Exercises — Reentry mechanics — ballistic coefficient β = m - (C_D A)

KYA: $\beta =m/(C_{D}A)$ mein substitute karo. KYUN: $\beta$ mass divided by "aerodynamic footprint" $C_{D}A$ hai — aur kuch compute nahi karna. $\beta =\frac{40}{0.2\times 0.01}=\frac{40}{0.002}=20000{\text{ kg/m}}^2.$ Ek chota, bhaari dart → bahut bada $\beta$ → ek "cannonball" jo gehraai mein ghus jaayega.

KYA: dono numbers compare karo. KYUN: chota $\beta$ = apne size ke hisaab se halka = "beach ball" jise hawa jaldi aur oonche rok leti hai; bada $\beta$ = "bowling ball" jo gehraai mein ghus jaata hai. Body P (chota $\beta =45$) feather hai — zyada oonche ruka, gentle heating. Body Q cannonball hai.

KYA: units track karo. $C_D$ dimensionless hai, isliye yeh kuch contribute nahi karta. $A$ ${\text{m}}^2$ mein hai, $m$ $\text{kg}$ mein. $[\beta ]=\frac{\text{kg}}{(\text{dimensionless})\cdot {\text{m}}^2}=\frac{\text{kg}}{{\text{m}}^2}.✓$

KYA: $v(h)=v_e\exp \left(-\frac{\rho H}{2\beta \sin \gamma }\right)$ use karo. Exponential KYUN? Kyunki density exponentially badhti hai jab tum neeche aate ho, aur ek exponential ka drag integral khud bhi exponential hota hai — yahi Allen–Eggers ne parent note mein integrate kiya tha. $\sin 90^{\circ }=1$. Exponent: $-\frac{\rho H}{2\beta \sin \gamma }=-\frac{1.0\times 7000}{2\times 6000\times 1}=-\frac{7000}{12000}=-\mathrm{0.5833.}$ $v=7000e^{-0.5833}=7000\times 0.558\approx 3906\text{ m/s}.$ Isne apni speed ka lagbhag 44% khoya hai — ek middling $\beta$, aadha feather aadha cannonball.

KYA: $a_{\max }=\frac{v_e^2\sin \gamma }{2eH}$ use karo. $\beta$ KYUN nahi hai? Parent note ne dikhaya tha ki jab tum density par maximise karte ho toh $\beta$ cancel ho jaata hai — peak-g ki magnitude sirf speed aur angle par depend karti hai. $a_{\max }=\frac{(6000{)}^2\times \sin 40^{\circ }}{2\times 2.71828\times 7000}=\frac{3.6\times 10^7\times 0.6428}{38056}\approx 608{\text{ m/s}}^2.$ G's mein: $608/9.81\approx 62g$.

KYA: parent derivation se peak condition ${\rho }^{\ast }=\frac{\beta \sin \gamma }{H}$ use karo. KYUN: $da/d\rho =0$ set karne par yeh ek special density milti hai jahan "zyada hawa" aur "speed bacha nahi" ka trade-off balance hota hai. ${\rho }^{\ast }=\frac{6000\times 1}{7000}=0.857{\text{ kg/m}}^3.$ Yeh sea-level density ka roughly $70\%$ hai — yeh bhaari body deep aur neeche peak karti hai. (Units: $\beta$ ${\text{kg/m}}^2$ mein hai aur $H$ $\text{m}$ mein, isliye $\beta /H$ ke units ${\text{kg/m}}^3$ hain — ek density, jaisa hona chahiye.)

KYA: dono speeds ka ratio $v_B/v_A$ compute karo. Setup KYUN: $v(h)$ ka exponent $-\rho H/(2\beta \sin \gamma )=-N/\beta$ hai, jahan $N=\rho H/(2\sin \gamma )$ $\beta$ ke alawa sab kuch collect karta hai. Kyunki dono bodies ek given altitude par same $\rho$, $H$ aur $\gamma$ dekhti hain, woh same $N$ share karti hain; sirf $\beta$ alag hai. Hum us altitude par evaluate karne ko keh rahe hain jahan $N=3500{\text{ kg/m}}^2$ hai. $\frac{v_A}{v_e}=e^{-N/{\beta }_A}=e^{-3500/500}=e^{-7}=9.12\times 10^{-4},$ $\frac{v_B}{v_e}=e^{-N/{\beta }_B}=e^{-3500/5000}=e^{-0.7}=\mathrm{0.497.}$ Ratio $v_B/v_A=e^{-0.7}/e^{-7}=e^{6.3}\approx 545$. Bhaari body B us altitude par lagbhag 545× faster hai — dramatic, kyunki $\beta$ ek exponent mein baitha hai.

