Exercises — Fairing separation — altitude, dynamic pressure requirements

Kya: parent note ke Step 1 se seedha definition $q=\frac{1}{2}\rho v^2$ apply karo. Yeh formula kyun: $q$ aane wali hawa ki kinetic-energy density hai; iske liye ek density aur ek speed chahiye, dono diye hain. $q=\frac{1}{2}(5\times 10^{-4})(2000{)}^2=0.5\times 5\times 10^{-4}\times 4\times 10^6=1000\text{Pa}.$ Answer: $q=1000\text{Pa}=1\text{kPa}$.

Kya: $\dot{q}=qv$ use karo (parent note ka Step 2). Kyun: heat flux force-flux times velocity hai; kyunki $q$ mein pehle se $\rho v^2$ hai, ek aur factor of $v$ lagane se $\rho v^3/2$ milta hai. $\dot{q}=qv=1000\times 2000=2\times 10^6{\text{W/m}}^2.$ Answer: $\dot{q}=2.0\times 10^6{\text{W/m}}^2$ — 1135 limit se kaafi upar, toh jettison karne ke liye bahut neeche/dense hai.

Kya: exponential density law $\rho ={\rho }_0{e}^{-h/H}$. Kyun: atmosphere har scale height par ek constant factor se patla hota hai, isliye altitude sirf exponent $-h/H$ ke zariye aata hai. Units check: $h=30$ km aur $H=7.5$ km unit km share karte hain, isliye $h/H=4$ ek pure number hai. $\rho =1.2e^{-30/7.5}=1.2e^{-4}=1.2\times 0.0183=0.0220{\text{kg/m}}^3.$ Answer: $\rho \approx 2.2\times 10^{-2}{\text{kg/m}}^3$.

Step 1 — density: altitude sirf hawa ke exponential thinning ke zariye aati hai, toh wahan se shuru karo. $h$ aur $H$ dono km mein hain, $\rho =1.2e^{-100/7.5}=1.2e^{-13.33}$. Kyunki $e^{-13.33}=1.62\times 10^{-6}$, $\rho =1.2\times 1.62\times 10^{-6}=1.94\times 10^{-6}{\text{kg/m}}^3.$ Step 2 — heat flux: $\dot{q}=\frac{1}{2}\rho v^3$ use karo. Yeh formula kyun aur $\frac{1}{2}\rho v^2$ kyun nahi: hum chahte hain ki payload kitni heat absorb kare, jo energy per second per area hai — aerodynamic force nahi. Free-molecular flow mein har molecule impact par apni kinetic energy $\frac{1}{2}{v}^2$ dump karta hai, aur mass arriving per second per area $\rho v$ hai; inhe multiply karo aur $\frac{1}{2}\rho v^3$ milta hai. Pressure $\frac{1}{2}\rho v^2$ par yeh extra factor of $v$ hi fast vehicles ko dangerous banata hai. $\dot{q}=0.5\times 1.94\times 10^{-6}\times (3500{)}^3=0.5\times 1.94\times 10^{-6}\times 4.29\times 10^{10}=4.16\times 10^4{\text{W/m}}^2.$ Decision: $4.16\times 10^4\gg 1135$ → abhi jettison mat karo; aur upar chhadho jahan $\rho$ girta rahe. Answer: $\dot{q}\approx 4.2\times 10^4{\text{W/m}}^2$, bahut zyada hai.

Step 1 — required density: $\dot{q}=\frac{1}{2}\rho v^3$ ko invert karo taaki woh density niklo jis par flux exactly limit ke barabar ho. Is tarah invert kyun: speed $v$ trajectory se set hai, aur flux ceiling mission se fix hai; sirf woh ek variable jo $\dot{q}$ ko neeche la sakta hai woh $\rho$ hai. Toh heating law ko $\rho$ ke liye solve karo $\dot{q}$ ke liye nahi. $\rho =\frac{2{\dot{q}}_{\text{limit}}}{v^3}=\frac{2\times 1135}{(3500{)}^3}=\frac{2270}{4.29\times 10^{10}}=5.29\times 10^{-8}{\text{kg/m}}^3.$ Step 2 — invert density law: $h=H\ln \frac{{\rho }_0}{\rho }$ (result km mein aata hai kyunki $H$ km mein hai). Log kyun: density altitude se $\rho ={\rho }_0{e}^{-h/H}$ ke zariye judi hai, jo ek exponential hai; $h$ ko exponent se nikaalane ka sirf ek hi tarika hai — natural log, jo exponential ko undo karta hai. Bonus mein yeh terrifying density ratio ($2\times 10^7$) ko tame number $16.9$ mein badal deta hai. $h=7.5\ln \frac{1.2}{5.29\times 10^{-8}}=7.5\ln (2.27\times 10^7)=7.5\times 16.94=127\text{km}.$ Answer: $h_{\text{jett}}\approx 127\text{km}$.

