Exercises — Aerobraking — gradual orbit lowering using atmospheric drag

Apoapsis girta hai; periapsis essentially fixed rehta hai. Kyun: brake $r_p$ par hota hai, toh woh radius wahi hai jahan tum already ho — us instant mein tum usse aur neeche nahi ja sakte. Wahan speed hatane ka matlab hai ki tumhare paas far side climb karne ke liye kam energy hai, toh opposite point ($r_a$) neecha aata hai. Relation $\Delta r_a\approx \frac{4a^2{v}_p}{\mu }\Delta v$ bilkul isi liye $r_a$ ke liye likha gaya hai.

$4$, $a^2$, $v_p$, $\mu$ sab positive hain. Toh $\Delta r_a$ ka sign $\Delta v$ jaisa hi hoga, yaani negative → apoapsis neecha jaata hai. Physically: har pass orbit ki far side ko shrink karta hai, bilkul wahi shrinking-oval wali picture.

$\Delta r_a=\frac{4a^2{v}_p}{\mu }\Delta v=\frac{4(2.0\times 10^7{)}^2(4000)}{4.28\times 10^{13}}(-1.0)$ Numerator: $4\times 4\times 10^{14}\times 4000=6.4\times 10^{18}$. $\Delta r_a=\frac{6.4\times 10^{18}}{4.28\times 10^{13}}(-1.0)\approx -1.50\times 10^5\text{m}=\boxed{-150\text{km}}$ Matlab: sirf 1 m/s ka loss far point ko ~150 km giraa deta hai. Free hai, aur bahut bada hai.

Vis-viva kyun, directly energy kyun nahi? Hume ek specific radius par speed chahiye; vis-viva wahi ek formula hai jo $(r,a)$ se seedha $v$ tak jaata hai. $v_p^2=\mu \left(\frac{2}{r_p}-\frac{1}{a}\right)=4.28\times 10^{13}\left(\frac{2}{3.8\times 10^6}-\frac{1}{2.0\times 10^7}\right)$ Andar: $\frac{2}{3.8\times 10^6}=5.263\times 10^{-7}$, $\frac{1}{2.0\times 10^7}=5.0\times 10^{-8}$; difference $=4.763\times 10^{-7}$. $v_p^2=4.28\times 10^{13}\times 4.763\times 10^{-7}=2.039\times 10^7\Rightarrow v_p\approx 4515\text{m/s}$ Quoted $4000$ m/s ke same order mein hai ✓ (parent note ne rounded value use ki thi).

Exponential kyun aur exponent ke liye kyun solve karein? Density double karna ek ratio question hai; exponentials ke ratios exponents ke differences mein convert ho jaate hain — aur wahi logarithms undo karte hain. Hum chahte hain $\frac{{\rho }_{\text{new}}}{{\rho }_{\text{old}}}=e^{-(h_{\text{new}}-h_{\text{old}})/H}=2$. Lower karne ka matlab $h_{\text{new}}<h_{\text{old}}$, toh drop $d=h_{\text{old}}-h_{\text{new}}>0$ lein: $e^{d/H}=2\Rightarrow d=H\ln 2=11\times 0.6931\approx \boxed{7.6\text{km}}$ Matlab: sirf 7.6 km ka chhota sa gehra dip double kar deta hai jo hawa tum feel karte ho — isliye corridor knife-edge hota hai.

Speed $v$ (aur isliye $v^3$) unchanged hai, sirf ${\rho }^{1/2}$ move karta hai: $\frac{{\dot{q}}_{\text{new}}}{{\dot{q}}_{\text{old}}}={\left(\frac{{\rho }_{\text{new}}}{{\rho }_{\text{old}}}\right)}^{1/2}=2^{1/2}\approx \boxed{1.41}$ 7.6 km ki slip ke liye 41% heating jump — altitude mein chhota, lekin watts mein khatarnak.

