Question bank — Gravity turn trajectory — pitch rate from aerodynamic angle of attack = 0

False. $\alpha =0$ ke saath thrust exactly velocity ke along point karta hai, isliye path ke perpendicular component zero hai — sirf gravity ka normal component $mg\cos \gamma$ trajectory ko bend karta hai.

True. Lift ek force hai jo velocity ke perpendicular $\alpha \ne 0$ se generate hoti hai; kyunki $\alpha =0$ hai, airflow vehicle se seedha takrata hai, toh lift vanish ho jaati hai aur sirf drag (tangential) bachta hai.

False. $\gamma =90^{\circ }$ par, $\cos 90^{\circ }=0$, toh $\dot{\gamma }=0$. Gravity velocity ke anti-parallel hai aur turn shuru karne ke liye koi sideways component nahi hai.

False. $\dot{\gamma }\propto 1/v$, toh $v$ double karne se $|\dot{\gamma }|$ half ho jaati hai — gravity ko path ka har metre bend karne ke liye kam time milta hai.

True sirf is trajectory ke liye, kyunki $\theta =\gamma +\alpha$ aur $\alpha =0$. Agar koi gust ya manoeuvre $\alpha \ne 0$ kar de, toh dono alag ho jayenge.

True. $\dot{\gamma }\propto \cos \gamma$, aur $\cos \gamma$ sabse bada ($=1$) $\gamma =0^{\circ }$ par hota hai, toh horizontal-flight instant mein given speed ke liye maximum turn rate hoti hai.

False. Drag velocity line ke along act karta hai (motion oppose karta hua), toh ye purely tangential hai — rocket ko slow karta hai lekin path curve karne ke liye zero normal component deta hai.

False. High $v$ se $\dot{\gamma }$ tiny ho jaata hai, toh ek fast upper stage bahut kam bend karta hai — zyaatar pitching pehle karni padti hai, jab slow aur low ho.

Error: ye thrust-vector steering describe karta hai, gravity turn nahi. Gravity turn thrust ko velocity ke along rakhta hai ($\alpha =0$); sideways force gravity ka normal component hai, jisme koi steering effort nahi chahiye.

Error: minus sign ka sirf matlab hai ki $\gamma$ decrease ho raha hai (nose vertical se horizontal ki taraf tip karta hai). $\gamma$ powered ascent ke dauran $0^{\circ }$ (horizontal) ki taraf jaata hai, negative territory mein nahi.

Error: $\alpha =0$ sirf nose aur velocity ke beech ka angle zero karta hai. Thrust, drag, aur gravity sab present hain aur generally cancel nahi karte; vehicle accelerate karta hai.

Koi galti nahi: curved motion ke liye radius satisfy karta hai $r=v/\dot{\gamma }$, toh $v^2/r=v^2\cdot (\dot{\gamma }/v)=v\dot{\gamma }$. Dono expressions identical hain.

Error: tower clear karne aur dense low atmosphere jaldi cross karne ke liye vertically launch karte hain, drag aur structural loads minimize karne ke liye. Phir ek tiny pitch-over turn seed karta hai; angled launch drag aur bending stress badhata hai. (Dekho Aerodynamic Drag and Max-Q.)

Error: $\sin \gamma$ term tangential hai — ye climb slow karta hai, speed change karta hai, direction nahi. Turning term normal component $-mg\cos \gamma$ hai.

Error: $\gamma$ horizon ke upar velocity ka angle hai; $\alpha$ nose aur velocity ke beech ka angle hai. Ye $\theta =\gamma +\alpha$ ke through related hain, equal nahi.

Bending loads lift se aate hain, ek side force jo $\alpha$ aur dynamic pressure ke saath scale karta hai. $\alpha =0$ ke saath lift zero hai, toh airframe ko dense air mein bhi koi aerodynamic side-load feel nahi hoti.

Kyunki $\dot{\gamma }\propto 1/v$, low speed se per second sabse badi turn rate milti hai, toh early mein ek small nudge efficiently poori trajectory set up kar deta hai. (Dekho Ascent Guidance and Pitch Program.)

Required force ($mv\dot{\gamma }$) aur available turning force ($mg\cos \gamma$) dono mein $m$ ka factor hai. Free-falling body ki tarah, acceleration mass-independent hai, toh $\dot{\gamma }$ sirf $g$, $\gamma$, aur $v$ par depend karta hai.

Gravity ka already ek tilted velocity ke perpendicular component hota hai, toh wo path bend karta hai bina kisi propellant cost ke. Thrust se vehicle turn karne ke liye $\alpha \ne 0$ chahiye, jo energy lift aur drag par waste karta hai.

Kyunki $\alpha =0$ aerodynamic lift completely eliminate kar deta hai; sirf curving agent gravity bachta hai, isliye "gravity turn" aur "zero-lift" dono same condition name karte hain.

Jaise $\gamma$ $90^{\circ }$ se drop karta hai, $\cos \gamma$ badhta hai, $\dot{\gamma }$ boost hota hai; lekin climbing speed $v$ bhi badhti hai, aur kuch point ke baad $1/v$ effect dominate karta hai aur turn rate girne lagti hai. Ye $\cos \gamma$ badhne aur $v$ badhne ke beech ek tug-of-war hai.

Ascent path ka instantaneous radius directly batana mushkil hai, lekin direction rate $\dot{\gamma }$ exactly wahi hai jo hum chahte hain. $a_{\perp }=v\dot{\gamma }$ likhne se equation directly us angle ke terms mein aa jaati hai jisko hum solve kar rahe hain. (Dekho Centripetal / Normal Acceleration.)

$\cos 90^{\circ }=0$, toh $\dot{\gamma }=0$. Rocket hamesha vertical rehta hai — ye degenerate case hi reason hai kyun symmetry todne ke liye deliberate initial pitch-over zaroori hai.

$\cos \gamma \to 1$, us speed ke liye maximum turn rate $|\dot{\gamma }|=g/v$ deta hai — gravity ab velocity ke fully perpendicular hai aur use sabse effectively bend karti hai.

$\dot{\gamma }=-g\cos \gamma /v$ blow up ho jaata hai jab $v\to 0$, infinitely fast turn predict karta hai. Physically isliye rocket ko pitch-over se pehle kuch vertical speed gain karni padti hai, aur isliye flat-plane model tabhi valid hai jab $v$ nonzero ho.

$\cos (\text{small negative})\approx 1>0$, toh $\dot{\gamma }$ negative rehta hai — path ground ki taraf neeche bend karta rahega, exactly woh diving behaviour jo aap expect karte; model consistent rehta hai.

Nahi. $\alpha \ne 0$ ke saath thrust aur lift normal components gain karte hain, turning equation mein terms add ho jaate hain, aur $\theta =\gamma +\alpha$ toh pitch rate aur flight-path rate alag ho jaate hain. Boxed formula ek zero-$\alpha$ special case hai.

$\dot{\gamma }\to 0$: gravity ke bina $\alpha =0$ par koi normal force nahi hai, toh path apne aap bend nahi ho sakta. Tab turning ke liye genuinely thrust vectoring ($\alpha \ne 0$) chahiye, unlike planet ki gravity se ascent.