Worked examples — Aerobraking — gradual orbit lowering using atmospheric drag

3.4.25 · D3 · Physics › Rocket Flight Mechanics › Aerobraking — gradual orbit lowering using atmospheric drag

Yeh page parent note ki "sab kuch dal do" companion hai. Hum har tarah ka case step by step dekhenge jo master relations produce kar sakti hain: ek normal brake, zero case, degenerate circular case, sign flip (braking at apoapsis), limiting deep-dive, ek real mission-style word problem, aur ek exam twist. Shuru karne se pehle, is page ke har symbol aur tool ko pin down karte hain.

Dekho bhi: Vis-Viva Equation, Orbital Energy and Semi-major Axis, Atmospheric Drag and Scale Height, Tsiolkovsky Rocket Equation, Aerocapture, Hohmann Transfer.

The scenario matrix

Is topic ke har problem ke liye aap in cells mein se kisi mein aayenge. "Applied at" column woh jagah batata hai jahan $\Delta v$ act karta hai — yeh sabse important fact hai yeh decide karne ke liye ki kaunsa orbital point move karta hai.

Poore time, jab tak bola na jaaye, hum Mars use karte hain: $\mu =4.28\times 10^{13}{\text{m}}^3/{\text{s}}^2$.

Example 1 — Cell A: normal periapsis brake

Forecast: Pehle guess karo — kya apoapsis $1$ km se zyada giregi ya kam? (Factor of $4$ aur badi $a^2$ tip deni chahiye.)

1. Sahi relation chuno. Yeh step kyun? Hume near point par ek chhoti speed loss se far point ki change chahiye — exactly yahi $\Delta r_a\approx \frac{4a^2{v}_p}{\mu }\Delta v$ ek line mein answer karta hai, aur koi doosri cheez ek line mein nahi karta.
2. Numbers substitute karo. Yeh step kyun? Plug in karo symbols ko metres mein turn karne ke liye. $\Delta r_a=\frac{4(2.0\times 10^7{)}^2(4000)}{4.28\times 10^{13}}(-1.0)$
3. Powers crunch karo. Yeh step kyun? $(2.0\times 10^7{)}^2=4.0\times 10^{14}$, toh numerator hai $4\cdot 4.0\times 10^{14}\cdot 4000=6.4\times 10^{18}$. $\Delta r_a=\frac{6.4\times 10^{18}}{4.28\times 10^{13}}(-1.0)\approx -1.50\times 10^5\text{m}=-150\text{km}$

Verify: Units: $\frac{({\text{m}}^2)(\text{m/s})(\text{m/s})}{{\text{m}}^3/{\text{s}}^2}=\frac{{\text{m}}^4/{\text{s}}^2}{{\text{m}}^3/{\text{s}}^2}=\text{m}$ ✓. Ek tiny $1$ m/s loss ne apoapsis $150$ km gira diya — yahi woh "leverage" hai jo aerobraking ko wait ke layak banata hai. Parent note ke estimate se match karta hai.

Example 2 — Cell B: zero case

Forecast: Agar atmosphere kuch nahi touch karta, kya orbit badlegi bilkul bhi?

1. $\Delta v=0$ substitute karo. Yeh step kyun? Yeh boundary case hai — yeh batata hai ki formula zero input par sensibly behave karta hai, jo kisi bhi acche formula mein hona chahiye. $\Delta a=\frac{2a^2{v}_p}{\mu }\cdot 0=0,\Delta r_a=\frac{4a^2{v}_p}{\mu }\cdot 0=0$
2. Interpret karo. Yeh step kyun? Koi energy nahi remove hui ($\Delta \epsilon =v_p\Delta v=0$) matlab $\epsilon$ unchanged, toh $a=-\mu /2\epsilon$ unchanged, toh poora ellipse identical hai.

Verify: Pure gravity energy conserve karti hai — drag ke bina ek pass must same orbit par return karna chahiye. Formula exactly $0$ deta hai. ✓ Isliye controllers "walk out" karte hain periapsis raise karke: ek baar jab dips empty space graze karne lagte hain, campaign apne aap progress karna band kar deta hai.

