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

3.4.13 · D3 · Physics › Rocket Flight Mechanics › Gravity turn trajectory — pitch rate from aerodynamic angle

Shuru karne se pehle, symbols ka ek do-second refresher, taaki kuch bhi bina explain ke use na ho:

Scenario matrix

Is topic se aane wala har question inhi cells mein se kisi ek mein fit hota hai. Example column batata hai kaun sa worked example ise cover karta hai.

Cell	Case class	Kya special hai	Example
A	Near-vertical ( $γ \approx 9 0^{\circ}$ )	$cos γ \approx 0$ → tiny turn rate	Ex 1
B	Mid-lean ( $γ \approx 4 5^{\circ}$ )	$cos γ$ moderate → brisk turn	Ex 2
C	Near-horizontal ( $γ \approx 0^{\circ}$ )	$cos γ \approx 1$ → fastest turn	Ex 3
D	Degenerate: exactly vertical ( $γ = 9 0^{\circ}$ )	$cos 9 0^{\circ} = 0$ → $\overset{γ}{˙} = 0$ , no turn	Ex 4
E	Degenerate: exactly horizontal ( $γ = 0^{\circ}$ )	$cos 0^{\circ} = 1$ → max possible $\overset{γ}{˙}$	Ex 4
F	Speed sensitivity ( $v$ vary karo, $γ$ fix karo)	$\overset{γ}{˙} \propto 1/ v$ test karta hai	Ex 5
G	Limiting behaviour ( $v \to \infty$ , $v \to 0$ )	equation ke asymptotes	Ex 6
H	Real-world word problem (turn radius nikalo)	Centripetal / Normal Acceleration se connect karo	Ex 7
I	Exam twist ( $α \neq = 0$ , toh $θ \neq = γ$ )	pitch angle ≠ path angle	Ex 8

Hum $g = 9.81 m/s^{2}$ lete hain har jagah, jab tak alag na bataya jaye.

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

Example 1 — Near-vertical (Cell A)

Worked example Pitch-over ke thodi der baad

Ek rocket ke paas $v = 100 m/s$ aur $γ = 8 9^{\circ}$ hai. Turn rate $\overset{γ}{˙}$ nikalo $rad/s$ aur $^{\circ} / s$ mein.

Forecast: nose almost seedha upar hai. Kya yeh tezi se bend karega ya bilkul nahi? (Padhne se pehle andaaza lagao.)

Step 1 — $cos γ$ nikalo. $cos 8 9^{\circ} = 0.01745$ . Yeh step kyun? Turning force gravity ka perpendicular slice hai, aur woh slice $m g cos γ$ hai. Near vertical, gravity almost exactly path ke saath ulti disha mein point karti hai, toh perpendicular mein almost kuch bachta nahi — isliye $cos γ$ tiny hai.

Step 2 — Boxed formula mein plug karo. $\overset{γ}{˙} = - \frac{9.81 \times 0.01745}{100} = - 1.712 \times 1 0^{- 3} rad/s$ Yeh step kyun? Direct substitution — formula already $\overset{γ}{˙}$ ke liye solve hai.

Step 3 — Degrees mein convert karo. $- 1.712 \times 1 0^{- 3} \times 57.30 = - 0.098 1^{\circ} / s$ . Yeh step kyun? $0.098$ degrees per second jaisi number slow lagti hai; radian value nahi lagti.

Verify: Rate negative hai (nose neeche tip ho raha hai, sahi hai) aur bahut chhoti — Forecast se match karta hai ki near-vertical rocket barely turn karta hai. Is rate par lagbhag $1^{\circ} /0.098 \approx 10$ seconds lagenge sirf ek degree drop karne mein. ✅

Example 2 — Mid-lean (Cell B)

Example 3 — Near-horizontal (Cell C)

Example 4 — Do degenerate endpoints (Cells D & E)

Worked example Exactly vertical aur exactly horizontal

$v = 100 m/s$ lo. $\overset{γ}{˙}$ compute karo (a) $γ = 9 0^{\circ}$ par aur (b) $γ = 0^{\circ}$ par.

Forecast: inme se ek zero turn deta hai. Kaun sa, aur kyun? (Andaaza lagao.)