Carry-through Δv: $\Delta v=I_{sp}{g}_0\ln \frac{m_0}{m_f}$. $\Delta v=460\times 9.81\times \ln \frac{25000}{7000}=4512.6\times \ln (3.571)=4512.6\times 1.2730=5745\text{m/s}.$ Drop-early Δv: dono masses se $m_{fr}$ ghataao: $\Delta v^{\prime }=4512.6\times \ln \frac{23800}{5800}=4512.6\times \ln (4.103)=4512.6\times 1.4117=6371\text{m/s}.$ Dono masses kyun giraate hain: fairing burn se pehle chali gayi, isliye woh na initial hai na final propellant mass — dono se use hataane par mass ratio $m_0/m_f$ badhti hai, aur bade ratio ka $\ln$ bada hota hai. Answer: gain $=6371-5745\approx 626\text{m/s}$.

Shared density: $\rho =1.2e^{-115/7.5}=1.2e^{-15.33}=1.2\times 2.19\times 10^{-7}=2.63\times 10^{-7}{\text{kg/m}}^3.$ Flux A: ${\dot{q}}_A=0.5\times 2.63\times 10^{-7}\times (3000{)}^3=0.5\times 2.63\times 10^{-7}\times 2.7\times 10^{10}=3.55\times 10^3{\text{W/m}}^2.$ Flux B: ${\dot{q}}_B=0.5\times 2.63\times 10^{-7}\times (5000{)}^3=0.5\times 2.63\times 10^{-7}\times 1.25\times 10^{11}=1.64\times 10^4{\text{W/m}}^2.$ Ratio: ${\dot{q}}_B/{\dot{q}}_A=(5000/3000{)}^3=(5/3{)}^3=\mathrm{4.63.}$ Interpretation: ussi 115 km par, B payload ko sirf $\dot{q}\propto v^3$ ki wajah se 4.6× zyada heat karta hai. Dono 1135 se upar hain, lekin B ko safe hone se pehle noticeably zyada upar jaana padega. Altitude sirf $\rho$ ka proxy hai; asli criterion $\rho v^3$ hai. Answer: ${\dot{q}}_A\approx 3.6\times 10^3$, ${\dot{q}}_B\approx 1.6\times 10^4{\text{W/m}}^2$, ratio $=4.63$.

Key trick: difference har constant term ko khatam kar deta hai. Kyunki $\ln \frac{{\rho }_0{v}^3/2}{\dot{q}}=\ln ({\rho }_0/2\dot{q})+3\ln v$, sirf $v$-dependence $3\ln v$ hai. $\Delta h=H\cdot 3(\ln v_2-\ln v_1)=3H\ln \frac{v_2}{v_1}=3\times 7.5\times \ln \frac{5000}{3000}.$ $\Delta h=22.5\times \ln (1.667)=22.5\times 0.5108=11.5\text{km}.$ $3\ln$ kyun: log ke andar $v^3$ ek coefficient $3$ bana deta hai $\ln v$ ke saamne — wohi wajah jisse L3.1 mein flux ratio cube tha. Answer: faster vehicle ko approximately 11.5 km zyada upar jettison karna hoga.

