Exercises — Atmospheric drag — exponential atmosphere model, orbit decay

KYA: $B$ drag-area per unit mass hai, $B=\frac{C_{D}A}{m}$. KYUN: yeh teen "kitna draggy hai" wale numbers ko ek mein bundle kar deta hai, taaki decay formulas ko sirf $B$ chahiye. $B=\frac{0.30}{15}=0.020{\text{m}}^2/\text{kg}$

Tool choice: Hume ek magnitude chahiye, isliye $a_{drag}=\frac{1}{2}B\rho v^2$ use karo (yeh $v^2$ wala form hai, koi vector ki zaroorat nahi). $a_{drag}=\frac{1}{2}(0.020)(5\times 10^{-12})(7670{)}^2$ $=\frac{1}{2}(0.020)(5\times 10^{-12})(5.883\times 10^7)\approx 2.94\times 10^{-6}{\text{m/s}}^2$ Yeh lagbhag $3\times 10^{-7}g$ hai — per second invisible, lekin mahino mein tabah kar deta hai.

Tool: $\rho (h)={\rho }_0{e}^{-(h-h_0)/H}$. Neeche jaane ka matlab $h<h_0$ hai, isliye $(h-h_0)$ negative hoga aur density badhegi.

• $50$ km neeche $=1H$: factor $e^{+1}\approx 2.718$ — density $\approx 2.72$ se multiply hoti hai.
• $150$ km neeche $=3H$: factor $e^{+3}\approx 20.09$ — density $\approx 20.1$ se multiply hoti hai. Reading: har scale height neeche jaane par $\rho$ ko $e$ se multiply karo. Teen scale heights = 20× ghana soup.

Tool: $\frac{da}{dt}=-B\rho \sqrt{\mu a}$ (orbit ke shrinkage ki rate). Pehle root: $\mu a=(3.986\times 10^{14})(6.778\times 10^6)=2.702\times 10^{21}$ $\sqrt{\mu a}=5.198\times 10^{10}{\text{m}}^2/\text{s}$ Phir: $\frac{da}{dt}=-(0.020)(5\times 10^{-12})(5.198\times 10^{10})=-5.198\times 10^{-3}\text{m/s}$ Per day convert karo ($86400$ s): $-5.198\times 10^{-3}\times 86400\approx -449\text{m/day}$ To orbit roughly $0.45$ km har roz drop hoti hai — is altitude par.

Tool: $H=\frac{k_{B}T}{m_{mol}g}$ — hydrostatic equilibrium plus ideal gas se derived (parent note dekho). $H=\frac{(1.380\times 10^{-23})(1000)}{(2.66\times 10^{-26})(8.7)}=\frac{1.380\times 10^{-20}}{2.314\times 10^{-25}}\approx 5.96\times 10^4\text{m}\approx 60\text{km}$ Reading: garam ($T$ bada) aur halki ($m_{mol}$ chhoti) air phuul jaati hai — mota scale height, exactly isliye thermosphere itni extended hai.

Tools: $\epsilon =-\frac{\mu }{2a}$ aur $v=\sqrt{\frac{\mu }{a}}$ (Two-Body Problem / Vis-Viva Equation se). $\epsilon =-\frac{3.986\times 10^{14}}{2(6.778\times 10^6)}=-2.941\times 10^7\text{J/kg}$ $v=\sqrt{\frac{3.986\times 10^{14}}{6.778\times 10^6}}=\sqrt{5.882\times 10^7}\approx 7670\text{m/s}$ Problem 1.2 mein humne jo speed assume ki thi usse consistent hai — good.

Step 1 — power = force · velocity. Drag acceleration hai ${\vec{a}}_{drag}=-\frac{1}{2}B\rho v\vec{v}$ (magnitude $\frac{1}{2}B\rho v^2$, direction $-\hat{v}$). Power per unit mass: $\frac{d\epsilon }{dt}={\vec{a}}_{drag}\cdot \vec{v}=-\frac{1}{2}B\rho v(\vec{v}\cdot \vec{v})=-\frac{1}{2}B\rho v\cdot v^2=-\frac{1}{2}B\rho v^3.$ KYUN negative: ${\vec{a}}_{drag}$, $\vec{v}$ ke anti-parallel hai, isliye dot product $-|\vec{a}||\vec{v}|$ hai — energy hamesha jaati hai. Step 2 — $a$ ke through cross-check. $\epsilon =-\mu /2a$ differentiate karo: $\frac{d\epsilon }{dt}=\frac{\mu }{2a^2}\frac{da}{dt}.$ $\frac{da}{dt}=-B\rho \sqrt{\mu a}$ substitute karo: $\frac{d\epsilon }{dt}=\frac{\mu }{2a^2}(-B\rho \sqrt{\mu a})=-\frac{B\rho }{2}\frac{\mu \sqrt{\mu a}}{a^2}=-\frac{1}{2}B\rho \frac{{\mu }^{3/2}}{a^{3/2}}.$ Kyunki $v=\sqrt{\mu /a}$ hai, hame milta hai $v^3=(\mu /a{)}^{3/2}={\mu }^{3/2}/a^{3/2}$, isliye yeh exactly $-\frac{1}{2}B\rho v^3$ hai. ✓ Dono formulas agree karte hain.

