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

3.4.13 · Physics › Rocket Flight Mechanics

WHAT is a gravity turn?

Pitch angle $θ$ : vehicle ki naak (thrust axis) ka horizon se upar ka angle.
Flight-path angle $γ$ : velocity vector ka horizon se upar ka angle.
$α = 0 \Rightarrow θ = γ$ . Naak aur velocity ek hi direction mein hain.

HOW: derive the pitch-rate equation from first principles

Hum vertical plane mein kaam karte hain, Earth ki curvature ko ignore karte hue (flat-Earth approximation, jo early ascent ke liye valid hai). Maano:

$v$ = speed (velocity ki magnitude),
$γ$ = horizon se upar flight-path angle,
$m$ = mass, $T$ = thrust, $g$ = gravity, $D$ = drag.

Kyunki $α = 0$ , thrust velocity ke saath hai aur drag velocity ke against hai — dono purely tangential hain. Gravity $m g$ seedha neeche point karti hai; hum ise velocity vector ke relative ek tangential part aur ek normal part mein split karte hain.

Gravity vs. velocity ki geometry. Velocity horizon se $γ$ angle banati hai. Gravity vertical (neeche ki taraf) hai. Resolve karo:

Velocity ke saath component (climb karte waqt forward motion ko oppose karta hai): $- m g sin γ$
Velocity ke perpendicular component (path ko neeche kheenchta hai): $- m g cos γ$

Ab turning. Ek particle jo speed $v$ se move kar raha hai aur jiski direction $γ$ change ho rahi hai uski normal acceleration $v \frac{d γ}{d t}$ hoti hai (yeh simply $a_{⊥} = v \overset{γ}{˙}$ hai, curved motion ka analogue $v^{2} / r$ ka jahan $r = v / \overset{γ}{˙}$ ).

Sirf ek hi normal force hai — gravity ka perpendicular component, $- m g cos γ$ (thrust aur drag ka koi normal component nahi kyunki $α = 0$ ). Path ke perpendicular Newton:

$m v \frac{d γ}{d t} = - m g cos γ$

$m$ cancel karo:

Yeh equation hume kya batati hai (Feynman-style reading)

$\overset{γ}{˙} \propto \frac{1}{v}$ : jitna tez jaoge, utna dheere mudo. Shuruaat mein, jab $v$ chhota hota hai, turn sabse tez hota hai — liftoff ke paas ek chhoti si kick poori trajectory set kar deti hai ("pitch-over manoeuvre").
$\overset{γ}{˙} \propto cos γ$ : turning sabse tez hoti hai jab horizontally fly kar rahe ho ( $γ = 0$ , $cos γ = 1$ ) aur zero hoti hai jab vertical ho ( $γ = 9 0^{\circ}$ , $cos 9 0^{\circ} = 0$ ).

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

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Test yourself (answers chhupaao)

$α = 0$ physically kya matlab hai? (naak bilkul velocity ke saath point karti hai ⇒ zero lift)
Koi thrust component rocket ko kyun nahi moda sakta? (thrust tangential hoti hai jab $α = 0$ )
Kaun si force path ko actually bend karti hai? (gravity ka normal component, $m g cos γ$ )
Ek perfectly vertical rocket gravity turn kyun shuru nahi kar sakta? ( $cos 9 0^{\circ} = 0$ )
Pitch-over shuruaat mein aur dheere kyun kiya jaata hai? ( $\overset{γ}{˙} \propto 1/ v$ ; chhota $v$ ⇒ har second bada turn)

Flashcards

Gravity turn condition on angle of attack

α = 0

(thrust along velocity, zero lift)

Gravity turn mein pitch angle aur flight-path angle ka relation

θ = γ

(kyunki

α = 0

)

Gravity-turn pitch-rate equation

\frac{d γ}{d t} = - \frac{g cos γ}{v}

Gravity-turn trajectory ko curve karne wali force

gravity ka normal component,

m g cos γ

Turn rate speed badhne par kyun girta hai?

\overset{γ}{˙} \propto 1/ v

; gravity ke paas path bend karne ke liye har metre par kam samay hota hai

Vertically fly karte waqt turn rate (

γ = 9 0^{\circ}

)

zero, kyunki

cos 9 0^{\circ} = 0

Horizontally fly karte waqt turn rate (

γ = 0^{\circ}

)

maximum, kyunki

cos 0^{\circ} = 1

Zero-lift ascent ka tangential (speed) equation

m \overset{v}{˙} = T - D - m g sin γ

Max-Q ke paas

α = 0

kyun rakhte hain?

aerodynamic side-loads (lift) aur structural bending minimize karne ke liye

Speed

v

par ghoomte body ki normal acceleration

a_{⊥} = v \overset{γ}{˙}

Recall Feynman: 12-saal ke bachche ko explain karo

Socho ek ball seedhi upar phenko. Gravity dheere dheere ise sideways-aur-neeche kheenchti hai toh uska path ek arc mein curve ho jaata hai. Ek rocket jaanboojhkar yahi karta hai: woh apni naak bilkul usi taraf point karta hai jis taraf woh pehle se ja raha hai aur bas gravity ko path upar se kheenchne deta hai, jaise ek dheema, andikhna haath uski flight ko bend kar raha ho. Jab rocket dheema ja raha hota hai, gravity ke paas ise bend karne ke liye bahut samay hota hai, toh woh tezi se muda karta hai. Jab woh bahut fast hota hai, gravity ise mushkil se nudge karti hai. Aur agar woh seedha upar ja raha ho, gravity seedha uske peeche kheenchti hai, sideways nahi — toh woh kabhi nahi muda. Isliye rockets liftoff ke turant baad thoda sa tip over karte hain, jab abhi dheeme hote hain.

Connections

Rocket Flight Mechanics
Tsiolkovsky Rocket Equation — $v (t)$ provide karta hai jo is $\overset{γ}{˙}$ mein feed hota hai
Flight-path angle and pitch angle — woh $θ = γ + α$ relationship
Aerodynamic Drag and Max-Q — kyun dense air mein zero lift matter karta hai
Centripetal / Normal Acceleration — $a_{⊥} = v \overset{γ}{˙}$ ki origin
Ascent Guidance and Pitch Program — initial pitch-over kaise command kiya jaata hai