Aerocapture — using atmosphere to decelerate into orbit: Exercises

3.4.24 · D4 · Physics › Rocket Flight Mechanics › Aerocapture — using atmosphere to decelerate into orbit

Shuru karne se pehle, ek reminder — har symbol ka plain words mein matlab:

Level 1 — Recognition

(Kya tum sahi formula pick kar sakte ho aur uska sign padh sakte ho?)

L1.1 Ek craft ek planet par $ε = + 2.0 km^{2} / s^{2}$ ke saath pahunchi. Kya yeh hyperbola, parabola, ya ellipse par hai? Kya bina kisi braking ke capture hogi?

Recall Solution

$ε$ ka sign hi poori kahani hai:

$ε > 0$ → hyperbola (escape kar jaata hai),
$ε = 0$ → parabola (marginal case),
$ε < 0$ → ellipse (captured).

Yahan $ε = + 2.0 > 0$ , toh yeh hyperbola par hai aur apne aap capture nahi hogi — yeh sidha nikal jaayegi. Yahi positive energy hai jo drag ko remove karni hoti hai.

L1.2 Ek captured ellipse ke liye energy hoti hai $ε = - μ / (2 a)$ . Ek Mars orbit ( $μ_{M} = 4.28 \times 1 0^{4} km^{3} / s^{2}$ ) ka semi-major axis $a = 12000$ km hai. $ε$ nikaalo.

Recall Solution

Seedha plug in karo: $ε = - \frac{μ}{2 a} = - \frac{4.28 \times 1 0 ^{4}}{2 \times 12000} = - \frac{4.28 \times 1 0 ^{4}}{2.4 \times 1 0 ^{4}} = - 1.783 km^{2} / s^{2} .$ Negative → captured. Theek hai, yeh ek bound orbit hai. Minus kyun: ek bound orbit ek "well" mein hoti hai; zero tak — yaani escape tak — pahunchne ke liye energy add karni padti hai.

L1.3 Drag deceleration per unit mass likho aur words mein batao ki $ρ$ , $v$ , $β$ mein se har ek usse kya karta hai.

Recall Solution

$a_{D} = \frac{ρ v ^{2}}{2 β} .$

$ρ ↑$ (thicken hawa, gehri dive) → zyada braking, linearly.
$v ↑$ → zyada braking, lekin $v^{2}$ ke hisaab se (speed double → drag chaar guna).
$β ↑$ (denser/chhota craft) → braking kam hoti hai, kyunki $β$ denominator mein hai; ek heavy compact craft usi hawa ko zyada resist karta hai.

Level 2 — Application

(Ek formula, real numbers, ek step.)

L2.1 Ek probe Venus par $v_{\infty} = 4 km/s$ hyperbolic excess speed ke saath pahunchi. Planet se door, uski arrival energy $ε_{in}$ kya hai?

Recall Solution

Door ( $r \to \infty$ ) potential term $- μ / r \to 0$ ho jaata hai, toh saari energy kinetic hai: $ε_{in} = \frac{v _{\infty}^{2}}{2} = \frac{4 ^{2}}{2} = 8.0 km^{2} / s^{2} .$ Positive, jaise har aane wali craft ki hoti hai — yahi $+ 8.0$ "excess" hai jise drag ko delete karna hai.

L2.2 Periapsis par pre-drag speed $v_{in} = 10.2$ km/s hai aur target captured orbit ko usi radius par $v_{o u t} = 9.4$ km/s chahiye. Atmosphere jo $Δ v$ free mein deti hai, woh nikaalo.

Recall Solution

Kyunki quick pass ke dauran periapsis radius $r_{p}$ almost change nahi hoti, sirf speed girti hai: $Δ v_{a er o} = v_{in} - v_{o u t} = 10.2 - 9.4 = 0.8 km/s .$ Yeh $0.8$ km/s ki braking hawa se aayi, fuel se nahi — yahi propellant hai jo tumne launch hi nahi kiya.