Example 3 — Cell C: degenerate circular orbit

Forecast: Ek circle mein koi distinct apoapsis nahi hota. Kya $\Delta r_a$ formula ab bhi kaam karta hai?

1. Circular speed nikalo. Yeh step kyun? Circle ke liye $a=r$, toh vis-viva deta hai $v^2=\mu (2/r-1/r)=\mu /r$. Yeh woh speed hai jahan se brake karte hain; ise hum $\Delta a$ relation mein apna $v$ use karenge. $v_c=\sqrt{\frac{\mu }{r}}=\sqrt{\frac{4.28\times 10^{13}}{2.0\times 10^7}}=\sqrt{2.14\times 10^6}\approx 1463\text{m/s}$
2. $\Delta a$ relation apply karo. Yeh step kyun? Helper 1, $\Delta a=\frac{2a^2{v}_c}{\mu }\Delta v$, ab bhi hold karta hai — yeh energy se aaya tha, jo eccentricity ki parwah nahi karta. $\Delta a=\frac{2(2.0\times 10^7{)}^2(1463)}{4.28\times 10^{13}}(-1.0)\approx -2.735\times 10^4\text{m}\approx -27.4\text{km}$
3. "Apoapsis" ka kya hota hai? Yeh step kyun? Yeh woh degenerate subtlety hai. Jis second aap brake karte ho, orbit circular nahi rahi ($e$ zero se upar jump karta hai): brake point naya apoapsis ban jaata hai (aapne us cheez ke top par slow kiya jo ab ek ellipse hai), aur ek naya periapsis $\frac{1}{2}$ orbit baad appear hota hai, neeche. Toh $r_a$ $\approx r$ rehta hai (aap usi par ho), aur periapsis girti hai. Naive $\Delta r_a$ formula, jisne assume kiya tha brake periapsis par hoti hai, apply nahi hota — uski assumption violate ho gayi.

Verify: $v_c^2=\mu /r=2.14\times 10^6$, aur $1463^2=2.14\times 10^6$ ✓. Lesson: $\Delta r_a$ formula sirf periapsis par brake ke liye hai; degenerate circle exactly woh jagah hai jahan woh assumption tootti hai, toh hum energy relation se $\Delta a$ ke liye wapas jaate hain.

Example 4 — Cell D: sign flip (apoapsis par brake)

Forecast: Parent note ke "forecast-then-verify" ne kaha tha ki apoapsis par braking periapsis ko girata hai. Example 1 ki comparison mein kitna?

1. $\Delta a$ relation $v=v_a$ ke saath use karo. Yeh step kyun? Helper 1 kisi bhi point par kaam karta hai; yahan brake apoapsis par hai, toh apoapsis speed $v_a$ use karte hain. $\Delta a=\frac{2a^2{v}_a}{\mu }\Delta v=\frac{2(2.0\times 10^7{)}^2(800)}{4.28\times 10^{13}}(-1.0)\approx -1.495\times 10^4\text{m}\approx -15.0\text{km}$
2. Pata karo kaunsa point move karta hai. Yeh step kyun? Hum apoapsis par hain, toh $r_a$ fixed hai. $2a=r_p+r_a$ se (semi-major-axis definition), $a$ mein change ke saath $r_a$ held fixed matlab sab kuch $r_p$ mein jaata hai: $\Delta r_p=2\Delta a$. $\Delta r_p=2\Delta a\approx -29.9\text{km}$
3. Example 1 se compare karo. Yeh step kyun? Periapsis par $v_p=4000$; apoapsis par $v_a=800$ — paanch guna slower. Kyunki $\Delta a\propto v$, apoapsis par braking 5× kam effective hai aur dangerous end ko girata hai.

Neeche wala figure (Figure 1) step 2 ko visual banata hai. Blue solid curve original ellipse hai; planet (orange) focus par hai. Red arrow woh brake $\Delta v<0$ hai jo apoapsis par apply kiya gaya (left, red dot). Kyunki hum apoapsis point ko fixed rakhte hain aur energy remove karte hain, ellipse sirf near side par andar shrink karti hai: gray dashed curve wahi apoapsis share karti hai lekin lower periapsis hai (green). Dhyan do ki near end (right) planet ki taraf khiich raha hai jabki far end barely move karta hai — periapsis brake ka mirror image.

