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

3.4.13 · D1 · Physics › Rocket Flight Mechanics › Gravity turn trajectory — pitch rate from aerodynamic angle

Is page mein kuch bhi assume nahi kiya gaya. Parent topic ek boxed equation par khatam hota hai, lekin woh equation symbols ki ek language mein likhi hui hai — $\gamma$, $t$, $g$, $v$, $\cos$, aur ek ajeeb fraction $\frac{d\gamma }{dt}$ — jo humne abhi tak seekhi nahi hai. Isliye woh equation hum yahan upar nahi likhenge. Iske bajaye hum uska har ek letter scratch se banayenge, ek-ek karke, ek doosre ke upar, har ek ko ek picture se anchor karke. Sirf aakhri line mein, jab har symbol earn ho jaayega, tab poori equation saamne aayegi.

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1. Horizon ke upar angle — "kitna steep hai" ki picture

Is topic mein sab kuch flat ground se upar maape gaye angles ke baare mein hai.

Figure dekho: gray line horizon hai, aur angle us arrow aur is line ke beech ki chhoti si wedge hai. Chhoti wedge matlab "almost flat"; quarter-turn ke paas ki wedge matlab "almost straight up". Yahi is topic mein sirf ek hi tarah ka angle hai, isliye picture ko yaad rakhein: flat ground se upar maapi gayi wedge.

Humein yeh kyun chahiye? Kyunki ek rocket ka poora kaam seedha upar (zameen aur gaadi haawa se nikalne ke liye) se almost sideways (orbit ke liye itni tezi se jaane ke liye) jaana hai. Woh safar sirf yahi wedge hai jo $90^{\circ }$ se $0^{\circ }$ ki taraf shrink karti hai.

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2. Do alag arrows: naak kahan point karti hai vs. kahan move ho raha hai

Yahi woh subtlety hai jis par poora topic tika hua hai. Ek rocket ke paas do directions hoti hain jo lagta hai same honi chahiye, lekin hamesha nahi hoti:

• Symbol $\theta$ Greek letter "theta" hai. Symbol $\gamma$ "gamma" hai. Yeh sirf names hain, jaise $x$ aur $y$, tradition se chosen.

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3. Angle of attack $\alpha$ — do arrows ke beech ka gap

Poore parent topic ka sabse important idea woh special case hai:

Topic $\alpha =0$ ki itni zyada parwah kyun karta hai? Kyunki haawa sirf tab rocket ko sideways push karti hai jab naak airflow se tilted away ho — us sideways push ko lift kehte hain, aur yeh structure ko bend aur stress karti hai. Agar $\alpha =0$ ho, toh haawa seedhe body ke along flow karti hai aur koi sideways air-push nahi hoti. Is fact ko apni pocket mein rakhein — Section 5 mein yeh do poori forces ko khatam kar dega. Thick air mein us sideways push ke kyun itna dangerous hai, yeh jaanne ke liye Aerodynamic Drag and Max-Q dekhein.

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4. Speed $v$ — kitni tezi, direction ignore karke

Figure 2 mein velocity arrow ki imagine karo: uski length $v$ hai, uska tilt $\gamma$ hai. Woh split — length vs. tilt — exactly wahi wajah hai ki final equation mein dono symbols honge.

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5. Forces: $T$, $D$, $m$, $g$ — aur kaunsi rocket ko turn kar sakti hain

Ab hum un chaar cheezoon ko naam dete hain jo rocket ko push aur pull karti hain.

Figure mein jo crucial geometry dikhti hai: thrust aur drag velocity line ke along point karte hain (kyunki $\alpha =0$ naak = velocity banata hai), lekin gravity uss line se perpendicular seedha neeche point karti rehti hai. Yahi mismatch poori kahaani hai.

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6. Ek force ko "along" aur "across" mein split karna — trigonometry earned

Gravity neeche point karti hai; rocket $\gamma$ tilt par move karta hai. Gravity use karne ke liye humein poochna hoga: iska kitna hissa path ke along backward push karta hai, aur kitna hissa path ke across sideways push karta hai? Ek arrow ko do perpendicular arrows mein split karna resolving into components kehlaata hai.

Is gravity $mg$ (hypotenuse) par apply karte hue, Figure 4 mein triangle padhte hue:

• Along-the-path part $mg\sin \gamma$: climb slow karta hai. Jab vertical ho ($\gamma =90^{\circ }$), $\sin 90^{\circ }=1$, isliye gravity climb se seedha fight karti hai — makes sense.
• Across-the-path part $mg\cos \gamma$: yahi turning force hai. Jab vertical ho, $\cos 90^{\circ }=0$, isliye koi sideways tug nahi — ek vertical rocket turning shuru nahi kar sakta. Jab flat ho ($\gamma =0$), $\cos 0^{\circ }=1$, gravity motion ke fully sideways kheenchti hai.

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7. Rate of change $\frac{d()}{dt}$ — "abhi yeh kitni tezi se change ho raha hai"

Humare paas $\gamma$ (tilt) hai aur gravity ka sideways pull hai. Lekin topic jaanna chahta hai $\gamma$ kitni tezi se shrink ho raha hai — ek rate. Humein ek symbol chahiye "koi quantity per second kitni tezi se change hoti hai" ke liye.