L2.3 $ρ = ρ_{0} e^{- h / H}$ use karo scale height $H = 15$ km ke saath — $h = 45$ km par hawa $h = 0$ ke mukable kitne factor se patli hai?

Recall Solution

$\frac{ρ ( 45 )}{ρ ( 0 )} = e^{- 45/15} = e^{- 3} \approx 0.0498.$ Toh lagbhag 20× patli ( $1/0.0498 \approx 20$ ). Yeh steep exponential hi wajah hai ki altitude mein thoda sa change drag ko dramatically badal deta hai — razor's-edge corridor (dekho Scale Height & Exponential Atmosphere).

Level 3 — Analysis

(Do ideas combine karo; behaviour ke baare mein soocho.)

L3.1 Ek probe Mars ( $μ_{M} = 4.28 \times 1 0^{4} km^{3} / s^{2}$ ) par $v_{\infty} = 3$ km/s ke saath pahunchi. Tum chahte ho ki exit energy $ε_{o u t} = - 1 km^{2} / s^{2}$ ho. (a) Drag ko kitni energy shed karni hogi? (b) Exit orbit ka semi-major axis $a$ kya hoga?

Recall Solution

(a) Arrival energy $ε_{in} = v_{\infty}^{2} /2 = 3^{2} /2 = 4.5 km^{2} / s^{2}$ . $∣Δ ε ∣ = ε_{in} - ε_{o u t} = 4.5 - (- 1) = 5.5 km^{2} / s^{2} .$ Tum $+ 4.5$ se neeche $- 1$ tak jaate ho. (b) $ε = - μ / (2 a)$ ko invert karo: $a = - \frac{μ}{2 ε _{o u t}} = - \frac{4.28 \times 1 0 ^{4}}{2 \times ( - 1 )} = 2.14 \times 1 0^{4} km .$ Ek loose, badi ellipse — exactly yahi chahiye taaki over-brake na ho.

L3.2 Do craft same altitude par same speed $v$ se dip karte hain. Craft A ka $β = 50$ kg/m², craft B ka $β = 200$ kg/m². Unki drag decelerations ka ratio $a_{D, A} / a_{D, B}$ kya hai, aur dono mein se kisko utni hi braking ke liye gehri dive karni hogi?

Recall Solution

$a_{D} = ρ v^{2} / (2 β)$ , aur $ρ, v$ dono same hain, toh $\frac{a _{D, A}}{a _{D, B}} = \frac{β _{B}}{β _{A}} = \frac{200}{50} = 4.$ A 4× zyada brake karta hai same altitude par — yeh "fluffier" hai (mass ke hisaab se zyada area). B ko utni hi braking ke liye $4 \times$ zyada density chahiye, yaani gehri dive — jo matlab hai zyada garam. Yahi wajah hai ki low- $β$ (bade blunt) shields safer hote hain. Dekho Ballistic Coefficient.

L3.3 Periapsis par $ρ = 2 \times 1 0^{- 5}$ kg/m³, $v = 6$ km/s $= 6000$ m/s, $β = 80$ kg/m² hai. Instantaneous drag deceleration m/s² mein aur "g" mein nikaalo ( $1 g = 9.81$ m/s²).

Recall Solution

Poora SI use karo (metres, seconds): $a_{D} = \frac{ρ v ^{2}}{2 β} = \frac{( 2 \times 1 0 ^{- 5} ) ( 6000 ) ^{2}}{2 \times 80} = \frac{( 2 \times 1 0 ^{- 5} ) ( 3.6 \times 1 0 ^{7} )}{160} = \frac{720}{160} = 4.5 m/s^{2} .$ g mein: $4.5/9.81 = 0.459 g \approx 0.46 g$ . Yahan gentle hai — lekin yeh sirf instantaneous peak hai; drag poore pass ke dauran km/s ka $Δ v$ remove karne ke liye kaam karta hai.

Level 4 — Synthesis

(Teen ya zyada ideas ko ek answer mein chain karo.)