$\Delta r_a$ ko constant kyun treat karein? Yeh pehla estimate hai; reality mein $v_p$ aur $a$ drift karte hain, lekin fixed step sahi order of magnitude deta hai. Total drop needed: $|\Delta r_a^{\text{total}}|=3.62\times 10^7-3.8\times 10^6=3.24\times 10^7\text{m}=32400\text{km}$ $N=\frac{32400\text{km}}{150\text{km/pass}}=\boxed{216\text{passes}}$ Matlab: ~ek useful pass per orbit aur multi-hour orbits ke saath, yeh weeks se months tak hoga — bilkul real MRO timescale. Design se slow hai.

Rocket equation kyun? Yahi ek law hai jo required speed change ko propellant mass mein convert karta hai, aur woh exponentially karta hai — isliye aerobraking faaydemand hoti hai. Dekho Tsiolkovsky Rocket Equation. $m_{\text{fuel}}=m_{\text{final}}\left(e^{\Delta v/v_e}-1\right)=1000\left(e^{1200/3000}-1\right)=1000\left(e^{0.4}-1\right)$ $e^{0.4}=1.4918$, toh $m_{\text{fuel}}=1000(0.4918)\approx \boxed{492\text{kg}}$ Matlab: aerobraking ne ~490 kg propellant ko patience se replace kar diya — wahi "hundreds of kg" jo parent note ne quote kiye the.

Sirf $\rho$ kyun change hota hai? Ek chhote se altitude nudge mein, periapsis par $v$ bahut kam hilta hai; fierce lever density hai. Toh $a_D$, $\rho$ track karta hai. $\frac{a_D^{\max }}{a_D}=\frac{{\rho }_{\text{new}}}{{\rho }_{\text{old}}}=e^{-\Delta h/H}=\frac{0.04}{0.05}=0.8$ Periapsis raise karna (bada $h$) $\rho$ shrink karta hai, jo ratio $<1$ se consistent hai. Solve karo: $-\frac{\Delta h}{H}=\ln 0.8\Rightarrow \Delta h=-H\ln 0.8=-11\times (-0.2231)\approx \boxed{2.45\text{km}}$ Matlab: periapsis sirf ~2.5 km utha lo (apoapsis par ek whisper jaisi burn) taaki drag safe box ke andar aa jaaye. Yahi aerobraking campaign ka real-time control loop hai.

Step — energy change. $\epsilon =\frac{1}{2}{v}^2-\mu /r$ se, fixed $r$ par ek impulse sirf kinetic term change karta hai: $d(\frac{1}{2}{v}^2)=vdv$. Toh $\Delta \epsilon =v_p\Delta v=4000\times (-1.0)=-4000\text{J/kg}.$ Step — energy se $a$. Kyunki $\epsilon =-\mu /2a$ hai, differentiate karein: $\Delta \epsilon =\frac{\mu }{2a^2}\Delta a$, toh $\Delta a=\frac{2a^2}{\mu }\Delta \epsilon =\frac{2(2.0\times 10^7{)}^2}{4.28\times 10^{13}}(-4000).$ Numerator $2\times 4\times 10^{14}=8\times 10^{14}$; divide karein: $\frac{8\times 10^{14}}{4.28\times 10^{13}}=18.69$; times $-4000$: $\Delta a\approx -7.48\times 10^4\text{m}=-74.8\text{km}.$ Step — apoapsis tak. $r_p$ fixed hai aur $2a=r_p+r_a$ hai, toh $\Delta r_a=2\Delta a\approx -150\text{km}$ ✓ — L2·Q1 se match karta hai. Dono raaste (direct $\Delta r_a$ formula aur energy chain) agree karte hain, loop close ho jaata hai.

Periapsis par apoapsis kyun lower karo, apoapsis par kyun nahi? ::: Periapsis par $v_p$ sabse bada hota hai, toh $\Delta \epsilon =v_p\Delta v$ har m/s mein sabse zyada energy remove karta hai, aur yeh far point ko safely lower karta hai (periapsis musibat se bahar rehta hai); apoapsis par braking periapsis ko hawa mein aur deep le jaayegi — khatarnak.