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

Turning gravity se hoti hai — specifically uske velocity vector ke perpendicular component se. Gravity ki magnitude $mg$ hai jo seedha neeche point karti hai. Isse velocity ke relative split karo (jo horizon ke upar angle $\gamma$ par hai): perpendicular slice ki size hai $mg\cos \gamma .$ Thrust aur drag tangential hain (kyunki $\alpha =0$), isliye woh sideways kuch nahi add karte. Figure s01 dekho: red arrow (labelled $mg\cos \gamma$) hi sirf ek sideways pull hai.

$\dot{\gamma }=0$. Vertical matlab $\gamma =90^{\circ }$, aur $\cos 90^{\circ }=0$, isliye perpendicular gravity component vanish ho jaata hai. Gravity seedha velocity line ke neeche point karti hai — yeh rocket ko slow kar sakti hai lekin isse turn karne ke liye koi sideways grip nahi hai. Ek vertical rocket hamesha vertical hi rehta hai.

Minus sign kehta hai ki $\gamma$ decrease ho raha hai: velocity arrow time ke saath horizon ki taraf neeche jhuk raha hai. Exactly yahi "turning over" ka matlab hai — rocket vertical se horizontal ki taraf lean karta hai. Agar sign $+$ hota, toh rocket upar uthta, jo gravity kabhi nahi karti.

Boxed law mein plug karo. $\cos 60^{\circ }=0.5$. $\dot{\gamma }=-\frac{9.81\times 0.5}{80}=-\frac{4.905}{80}=-0.06131\text{rad/s}.$ Convert karo: $180/\pi \approx 57.30$ se multiply karo: $-0.06131\times 57.30\approx -3.51^{\circ }/\text{s}.$ Padhna: nose lagbhag saadhe teen degree har second tip ho raha hai — ek brisk turn kyunki $\gamma$ vertical se kaafi neeche hai aur $v$ abhi bhi chhota hai.

$v$ ke liye law rearrange karo. Shuru karo $|\dot{\gamma }|=\frac{g\cos \gamma }{v}\Rightarrow v=\frac{g\cos \gamma }{|\dot{\gamma }|}.$ $\cos 50^{\circ }=0.6428$ ke saath: $v=\frac{9.81\times 0.6428}{0.02}=\frac{6.306}{0.02}=315.3\text{m/s}.$ $v$ ke liye solve kyun kiya? Hume desired rate diya gaya tha aur pucha gaya tha ki kaunsi condition isse produce karti hai — isliye $v$ unknown hai, aur algebra ek clean division hai.

Wahi formula, bas $g$ swap karo. $\cos 30^{\circ }=0.8660$. $\dot{\gamma }=-\frac{1.62\times 0.8660}{40}=-\frac{1.4029}{40}=-0.03507\text{rad/s}.$ Degrees mein: $-0.03507\times 57.30\approx -2.01^{\circ }/\text{s}$. Lesson: kamzor gravity ⇒ kamzor sideways pull ⇒ slower gravity turn. Moon par tumhe zyada deliberately lean karna padta hai kyunki gravity kam help karti hai.

Kyunki $\gamma$ (aur isliye $\cos \gamma$) aur $g$ identical hain, sirf $1/v$ alag hai. Isliye $\frac{{\dot{\gamma }}_A}{{\dot{\gamma }}_B}=\frac{1/v_A}{1/v_B}=\frac{v_B}{v_A}=\frac{360}{120}=3.$ Rocket A teen guna tez turn karta hai. Numeric check: $\cos 45^{\circ }=0.7071$. ${\dot{\gamma }}_A=-9.81(0.7071)/120=-0.05781$ rad/s; ${\dot{\gamma }}_B=-9.81(0.7071)/360=-0.01927$ rad/s; ratio $=3.00$. ✅ Insight: speed turning par ek brake hai. Slow rocket apna path har second kahin zyada bend karta hai — exactly yahi wajah hai ki critical pitch-over jaldi kiya jaata hai, jab $v$ chhota ho.

$\gamma =0$ par: $|{\dot{\gamma }}_0|=g\cos 0/v=g/v$ (kyunki $\cos 0=1$). Hume chahiye $\frac{g\cos \gamma }{v}=\frac{1}{2}\cdot \frac{g}{v}\Rightarrow \cos \gamma =\frac{1}{2}.$ $g$ aur $v$ cancel ho jaate hain — isliye answer ek pure angle hai: $\gamma =\arccos (0.5)=60^{\circ }$. $\arccos$ kyun? Hume cosine ($0.5$) pata hai aur angle chahiye. $\arccos$ woh tool hai jo cosine ko undo karta hai — yeh answer deta hai "kis angle ka yeh cosine hai?" $[0^{\circ },90^{\circ }]$ mein unique answer $60^{\circ }$ hai.

