Aerocapture — using atmosphere to decelerate into orbit: Worked examples

3.4.24 · D3 · Physics › Rocket Flight Mechanics › Aerocapture — using atmosphere to decelerate into orbit

Agar koi symbol yahan ajaib lage, woh parent mein build hua tha — dekho Aerocapture — using atmosphere to decelerate into orbit (index 3.4.24) aur prerequisites pages jo end mein linked hain.

Scenario matrix

Har aerocapture problem in cells mein se kisi ek mein rehti hai. "Sign / regime" column woh key knob hai — woh cheez jiska sign ya size poora answer decide karta hai.

#	Cell (case class)	Key knob & uska regime	Covered by
A	Capture check — kya hum bound end hue?	Sign of $ε_{o u t}$ : $+$ / $0$ / $-$	Ex 1, Ex 2
B	Marginal / degenerate — exactly parabolic	$ε = 0$ (the boundary)	Ex 2
C	Over-braking (crash) — bahut deep	$ε_{o u t}$ far negative, $r_{p} < R_{surface}$	Ex 3
D	Under-braking (skip-out) — bahut shallow	$Δ ε$ too small	Ex 4
E	Ballistic coefficient extremes — $β$ small vs large	$a_{D} \propto 1/ β$	Ex 5
F	Density / altitude limit — kitna deep hai deep?	$ρ = ρ_{0} e^{- h / H}$ , $h \to 0$ aur $h \to \infty$	Ex 6
G	Real-world word problem — Δv & fuel saved	mass fraction $1 - e^{- Δ v / v_{e}}$	Ex 7
H	Exam twist — heating vs braking trade	$\overset{q}{˙} \propto v^{3}$ vs $Δ ε \propto v^{2}$	Ex 8

Hum do planets use karte hain taaki numbers concrete rahein:

Cell A + energy ka sign — poora game

Figure — Aerocapture — using atmosphere to decelerate into orbit

Upar wali picture $ε$ ki number line hai. Is page par sab kuch yahi decide karna hai ki spacecraft zero ke kis taraf land karta hai.

$ε > 0$ : hyperbola — craft ke paas infinity par bhi spare kinetic energy hai, woh nikal jaata hai. (Red, right side.)
$ε = 0$ : parabola — exact boundary, "bas abhi abhi escape karta hai". (Amber, the wall.)
$ε < 0$ : ellipse — bound, captured, hamesha loop karta hai. (Cyan, left side.)

Worked example Ex 1 — Cell A: ek clean capture

Ek probe Mars par periapsis radius $r_{p} = 3490 km$ (100 km upar) par pahunchti hai. Drag se pehle wahan uski speed $v_{in} = 5.5 km/s$ hai. Drag phir $Δ v = 0.8 km/s$ remove karta hai to woh $v_{o u t} = 4.7 km/s$ par exit karti hai. Kya yeh captured hai?

Forecast: Compute karne se pehle $ε_{o u t}$ ka sign guess karo. Usi radius par slower hone ka matlab hai kam kinetic energy — kya woh zero se neeche jayegi?

Incoming energy $ε_{in} = \frac{v _{in}^{2}}{2} - \frac{μ _{M}}{r _{p}} = \frac{5. 5 ^{2}}{2} - \frac{4.28 \times 1 0 ^{4}}{3490}$ . Yeh step kyun? $ε$ woh ek number hai jo capture decide karta hai; ise pass ke pehle aur baad compute karo. $= 15.125 - 12.264 = 2.861 km^{2} / s^{2}$ . Positive → abhi bhi hyperbola aate waqt. Good, ise capture karna hai.
Outgoing energy $ε_{o u t} = \frac{4. 7 ^{2}}{2} - \frac{4.28 \times 1 0 ^{4}}{3490} = 11.045 - 12.264 = - 1.219 km^{2} / s^{2}$ . Yeh step kyun? Same $r_{p}$ (fast pass, radius barely changes), sirf $v$ giri. Potential term identical hai, isliye sirf kinetic energy change hua.
Sign padho. $ε_{o u t} < 0$ → captured (ellipse). ✅

Verify: Change $Δ ε = ε_{o u t} - ε_{in} = - 1.219 - 2.861 = - 4.08 km^{2} / s^{2}$ . Speed form se cross-check karo $Δ ε = \frac{1}{2} (v_{o u t}^{2} - v_{in}^{2}) = \frac{1}{2} (22.09 - 30.25) = - 4.08$ . Same number — theek hai, potential term exactly waisa cancel hua jaisa claim kiya gaya tha.