• Isi tarah $\frac{dv}{dt}$ hai ki speed kitni tezi se change hoti hai (path ke along acceleration).
• Dot notation $\dot{\gamma }$ ka matlab bilkul wahi hai jo $\frac{d\gamma }{dt}$ ka hai — upar ek dot sirf "rate of change per second" ka shorthand hai. Hum $\dot{\gamma }$ sirf is definition ke baad use karte hain, pehle kabhi nahi.
• Units caution: jab yeh rate physics formula ke andar ho, toh $\dot{\gamma }$ radians per second mein nikalta hai. Hum ise feel ke liye degrees per second mein report kar sakte hain, lekin sirf radians mein compute karne ke baad.

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8. Normal (turning) acceleration $v\dot{\gamma }$ — derived, asserted nahi

Jab koi moving cheez turn karti hai, uski direction change hoti hai chahe speed na bhi bade. Direction change karna khud ek acceleration hai, aur Newton ke anusaar iske liye ek sideways force chahiye. Aao hum formula banayein rather than quote karein.

Toh jitni tezi se ja rahe ho ($v$) ya jitni tezi se arrow swing kare ($\dot{\gamma }$), utni badi sideways acceleration. Yeh familiar $v^2/r$ ka curved-motion cousin hai; yahan turn radius $r=v/\dot{\gamma }$ hai. Deeper detail Centripetal / Normal Acceleration mein milega.

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Sab kuch ek saath (parent result ka preview)

Ab — aur sirf ab — har symbol earn ho chuka hai, isliye hum finally parent equation likh sakte hain. Newton across the path kehta hai: (mass)$\times$(sideways acceleration) = (sideways force). Sideways acceleration $v\dot{\gamma }$ hai (Section 8); sideways force gravity ka across-path part hai, $-mg\cos \gamma$ (Section 6), negative kyunki yeh $\gamma$ ko neeche kheenchti hai: $mv\dot{\gamma }=-mg\cos \gamma .$ Dono sides se mass $m$ cancel karo: $\boxed{\frac{d\gamma }{dt}=-\frac{g\cos \gamma }{v}}$ Ise apni nazar se wapas padho: gravity ka sideways fraction ($\cos \gamma$), gravity ki strength ($g$) se scale hoke, aap already kitni tezi se ja rahe ho ($v$) se divide — yahi aapka turn rate hai. (Aur yaad rakhein: compute karne ke liye $\gamma$ radians mein daalo, aur answer $\dot{\gamma }$ radians per second mein aayega.) Poora derivation parent note mein hai the parent topic Rocket Flight Mechanics ke andar; $v$ aur $m$ flight ke dauran kya hota hai yeh jaanne ke liye Ascent Guidance and Pitch Program aur Tsiolkovsky Rocket Equation bhi dekhein.

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Prerequisite map

Neeche ka diagram dikhata hai ki aapne jo har foundation abhi banaya woh final pitch-rate equation mein kaise feed karta hai. Arrows ko "is ke liye zaroorat hai" ki tarah padho. Do streams notice karo: left stream (angles → components → sideways gravity pull) force provide karta hai; right stream (speed aur rate of change → normal acceleration) turning provide karta hai. Woh boxed equation par milte hain.

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Equipment checklist

Khud test karo: dash ke baad answer cover karo aur zor se bolo. Agar koi bhi answer fuzzy lage, toh us section ko dobara padho.

• "Angle above the horizon" kya measure karta hai, aur "below" ka sign kya hai? — Flat ground line se upward tilt; $0^{\circ }$ = sideways, $+90^{\circ }$ = seedha upar, negative = horizon ke neeche point karna (descending).
• Formula ke andar $\gamma$ radians mein kyun hona chahiye? — Kyunki physics equations mein $\sin$ aur $\cos$ radians expect karte hain; degrees galat numbers dete hain jab tak convert nahi karo ($180^{\circ }=\pi$).
• Pitch angle $\theta$ kya hai? — Woh direction jisme rocket ki naak (aur thrust) point karti hai, horizon ke upar maapi gayi.
• Flight-path angle $\gamma$ kya hai? — Woh direction jisme rocket actually travel karta hai (uski velocity), horizon ke upar maapi gayi.
• Angle of attack $\alpha$ kya hai? — Naak aur travel directions ke beech signed gap: $\alpha =\theta -\gamma$.
• $\alpha =0$ kya true hone par majboor karta hai, aur kaunsi forces disable hoti hain? — $\theta =\gamma$; aur thrust aur drag ka tab koi across-path component nahi hota, isliye woh rocket ko turn nahi kar sakte.
• Symbol $v$ kya represent karta hai? — Speed — velocity arrow ki length, direction strip karke.
• Gravity triangle par kaunsa hissa rocket ko turn karta hai? — Across-path part $mg\cos \gamma$; along-path part $mg\sin \gamma$ sirf climb slow karta hai.
• Jab vertical ho toh sideways tug kyun nahi hota? — $\cos 90^{\circ }=0$, isliye gravity ka across-path component vanish ho jaata hai.
• $\frac{d\gamma }{dt}$ (same as $\dot{\gamma }$) ka kya matlab hai? — Flight-path angle abhi per second kitni tezi se change ho raha hai (negative = tip over ho raha hai).
• Normal acceleration $v\dot{\gamma }$ kahan se aata hai? — Velocity arrow (length $v$) $d\gamma$ se swing karne par uski tip $vd\gamma$ move karti hai; $dt$ se divide karo toh $v\dot{\gamma }$ milta hai.