$r=v/\dot{\gamma }$ kyun? Curved motion mein normal (centre ki taraf) acceleration $a_{\perp }=v\dot{\gamma }$ hai, aur saath hi $a_{\perp }=v^2/r$ bhi hai (dekho Centripetal / Normal Acceleration). Inhe equal set karo: $v^2/r=v\dot{\gamma }\Rightarrow r=v/\dot{\gamma }$. Ab gravity-turn law $|\dot{\gamma }|=g\cos \gamma /v$ substitute karo: $r=\frac{v}{|\dot{\gamma }|}=\frac{v}{g\cos \gamma /v}=\frac{v^2}{g\cos \gamma }.$ Numbers: $\cos 40^{\circ }=0.7660$. $r=\frac{150^2}{9.81\times 0.7660}=\frac{22500}{7.515}\approx 2994\text{m}\approx 3.0\text{km}.$ Padhna: path us instant par $\sim 3$ km radius ke circle ke along curve karta hai. Jaise $v$ badhta hai, $r$ $v^2$ ki tarah badhta hai — path tezi se seedha hota jaata hai.

Step 1 — separate. Saare $\gamma$ ek side par group karo, $t$ doosri side par: $\frac{d\gamma }{\cos \gamma }=-\frac{g}{\bar{v}}dt.$ Kyun? Left side par time sirf $\gamma$ ke through appear karta hai, isliye $\gamma$ isolate karne se har side ko apne aap integrate kiya ja sakta hai. Step 2 — $\int \frac{d\gamma }{\cos \gamma }=\ln |\sec \gamma +\tan \gamma |$ kahan se aata hai? Yeh koi jaadu nahi hai — yahan ek-line reason hai. $\sec \gamma =1/\cos \gamma$ ke top aur bottom ko clever factor $(\sec \gamma +\tan \gamma )$ se multiply karo: $\sec \gamma =\sec \gamma \cdot \frac{\sec \gamma +\tan \gamma }{\sec \gamma +\tan \gamma }=\frac{{\sec }^2\gamma +\sec \gamma \tan \gamma }{\sec \gamma +\tan \gamma }.$ Ab denominator $u=\sec \gamma +\tan \gamma$ dekho. Iska derivative hai $\frac{du}{d\gamma }=\sec \gamma \tan \gamma +{\sec }^2\gamma$ — jo exactly numerator hai! Isliye integrand $u^{\prime }/u$ form ka hai, jiska integral $\ln |u|$ hai: $\int \sec \gamma d\gamma =\int \frac{u^{\prime }}{u}d\gamma =\ln |u|=\ln |\sec \gamma +\tan \gamma |.$ (Geometrically, $\sec \gamma +\tan \gamma$ woh length hai jo tangent line ke along sweep hoti hai jaise $\gamma$ khulta hai — log isliye appear hota hai kyunki woh length multiplicatively badhti hai, usi tarah jaise $\ln$ tab dikhta hai jab koi quantity khud ke proportion mein barhti hai.) Step 3 — bounds apply karo. $[\ln |\sec \gamma +\tan \gamma |{]}_{{\gamma }_1}^{{\gamma }_2}=-\frac{g}{\bar{v}}\Delta t.$ ${\gamma }_2=70^{\circ }$ par: $\sec 70^{\circ }=2.924$, $\tan 70^{\circ }=2.747$, sum $=5.671$, $\ln =1.7354$. ${\gamma }_1=80^{\circ }$ par: $\sec 80^{\circ }=5.759$, $\tan 80^{\circ }=5.671$, sum $=11.430$, $\ln =2.4362$. Left side $=1.7354-2.4362=-0.7008$. Step 4 — $\Delta t$ ke liye solve karo. $-0.7008=-\frac{9.81}{200}\Delta t=-0.04905\Delta t\Rightarrow \Delta t=\frac{0.7008}{0.04905}\approx 14.29\text{s}.$ Sanity check: vertical ke paas turn slow hoti hai (chhota $\cos \gamma$), isliye sirf $10^{\circ }$ tip karne mein kaafi time lagta hai — lagbhag 14 s reasonable hai.

$\gamma$ (isliye $\cos \gamma$) aur $g$ fixed hain, isliye $|\dot{\gamma }|\propto 1/v$. End-to-start ratio: $\frac{|{\dot{\gamma }}_{\text{end}}|}{|{\dot{\gamma }}_{\text{start}}|}=\frac{v_{\text{start}}}{v_{\text{end}}}=\frac{100}{400}=\mathrm{0.25.}$ Turn rate quarter ho jaati hai jaise engine vehicle ko accelerate karta hai. Guidance matlab: trajectory ki almost saari shaping tab hoti hai jab rocket slow ho. Jab tak woh fast ho jaata hai, gravity path ko barely bend karti hai, isliye pitch program essentially "initial lean sahi se set karo aur phir physics ko coast karne do" ban jaata hai. Yahi wajah hai ki early mein ek chhoti pitch-over error late wali error se kahin zyada costly hai.