L4.1 Ek craft Earth ( $μ_{E} = 3.986 \times 1 0^{5} km^{3} / s^{2}$ ) par $v_{\infty} = 3$ km/s ke saath pahunchi aur periapsis radius $r_{p} = 6500$ km tak dip karti hai. (a) Periapsis par uski pre-drag speed $v_{in}$ nikaalo. (b) Tum chahte ho ki exit orbit ki $ε_{o u t} = - 8 km^{2} / s^{2}$ ho. Same $r_{p}$ par required post-drag speed $v_{o u t}$ nikaalo. (c) $Δ v_{a er o}$ nikaalo.

Recall Solution

(a) Infinity se periapsis tak energy conserved hai (atmosphere mein entry ke waqt drag abhi nahi lagi hai; pre-drag energy $ε_{in} = v_{\infty}^{2} /2$ use karo). $ε = v^{2} /2 - μ / r$ se: $v_{in} = 2 (ε_{in} + \frac{μ}{r _{p}}) = 2 (\frac{3 ^{2}}{2} + \frac{3.986 \times 1 0 ^{5}}{6500}) .$ Andar: $\frac{9}{2} = 4.5$ aur $\frac{3.986 \times 1 0 ^{5}}{6500} = 61.32$ . Sum $= 65.82$ , times 2 $= 131.65$ . $v_{in} = 131.65 = 11.47 km/s .$ (b) Same $r_{p}$ , lekin ab $ε_{o u t} = - 8$ ke saath: $v_{o u t} = 2 (ε_{o u t} + \frac{μ}{r _{p}}) = 2 (- 8 + 61.32) = 2 \times 53.32 = 106.65 = 10.33 km/s .$ (c) $Δ v_{a er o} = v_{in} - v_{o u t} = 11.47 - 10.33 = 1.15 km/s$ ki free braking. (Dekho Vis-viva Equation $v (r)$ relation ke liye jo yahan use hua.)

L4.2 L4.1 continue karte hue, agar ship ke paas $v_{e} = 3$ km/s exhaust speed wala engine hota, toh aerocapture kitna propellant mass fraction bachata hai us $1.15$ km/s ko fuel ki jagah hawa se supply karke?

Recall Solution

Tsiolkovsky Rocket Equation se, $Δ v$ ke burn ke liye mass fraction chahiye: $\frac{Δ m}{m} = 1 - e^{- Δ v / v_{e}} = 1 - e^{- 1.15/3} = 1 - e^{- 0.3833} .$ $e^{- 0.3833} = 0.6816$ , toh $\frac{Δ m}{m} = 1 - 0.6816 = 0.318 \approx 32%.$ Arrival mass ka lagbhag ek tehaai propellant ke roop mein nahi laana pada — yahi aerocapture ka poora economic case ek number mein hai.

Level 5 — Mastery

(Design-level: multiple constraints, exponential atmosphere, aur ek judgement call.)

L5.1 — Entry corridor. Drag deceleration hai $a_{D} = ρ v^{2} / (2 β)$ jahan $ρ = ρ_{0} e^{- h / H}$ , toh fixed speed $v$ par, $a_{D} \propto e^{- h / H}$ . Maano mission ke liye peak deceleration $a_{m i n} = 3 g$ aur $a_{m a x} = 6 g$ ke beech honi chahiye (zyada kam → skip out aur escape; zyada zyada → crash/overheat). $H = 8$ km lo. Altitude window $Δ h = h_{m i n} - h_{m a x}$ (corridor width) kya hogi jo tumhe is band ke andar rakhe?

Recall Solution

Jitna upar jao (bada $h$ ), utni patli hawa, utna chhota $a_{D}$ . Toh:

$h_{m a x}$ (sabse upar allowed) correspond karta hai sabse chhoti allowed drag $a_{m i n}$ se.
$h_{m i n}$ (sabse neeche allowed) correspond karta hai sabse badi allowed drag $a_{m a x}$ se.

