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

False. Δv ek change in velocity hai — ek budget jo tumhari starting velocity mein add hoti hai (aur jisme se real gravity/drag losses subtract hoti hain). Ek rocket jo pehle se 2000 m/s pe chal raha hai aur Δv = 3000 m/s kamaata hai, woh 5000 m/s ke paas pahunchta hai, 3000 nahi.

False. Δv mass ratio ke log ke saath scale karta hai, propellant mass ke saath nahi. Δv double karne ke liye mass ratio ko square karna padta hai ($\ln R^2=2\ln R$), jiske liye double se kahin zyada fuel chahiye.

True. Koi propellant nahi jala, toh mass ratio 1 hai aur $\ln 1=0$, jisse Δv = 0 milta hai. Peechhe throw karne ke liye kuch nahi hai, isliye koi kick nahi — dekho Newton's Third Law.

True. Kyunki $\Delta v=v_e\ln R$ hai, Δv seedha $v_e$ ke proportional hai. Exhaust ko tez throw karne se har kilogram throw pe bada forward kick milta hai — isliye engineers high $v_e$ ke peeche bhaagte hain (ion engines, hot gases).

False. Derivation mein assume kiya jaata hai ki vacuum hai aur koi external force nahi; thrust purely mass eject karne se aata hai aur Conservation of Momentum se. Air actually hurt karta hai (drag), push karne mein help nahi karta.

False. Ye do masses ka ratio hai, toh units cancel ho jaate hain — $R$ dimensionless hai. Ye zaroori hai, kyunki tum sirf ek pure number ka $\ln$ le sakte ho.

True. Δv sirf ratio $m_0/m_f$ par depend karta hai; dono ko same factor se scale karne se ratio (aur uska log) fixed rehta hai. Ek bada rocket aur ek chhota rocket, jinka proportion same ho, same Δv kamaate hain.

False. Ye ideal case hai — inertial frame, no gravity, no drag, constant $v_e$. Reality mein is ideal Δv se gravity-loss aur drag-loss terms subtract karne padte hain.

Galat. $m_f$ woh mass hai jo burn ke baad bachi rehti hai — dry structure + engines + payload. Propellant jo jala woh hai $m_0-m_f$, woh mass jo exhaust ke roop mein nikal gayi.

Galat. Log ke andar ek ratio hota hai: $\ln (m_0/m_f)$. Subtraction se galat units milenge (log ke andar kg meaningless hai) aur galat physics — $dv=-v_{e}dm/m$ ko integrate karne se ratio milta hai, difference nahi.

Galat. Bade tanks dry mass bhi badhate hain, $m_f$ ko raise karte hain aur ratio ka faayda kam karte hain. Δv sirf logarithmically badhta hai jabki structural mass tezi se badhta hai — isliye hum Multistage Rockets use karte hain.

Galat. $dmdv$ do infinitesimals ka product hai (second-order), first-order terms se infinitely chhota, isliye ise sahi tarike se drop kiya jaata hai. Ise rakhne se final equation mein koi fark nahi padta.

Galat. $v_e$ exhaust speed hai rocket ke relative. Derivation mein ejected chunk ground frame mein $v-v_e$ pe move karta hai, precisely isliye kyunki $v_e$ relative speed hai.

Galat. Ye $v_e=g_0{I}_{sp}$ se linked hain, jahan $g_0\approx 9.81{\text{ m/s}}^2$. $I_{sp}$ ke seconds $g_0$ se multiply hone par m/s ban jaate hain — dekho Specific Impulse (Isp).

Galat. Log mathematics se forced hai: velocity gain $dv=-v_{e}dm/m$ follow karta hai, aur $1/m$ ko integrate karne se zaroor $\ln m$ milta hai. Dekho Natural Logarithm and Integration of 1/x.

Kyunki exhaust ka same chunk ek halke rocket ko heavy rocket se zyada accelerate karta hai. Momentum conservation se $dv=-v_{e}dm/m$ milta hai, toh jo matter karta hai woh hai current mass ke relative throw ki gayi mass.

Fractional mass change $dm/m$ chhotha hota hai jab $m$ bada ho, toh har kilogram throw se kam $dv$ milta hai. Bade rockets apna zyada fuel sirf us fuel ko move karne mein waste karte hain jo abhi bhi board pe hai.

Staging mid-flight empty dry mass drop karta hai, remaining journey ke liye $m_f$ kam karta hai aur effective mass ratio boost karta hai. Ye logarithm ki us punishing demand se bachata hai jo exponentially zyada fuel maangti hai — dekho Multistage Rockets.

Dono same momentum balance se aate hain: thrust instantaneous force describe karta hai mass flow se, jabki rocket equation wahi relation hai jo poore burn par integrate kiya gaya ho — dekho Thrust and Mass Flow Rate.

Kyunki propellant khud woh mass hai jise carry aur accelerate karna padta hai. Fuel add karne se $m_0$ badhta hai, lekin log itna dheere badhta hai ki returns collapse ho jaate hain — Δv double karne ke liye ratio ko square karna padta hai.

Ek mission kai maneuvers ka sum hota hai (launch, transfers, landing), har ek ek speed change ka chunk maangta hai. Δv direction se independent hokar maneuvers mein cleanly add hota hai, toh planners ek total Δv budget track karte hain.

Zero. Koi propellant eject nahi hua, toh $m_f=m_0$, ratio 1 hai, aur $\ln 1=0$. Unused fuel sirf dead weight hai, speed source nahi.

Δv $\to 0$, kyunki $\ln R\to \ln 1=0$. Chhote fuel fraction ke liye equation almost linear hai ($\ln R\approx R-1$), toh thoda fuel thoda Δv deta hai.

Ratio $R=m_0/m_f\to \infty$, toh $\ln R\to \infty$ aur Δv bina bound ke badhta jaata. Ye unphysical hai — tum kabhi $m_f=0$ nahi reach kar sakte kyunki structure, engines, aur payload hamesha rehte hain.

Nahi. Kyunki $m_0\ge m_f$ (tum sirf mass lose karte ho), $R\ge 1$ aur $\ln R\ge 0$, toh Δv hamesha non-negative hota hai. Deceleration thrust ko backward point karne se aati hai, negative Δv se nahi.

Zero, kyunki $\Delta v=v_e\ln R=0$ jab $v_e=0$ ho. Zero relative speed par mass eject karne se koi momentum nahi jaata, toh koi forward kick nahi — dekho Conservation of Momentum.

Forward Δv chhhota ho jaayega, kyunki sirf exhaust velocity ka backward component forward thrust produce karta hai. Clean equation assume karta hai ki exhaust seedha motion ke opposite eject hota hai.