Kya hota hai: log ke andar, ${\rho }_0{v}^3/2\to 0$, toh $\ln (\text{kuch}\to 0^{+})\to -\infty$, jo $h_{\text{jett}}\to -\infty$ deta hai. Physical reading: ek vehicle jo barely move kar raha hai woh har molecule par almost koi kinetic energy deposit nahi karta, isliye uska heat flux limit se har altitude par — yahan tak ki ground par bhi neeche hoga. Ek negative "required" altitude ka matlab bas yeh hai ki "constraint pehle se satisfy hai; koi altitude margin ki zaroorat nahi." Formula sahi hai lekin iska output clamp karna hoga: $h_{\text{jett}}=\max (0,\dots )$. Answer: $h_{\text{jett}}\to -\infty$; interpret karo "har jagah safe hai," $h\ge 0$ tak clamp karo.

(a) Altitude. Required density: Pehle density kyun: flux limit aur speed dono fix hain, toh — bilkul L2.2 ki tarah — sirf woh knob jo $\dot{q}$ ko limit tak la sakta hai woh density hai. Heating law ko $\rho$ ke liye solve karo, phir us density ko altitude mein convert karo. $\rho =\frac{2\times 1135}{(4200{)}^3}=\frac{2270}{7.409\times 10^{10}}=3.064\times 10^{-8}{\text{kg/m}}^3.$ $h=7.5\ln \frac{1.2}{3.064\times 10^{-8}}=7.5\ln (3.916\times 10^7)=7.5\times 17.483=131.1\text{km}.$ Log phir exponential atmosphere ko undo karta hai taaki hum seedha altitude padh sakein (km mein, kyunki $H$ km mein hai). (b) Δv gain. $I_{sp}{g}_0=455\times 9.81=4463.6\text{m/s}$. Do mass ratios compare kyun: rocket equation kehta hai Δv sirf ratio $m_0/m_f$ par depend karta hai. Fairing carry karne par dono masses high rehti hain; burn se pehle drop karne par dono $m_{fr}$ se girte hain, jo ratio badhata hai aur isliye achievable Δv bhi. Gain dono ke beech ka difference hai. Carry-through: $\Delta v=4463.6\ln \frac{22000}{6500}=4463.6\times 1.2192=5442\text{m/s}.$ Drop early: $\Delta v^{\prime }=4463.6\ln \frac{20900}{5400}=4463.6\times 1.3527=6039\text{m/s}.$ Gain: $6039-5442\approx 597\text{m/s}$. Answer: $\approx 131\text{km}$ par jettison karo; Δv gain $\approx 597\text{m/s}$.

Flux $\propto \rho$ yahan kyun scale karta hai: heating law $\dot{q}=\frac{1}{2}\rho v^3$ hai. Fixed speed par, $v^3$ ek constant multiplier hai, isliye $\dot{q}$ directly $\rho$ ke proportional hai — density aadhi karo, flux aadhi ho jaati hai. Yeh hume sirf unki densities compare karke do altitudes compare karne deta hai. 121 km par density: 131.1 km case se ratio sirf altitude drop use karta hai (dono altitudes km mein, isliye exponent unitless hai): $\frac{\rho (121)}{\rho (131.1)}=e^{(131.1-121)/7.5}=e^{10.1/7.5}=e^{1.347}=\mathrm{3.85.}$ Kyunki $\dot{q}\propto \rho$ at fixed $v$, flux limit ka 3.85× hai, yaani $\approx 4370{\text{W/m}}^2$. Reading: sirf 10 km pehle drop karne par heating almost chaar guna ho jaati hai — exponential atmosphere constraint ko sharp banata hai. Extra Δv payload ko cook karne ki keemat par nahi aata. Answer: flux $\approx 3.85\times$ limit $\approx 4.4\times 10^3{\text{W/m}}^2$ — forbidden.

Do competing effects ko jettison altitude $h$ ke functions ki tarah set up karo:

1. Δv benefit $h$ mein monotonically decreasing hai. Δv $m_{fr}$ shed karne se save hota hai: jitna pehle (lower $h$) drop karo, utni der tak halka fly karte ho, isliye $\Delta v_{\text{saved}}(h)$ $h$ badhne par decreases karta hai. Koi interior maximum nahi hai — benefit sirf chhota $h$ chahta hai.
2. Heating constraint ek hard floor hai. $\dot{q}(h)=\frac{1}{2}{\rho }_0{e}^{-h/H}{v}^3$ $h$ ke saath decreases karta hai. Payload sirf wahan survive karta hai jahan $\dot{q}(h)\le {\dot{q}}_{\text{limit}}$, yaani $h\ge h_{\text{jett}}$.

