Exercises — Shock waves — Mach number, Mach cone — - CRITICAL for rockets -

KYA karte hain: ratio $M$ banao — sirf yahi regime decide karta hai. $M=\frac{v_{\text{obj}}}{v_s}=\frac{270}{340}=0.794$ KYU: raw speed akela nahi bata sakta — 270 m/s fast hai, lekin sawaal yeh hai ki kya usne apni ripples ko beat kiya? Kyunki $M<1$ hai, usne nahi kiya. Answer: $M\approx 0.79$ → subsonic. Kyunki $\sin \theta =1/M=1.26>1$ ka koi real angle nahi hota, koi Mach cone exist nahi karta. Object apne khud ke expanding wavefronts ke andar hi rehta hai.

KYA/KYU: ek formula ko invert karo. $\sin 90^{\circ }=1$, toh $\sin \theta =\frac{1}{M}=1\Rightarrow M=1.$ KAISA DIKHTA HAI: cone poori tarah khul ke flight path ke perpendicular ek flat plane ban gaya hai — wavefronts seedha aage ek single wall mein pile ho jaate hain. Yeh transonic case hai, $M=1$, yaani "sound barrier."

$\sin \theta =\frac{1}{M}=\frac{1}{2.5}=0.4\Rightarrow \theta ={\sin }^{-1}(0.4)=23.6^{\circ }.$ KYU ${\sin }^{-1}$: hume ratio $\sin \theta$ pata hai aur angle chahiye — ${\sin }^{-1}$ ("kis angle ka yeh sine hai?") $\sin$ ko undo karta hai. Ek faster rocket ($M=2.5>2$) $M=2$ ke $30^{\circ }$ se thinner cone deta hai.

Step 1 — $M$ nikalo. $\sin \theta =1/M$ ko invert karo: $M=\frac{1}{\sin 25^{\circ }}=\frac{1}{0.4226}=\mathrm{2.366.}$ Step 2 — speed nikalo. Definition se $M=v_{\text{obj}}/v_s$ hai, toh wapas multiply karo: $v_{\text{obj}}=M{v}_s=2.366\times 340\approx 804\text{m/s}.$ KYU do steps: angle sirf ratio $M$ encode karta hai; actual speed recover karne ke liye tumhe $v_s$ wapas introduce karni hogi, woh physical scale jo $M$ ne throw away ki thi.

Cone angle sirf $M$ par depend karta hai, toh $M=1/\sin 25^{\circ }=2.366$ phir se aaya. Lekin $v_{\text{obj}}=M{v}_s=2.366\times 295\approx 698\text{m/s}.$ Lesson: identical cone shape ≠ identical speed. Thandi air → slower sound → wahi $M$ matlab ek lower actual speed.

Step 1 — Mach angle. $\sin \theta =1/1.8=0.5556\Rightarrow \theta =33.75^{\circ }$. Step 2 — cone wall zameen se kahan milta hai. Cone surface flight path se angle $\theta$ banata hai. Right triangle dekho (height $h$ opposite $\theta$ ke, horizontal lag $x$ adjacent) — wall zameen se horizontal distance par milti hai $x=\frac{h}{\tan \theta }=\frac{3000}{\tan 33.75^{\circ }}=\frac{3000}{0.6682}\approx 4490\text{m}$ jet ke peeche. KYU $\tan$: hum opposite side ($h$) ko adjacent side ($x$) se same angle par relate kar rahe hain, aur opposite-over-adjacent $\tan$ hi hota hai — yeh natural tool tab hai jab dono legs matter karein aur hypotenuse na kare. Step 3 — time. Jet ki ground speed $v_{\text{obj}}=M{v}_s=1.8\times 340=612\text{m/s}$ hai. Cone jet se $x$ door trail karta hai; wall tumhak reach karne se pehle jet ko utni door fly karni hogi: $t=\frac{x}{v_{\text{obj}}}=\frac{4490}{612}\approx 7.3\text{s}.$

Jab $M\to 1^{+}$, $\sin \theta =1/M\to 1$, toh $\theta \to 90^{\circ }$ aur $\tan \theta \to \infty$. Tab $t=\frac{h}{v_{s}M\tan \theta }\to \frac{h}{v_s\cdot 1\cdot \infty }=0.$ Ruko — direction check karo: horizontal lag $x=h/\tan \theta \to 0$ hai, toh cone wall essentially vertical, seedha jet ke neeche hai. Boom almost us waqt hi aa jaati hai jab jet seedha upar hota hai. Physical meaning: exactly $M=1$ par "cone" ground track ke perpendicular ek flat wall hai — koi trailing lag nahi hota. (Jis blow-up ki tumhe expect hogi woh $M\to \infty$ par hoti hai: $\theta \to 0$, $\tan \theta \to 0$, $x\to \infty$ — wall infinitely far peeche trail karti hai.)