Worked example Ex 2 — Cells A & B: kitna braking

exactly parabolic deta hai (the boundary)? Same $r_{p}$ , same $v_{in} = 5.5$ km/s. Kaun sa exit speed $v_{o u t}^{⋆}$ craft ko parabola par chhod deta hai ( $ε = 0$ )? Isse kam koi bhi braking capture fail karta hai.

Forecast: $ε = 0$ ka matlab hai poori kinetic exactly potential balance karti hai. Guess karo $v_{o u t}^{⋆}$ $4.7$ km/s se upar hai ya neeche.

$ε = 0$ set karo: $\frac{v _{o u t}^{⋆} ^{2}}{2} = \frac{μ _{M}}{r _{p}}$ . Kyun? Parabola escape aur capture ke beech razor's edge hai — yeh defined hai $ε = 0$ se.
Solve karo $v_{o u t}^{⋆} = 2 μ_{M} / r_{p} = 2 \cdot 12.264 = 24.528 = 4.953 km/s$ . Kyun? Yeh exactly escape speed hai $r_{p}$ par — koi coincidence nahi: escape speed woh speed hai jis par $ε = 0$ hota hai.
Interpret karo: Captured hone ke liye tumhe $4.953$ km/s se neeche exit karna hoga. Hamara Ex 1 exit ( $4.7$ ) usse neeche hai → captured. Agar drag weaker hota aur $v_{o u t} = 5.0 > 4.953$ chhod deta, toh woh abhi bhi escape karta (Cell D preview).

Verify: $v_{o u t}^{⋆}$ wapas plug karo: $ε = \frac{1}{2} (4.953)^{2} - 12.264 = 12.264 - 12.264 = 0$ . Exactly the boundary. ✅

Cell C — over-braking: crash floor

Figure dikhata hai kyun "safety ke liye deeper aim karo" ek trap hai: jo ellipse tum create karte ho uska ek periapsis hai; agar tum bahut zyada energy bleed karo, woh periapsis surface ke neeche sink ho jaata hai aur craft zameen se takraata hai.

Worked example Ex 3 — Cell C: itna hard braking ki orbit planet intersect kare

Mars par ek craft $r_{p} = 3490$ km par $v_{in} = 5.5$ km/s se pass karta hai. Drag enormous hai aur woh sirf $v_{o u t} = 3.0$ km/s par exit karta hai. Kya resulting orbit safe hai, ya crash hoga?

Forecast: Bahut slow exit = bahut negative energy = ek tight ellipse. Guess karo uska far point (apoapsis) ya near point crash ke liye matter karta hai.

Exit energy $ε_{o u t} = \frac{3. 0 ^{2}}{2} - 12.264 = 4.5 - 12.264 = - 7.764 km^{2} / s^{2}$ . Kyun? Establish karta hai ki hum bound hain (bahut negative → tight ellipse).
Semi-major axis from $ε = - μ / (2 a)$ : $a = - \frac{μ _{M}}{2 ε _{o u t}} = - \frac{4.28 \times 1 0 ^{4}}{2 ( - 7.764 )} = 2756 km$ . Kyun? $a$ poori ellipse ki size set karta hai; chhota $a$ matlab chhoti orbit.
Danger check: current point $r_{p} = 3490$ km wahan hai jahan hum hain, lekin kya yeh periapsis hai? Kyunki $r_{p} = 3490 > a = 2756$ , jis point par hain woh semi-major axis se beyond hai — yeh orbit $r_{p}$ ko apoapsis tak reach bhi nahi kar sakti without... actually $r_{p} > 2 a = 5512$ ? Nahi, $r_{p} < 2 a$ . Physical impossibility flag: apoapsis $r_{a} \leq 2 a = 5512$ km aur periapsis $r_{p, orbit} = 2 a - r_{a}$ . Kyunki $a < r_{p}$ , ellipse ka periapsis $R_{M} = 3390$ se chhota hona chahiye. Kyun? Ek ellipse ke do radii ka average $a$ hota hai: $r_{p, orbit} + r_{a} = 2 a$ . Agar $a = 2756 < R_{M} = 3390$ , toh larger radius side bhi andar ki taraf khich raha hai — near side zameen ke andar gahri hai.
Verdict: $2 a = 5512 < 2 R_{M} = 6780$ surface-clearing orbit ke liye impossible hai — crash. Over-braking. ❌

