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

3.3.2 · D3 · Physics › Rocket Propulsion › Δv = v_e · ln(m₀ - m_f) — understanding each term

Is page mein parent equation $Δ v = v_{e} ln (m_{0} / m_{f})$ ko har us situation se guzara gaya hai jo exam mein aa sakti hai. Pehle hum sabhi cases ka ek map banate hain, phir har cell ke liye ek example karte hain. Agar koi scenario exam mein mile, toh woh neeche ki table mein zaroor milega.

Shuru karne se pehle, ek baar har symbol ka matlab yaad kar lo, kyunki yeh har line mein use hote hain:

$Δ v$ = change in velocity jo rocket khareed sakta hai (m/s), yeh ek budget hai, final speed nahi.
$v_{e}$ = effective exhaust velocity (m/s), gas kitni tez rocket ke relative se nikalta hai.
$m_{0}$ = wet mass (kg) = shuruaat mein sab kuch, poora propellant include karke.
$m_{f}$ = dry mass (kg) = burn ke baad jo kuch bachta hai sab kuch.
$R = m_{0} / m_{f}$ = mass ratio, ek pure number jiska koi unit nahi.
$ln$ = natural logarithm, "shrinking machine" — dekho Natural Logarithm and Integration of 1/x.
$e \approx 2.718$ = us logarithm ka base; $e^{x}$ $ln x$ ko undo karta hai.

The scenario matrix

Equation mein sirf teen knobs hain ( $v_{e}$ , $m_{0}$ , $m_{f}$ ), lekin iske baare mein jo sawaal pooche jaate hain woh in classes mein aate hain:

Cell	Case class	Kya unknown hai	Degenerate / edge check
A	Forward: saari masses + $v_{e}$ diye hain, $Δ v$ nikalo	$Δ v$	—
B	Degenerate: koi fuel nahi jala, $m_{0} = m_{f}$	$Δ v$	$ln 1 = 0$
C	Fuel ke liye inverse: target $Δ v$ diya hai, $R$ nikalo	$R$ , propellant mass	$e^{x}$ chahiye
D	Specific impulse ke zariye: $v_{e}$ nahi diya, $I_{s p}$ diya hai	$Δ v$	$v_{e} = g_{0} I_{s p}$
E	Limiting behaviour: $R \to \infty$ (sab fuel, koi dry mass nahi)	$Δ v$ ki ceiling	$ln R \to \infty$ slowly
F	Two-stage: do burns chain karo	total $Δ v$	$Δ v$ add hote hain, ratios multiply hote hain
G	Word problem: losses ke saath real vehicle	actual speed gain	gravity loss subtract karo
H	Exam twist: $Δ v$ , $m_{0}$ , $v_{e}$ diye hain, $m_{f}$ nikalo	$m_{f}$	log ko do baar invert karo

Neeche ka har example apne cell letter se tagged hai. Milake yeh har row ko cover karte hain.

Cell A — seedha-saadha forecast

Worked example A · Masses aur

$v_{e}$ diye hain, $Δ v$ nikalo Ek rocket ka $m_{0} = 500, 000$ kg, $m_{f} = 100, 000$ kg, $v_{e} = 3000$ m/s hai. $Δ v$ nikalo.

Forecast: ratio 5 hai, aur $ln 5 \approx 1.6$ , toh andaaza lagao thoda $3000 \times 1.6 \approx 4800$ m/s se kam. Yeh number yaad rakho.

Mass ratio compute karo $R = m_{0} / m_{f} = 500000/100000 = 5$ . Yeh step kyun? $Δ v$ masses ke ratio par depend karta hai, unke difference par nahi, isliye ratio pehli cheez hai jo fix karni hai.
Logarithm lo: $ln 5 = 1.609$ . Yeh step kyun? Equation $R$ ko $ln$ se guzarti hai kyunki velocity gain $d v = - v_{e} d m / m$ ki tarah accumulate hoti hai; $1/ m$ integrate hokar log ban jaata hai.
$v_{e}$ se multiply karo: $Δ v = 3000 \times 1.609 = 4828$ m/s. Yeh step kyun? Mass ka har fractional bit jo lose hota hai, usse $v_{e}$ times us fraction ki speed milti hai.