KYA dominate karta hai: $\left|\frac{da}{dt}\right|=B\rho \sqrt{\mu a}$ mein, density $\rho$ exponentially change hoti hai jabki $\sqrt{\mu a}$ sirf ek gentle square root ki tarah change hota hai. Isliye do altitudes par ratio essentially density ratio hai. KYUN: altitude $\Delta h$ drop karne par density $e^{+\Delta h/H}$ se multiply hoti hai (Problem 1.3 ke sign logic se).

• Ek full scale height ($\Delta h=H$): factor $e^1\approx 2.72$.
• $\Delta h=30$ km, $H=60$ km: factor $e^{30/60}=e^{0.5}\approx 1.65$. Reading: har $60$ km neeche jaane par shrink rate lagbhag triple ho jaati hai, aur thirty km mein hi yeh $65\%$ boost ho jaati hai. Yahi compounding "runaway" hai — jitna neeche girte ho, utni tez girte ho.

KAHAN density sabse badi hai: perigee sabse neeche ka point hai, isliye $\rho (r_p)\gg \rho (r_a)$ — exponentially bada. Drag magnitude $\propto \rho v^2$ hai, aur $v$ bhi perigee par sabse bada hota hai, isliye drag "kick" overwhelmingly brief perigee pass mein concentrated hai. SHAPE ke saath kya hota hai: perigee par ek drag impulse energy wahan remove karta hai jahan satellite sabse gehra hai, jo sabse strongly apogee (door wala side) ko neecha karta hai, jabki perigee barely move karta hai. Kai orbits mein apogee perigee ki taraf walk down karta hai — ellipse circularize ho jaata hai. Dekho Perturbations in Orbital Mechanics. Phir: jab lagbhag circular aur low ho jaata hai, har point thick air mein hota hai, $\rho$ uniformly high hoti hai, aur final plunge follow karta hai.

$\epsilon =-\mu /2a$ se start karo. Magnitude ka natural log lo: $\ln |\epsilon |=\ln (\mu /2)-\ln a$. Time mein differentiate karo: $\frac{1}{\epsilon }\frac{d\epsilon }{dt}=-\frac{1}{a}\frac{da}{dt}.$ Interpretation: $\frac{da}{dt}<0$ (orbit shrink hoti hai), isliye right side positive hai; lekin $\epsilon <0$ hai, isliye positive $\frac{1}{\epsilon }\frac{d\epsilon }{dt}$ ka matlab hai $\frac{d\epsilon }{dt}<0$ — energy decrease hoti hai. Consistent: jab $a$ girta hai, $\epsilon$ aur negative hoti jaati hai (gehra well), aur speed badhti hai. Yeh drag paradox ek clean identity ke roop mein stated hai. Dekho Orbital Energy & Specific Mechanical Energy.

Local estimate: constant $449$ m/day par, $t=\frac{\Delta a}{|da/dt|}=\frac{50000\text{m}}{449\text{m/day}}\approx 111\text{days}$ pehle $50$ km girne mein. KYUN true lifetime chhoti hai: rate constant nahi hai. Jab $a$ girta hai, altitude girta hai, $\rho$ exponentially badhta hai (Problem 3.2), isliye $|da/dt|$ badhta hai aur har successive $50$ km kam time leta hai. Shrinking times ki sequence ko sum karne se total $111\text{days}\times (\text{50-km steps ki number})$ se bahut kam hota hai. Constant-rate number sirf pehle step ke liye upper-bound feel deta hai.

