ Compare: honest single stage sirf $\approx 6002$ m/s deta hai jabki staged $7144$ m/s deta hai — staging $1100$ m/s se zyada se jeetti hai. Kyun: single stage ko apna sara dead structure ( $90, 000$ kg) akhir tak dhaona padta hai, $R$ low rakhta hai. Staging drop karta hai bhaari Stage-1 shell ko beech mein, toh final push ek bahut halke rocket par kaam karta hai — aakhri leg ke liye zyada effective ratio. Dry mass dushman hai, aur log use carry karne ki saza deta hai — dekho Multistage Rockets.

Level 5 — Mastery

L5.1 — Budget se backward design karna

Ek Mars-transfer stage ko $Δ v = 5700$ m/s deliver karna hai. Engine $I_{s p} = 340$ s deta hai ( $g_{0} = 9.81$ m/s²). Payload + structure jo survive karna chahiye woh $m_{f} = 25, 000$ kg hai. (a) $v_{e}$ nikalo. (b) Required mass ratio $R$ nikalo. (c) Required wet mass $m_{0}$ aur zaroori propellant mass nikalo.

Recall Solution

(a) $v_{e} = g_{0} I_{s p} = 9.81 \times 340 = 3335.4$ m/s. (b) Ulta chalao: $ln R = Δ v / v_{e} = 5700/3335.4 = 1.70880$ , toh $R = e^{1.70880} \approx 5.522.$ (c) $m_{0} = R \cdot m_{f} = 5.522 \times 25, 000 \approx 138, 050$ kg. Propellant $= m_{0} - m_{f} \approx 138, 050 - 25, 000 = 113, 050$ kg. Sense check: propellant wet mass ka $\sim 82%$ hai — bhaari, lekin parent ke mushkil LEO example ke $94%$ se bahut dur. Kam $Δ v$ demand ⇒ gentler ratio. ✔

L5.2 — Design ki sensitivity

L5.1 ka engine use karo ( $v_{e} = 3335.4$ m/s), budget $5700$ se $6300$ m/s tak creep karta hai (ek $600$ m/s increase). $m_{f} = 25, 000$ kg rakho, extra propellant kitna chahiye? Comment karo ki ek modest $Δ v$ bump itna zyada kyun cost karta hai.

Recall Solution

Naya $ln R^{'} = 6300/3335.4 = 1.88872$ , toh $R^{'} = e^{1.88872} \approx 6.611$ . Naya $m_{0}^{'} = 6.611 \times 25, 000 \approx 165, 280$ kg. Naya propellant $= 165, 280 - 25, 000 = 140, 280$ kg. Extra propellant $= 140, 280 - 113, 050 \approx 27, 230$ kg. Itna costly kyun: kyunki $R = e^{Δ v / v_{e}}$ , $Δ v$ mein exponential hai. Budget mein linear $+ 600$ m/s ratio ko $e^{600/3335.4} \approx 1.197$ se multiply karta hai — wet mass mein $\sim 20%$ jump. Har extra m/s tumhe exponentially tax karta hai. Yeh rocket equation ki tyranny ka mastery-level statement hai.

Wrap-up recall

Recall Cover and answer

Jab $Δ v$ aur $v_{e}$ diya ho toh $R$ kaise nikaalte hain? ::: $R = e^{Δ v / v_{e}}$ (exponential log undo karta hai).
Multi-stage $Δ v$ values add kyun hoti hain? ::: Har stage $d v$ alag integrate karta hai; same-direction velocity changes sum hoti hain.
$+ 600$ m/s budget bump itna fuel kyun cost karta hai? ::: Mass ratio $Δ v$ mein exponential hai, toh fuel compounds karta hai.
$I_{s p}$ ko $v_{e}$ mein kya convert karta hai? ::: $g_{0}$ se multiply karo: $v_{e} = g_{0} I_{s p}$ .

Connections

Parent topic (Hinglish)
Conservation of Momentum — derivation ki foundation.
Specific Impulse (Isp) — $v_{e} = g_{0} I_{s p}$ conversion jo L2.3, L5 mein use hota hai.
Multistage Rockets — L4 staging results.
Natural Logarithm and Integration of 1/x — kyun $ln$ aur $e^{x}$ aate hain.
Thrust and Mass Flow Rate — differential cousin $F = v_{e} \overset{m}{˙}$ .