Verify: units $(m/s) \times (dimensionless) = m/s$ hain ✔. Hamara forecast (~4800) match karta hai. Propellant burned $= m_{0} - m_{f} = 400, 000$ kg, total se kam — sahi lag raha hai ✔.

Cell C — required fuel ke liye invert karna

Worked example C · Target

$Δ v$ diya hai, mass ratio nikalo Tumhe $Δ v = 9400$ m/s (ek typical low-Earth-orbit budget) $v_{e} = 3400$ m/s ke saath achieve karni hai. Tumhe kaun sa mass ratio $R$ chahiye, aur rocket ka kitna fraction propellant hai?

Forecast: $9400/3400 \approx 2.8$ , aur $e^{2.8}$ ek bada number hai — guess karo $R$ around 15, yaani rocket ka zyaadatar hissa fuel hai.

Log ko isolate karo: dono sides ko $v_{e}$ se divide karo. $ln R = \frac{Δ v}{v _{e}} = \frac{9400}{3400} = 2.765$ Yeh step kyun? Hum $R$ chahte hain, aur woh $ln$ ke andar band hai; pehle $v_{e}$ hatao toh log expose hota hai.
Log ko $e^{x}$ se undo karo: $R = e^{2.765} = 15.88$ Yeh step kyun? Kyunki $e^{l n R} = R$ — exponentiation natural log ka exact inverse hai.
Propellant fraction: agar $R = m_{0} / m_{f} = 15.88$ , toh $m_{f} / m_{0} = 1/15.88 = 0.063$ , toh propellant $1 - 0.063 = 0.937$ hai, yaani lagbhag $93.7%$ . Yeh step kyun? Dry fraction ratio ka reciprocal hai; baaki sab fuel hai.

Verify: plug back karo — $3400 \times ln (15.88) = 3400 \times 2.765 = 9401$ m/s ≈ target ✔. Lagbhag 94% propellant rocket equation ki famous "tyranny" hai.

Cell D — specific impulse se jaana

Worked example D ·

$v_{e}$ nahi diya, sirf $I_{s p}$ diya hai Ek engine $I_{s p} = 450$ s advertise karta hai. Dry mass (structure + payload) $m_{f} = 20, 000$ kg hai aur yeh $300, 000$ kg propellant le jaata hai. $Δ v$ nikalo. $g_{0} = 9.81$ m/s² use karo.

Forecast: $450$ s ek high-performance hydrogen engine hai, toh $v_{e}$ around 4400 m/s hoga; itne zyaada fuel ke saath expect karo $Δ v$ 10 km/s se zyaada.

$I_{s p}$ ko exhaust velocity mein convert karo: $v_{e} = g_{0} I_{s p} = 9.81 \times 450 = 4414.5$ m/s. Yeh step kyun? Equation ko ek speed chahiye, lekin $I_{s p}$ seconds mein measure hota hai; dekho Specific Impulse (Isp). Bridge hai $v_{e} = g_{0} I_{s p}$ .
Wet mass banao: $m_{0} = 300000 + 20000 = 320000$ kg. Yeh step kyun? $m_{0}$ mein saara propellant plus sab kuch jo bachta hai, yeh sab include hona chahiye.
Mass ratio $R = 320000/20000 = 16$ , aur $ln 16 = 2.773$ . Yeh step kyun? Wahi core computation jaise hamesha — ratio ko log se guzaro.
$Δ v = 4414.5 \times 2.773 = 12, 241$ m/s.

Verify: units — $v_{e}$ ke liye $(m/s^{2}) (s) = m/s$ ✔, phir $\times$ dimensionless ✔. Value LEO budget se aaram se zyaada hai, jaise forecast tha ✔.

Cell E — limiting ceiling

Worked example E · Ek stage zyaada se zyaada kitna

$Δ v$ de sakta hai? Maano $v_{e} = 3400$ m/s fixed hai. Agar tum dry mass ko jaadu se zero ki taraf shrink kar sako (sab fuel, koi structure nahi), toh $R \to \infty$ hone par $Δ v$ ka kya hoga?