Combine karo: hum "wasted Δv" minimize karte hain (equivalently Δv saved maximize karte hain) subject to $h\ge h_{\text{jett}}$. Objective $h$ ko neeche push karta hai; constraint use $h=h_{\text{jett}}$ par rokti hai. Ek half-line $[h_{\text{jett}},\infty )$ par strictly monotone objective ka minimiser endpoint par baitha hota hai. Isliye optimum exactly $h_{\text{opt}}=h_{\text{jett}}$ hai, jahan $\dot{q}={\dot{q}}_{\text{limit}}$. Boundary kyun: jab bhi koi benefit strictly ek direction favour kare aur safety use block kare, sabse best yahi hai ki seedha wall se lag jaao. Pehle drop karna safety violate karta hai; baad mein drop karna bina wajah Δv throw away karta hai. Equality $\dot{q}={\dot{q}}_{\text{limit}}$ hi optimum hai. Yeh Ascent Trajectory Optimization se connect karta hai: poora trajectory is tarah shape kiya jaata hai taaki yeh boundary jitni jaldi climb allow kare utni jaldi reach ho.

Required density fixed hai (yeh sirf $v$ aur flux limit par depend karta hai): $\rho =3.064\times 10^{-8}{\text{kg/m}}^3$ L4.1 se. Sirf inversion $h=H\ln ({\rho }_0/\rho )$ shift karti hai, aur pure-number log $\ln ({\rho }_0/\rho )=17.483$ constant hai. $h_{\min }=7.0\times 17.483=122.4\text{km},h_{\max }=8.0\times 17.483=139.9\text{km}.$ Spread: approximately $122\text{km}$ se $140\text{km}$ tak — ek $17.5\text{km}$ band, jo parent note mein quoted "110–140 km" range se match karta hai. Conservative choice: bada $H$ matlab atmosphere zyada altitude tak denser rehta hai, isliye usi low density tak pahunchne ke liye zyada upar wait karna padta hai. $H=8.0$ km ke liye design karna (jettison $\approx 140$ km par) "puffy" atmosphere wale din bhi safety guarantee karta hai; payload kabhi over-heated nahi hoga, thodi extra Δv ki keemat par. Safety dominate karti hai → high end choose karo. Answer: $h_{\text{jett}}\in [122,140]\text{km}$; conservatively $\approx 140$ km ke liye design karo.

(a) Altitude. Required density: $\rho =\frac{2\times 1135}{(4600{)}^3}=\frac{2270}{9.7336\times 10^{10}}=2.332\times 10^{-8}{\text{kg/m}}^3.$ Density law invert karo (result km mein): $h=7.5\ln \frac{1.2}{2.332\times 10^{-8}}=7.5\ln (5.146\times 10^7)=7.5\times 17.756=133.2\text{km}.$ (b) Δv gain. $I_{sp}{g}_0=462\times 9.81=4532.2\text{m/s}$. Carry-through: $\Delta v=4532.2\ln \frac{30000}{8000}=4532.2\times 1.3218=5991\text{m/s}.$ Drop early: $\Delta v^{\prime }=4532.2\ln \frac{28500}{6500}=4532.2\times 1.4784=6701\text{m/s}.$ Gain: $6701-5991\approx 710\text{m/s}$. (c) Decision: fairing exactly $\approx 133$ km par jettison karo, jis pal $\dot{q}$ 1135 W/m² tak girta hai — woh earliest altitude jo thermally safe hai, jo L5.1 ke hisaab se Δv-optimal point bhi hai. Answer: $h\approx 133\text{km}$; Δv gain $\approx 710\text{m/s}$; flux-limit boundary par drop karo.

L1: $q=\frac{1}{2}\rho v^2$ aur $\dot{q}=qv$ mein plug in karo. L2: altitude ke liye rearrange karo, rocket equation apply karo. L3: $\dot{q}\propto v^3$ speed ko dominate karta hai; $v\to 0$ limit handle karo. L4: exponential density constraint ko sharp banata hai. L5: optimum exactly heating boundary par hai aur tum conservative (large-$H$) extreme ke liye design karte ho.