Kyunki $a_{D} \propto e^{- h / H}$ , do drag limits ka ratio ek altitude difference mein map hota hai: $\frac{a _{m a x}}{a _{m i n}} = \frac{e ^{- h_{m i n} / H}}{e ^{- h_{m a x} / H}} = e^{(h_{m a x} - h_{m i n}) / H} = e^{- Δ h / H} .$ Corridor width $∣Δ h ∣ = h_{m i n} - h_{m a x}$ ka magnitude solve karo: $∣Δ h ∣ = H ln \frac{a _{m a x}}{a _{m i n}} = 8 ln \frac{6}{3} = 8 ln 2 = 8 \times 0.6931 = 5.55 km .$ Poora safe corridor sirf ≈ 5.5 km tall hai. Couple of km chook gaye toh ya toh skip back ho ke escape, ya overshoot ho ke crash — yahi razor's-edge hai jiski wajah se aerocapture mein precise navigation aur lift steering (bank-angle control) chahiye, sirf "gehri aim" nahi. Figure dekho: drag band narrow hai kyunki density curve itni steep hai.

Figure — Aerocapture — using atmosphere to decelerate into orbit

L5.2 — Heating vs braking, design trade. Peak stagnation heating scale hoti hai $\overset{q}{˙} \propto ρ v^{3}$ jabki braking scale hoti hai $a_{D} \propto ρ v^{2}$ . Do arrival scenarios same required drag $a_{D}$ tak pahunche: scenario X $(ρ_{X}, v_{X}) = (ρ_{0}, v_{0})$ par, aur scenario Y aadhi speed par $v_{Y} = v_{0} /2$ . (a) Same $a_{D}$ ke liye Y ko kitni density $ρ_{Y}$ chahiye? (b) X se Y mein peak heating kis factor se change hoti hai?

Recall Solution

(a) Same drag: $ρ_{X} v_{X}^{2} = ρ_{Y} v_{Y}^{2}$ , toh $ρ_{Y} = ρ_{X} \frac{v _{X}^{2}}{v _{Y}^{2}} = ρ_{0} \frac{v _{0}^{2}}{( v _{0} /2 ) ^{2}} = ρ_{0} \times 4 = 4 ρ_{0} .$ Dheemi Y ko utni hi braking ke liye 4× ghani hawa mein dive karna hoga. (b) Heating ratio: $\frac{q ˙ _{Y}}{q ˙ _{X}} = \frac{ρ _{Y} v _{Y}^{3}}{ρ _{X} v _{X}^{3}} = \frac{4 ρ _{0}}{ρ _{0}} (\frac{v _{0} /2}{v _{0}})^{3} = 2 \times \frac{1}{8} = 0.25.$ Y sirf ek chauthaai rate par heat karta hai. Lesson: speed par cube, density par half-power se zyada dominant hai — kam speed par brake karna (chahe ghani hawa mein) thermally kaafi gentle hota hai. Yahi wajah hai ki tezi se enter karna hard case hai (dekho Atmospheric Entry & Heating): heating, braking nahi, binding constraint hota hai.

Recall Master checklist (hidden — page close karne se pehle self-test karo)

$ε$ ka sign decide karta hai captured (−) vs escaping (+). ::: Sahi — magnitude sirf orbit tightness batata hai.
$v_{\infty}$ vs $v_{in}$ ? ::: $v_{in} = 2 (v_{\infty}^{2} /2 + μ / r_{p})$ — gravity periapsis se pehle speed badha deti hai.
$a_{D}$ mein $v$ ki kaunsi units? ::: m/s (SI), kyunki $ρ$ aur $β$ metric mein hain.
Air density ke liye kaunsa radius? ::: altitude $h = r - R_{planet}$ , $r$ nahi.
Heating limit kyun hai, braking nahi? ::: $\overset{q}{˙} \propto v^{3}$ , $a_{D} \propto v^{2}$ se zyada dominant hai — speed heat ko sabse zyada hurt karta hai.