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

3.4.13 · D2 · Physics › Rocket Flight Mechanics › Gravity turn trajectory — pitch rate from aerodynamic angle

Yeh page parent topic ka picture-first companion hai, jo Rocket Flight Mechanics ke andar hai.

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Step 1 — Rocket ko ek moving dot aur ek arrow ki tarah draw karo

KYA. Woh lamba chamkila cylinder bhool jaao. Physics ke liye ek rocket sirf ek point hai (uski mass, sab kuch ek jagah squish ho gayi) jiske saath ek velocity arrow laga hua hai — ek aisa arrow jo kehta hai "yeh woh direction hai jisme main move kar raha hoon, aur iski length batati hai kitni tezi se."

KYUN. Turn karna us arrow ke direction mein badlaav hai. Agar hum baat karna chahte hain "rocket kitni tezi se turn karta hai," toh pehle hamare paas ek single clean arrow hona chahiye jiske swinging ko hum measure kar sakein. Poori rocket ki shape hamen distract kar degi.

PICTURE. Burnt-orange arrow dekho. Uski tail rocket ki current position hai; uski length woh speed hai jise hum $v$ (metres per second) bolenge. Woh horizon ke upar jitna angle banata hai woh teal mein drawn hai — woh angle poore show ka star hai.

Kyunki rocket angle of attack ==angle of attack $\alpha =0$== ke saath fly karta hai (nose bilkul velocity ke saath), nose jis direction mein point karta hai ($\theta$) woh us direction ke barabar hai jisme woh move karta hai ($\gamma$). Toh "nose kitni tezi se pitch karta hai" yahi hai "$\gamma$ kitni tezi se change hota hai." Ek arrow sab kuch bata deta hai.

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Step 2 — "Turning" ka matlab kya hai? Arrow ko swing karte hue dekho

KYA. Thoda sa time guzarne do — use $dt$ kaho ("thoda sa time," itna chhota ki numbers barely change hon). Us instant mein arrow ek tiny angle se swing karta hai, jise hum $d\gamma$ kehte hain ("$\gamma$ mein change ka thoda sa hissa").

KYUN. Hum ek rate chahte hain: angle change per second. Woh exactly $\frac{d\gamma }{dt}$ hai — ise "d-gamma d-t" padhte hain, matlab "har guzarte second ke liye $\gamma$ kitna change hota hai." Yeh fraction hi woh poori quantity hai jise hum dhundh rahe hain.

PICTURE. Purana arrow (faint) aur naya arrow (solid) almost ek doosre ke upar baithe hain; unke beech ka plum wedge $d\gamma$ hai. Dhyan do yeh neeche ki taraf swing karta hai — horizon ki taraf. Yeh neeche wala swing hi wajah hai ki hum baad mein ek minus sign dekhenge.

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Step 3 — Arrow ki tip ek curve par travel karti hai; uska sideways nudge dhundho

KYA. Agar arrow swing karta rahe, toh rocket ka path ek curve hai, seedhi line nahi. Path ko bend karne ke liye, kuch cheez rocket ko sideways — jahan woh already ja raha hai uske perpendicular — push karni chahiye. Us sideways acceleration ko $a_{\perp }$ kaho ("a-perp," perpendicular acceleration).

KYUN. Newton kehta hai velocity ka direction change karne ke liye ek force chahiye, sirf speed nahi. Sideways push hi woh cheez hai jo path ko curve karti hai. Hume ek formula chahiye ki kitna bada sideways push kitna tez turn produce karta hai.

PICTURE. Rocket curve ke saath thodi si distance $vdt$ move karta hai (orange). Us step mein direction $d\gamma$ se swing hui. Ek chhote arc ki geometry kehti hai sideways displacement $vdt\times d\gamma$ hai, toh sideways acceleration speed times turn-rate hai:

Ise ulta padho: agar hum sideways acceleration jaante hain, toh hum turn rate jaante hain — $\dot{\gamma }=a_{\perp }/v$. Toh hunt narrow ho jaati hai: sideways acceleration kya provide karta hai?

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Step 4 — Sirf gravity sideways push kar sakti hai (kyunki $\alpha =0$)

KYA. Har force list karo. Thrust $T$ aage push karta hai. Drag $D$ peeche push karta hai. Gravity $mg$ seedha neeche pull karti hai. Kyunki nose exactly velocity ke saath hai ($\alpha =0$), thrust aur drag purely arrow ke saath hain — woh hamen tez ya dhima kar sakte hain lekin woh sideways push nahi kar sakte.

KYUN. Turning ke liye ek sideways force chahiye. Thrust aur drag ka yahan exactly zero sideways part hai, toh woh rule out ho jaate hain. Sirf gravity baaki bachhti hai path ko bend karne ke liye. Yahi wajah hai ki is manoeuvre ko gravity turn kaha jaata hai.

PICTURE. Orange (thrust) aur uska opposing drag arrow ke saath flat lie karte hain — koi sideways reach nahi. Sirf grey gravity arrow, seedha neeche point karta hua, direction of motion ke across lean karta hai.

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Step 5 — Gravity ko "along" aur "sideways" mein split karo ek right triangle use karke

KYA. Gravity seedha neeche point karti hai size $mg$ ke saath. Hum ise do arrows mein todenge: ek velocity ke saath (climbing karte waqt backward) aur ek perpendicular (woh sideways wala jis par hamein dhyan dena hai). Is splitting ko components mein resolve karna kehte hain.

KYUN. Sirf perpendicular part rocket ko turn karta hai. Use karne ke liye hume exactly measure karna hoga ki woh kitna bada hai. Ek right triangle "seedha-neeche gravity + tilted velocity" ko do clean numbers mein badal deta hai.