Verify: $r_{p}$ par escape speed $4.953$ km/s thi (Ex 2). Hum $3.0 ≪ 4.953$ par exit kiye, yaani hugely sub-escape → deeply bound → dangerously tight. Consistent. ✅

Cell D — under-braking: skip-out

Worked example Ex 4 — Cell D: bahut shallow, phir bhi escape

Ek craft corridor mein high aim karta hai. Woh $v_{in} = 5.5$ km/s ( $r_{p} = 3490$ km) par enter karta hai lekin thin air sirf $Δ v = 0.3$ km/s remove karti hai, to $v_{o u t} = 5.2$ km/s. Captured?

Forecast: Yaad karo boundary exit speed $4.953$ km/s thi. $5.2$ usse upar hai ya neeche?

Exit energy $ε_{o u t} = \frac{5. 2 ^{2}}{2} - 12.264 = 13.52 - 12.264 = 1.256 km^{2} / s^{2}$ . Kyun? Direct sign test.
Sign: $ε_{o u t} > 0$ → abhi bhi hyperbolic → craft wapas skip karke escape karta hai. ❌
Kitna aur braking chahiye tha? Chahiye $v_{o u t} \leq v_{o u t}^{⋆} = 4.953$ , yaani kam se kam $Δ v = 5.5 - 4.953 = 0.547$ km/s. Hume sirf $0.3$ mila. Shortfall $0.247$ km/s. Kyun? Quantify karta hai "kitna shallow bahut zyada shallow hai" — corridor ki upper wall.

Verify: $ε_{o u t} = 1.256 > 0$ aur $v_{o u t} = 5.2 > 4.953 = v_{o u t}^{⋆}$ : dono tests agree karte hain ki woh escape kar gaya. ✅

Cell E — ballistic coefficient extremes

Worked example Ex 5 — Cell E: same air,

$β$ se opposite fates Do craft Earth ke atmosphere mein $h$ par hit karte hain jahan $ρ = 1.0 \times 1 0^{- 4} kg/m^{3}$ hai, dono $v = 7.0$ km/s $= 7000$ m/s par. Craft A (blunt): $β_{A} = 50 kg/m^{2}$ . Craft B (dense): $β_{B} = 200 kg/m^{2}$ . Unke decelerations compare karo.

Forecast: $a_{D} \propto 1/ β$ . Compute karne se pehle ratio $a_{D, A} / a_{D, B}$ guess karo.

A ka decel: $a_{D, A} = \frac{ρ v ^{2}}{2 β _{A}} = \frac{( 1 0 ^{- 4} ) ( 7000 ) ^{2}}{2 ( 50 )} = \frac{4900}{100} = 49 m/s^{2}$ . Kyun? SI throughout use karo ( $ρ$ kg/m³ mein, $v$ m/s mein, $β$ kg/m² mein) taaki answer m/s² mein aaye.
B ka decel: $a_{D, B} = \frac{( 1 0 ^{- 4} ) ( 7000 ) ^{2}}{2 ( 200 )} = \frac{4900}{400} = 12.25 m/s^{2}$ .
Ratio: $a_{D, A} / a_{D, B} = 49/12.25 = 4$ . Exactly $β_{B} / β_{A} = 200/50 = 4$ . Kyun? Confirm karta hai $a_{D} \propto 1/ β$ : fluffy craft same air mein $4 \times$ harder brake karta hai, isliye woh upar hi capture ho sakta hai, jahan thanda hota hai.