Jaise $\gamma \to 0$, $\cos \gamma \to 1$ (iska sabse bada value) hota hai, isliye per-radian pull sabse strong hoti hai wahan — yeh help karta hai. Lekin do cheezein isse fight karti hain:

1. Real ascent mein $v$ large ho jaata hai (hazaaron m/s), aur $|\dot{\gamma }|=g\cos \gamma /v\to$ small kyunki $v$ bahut bada hai. Isliye turn rate shrink ho jaati hai even though $\cos \gamma$ apne max ke paas hai.
2. Flat-Earth model mein gravity ka hamesha ek downward component hota hai jo $\gamma$ ko $0^{\circ }$ ke horizontal ho jaane ke baad below kheenchta hai, isliye pure gravity nose ko horizon ke neeche tipate rehti — tum descend karne lagte. Conclusion: ek pure flat-Earth gravity turn $\gamma =0$ se guzar sakti hai, lekin wahan rukti nahi — gravity $\gamma$ ko negative kheenchti rehti hai. Real orbital insertion ko Earth ki curvature chahiye: jaise rocket itni tezi se jaata hai ki ground uske neeche curve away ho jaata hai aur "level" ek moving target ban jaata hai, $\gamma$ ko diving ke bina $0$ ke paas settle karne deta hai. Flat-Earth law sirf early-ascent tool hai; burnout ke paas tumhe curved-Earth model par switch karna padega.

Constraint hai $|\dot{\gamma }|\le 0.05$, yaani $\frac{g\cos \gamma }{v}\le 0.05\Rightarrow v\ge \frac{g\cos \gamma }{0.05}.$ $\cos 85^{\circ }=0.08716$ ke saath: $v_{\min }=\frac{9.81\times 0.08716}{0.05}=\frac{0.8550}{0.05}=17.1\text{m/s}.$ Isliye $\sim 17\text{m/s}$ se upar koi bhi speed $\gamma =85^{\circ }$ par constraint satisfy karta hai — comfortably easy, kyunki near-vertical flight mein $\cos \gamma$ tiny hota hai. Asli khatara baad mein aata hai, abhi nahi. Jaise $\gamma$ level ki taraf girta hai, $\cos \gamma$ $1$ ki taraf chadhta hai, isliye required minimum speed bhi usके saath chadhti hai. Extreme horizontal case $\gamma =0^{\circ }$ par: $v_{\min }(\gamma =0)=\frac{9.81\times \cos 0}{0.05}=\frac{9.81}{0.05}=196.2\text{m/s}.$ Isliye constraint near vertical par sabse loose hai ($17\text{m/s}$ kaafi hai) lekin near horizontal par sabse tight hai ($196\text{m/s}$ chahiye). Ek rocket jo abhi bhi $\sim 196\text{m/s}$ se slower hai jab uska path level hota hai woh payload ki angular-rate limit exceed kar dega aur use damage kar sakta hai. Design fix: ensure karo ki vehicle $\approx 196\text{m/s}$ se past accelerate kar le $\gamma$ ke $0$ ke paas aane se pehle — yani pehle speed gain karo taaki turn ka late, low-$\gamma$ part gentle rahe. Bahut slowly launch karna exactly isliye dangerous hai kyunki singular-worst moment low-$\gamma$ end par hai, loud, slow lift-off par nahi.

Constraint se speed (L5.2 ki tarah), $\cos 40^{\circ }=0.7660$: $v=\frac{g\cos \gamma }{0.05}=\frac{9.81\times 0.7660}{0.05}=\frac{7.515}{0.05}=150.3\text{m/s}.$ L3.3 ke formula se radius (ya directly $r=v/|\dot{\gamma }|$): $r=\frac{v}{|\dot{\gamma }|}=\frac{150.3}{0.05}=3006\text{m}\approx 3.0\text{km}.$ (Check karo $r=v^2/(g\cos \gamma )=150.3^2/7.515=3006$ m se. ✅) Padhna: max turn rate enforce karna fix karta hai ki tum kitni fast chal rahe ho aur path kitni gently curve karta hai — ek payload spec ek saath do trajectory numbers lock kar deti hai. Notice karo $r=v/|\dot{\gamma }|$ sabse clean route hai: constraint limit par, radius sirf speed divided by allowed rate hai.

• Rocket ko kaunsi force turn karti hai? ::: gravity ka perpendicular component $mg\cos \gamma$.
• Turn rate vs speed? ::: $|\dot{\gamma }|\propto 1/v$ — faster = slower turn.
• Turn radius formula? ::: $r=v^2/(g\cos \gamma )$.
• $\sec \gamma$ ka antiderivative? ::: $\ln |\sec \gamma +\tan \gamma |$.
• Turn-rate limit sabse zyada kahan bite karta hai? ::: sabse chhote $\gamma$ par (near horizontal), jahan $\cos \gamma \to 1$.
• Liftoff par law ko kya todta hai? ::: $v\to 0$ se $1/v$ diverge karta hai, aur $\gamma =90^{\circ },v=0$ ek $0/0$ indeterminate hai — isliye hum $v>0$ hone par pitch-over seed karte hain.