KYA: dono peak altitudes subtract karo. KYUN: $h^{\ast }$ $\beta$ par sirf $\ln (1/\beta )$ ke through depend karta hai, isliye factor-of-10 change ek fixed additive shift hai. $h_A^{\ast }-h_B^{\ast }=H\ln \frac{{\rho }_0}{{\rho }_A^{\ast }}-H\ln \frac{{\rho }_0}{{\rho }_B^{\ast }}=H\ln \frac{{\rho }_B^{\ast }}{{\rho }_A^{\ast }}=H\ln \frac{{\beta }_B}{{\beta }_A}.$ $=7000\times \ln 10=7000\times 2.3026\approx 16118\text{ m}.$ B lagbhag 16.1 km neeche A se peak karta hai.

Neeche ka chalkboard sketch altitude $h$ (km, vertical axis) ko air density $\rho$ (kg/m³, horizontal axis) ke against plot karta hai. Pale-yellow curve exponential atmosphere $\rho ={\rho }_0{e}^{-h/H}$ hai: ground ke paas dense (right), upar thin (left). Do dashed lines har body ki peak-deceleration density ${\rho }^{\ast }=\beta \sin \gamma /H$ mark karti hain — body A (blue, chota $\beta$) oonche low density par peak karta hai; body B (pink, 10× $\beta$) 10× density par, kaafi neeche peak karta hai. Dono ke beech ka vertical double-arrow exactly $H\ln 10=16.1\text{ km}$ shift hai jo humne abhi compute kiya, yeh reinforce karta hai ki $\beta$ ka ek decade ek "log-decade" altitude cost karta hai.

KYA: peak-g ko heating se alag karo. KYUN: $a_{\max }=v_e^2\sin \gamma /(2eH)$ mein $\beta$ nahi hai, isliye peak-g identical hai. Lekin heating deceleration point par air density ke saath scale karti hai, aur peak ki altitude, $h^{\ast }=H\ln ({\rho }_0/{\rho }^{\ast })$, $\beta$ par DEPEND KARTI HAI. High-$\beta$ warhead dense air mein deep peak karta hai ⇒ equal g ke bawajood kaafi zyada heating. Jhooth — same g, bahut alag heat load. (Dekho Aerodynamic heating and thermal protection systems.)

KYA: force balance rearrange karo aur $\beta$ identify karo. KYUN: wahi combination $m/(C_{D}A)=\beta$ jo reentry mein aaya tha yahan bhi fir se aata hai — parent note ne Terminal velocity ke saath is link ko flag kiya tha. $\frac{1}{2}\rho v_t^2{C}_{D}A=mg$ se shuru karo. $C_{D}A$ se divide karo: $\frac{1}{2}\rho v_t^2=\frac{mg}{C_{D}A}=\beta g\Rightarrow v_t^2=\frac{2\beta g}{\rho }\Rightarrow v_t=\sqrt{\frac{2\beta g}{\rho }}.$ Numbers: $v_t=\sqrt{\frac{2\times 45\times 9.81}{1.0}}=\sqrt{882.9}\approx 29.7\text{ m/s}.$ Ek halki body ~30 m/s par settle karti hai — parachute jaisi. Isliye low-$\beta$ = gentle.

KYA: zaroori $\beta$ ke liye ${\rho }^{\ast }=\beta \sin \gamma /H$ ko invert karo, phir $A$ ke liye $\beta =m/(C_{D}A)$ ko invert karo. KYUN: hum physics ulta chala rahe hain — ek desired outcome (peak altitude) se ek design variable (area) tak. Step 1 — zaroori $\beta$: $\beta =\frac{{\rho }^{\ast }H}{\sin \gamma }=\frac{0.05\times 7000}{\sin 30^{\circ }}=\frac{350}{0.5}=700{\text{ kg/m}}^2.$ Step 2 — zaroori $A$: $A=\frac{m}{C_D\beta }=\frac{4000}{1.4\times 700}=\frac{4000}{980}\approx 4.08{\text{ m}}^2.$ Ek ~4 m² heat shield peak ko achhi tarah se thin ${\rho }^{\ast }=0.05$ par, oonche rakhta hai.

KYA: pehle aur baad ka $\beta$ compute karo; ratio lo. KYUN: fixed mass par $\beta \propto 1/(C_{D}A)$, isliye ratio sirf footprints ka inverse ratio hai. ${\beta }_{\text{before}}=\frac{5200}{16.9}=307.7,{\beta }_{\text{after}}=\frac{5200}{2.6}=2000.$ $\frac{{\beta }_{\text{after}}}{{\beta }_{\text{before}}}=\frac{16.9}{2.6}=\mathrm{6.5.}$ $\beta$ 6.5× jump karta hai — body achanak kaafi zyada "cannonball" ban jaati hai aur plunge karti hai. (Yeh Skip vs ballistic vs lifting reentry ki ballistic side hai.)