Verify: $g$ 's mein: A feels $49/9.81 = 5.0 g$ , B feels $12.25/9.81 = 1.25 g$ . Ratio $4$ , aur dono plausible aerocapture loads hain. ✅

Cell F — density aur altitude limits

Worked example Ex 6 — Cell F: target density ke liye kitna deep? (aur do limits)

Earth: $ρ (h) = ρ_{0} e^{- h / H}$ , $ρ_{0} = 1.2$ , $H = 8$ km. (a) Kis altitude $h$ par $ρ = 1.0 \times 1 0^{- 4} kg/m^{3}$ hai (Ex 5 mein use ki gayi density)? (b) Do degenerate limits $h \to 0$ aur $h \to \infty$ check karo.

Forecast: Ratio $ρ / ρ_{0} = 1 0^{- 4} /1.2 \approx 8.3 \times 1 0^{- 5}$ tiny hai — guess karo $h$ tens mein hoga ya hundreds of km mein.

Exponential ko invert karo. $ρ = ρ_{0} e^{- h / H}$ se, natural log lo: $ln (ρ / ρ_{0}) = - h / H$ , to $h = - H ln (ρ / ρ_{0})$ . Yeh tool — logarithm — kyun? Unknown $h$ exponent mein baitha hai. Woh ek operation jo variable ko exponent se bahar kheechta hai woh uska inverse hai, $ln$ . Bas yahi wajah hai ki hum isse use karte hain.
Plug in karo: $h = - 8 ln (\frac{1 0 ^{- 4}}{1.2}) = - 8 ln (8.333 \times 1 0^{- 5}) = - 8 (- 9.393) = 75.1 km$ . Kyun? Confirm karta hai ki Earth par upper-atmosphere aerocapture ~75 km ke paas hoti hai.
Limit $h \to 0$ : $ρ \to ρ_{0} e^{0} = ρ_{0} = 1.2$ — full sea-level density (dense, craft ko taar dega). Limit $h \to \infty$ : $ρ \to ρ_{0} e^{- \infty} = 0$ — vacuum, bilkul drag nahi. Limits kyun check karein? Woh physics bracket karte hain: "solid wall" ( $h = 0$ ) aur "no brake" ( $h = \infty$ ) ke beech thin usable band hai. Kuch $H$ (few tens of km) poora action cover karta hai.

Verify: $h = 75.1$ km ko $ρ_{0} e^{- h / H} = 1.2 e^{- 75.1/8} = 1.2 e^{- 9.39} = 1.2 (8.33 \times 1 0^{- 5}) = 1.0 \times 1 0^{- 4}$ se wapas push karo. ✅ Input density se match karta hai.

Cell G — real-world word problem: fuel saved

Worked example Ex 7 — Cell G: saved

$Δ v$ ko saved mass mein convert karo Ek Mars orbiter ko insertion burn ka $Δ v = 0.8$ km/s chahiye hoga. Iske bajaye woh aerocapture karta hai, woh $0.8$ km/s drag se free mein leta hai. Engine exhaust speed $v_{e} = 3$ km/s. Arrival mass ka kitna fraction propellant ke roop mein bachta hai?

Forecast: Tsiolkovsky Rocket Equation mass fraction $1 - e^{- Δ v / v_{e}}$ deta hai. $Δ v ≪ v_{e}$ ke saath, guess karo: 25% ke kareeb ya 5% ke kareeb?

Exponent: $Δ v / v_{e} = 0.8/3 = 0.2667$ . Kyun? Rocket equation is ratio mein exponential hai — "kitna hard" burn hai uska natural unit.
Mass fraction saved: $f = 1 - e^{- 0.2667} = 1 - 0.7659 = 0.2341$ . Kyun? $e^{- Δ v / v_{e}}$ woh mass ka fraction hai jo tum burn ke baad rakhoge; $1$ minus woh propellant hai jo tum spend karte — aur ab nahi karte.
Interpret karo: Arrival mass ka ~ $23.4%$ ab propellant nahi hona chahiye — woh payload ho sakta hai, ya kabhi launch hi nahi kiya gaya. Enormous.