Step 1 — Case (a), $γ = 9 0^{\circ}$ . $cos 9 0^{\circ} = 0$ , toh $\overset{γ}{˙} = - \frac{9.81 \times 0}{100} = 0 rad/s .$ Yeh step kyun? Seedha upar, gravity exactly velocity ke anti-parallel hai — pure braking, koi perpendicular slice nahi. Ek perfectly vertical rocket kabhi turning shuru nahi karta. Isliye launches deliberately ek tiny initial pitch-over inject karte hain symmetry todne ke liye.

Step 2 — Case (b), $γ = 0^{\circ}$ . $cos 0^{\circ} = 1$ , toh $\overset{γ}{˙} = - \frac{9.81 \times 1}{100} = - 0.0981 rad/s = - 5.62 2^{\circ} / s .$ Yeh step kyun? Flat fly karte waqt, gravity entirely perpendicular hai — is speed par maximum possible turn rate.

Verify: Case (a) $= 0$ (no-turn endpoint, Forecast se match karta hai) aur case (b) $= - g / v = - 9.81/100$ exactly (sabse bada magnitude, kyunki $cos γ \leq 1$ ). Dono endpoints physics ki demand ke mutabik behave karte hain. ✅

Example 5 — Speed sensitivity (Cell F)

Example 6 — Limiting behaviour (Cell G)

Worked example Speed ke extremes par kya hota hai?

$γ = 4 5^{\circ}$ fixed ke saath, $\overset{γ}{˙}$ describe karo jab $v \to \infty$ aur jab $v \to 0^{+}$ .

Forecast: bahut tez $v$ ke liye, kya turn rate bada number approach karta hai ya vanish ho jaata hai? (Andaaza lagao.)

Step 1 — Large $v$ . $\overset{γ}{˙} = - g cos γ / v$ . Jab $v \to \infty$ toh fraction $\to 0$ . Yeh step kyun? Jitna tez udo, gravity ko path ke har metre ko nudge karne ka utna kam time milta hai — trajectory nearly straight ho jaati hai. Mathematically $1/ v \to 0$ .

Step 2 — Tiny $v$ . Jab $v \to 0^{+}$ , $\overset{γ}{˙} = - g cos γ / v \to - \infty$ . Yeh step kyun? Vanishing speed par, gravity ko "baar baar mauka" milta hai har metre ke liye — direction whip around kar jaati hai. Reality mein rocket kabhi $v = 0$ par sideways pointed nahi hota, lekin equation warn karta hai ki low-speed turning violently fast hai, yahi wajah hai ki pitch-over ek tiny gentle kick hota hai, sharp nahi.

Verify: $γ = 4 5^{\circ}$ par numbers se check karo: $v = 1000$ par, $\overset{γ}{˙} = - 9.81 (0.7071) /1000 = - 0.006936 rad/s$ (chhota, → 0 trend); $v = 5$ par, $\overset{γ}{˙} = - 9.81 (0.7071) /5 = - 1.387 rad/s$ (bada, → −∞ trend). Dono limits confirm hote hain. ✅

Example 7 — Real-world word problem: turn radius (Cell H)

Worked example Rocket jo curve carve kar raha hai woh kitna tight hai?

Ek rocket $v = 200 m/s$ , $γ = 3 0^{\circ}$ par fly karta hai. Uski curving path ko locally ek circle ke arc ki tarah treat karo. Nikalo (a) turn rate $\overset{γ}{˙}$ , aur (b) flight path ki radius of curvature $R$ is instant par.

Forecast: kya radius kuch metres hogi, ya kilometres? (Andaaza lagao — ek fast rocket gently bend karta hai.)

Step 1 — Turn rate. $cos 3 0^{\circ} = 0.86603$ , $\overset{γ}{˙} = - \frac{9.81 \times 0.86603}{200} = - 0.042479 rad/s .$

Step 2 — Radius se relate karo. Radius $R$ ke circle par speed $v$ se move karne wala ek point angle at rate $∣ \overset{γ}{˙} ∣ = v / R$ sweep karta hai (yeh hai Centripetal / Normal Acceleration: $a_{⊥} = v \overset{γ}{˙} = v^{2} / R$ ). Toh $R = \frac{v}{∣ γ ˙ ∣} .$ Yeh step kyun? Parent note ne note kiya tha $r = v / \overset{γ}{˙}$ ; perpendicular gravity slice exactly centripetal turning supply karta hai, toh instantaneous circle ka radius $v /∣ \overset{γ}{˙} ∣$ hai.