Forecast: $ln$ bina bound ke badhta hai, toh koi hard ceiling nahi — lekin yeh bahut dhire badhta hai. Guess: $R$ double karne se sirf ek fixed lump of $Δ v$ add hota hai.

Dekho $ln R$ kaise badhta hai jab $R$ increase hota hai. $R$ ko ek fixed factor se har multiplication mein $ln$ ki ek fixed amount add hoti hai. Yeh step kyun? Kyunki $ln (k R) = ln k + ln R$ — ek constant $ln k$ add hota hai chahe $R$ pehle se kitna bhi bada ho. Figure dekho.
Concretely, jitni baar tum $R$ double karo, $Δ v$ $v_{e} ln 2 = 3400 \times 0.693 = 2356$ m/s badhta hai — har doubling mein same jump. Yeh step kyun? $ln (2 R) - ln R = ln 2$ , ek constant hai, toh har doubling per $Δ v$ step constant hai.
Conclusion: $Δ v \to \infty$ jab $R \to \infty$ , lekin sirf logarithmically. Aur 2356 m/s paane ke liye poora mass ratio fir se double karna padega — cost explode ho jaati hai.

Verify: doubling law numerically check karo. $R = 8$ se $R = 16$ tak: $3400 (ln 16 - ln 8) = 3400 \times 0.693 = 2356$ m/s ✔. $R = 16$ se $R = 32$ tak: same $2356$ m/s ✔. Equal steps log growth confirm karte hain.

Cell F — do stages chain karna

Worked example F · Two-stage rocket,

$Δ v$ s add karo Stage 1: $m_{0} = 100, 000$ kg, $m_{f} = 40, 000$ kg tak burn hota hai, $v_{e} = 3000$ m/s. Stage 2 (khaali stage-1 hardware drop karne ke baad): $m_{0} = 30, 000$ kg, $m_{f} = 8, 000$ kg, $v_{e} = 4000$ m/s. Total $Δ v$ nikalo.

Forecast: har stage ka ratio modest hai (2.5 aur 3.75), toh har ek couple of km/s dega; total shayad 6–8 km/s.

Stage 1: $R_{1} = 100000/40000 = 2.5$ , $Δ v_{1} = 3000 ln 2.5 = 3000 \times 0.9163 = 2749$ m/s. Yeh step kyun? Har stage independently wahi rocket equation follow karta hai.
Stage 2: $R_{2} = 30000/8000 = 3.75$ , $Δ v_{2} = 4000 ln 3.75 = 4000 \times 1.3218 = 5287$ m/s. Yeh step kyun? Separation ke baad khaali tank gone ho jaata hai, toh stage 2 ek nayi, halki wet mass se start karta hai — exactly issi liye staging log ko beat karta hai; dekho Multistage Rockets.
Add karo: $Δ v_{total} = 2749 + 5287 = 8036$ m/s. Yeh step kyun? $Δ v$ velocity mein change hai; sequential changes seedha add ho jaate hain.

Verify: ratios ek effective $2.5 \times 3.75 = 9.375$ mein multiply ho jaate hain, lekin kyunki do stages ke alag-alag $v_{e}$ hain tum ek log use nahi kar sakte — tumhe do alag add karne padte hain, jo humne kiya ✔. Total km/s range mein forecast ke anusaar ✔.

Cell G — losses ke saath real-world word problem

Worked example G · Real launch, gravity loss subtract karo

Ek booster ka $m_{0} = 550, 000$ kg, $m_{f} = 150, 000$ kg, $v_{e} = 3100$ m/s hai. 150-second vertical climb ke dauran roughly $150 \times 9.81 = 1472$ m/s gravity drag se loss hota hai. Rocket ko actual mein kitna speed gain milta hai?

Forecast: ideal $Δ v$ around 4000 m/s, minus ~1500 loss, toh around 2500 m/s real speed.

