Question bank — Trajectory optimization — minimum gravity loss, minimum drag loss

Galat. Gravity loss $\int g\sin \gamma dt$ $\gamma =0^{\circ }$ par zero ho jaata hai, lekin dense low air mein flat aur fast urne se $\frac{1}{2}\rho v^2$ bahut bada ho jaata hai, isliye drag loss explode kar jaata hai — tumhe sum optimize karna hoga, sirf ek term nahi.

Galat. Gravity loss hai $\int g\sin \gamma dt$ — yeh flight-path angle aur burn time par depend karta hai, na ki reach ki gayi altitude par. Isse potential energy $mgh$ se confuse karna hi trap hai; ek slow rocket jo vertically loiter karta hai woh uss fast rocket se kahin zyada lose karta hai jo same height tak pahunchta hai.

Sahi. Ek gravity turn mein thrust velocity vector ke saath aligned rakha jaata hai ($\alpha =0$), isliye koi bhi thrust sideways waste nahi hota; trajectory sirf gravity ke free perpendicular component se bend hoti hai.

Galat. Drag loss typically 30–150 m/s hota hai, jabki gravity loss 1–2 km/s tak hota hai. Gravity loss optimization ko dominate karta hai — isliye hi pitch program mostly horizontal turn karne ki chinta karta hai.

Galat. Off-axis thrust angle $\alpha$ par sirf $\cos \alpha$ forward deliver karta hai; shortfall $\int \frac{T}{m}(1-\cos \alpha )dt$ steering loss hai (dekho Steering losses and thrust vectoring). Gravity turn isse $\alpha =0$ rakh ke avoid karta hai.

Sahi. Zyada thrust-to-weight ka matlab hai shorter burn, aur gravity loss $\int g\sin \gamma dt$ tab shrink hota hai jab burn time $t_b$ shrink hota hai — tum gravity se ladte hue kam time spend karte ho.

Sahi. $\sin 90^{\circ }=1$ hone se poora $g$ velocity ko oppose karta hai, isliye vertical burn ka har second maximum possible gravity loss $g\Delta t$ costat hai.

Galat. Gravity abhi bhi act karti hai — woh sirf path ke saath koi work nahi karti. Uska poora perpendicular component $g\cos \gamma$ ab trajectory ko bend karta hai (velocity ko curve karta hai), jo exactly woh turning force hai jise gravity turn exploit karta hai.

Galat. $L(\gamma )=\Delta v_{\text{grav}}+\Delta v_{\text{drag}}$ convex hai: $\gamma =90^{\circ }$ par high (gravity) aur $\gamma =0^{\circ }$ par high (drag), isliye minimum beech mein hota hai — woh interior compromise hi gravity turn hai.

Sine, cosine nahi, gravity ko velocity par project karta hai. Gravity motion ko $g\sin \gamma$ se oppose karti hai; $g\cos \gamma$ wala part perpendicular part hai jo path turn karta hai, woh part nahi jo speed ka cost karta hai.

Galat trig function hai. Sirf perpendicular gravity component hi velocity ko turn karta hai, isliye $\frac{d\gamma }{dt}=-\frac{g\cos \gamma }{v}$. $\sin$ use karna (galat tarike se) turn ko tab fastest dikhata jab already vertical ho.

Ulta hai. Turn rate $v$ ke inversely proportional hai: high speed par velocity vector ko bend karna mushkil hota hai, isliye pitch kick tab lagaao jab $v$ abhi bhi small ho.

Sign error hai. Losses ideal side par add hote hain: $\Delta v_{\text{ideal}}=\Delta v_{\text{orbit}}+\Delta v_{\text{grav}}+\Delta v_{\text{drag}}+\Delta v_{\text{steer}}$. Tsiolkovsky se ideal $\Delta v$ ko orbit plus saare losses cover karne chahiye.

Woh integrand ek force hai, acceleration nahi. Drag loss hai $\int \frac{D}{m}dt$ — tumhe time ke upar integrate karne se pehle mass se divide karke deceleration $D/m$ nikalna hoga (dekho Aerodynamic Drag and Max-Q).

Throttling up se $v$ low neeche badhti hai jahan $\rho$ bada hota hai, jisse $\frac{1}{2}\rho v^2$ aur bura ho jaata hai. Real vehicles max-Q ke through down throttle karte hain taaki $\rho v^2$ (dynamic pressure) manageable rahe.