Step 3 — Compute karo. $R = \frac{200}{0.042479} = 4708 m \approx 4.71 km .$

Verify (do tarike se).

Formula directly substitute karo: $R = v /∣ \overset{γ}{˙} ∣ = v \cdot v / (g cos γ) = v^{2} / (g cos γ) = 20 0^{2} / (9.81 \times 0.86603) = 4708 m$ . ✅
Units: $\frac{m/s}{rad/s} = m$ (radians dimensionless hote hain). ✅ Multi-kilometre radius Forecast se match karta hai ki fast rockets gently bend karte hain.

Example 8 — Exam twist: pitch angle jab $α \neq = 0$ (Cell I)

Worked example Trap: nose ≠ velocity

Ek gust briefly rocket ko angle of attack $α = 4^{\circ}$ deta hai jabki uska flight-path angle $γ = 5 5^{\circ}$ hai. Pitch angle $θ$ kya hai (woh direction jis taraf nose point karta hai)? Kya yeh still ek pure gravity turn hai?

Forecast: ideal gravity turn mein $θ = γ$ . Gust ke saath, kya $θ$ bada hoga ya chhota $γ$ se? (Andaaza lagao.)

Step 1 — General relation yaad karo. Flight-path angle and pitch angle se, pitch angle, flight-path angle aur angle of attack hamesha satisfy karte hain $θ = γ + α .$ Yeh step kyun? $α$ define hai us angle ke roop mein jo nose point karta hai ( $θ$ ) aur velocity point karti hai ( $γ$ ) ke beech. Toh nose $α$ upar hai velocity se.

Step 2 — Plug in. $θ = 5 5^{\circ} + 4^{\circ} = 5 9^{\circ}$ .

Step 3 — Kya yeh abhi bhi gravity turn hai? Nahi. Gravity turn require karta hai $α = 0$ . $α = 4^{\circ}$ ke saath wings/body lift generate karte hain, thrust ka ek small normal component aa jaata hai, aur extra structural side-loads appear hote hain Aerodynamic Drag and Max-Q ke paas. Pitch-rate equation $\overset{γ}{˙} = - g cos γ / v$ ab poori turning capture nahi karta — ab ek aerodynamic contribution bhi hai.

Verify: Special case sanity check: agar gust vanish ho jaaye ( $α \to 0$ ), $θ = γ + 0 = γ = 5 5^{\circ}$ , gravity-turn condition $θ = γ$ recover ho jaati hai. Formula gracefully ideal case mein degrade ho jaata hai. ✅

Active recall

Recall Khud ko test karo (answers hide karo)

Near vertical ( $γ \to 9 0^{\circ}$ ) par, turn rate tiny kyun hoti hai? ::: $cos γ \to 0$ , toh gravity ka perpendicular slice vanish ho jaata hai. Kaun sa endpoint $\overset{γ}{˙} = 0$ exactly deta hai? ::: $γ = 9 0^{\circ}$ (seedha upar). Fixed $γ$ par $v$ double karne par $∣ \overset{γ}{˙} ∣$ ka kya hota hai? ::: Halve ho jaata hai, kyunki $\overset{γ}{˙} \propto 1/ v$ . Instantaneous turn radius ka formula? ::: $R = v /∣ \overset{γ}{˙} ∣ = v^{2} / (g cos γ)$ . Agar $α = 4^{\circ}$ aur $γ = 5 5^{\circ}$ hai, toh $θ$ kya hai? ::: $θ = γ + α = 5 9^{\circ}$ . Fixed $γ$ par $v \to \infty$ hone par $\overset{γ}{˙}$ ka kya hota hai? ::: Yeh $\to 0$ ho jaata hai; path straight ho jaata hai.