Verify: $e^{- 0.2667} = 0.7659$ , aur $0.7659 \cdot e^{+ 0.2667} = 1.000$ . Round-trip consistent, aur $0.234$ parent note ke stated ≈0.23 se match karta hai. ✅

Cell H — exam twist: heating vs braking

Worked example Ex 8 — Cell H: "faster entry harder brake karta hai, to easier hai." Sach?

$v$ vs $2 v$ par enter karne ki comparison karo (same $ρ$ , same craft). Braking force $v^{2}$ ke saath scale karta hai; energy to remove $v^{2}$ ke saath scale karti hai; peak heating $v^{3}$ ke saath scale karti hai (Sutton–Graves, dekho Atmospheric Entry & Heating). Jab tum $v$ double karte ho toh har cheez kitne factor se badhti hai?

Forecast: Teeno badhte hain — lekin kaun sabse tez badhta hai, aur kya woh help karta hai ya hurt?

Drag force factor: $F_{D} \propto v^{2} \Rightarrow (2)^{2} = 4 \times$ . Kyun? $F_{D} = \frac{1}{2} ρ v^{2} C_{D} A$ se.
Energy-to-shed factor: excess $ε_{in} \approx \frac{1}{2} v_{\infty}^{2} \propto v^{2} \Rightarrow 4 \times$ . Kyun? Braking $4 \times$ stronger hua, lekin kaam bhi $4 \times$ bada hua — yeh cancel ho jaate hain. Speed se koi net advantage nahi.
Heating factor: $\overset{q}{˙} \propto v^{3} \Rightarrow (2)^{3} = 8 \times$ . Kyun? Heating dono braking aur energy ko ek extra factor of $v$ se outpace karta hai. To speed double karne se braking-vs-work balance unchanged rehta hai lekin thermal problem 8× worse ho jaata hai.
Verdict: faster entry harder hai, kyunki heating (sach mein limiting constraint) cubic hai jabki sab helpful cheezein sirf quadratic hain. "Faster harder brake karta hai" wali intuition ek trap hai. ✅ (Parent ke mistakes list mein steel-manned.)

Verify: Ratios $4 : 4 : 8$ . Heating ko braking se divide karo: $8/4 = 2 =$ ek extra power of $v$ — exactly cubic-vs-quadratic gap. ✅

Recall Self-test: yeh kaun sa cell hai?

Ek craft apne single atmospheric pass ke baad $ε_{o u t} = + 0.4 km^{2} / s^{2}$ ke saath exit karta hai. Captured? ::: Nahi — $ε > 0$ abhi bhi hyperbolic hai (Cell D, skip-out). Ise zyada braking chahiye. Tum exact braking chahte ho jo parabola de. $ε_{o u t}$ kya hai? ::: Exactly $0$ (Cell B, the boundary = escape speed at $r_{p}$ ). Craft A ka $β = 50$ hai, craft B ka $β = 200$ , same air & speed. Kaun harder brake karta hai aur kitna? ::: A, $4 \times$ se, kyunki $a_{D} \propto 1/ β$ (Cell E). Entry speed double karne se peak heating kitne se multiply hoti hai? ::: $8$ ( $\overset{q}{˙} \propto v^{3}$ ) — Cell H. Target density dene wala altitude find karne ke liye, kaun sa function $ρ_{0} e^{- h / H}$ ko invert karta hai? ::: Natural log: $h = - H ln (ρ / ρ_{0})$ (Cell F).

Prerequisites & neighbours: Vis-viva Equation · Hyperbolic Trajectories & Hyperbolic Excess Velocity · Orbit Insertion Burns · Aerobraking · Atmospheric Entry & Heating · Ballistic Coefficient · Tsiolkovsky Rocket Equation · Scale Height & Exponential Atmosphere