Tool: $\frac{da}{dt}=-B\rho \sqrt{\mu a}$. Same orbit ⇒ same $\rho$, $\mu$, $a$. Isliye decay rates ka ratio sirf $B$ ka ratio hai: $\frac{(da/dt{)}_Y}{(da/dt{)}_X}=\frac{B_Y}{B_X}=\frac{0.5}{0.005}=100.$ Y instantaneously $100\times$ tez decay karta hai, isliye (first order mein) Y ki lifetime roughly X ki $1/100$ hogi. Reading: big-area/light-mass bodies (sails, balloons) natural deorbit devices hain — yeh exactly drag-sail deorbit principle hai. Dekho Reentry Aerodynamics & Ballistic Coefficient.

$\frac{da}{dt}=-B\rho (a)\sqrt{\mu a}$ se, variables separate karo: $dt=\frac{da}{-B\rho (a)\sqrt{\mu a}}\Rightarrow t={\int }_{a_1}^{a_2}\frac{-da}{B{\rho }_1{e}^{-(a-a_1)/H}\sqrt{\mu a}}.$ Kyunki $a_2<a_1$ (decaying) hai, limits flip karo taaki positive ho: $t={\int }_{a_2}^{a_1}\frac{e^{+(a-a_1)/H}}{B{\rho }_1\sqrt{\mu a}}da.$ Kaunsa end dominate karta hai: integrand $e^{+(a-a_1)/H}$ carry karta hai, jo $a=a_1$ ke paas sabse bada hota hai (uupar, thin air wali start) jahan $\rho$ sabse chhota hota hai, isliye $da$ per unit $dt$ wahan sabse bada hota hai. Lagbhag poori lifetime top few scale heights mein slowly bleed karne mein spend hoti hai. Jab tak satellite deep hota hai, $\rho$ huge hai, $dt$ tiny hai, aur final kilometres flash ho jaate hain. Yeh "long quiet decay, sudden fiery end" picture se match karta hai.

• (a) $\rho \to 0$: $\frac{da}{dt}\to 0$ — koi air nahi, koi decay nahi. Orbit eternal hai, exactly frictionless Two-Body Problem limit. ✓
• (b) $B\to 0$: $\frac{da}{dt}\to 0$ — zero drag-area-per-mass wala body (infinite density ka cannonball) atmosphere ignore karta hai. ✓ Isliye heavy compact objects reentry survive karte hain: tiny $B$.
• (c) $a\to R_{\oplus }$: $\sqrt{\mu a}$ finite rehta hai, lekin model ka $\rho (a)={\rho }_1{e}^{+(a_1-a)/H}$ blow up ho jaata hai jab $a$ girta hai — decay rate diverge ho jaata hai. Physically: satellite ab orbit nahi kar raha, yeh dense air mein plow kar raha hai aur reenter karta hai. Circular-orbit assumption ($v=\sqrt{\mu /a}$) toot gayi hai; tumhe ballistic-reentry model par switch karna hoga. Divergence model ka polite tarika hai tumhe batane ka ki yeh expire ho gaya. Dekho Reentry Aerodynamics & Ballistic Coefficient.

(i) Speed: $v=\sqrt{\mu /a_1}=\sqrt{3.986\times 10^{14}/6.778\times 10^6}\approx 7670\text{m/s}$ (jaisa 2.3 mein hai). (ii) Initial rate: $|da/dt|=B{\rho }_1\sqrt{\mu a_1}=(0.02)(5\times 10^{-12})(5.198\times 10^{10})\approx 5.20\times 10^{-3}\text{m/s}\approx 449\text{m/day}$ (jaisa 2.1 mein hai). (iii) $\Delta h=120$ km $=2H$ ke baad: density $e^2\approx 7.389$ se badhti hai: $\rho =5\times 10^{-12}\times e^2\approx 3.69\times 10^{-11}{\text{kg/m}}^3.$ Naya $a\approx 6.658\times 10^6$ m, isliye $\sqrt{\mu a}\approx \sqrt{(3.986\times 10^{14})(6.658\times 10^6)}=\sqrt{2.654\times 10^{21}}\approx 5.152\times 10^{10}$. Phir: $|da/dt|=(0.02)(3.69\times 10^{-11})(5.152\times 10^{10})\approx 3.80\times 10^{-2}\text{m/s}\approx 3287\text{m/day}.$ Rate $\sim 0.45$ km/day se $\sim 3.3$ km/day tak jump kar gayi — roughly $7\times$ tez, essentially $e^2$ density factor ($\sqrt{\mu a}$ barely moved). (iv) Ek sentence: kyunki decay rate khud exponentially accelerate hota hai jab orbit sink hoti hai, ek single boost sirf clock reset karta hai — repeated reboosts ke bina satellite hamesha thicken hoti air mein slide back ho jaata hai aur reenter karta hai. Dekho Perturbations in Orbital Mechanics.