PICTURE. Right triangle banao: neeche wala gravity arrow $mg$ hypotenuse hai; velocity direction horizon se angle $\gamma$ par baithe hai. Ek perpendicular daalo. Do legs hain:

• velocity ke saath: $mg\sin \gamma$
• velocity ke perpendicular: $mg\cos \gamma$

$\cos$ perpendicular leg par kyun aata hai? Kyunki "seedha neeche" aur "perpendicular-to-velocity" ke beech ka angle $\gamma$ hi hai, aur $\cos$ hai "adjacent over hypotenuse" — woh leg jo us angle ko hug karti hai. Figure mein plum leg trace karo: woh $mg\cos \gamma$ hai, hamara sideways push.

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Step 6 — Sideways force ko sideways acceleration ke barabar set karo (Newton)

KYA. Perpendicular direction mein Newton ka law: (sideways force) $=$ (mass) $\times$ (sideways acceleration). Sideways force gravity ka perpendicular leg hai jo arrow ko neeche pull karti hai — toh woh negative hai (woh $\gamma$ ko reduce karti hai). Sideways acceleration Step 3 se $v\dot{\gamma }$ hai.

KYUN. Yeh woh ek honest equation hai jo cause (gravity ka perpendicular leg) ko effect (turn) se connect karti hai. Pehle sab kuch dono sides build kar raha tha; ab hum unhe glue karte hain.

PICTURE. Balance ki left side: plum sideways-gravity leg $-mg\cos \gamma$. Right side: $m$ times sideways acceleration $v\dot{\gamma }$. Scale balance ho jaata hai.

Ab dono sides se mass $m$ cancel karo (woh har term mein appear hota hai — ek halka rocket aur ek bhaari rocket same tarah turn karte hain, kyunki gravity saari masses ko equally accelerate karti hai), phir $v$ se divide karo:

Kyunki $\alpha =0\Rightarrow \theta =\gamma$, yeh pitch rate $\dot{\theta }$ bhi hai.

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Step 7 — Woh do edge cases jo equation quietly handle karta hai

KYA. Extremes check karo taaki koi scenario surprise na kare.

KYUN. Jis formula par aap trust karte ho woh apni boundaries par survive karna chahiye. Do matter karte hain: bilkul vertical ($\gamma =90^{\circ }$) aur bilkul level ($\gamma =0^{\circ }$).

PICTURE. Do mini-scenes.

• Vertical, $\gamma =90^{\circ }$: gravity exactly opposite arrow point karti hai. Uska sideways leg $mg\cos 90^{\circ }=0$ length ka hai. Koi sideways push nahi → $\dot{\gamma }=0$. Ek perfectly vertical rocket apne aap kabhi nahi turns — isliye hum ek tiny deliberate pitch-over dete hain cheezein start karne ke liye (dekho Ascent Guidance and Pitch Program).
• Level, $\gamma =0^{\circ }$: gravity arrow ke entirely perpendicular hai; $mg\cos 0^{\circ }=mg$, poora weight path ko bend karta hai. Turning sabse tez hoti hai yahan (diye gaye speed ke liye).

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Ek-picture summary

Sab kuch ek canvas par: $\gamma$ angle par arrow, neeche-gravity apne along-leg $mg\sin \gamma$ aur apne sab-se-zaroori perpendicular leg $mg\cos \gamma$ mein split hoti hui, woh leg sideways acceleration $v\dot{\gamma }$ ko feed karti hui, aur resulting curved path horizon ki taraf bend hoti hui — final boxed law ke saath labelled.

Recall Feynman retelling — ise apne saadhe shabdon mein bol ke dikhao

Ek rocket ek dot hai jiske saath ek speed-arrow hai jo horizon ke upar $\gamma$ angle par tika hua hai. Turn karne ke liye, uska arrow swing karna chahiye, aur swing karne ke liye use ek sideways push chahiye. Nose ko velocity ke saath glued rakhe ($\alpha =0$), engine aur drag dono straight arrow ke saath point karte hain — woh sirf change kar sakte hain kitni tezi se, kabhi nahi kis taraf. Toh steer karne ke liye sirf ek cheez bacha hai gravity. Gravity seedha neeche point karti hai; sirf uska woh slice jo arrow ke across lie karta hai path ko turn kar sakta hai, aur ek right triangle kehta hai woh slice $mg\cos \gamma$ hai. Newton ka sideways law, "sideways force $=$ mass times sideways acceleration," padhta hai $mv\dot{\gamma }=-mg\cos \gamma$. Mass cancel karo, speed se divide karo, aur nikal aata hai $\dot{\gamma }=-g\cos \gamma /v$: turn neeche hai (minus), gravity se driven ($g$), sirf uske across-slice ka use karke ($\cos \gamma$), aur speed se slow ($1/v$). Seedha upar, $\cos 90^{\circ }=0$, toh kuch nahi turns — isliye launch par gentle nudge. Level flight, $\cos 0^{\circ }=1$, sabse hardest turns.

Recall Quick self-test

Rocket ko kaun sa force turn karta hai? ::: Gravity ka perpendicular slice, $mg\cos \gamma$. Thrust aur drag use kyun nahi turn karte? ::: $\alpha =0$ ke saath woh velocity ke saath lie karte hain, toh unka sideways part zero hai. Ek turning body ka sideways acceleration kya hota hai? ::: $v\dot{\gamma }$ (speed times turn rate). Launch par $\dot{\gamma }=0$ kyun hota hai? ::: $\gamma =90^{\circ }$ deta hai $\cos 90^{\circ }=0$, koi sideways gravity nahi. $v$ double karna turn speed up karta hai ya slow? ::: Slow karta hai — $\dot{\gamma }\propto 1/v$, toh woh aadha ho jaata hai.