Warm layer: $M_1=700/340=2.059$, toh $\sin {\theta }_1=1/2.059=0.4857$, ${\theta }_1=29.06^{\circ }$. Cold layer: $M_2=700/295=2.373$, toh $\sin {\theta }_2=1/2.373=0.4214$, ${\theta }_2=24.93^{\circ }$. Reasoning: thandi air mein climb karne se $v_s$ kam hoti hai, toh fixed true speed par $M$ badhta hai, jo $\sin \theta$ kam karta hai, toh cone narrower hoti hai ($29.06^{\circ }\to 24.93^{\circ }$). Rocket ne bilkul speed up nahi ki — medium change hua. Exactly isliye engineers wave drag aur heating predict karte waqt raw speed ki jagah $M$ track karte hain.

Same cone angle ⇒ same $M$: $M_Q=M_P=3.0$ (angle sirf $M$ par depend karta hai). Maan lo P ki sound speed $v_s$ hai; tab $v_{s,Q}=1.10v_s$. $\frac{v_Q}{v_P}=\frac{M_Q{v}_{s,Q}}{M_P{v}_s}=\frac{3.0\cdot 1.10v_s}{3.0\cdot v_s}=\mathrm{1.10.}$ Toh Q ko identical cone trace karne ke liye P se $10\%$ faster fly karna hoga. Insight: shape match karna force karta hai $M$ match karo; faster-sound air mein $M$ match karna force karta hai proportionally higher true speed.

(a) Ek proper trailing cone ko $M>1$ chahiye, yaani $\sin \theta =1/M<1$. $t=0$ par, $M=1.2>1$ already hai, toh cone shuru se hi ek trailing cone ke roop mein exist karta hai (sabhi $t\ge 0$ ke liye $\theta <90^{\circ }$). (b) $t=0$ par: $M=1.2$, $\sin \theta =1/1.2=0.8333$, $\theta =56.44^{\circ }$. $t=4$ par: $M=1.2+0.5(4)=3.2$, $\sin \theta =1/3.2=0.3125$, $\theta =18.21^{\circ }$. (c) Jab $t\to \infty$, $M\to \infty$, toh $\sin \theta \to 0$ aur $\theta \to 0^{\circ }$. Kyunki $M$ monotonically increase karta hai, $\theta$ monotonically shrink karta hai — cone continuously $56.4^{\circ }$ se ek needle ki taraf narrow hoti hai. Yeh $t=0$ par widest hai (slowest $M$) aur kabhi re-widen nahi karti.

Smaller $\theta$ larger $M$ correspond karta hai, toh limiting case $\theta ={\theta }_{\min }=15^{\circ }$ hai: $\sin 15^{\circ }=\frac{1}{M_{\max }}\Rightarrow M_{\max }=\frac{1}{\sin 15^{\circ }}=\frac{1}{0.2588}=\mathrm{3.86.}$ Interpretation: capsule ko $\theta >15^{\circ }$ rakhne ke liye $M\approx 3.86$ se neeche rehna hoga. Us speed se upar Mach cone bahut flat ho jaata hai, exactly woh heating regime jise shield survive nahi kar sakti. (Real re-entry bahut zyada extreme, hypersonic hoti hai — yeh ek simplified design gate hai.)

(a) $M=\frac{1}{\sin 28^{\circ }}=\frac{1}{0.4695}=\mathrm{2.130.}$ (b) $v_{\text{obj}}=M{v}_s=2.130\times 330=703\text{m/s}.$ (c) $x=\frac{h}{\tan \theta }=\frac{1500}{\tan 28^{\circ }}=\frac{1500}{0.5317}=2821\text{m}.$ (d) $t=\frac{x}{v_{\text{obj}}}=\frac{2821}{703}=4.01\text{s}.$ Yeh order kyun: angle → $M$ ($\sin$ invert karo), $M$ → speed ($v_s$ se multiply karo), speed + geometry → lag ($\tan$), lag ÷ speed → time. Har arrow exactly ek earned tool use karta hai.