Figure 1 — Apoapsis par braking periapsis ko girata hai (Cell D).

Verify: Ratio of effectiveness $=v_p/v_a=4000/800=5$, aur size changes compare karo: Example 1 ka $\Delta a=\frac{1}{2}\Delta r_a=-75$ km versus is case ka $-15$ km, same $5\times$ deta hai ✓. Exactly isliye real drag periapsis par rehti hai: woh dono safer aur zyada efficient hai.

Example 5 — Cell E: limiting deep-dive

Forecast: Sirf $11$ km girana — surely modest change? Exponential dekho.

1. Density factor. Yeh step kyun? Altitude $\Delta h=-H$ se kam karne par density $e^{-(h-H)/H}/e^{-h/H}=e^{+1}$ se multiply hoti hai. Exponential hi poori kahani hai. $\frac{{\rho }_{\text{new}}}{{\rho }_{\text{old}}}=e^1\approx 2.718$
2. Speed loss per pass. Yeh step kyun? Parent ke drag deceleration $a_D=\frac{\rho v^2{C}_{D}A}{2m}$ se (symbols in Helper 3), hume milta hai $a_D\propto \rho$, toh $\Delta v$ per pass $\rho$ ke saath scale karta hai (speed $v_p$ roughly unchanged chhote $r_p$ move ke liye). $\Delta v_{\text{new}}\approx 2.718\times (-1.0)=-2.72\text{m/s}$
3. Dynamic pressure aur heating. Yeh step kyun? Dynamic pressure hai $q=\frac{1}{2}\rho v^2$, toh fixed $v$ par yeh $\rho$ ke saath scale karta hai ($\times e=2.72$). Heating-rate law $\dot{q}\propto {\rho }^{1/2}{v}^3$ standard convective stagnation-point (Sutton–Graves) model hai jo Aerocapture mein use hota hai: heat delivered per second grow karta hai air ke mass flux ki tarah jo nose par lagti hai ($\propto \rho v$) times energy per unit mass jo woh carry karta hai ($\propto v^2$), ek ${\rho }^{-1/2}$ correction ke saath boundary layer se — net ${\rho }^{1/2}{v}^3$. Fixed $v$ par yeh ${\rho }^{1/2}$ ke saath scale karta hai ($\times e^{1/2}=1.65$). $q:\times 2.72,\dot{q}:\times 1.65$

Verify: $e^1=2.718$, $e^{0.5}=1.649$ ✓. Sirf $11$ km deeper jaane se braking lagbhag triple ho jaata hai lekin dynamic pressure bhi lagbhag triple ho jaata hai — isliye "bas thoda aur deep dip karo jaldi finish karne ke liye" dangerous hai, aur safe corridor narrow kyun hota hai. Limiting behaviour: jaise jaise $r_p\to$ dense atmosphere, $\rho$ without bound grow karta hai aur craft overheat ho jaata hai.

Example 6 — Cell F: real word problem

Forecast: Mahino-long campaign — hundreds of passes? Tens? Ek order of magnitude guess karo.

1. Total apoapsis remove karna hai. Yeh step kyun? Cover karne ki distance workload set karti hai. $\Delta r_a^{\text{total}}=3.35\times 10^7-6.9\times 10^6=2.66\times 10^7\text{m}$
2. Per-pass drop se divide karo. Yeh step kyun? Constant-drop approximation pehla estimate deta hai (real drops $a$ shrink hone ke saath kam hoti hain, toh yeh lower bound hai). $N\approx \frac{2.66\times 10^7}{1.5\times 10^5}\approx 177\text{passes}$
3. Passes ko calendar time mein convert karo. Yeh step kyun? Helper 2 use karo, period $T=2\pi \sqrt{a^3/\mu }$, early (sabse badi) orbit par: $T=2\pi \sqrt{\frac{(2.0\times 10^7{)}^3}{4.28\times 10^{13}}}=2\pi \sqrt{\frac{8.0\times 10^{21}}{4.28\times 10^{13}}}=2\pi \sqrt{1.869\times 10^8}\approx 8.59\times 10^4\text{s}\approx 23.9\text{h}$ Toh $177$ passes roughly ek din each par $\Rightarrow$ kai mahine (early passes sabse lambe hote hain; orbit period $a$ fall hone ke saath shrink karti hai, toh real campaign thoda shorter hota hai).