Ideal $Δ v$ : $R = 550000/150000 = 3.667$ , toh $Δ v_{ideal} = 3100 ln 3.667 = 3100 \times 1.2993 = 4028$ m/s. Yeh step kyun? Tsiolkovsky formula budget deta hai yeh maankar ki koi external forces nahi hain.
Gravity loss subtract karo: real speed gain $= 4028 - 1472 = 2556$ m/s. Yeh step kyun? Gravity puri climb mein peeche kheenchti hai; woh momentum debt seedha ideal budget mein se aati hai. $Δ v$ ek budget hai, final speed nahi — yeh classic mistake hai jo parent note warn karta hai.

Verify: real gain (2556) ideal (4028) se kam hai ✔ — losses budget ko hamesha shrink karte hain, kabhi grow nahi karte. Order-of-magnitude forecast se match karta hai ✔.

Cell H — exam twist: dry mass nikalo

Worked example H ·

$Δ v$ , $m_{0}$ , $v_{e}$ diye hain, $m_{f}$ nikalo Ek design ko $v_{e} = 3200$ m/s aur fixed wet mass $m_{0} = 80, 000$ kg ke saath $Δ v = 6000$ m/s hit karni hai. Kitna dry mass $m_{f}$ allowed hai, aur isse kitna propellant imply hota hai?

Forecast: $6000/3200 \approx 1.9$ , toh $R = e^{1.9} \approx 6.6$ ; dry mass $m_{0}$ ko isse divide karne par roughly 12,000 kg hoga.

Required log nikalo: $ln R = Δ v / v_{e} = 6000/3200 = 1.875$ . Yeh step kyun? Cell C jaisa hi isolation — log expose karne ke liye $v_{e}$ hatao.
Ratio ke liye exponentiate karo: $R = e^{1.875} = 6.521$ . Yeh step kyun? $e^{x}$ $ln$ ko undo karta hai, pure mass ratio deta hai.
Dry mass solve karo: $m_{f} = m_{0} / R = 80000/6.521 = 12, 268$ kg. Yeh step kyun? Kyunki $R = m_{0} / m_{f}$ hai, rearrange karne par $m_{f} = m_{0} / R$ milta hai.
Propellant needed: $m_{0} - m_{f} = 80000 - 12268 = 67, 732$ kg. Yeh step kyun? Propellant burned hamesha wet minus dry hota hai.

Verify: plug back karo — $3200 ln (80000/12268) = 3200 ln (6.521) = 3200 \times 1.875 = 6000$ m/s ✔. Dry mass forecast se match karta hai (~12,000 kg) ✔.

Wrap-up

Recall Har exam question kaun sa cell hai?

"Masses se Δv nikalo" ::: Cell A.
"Engine off, mass unchanged" ::: Cell B, answer 0.
"Is Δv ke liye kitna fuel chahiye?" ::: Cell C ( $e^{x}$ se invert karo).
"Sirf $I_{s p}$ diya hai" ::: Cell D ( $v_{e} = g_{0} I_{s p}$ use karo).
"Zyaada fuel itna kam kyun help karta hai?" ::: Cell E (log slowly badhta hai).
"Do stages, total Δv?" ::: Cell F (Δv's add karo).
"Real launch speed?" ::: Cell G (gravity loss subtract karo).
"Allowed dry mass nikalo" ::: Cell H ( $m_{f} = m_{0} / e^{Δ v / v_{e}}$ solve karo).

Connections

Parent topic — equation khud.
Conservation of Momentum — woh derivation jis par yeh examples based hain.
Newton's Third Law — mass fenkne se thrust kyun milta hai.
Specific Impulse (Isp) — Cell D mein use hua.
Multistage Rockets — Cell F ke peeche physics.
Thrust and Mass Flow Rate — Cell G ki climb ke peeche ka differential cousin.
Natural Logarithm and Integration of 1/x — $ln$ aur $e$ kyun pair up karte hain.

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

The scenario matrix

Cell A — seedha-saadha forecast

Cell B — degenerate "no burn"

Cell C — required fuel ke liye invert karna

Cell D — specific impulse se jaana

Cell E — limiting ceiling

Cell F — do stages chain karna

Cell G — losses ke saath real-world word problem

Cell H — exam twist: dry mass nikalo

Wrap-up

Connections