Yeh $1-\cos \alpha$ ke scale par hota hai, jo small angles ke liye ${\alpha }^2$ ki tarah badhta hai — ek $2^{\circ }$ offset ek $1^{\circ }$ offset ka roughly chaar guna cost karta hai, double nahi.

Dense low atmosphere se jaldi punch karne ke liye speed build hone se pehle, $\frac{1}{2}\rho v^2$ (aur heating aur structural load) ko low rakhte hue. Sirf thick air clear karne ke baad hi woh pitch over afford kar sakta hai.

Kyunki thrust velocity ke saath aligned rehta hai aur path bend karne ke liye koi aerodynamic lift (ya side thrust) use nahi hota — sirf turning force gravity ka perpendicular component hai, jo free hai.

Turn rate $g\cos \gamma /v$ $v$ ke inversely proportional hai. Jab $v$ small hota hai tab rate bada hota hai, isliye ek chota early nudge ek aisa turn seed karta hai jo naturally badhta hai; bahut der karo toh $v$ high hogi, path stiff hogi, aur fuel khatam hone se pehle horizontal nahi pahunch sakte.

Dono ek same knob se control hote hain, pitch angle $\gamma$ se, aur woh opposite directions mein respond karte hain. $\gamma$ ko ek ki help ke liye push karna hamesha doosre ko hurt karta hai, isliye sirf sum ka ek meaningful minimum hota hai.

Gravity loss $\int g\sin \gamma dt$ isliye girta hai kyunki burn time mein shorter hai; lekin ek shorter, harder burn $v$ ko low neeche faster push karta hai, jo drag loss raise kar sakta hai. Benefit gravity ke time-integral par hai, drag par nahi.

Orbit ke liye badi horizontal velocity chahiye planet ke around girte rehne ke liye; vertical speed sirf upar jaati hai aur wapas aati hai. Isliye thrust eventually sideways redirect karna hota hai — yahi kaam gravity turn gradually karta hai.

Along-track (tangential) equation sirf velocity ke parallel force components dekhti hai. Gravity ka parallel part $g\sin \gamma$ hai; uska perpendicular part $g\cos \gamma$ normal equation mein rehta hai, jahan woh path turn karta hai instead of use slow karne ke.

$\Delta v_{\text{grav}}=\int gdt$ (kyunki $\sin 90^{\circ }=1$) — maximum possible, kyunki poora weight har instant motion ko oppose karta hai. Koi bhi orbital (horizontal) speed build nahi hoti.

Gravity loss vanish ho jaata hai ($\int g\sin \gamma dt=0$) aur turn rate $g\cos \gamma /v\to 0$, isliye velocity vector khud kabhi rotate nahi karta — koi gravity turn nahi hota, aur koi bhi turning steering thrust se pay karni padti hai.

$\frac{d\gamma }{dt}=-\frac{g\cos 90^{\circ }}{v}=0$ — ek perfectly vertical rocket kabhi bhi khud se turn shuru nahi karega. Symmetry break karne aur gravity ko take over karne ke liye ek deliberate small pitch kick off vertical zaroori hai.

Nahi — liftoff par rocket near vertical rakha jaata hai ($\cos \gamma \approx 0$) aur active control mein hota hai, isliye product finite rehta hai. Formula warn karta hai ki turning easiest tab hoti hai jab slow ho, na ki yeh uncontrollably hoti hai.

Drag loss $\int \frac{\frac{1}{2}\rho v^2{C}_{D}A}{m}dt\to 0$ kyunki $\rho \to 0$, chahe ab $v$ bahut bada ho. Isliye poora drag-vs-gravity trade sirf neeche matter karta hai; ek baar air se bahar nikal gaye, sirf gravity loss se ladna bacha.

Drag product $\rho v^2$ par depend karta hai; max-Q ke baad, $\rho$ $v^2$ ke badhne se zyada tezi se girta hai, isliye dynamic pressure — aur isliye drag deceleration $D/m$ — speed badhne ke bawajood girta hai.

$\Delta v_{\text{grav}}={\int }_0^{t_b}g\sin \gamma dt\to 0$ — burn mein koi time spend nahi hua, gravity ko motion ke against act karne ka koi chance nahi mila. Yeh woh idealized limit hai jo dikhata hai kyun high thrust-to-weight prized hai.