Verify: $2.66\times 10^7/1.5\times 10^5=177.3$ ✓. $T=2\pi \sqrt{8.0\times 10^{21}/4.28\times 10^{13}}=2\pi \sqrt{1.869\times 10^8}\approx 8.59\times 10^4$ s ✓. Order of magnitude "hundreds of passes over months" real MGS/MRO campaigns se match karta hai. ✓

Example 7 — Cell G: exam twist (drag force se)

Forecast: Chhoti deceleration, short time — kya $\Delta v$ $\approx 1$ m/s hoga (Example 1 ki tarah)?

1. Deceleration ko $\Delta v$ mein convert karo. Yeh step kyun? $\Delta v$ deceleration times woh time hai jab tak woh act karta hai (impulse per unit mass), aur yeh negative hai kyunki craft slow down hota hai. $\Delta v=-a_{D}t=-(0.02)(50)=-1.0\text{m/s}$
2. $\Delta a$ compute karo. Yeh step kyun? Helper 1 speed loss (periapsis par, toh $v=v_p$) ko orbit-size change mein convert karta hai. $\Delta a=\frac{2a^2{v}_p}{\mu }\Delta v=\frac{2(4.0\times 10^{14})(4000)}{4.28\times 10^{13}}(-1.0)\approx -7.48\times 10^4\text{m}\approx -74.8\text{km}$
3. $\Delta r_a$ compute karo. Yeh step kyun? Kyunki brake periapsis par hai, $r_p$ fixed hai aur $2a=r_p+r_a$ deta hai $\Delta r_a=2\Delta a$. $\Delta r_a=2\Delta a\approx -1.50\times 10^5\text{m}=-150\text{km}$

Verify: $\Delta v=-1.0$ m/s Example 1 ka $-150$ km reproduce karta hai ✓ — accha, dono routes (given $\Delta v$ vs given drag force) agree karte hain. $a_{D}t$ ke units: $({\text{m/s}}^2)(\text{s})=\text{m/s}$ ✓.

Example 8 — Cell H: campaign ka ant

Forecast: Ek positive (prograde) burn — kya yeh near end uthata hai ya far end?

1. Apoapsis radius nikalo, phir vis-viva se $v_a$. Yeh step kyun? Hume un point par sach mein speed chahiye jahan hum burn karte hain, aur vis-viva kahin bhi speed deta hai jab hum $r$ aur $a$ jaante hain. Apoapsis: $r_a=2a-r_p=2(6.95\times 10^6)-3.5\times 10^6=1.04\times 10^7$ m. $v_a=\sqrt{\mu \left(\frac{2}{r_a}-\frac{1}{a}\right)}=\sqrt{4.28\times 10^{13}\left(\frac{2}{1.04\times 10^7}-\frac{1}{6.95\times 10^6}\right)}\approx 1.63\times 10^3\text{m/s}$
2. Helper 1 apoapsis par apply karo. Yeh step kyun? Hum $r_a$ par burn karte hain, toh $v=v_a$ use karo; positive $\Delta v$ energy raise karta hai, toh $a$ grow karta hai. $\Delta a=\frac{2a^2{v}_a}{\mu }\Delta v=\frac{2(6.95\times 10^6{)}^2(1633)}{4.28\times 10^{13}}(+2.0)\approx +7.37\times 10^3\text{m}\approx +7.4\text{km}$
3. Kaunsa point uthta hai? Yeh step kyun? Hum apoapsis par hain, toh $r_a$ fixed hai; $2a=r_p+r_a\Rightarrow \Delta r_p=2\Delta a$. $\Delta r_p=2\Delta a\approx +14.7\text{km}$

Verify: $v_a\approx 1633$ m/s vis-viva se (neeche check kiya gaya). Positive $\Delta v$ → positive $\Delta r_p$: periapsis atmosphere se bahar uthti hai, exactly woh "walk-out" jo parent note describe karta hai. Sign correct hai: periapsis uthana aerobraking band karta hai.

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