KYA: $a$ ko $\rho$ ke respect mein differentiate karo, zero set karo, back-substitute karo. Differentiate KYUN? Smooth curve ka peak wahan hai jahan uski slope zero hai — calculus woh tool hai jo "sabse bura instant" bina guess kiye dhundh leta hai (dekho Newton's second law jahan se $a$ aaya). Write $a(\rho )=C\rho e^{-k\rho }$ jahan $C=v_e^2/(2\beta )$ aur $k=H/(\beta \sin \gamma )$ hai. Product rule: $\frac{da}{d\rho }=C(e^{-k\rho }-k\rho e^{-k\rho })=C{e}^{-k\rho }(1-k\rho ).$ Exponential kabhi zero nahi hota, isliye $\frac{da}{d\rho }=0\Rightarrow 1-k{\rho }^{\ast }=0\Rightarrow {\rho }^{\ast }=1/k=\frac{\beta \sin \gamma }{H}.$ Back-substitute karo. Note karo $k{\rho }^{\ast }=1$, isliye $e^{-k{\rho }^{\ast }}=e^{-1}$: $a_{\max }=C{\rho }^{\ast }{e}^{-1}=\frac{v_e^2}{2\beta }\cdot \frac{\beta \sin \gamma }{H}\cdot \frac{1}{e}=\frac{v_e^2\sin \gamma }{2eH}.$ $C$ mein $\beta$ cancel ho jaata hai ${\rho }^{\ast }$ wale $\beta$ se. Peak-g ko $\beta$ nahi dikhta. $■$

KYA: $\sin \gamma$ factor ko apne poore domain $0^{\circ }\le \gamma \le 90^{\circ }$ mein evaluate karo. KYUN: contract kehta hai har case cover karo — $\sin \gamma$ 0 se 1 tak jaata hai, isliye peak-g 0 se apne max tak jaata hai.

• $\gamma =90^{\circ }$ (vertical dive): $\sin \gamma =1$, sabse bada peak-g $=v_e^2/(2eH)$. Brutal, brief, deep. Warhead territory.
• $0^{\circ }<\gamma <90^{\circ }$ (typical entry): $\sin \gamma$ 0 aur 1 ke beech proportionally peak-g scale karta hai. Capsules ek chota $\gamma$ (~5–7°) choose karte hain taaki g's survivable rahein.
• $\gamma \to 0^{\circ }$ (grazing, upar se approach karte hue): $\sin \gamma \to 0$, isliye $a_{\max }\to 0$. Physically body barely atmosphere mein dip karti hai aur bahut gently slow hoti hai — lekin yeh skip regime hai: bahut shallow aur vehicle atmosphere se wapas bounce ho jaata hai descent commit karne ki jagah (dekho Skip vs ballistic vs lifting reentry). Allen–Eggers straight-line assumption bhi yahan toot jaati hai, kyunki path ab approximately straight nahi hai — flight path curve karti hai aur vehicle re-ascend kar sakta hai. Isliye formula ka "$a_{\max }\to 0$" limit ke taur par sahi hai lekin underlying model trustworthy rehna band ho jaata hai.
• $\gamma =0$ exactly (fully degenerate): descent rate $dh/dt=-v\sin \gamma =0$ hai, isliye altitude kabhi kam nahi hoti — vehicle horizontally skim karta hai aur actually reenter hi nahi karta. Formula sahi taur par zero deceleration return karta hai ($a_{\max }=0$), jo honest answer hai: koi downward motion nahi toh dense air mein koi plunge nahi jo tumhe decelerate kare. Model simply ek aisi body par applicable nahi hai jo kabhi descend hi nahi karti.

KYA: $\beta$ ko uske dono extremes par bhejo, $\sin \gamma >0$ fixed aur positive rakho. KYUN: limits woh "ideal" endpoints reveal karte hain jinke beech design rehta hai.

• $\beta \to \infty$ (infinitely dense cannonball): exponent $=-\rho H/(2\beta \sin \gamma )\to 0^{-}$, isliye $v\to v_e{e}^0=v_e$. Body hawa ko bilkul ignore karti hai aur apni poori entry speed neeche tak maintain karti hai.
• $\beta \to 0^{+}$ (infinitely light feather): $\sin \gamma >0$ ke saath exponent $\to -\infty$, isliye $v\to v_e{e}^{-\infty }=0$. Body atmosphere ki pehli hint se hi practically rok li jaati hai. Real vehicles in dono poles ke beech hote hain; $\beta$ choose karna yeh choose karna hai ki tum is spectrum par kahan land karte ho.

Peak-g formula mein $\beta$ nahi hai; $\beta$ sirf altitude/heating set karta hai; $\beta$ ek exponent mein rehta hai isliye retained speed par uska effect exponential hai, linear nahi. In teeno mein se kaun sa tumne is page par galat kiya? ::: Us level ka mistake callout dobara dekho.