Terminal landing — propulsive descent, suicide burn: Exercises

3.4.26 · D4 · Physics › Rocket Flight Mechanics › Terminal landing — propulsive descent, suicide burn

Definition Is poore page par use hone wala sign convention

Hum 1-D vertical motion mein kaam karte hain aur up = positive, down = negative choose karte hain (wahi convention jo parent note mein hai).

Altitude $h \geq 0$ : ground ke upar ki height.
Girte waqt Velocity negative hoti hai (neeche ki taraf); iska magnitude — seedha "speed" — hum $v > 0$ kehte hain. To ek lander "falling at $v = 200 m/s$ " ka matlab velocity $- 200 m/s$ hai.
$g > 0$ gravity ka magnitude hai; gravity ka acceleration $- g$ hai (neeche ki taraf point karta hai).
$a_{T} = F / m > 0$ woh upar ki taraf ka acceleration hai jo engine akela produce karta. Yahan $F$ engine thrust force (newtons mein) hai aur $m$ vehicle ki mass (kg mein) hai; force ko mass se divide karne par acceleration milta hai, Newton ke doosre law $F = ma$ se (dekho Thrust and Specific Impulse jahan se $F$ aata hai).
Burn ke dauran net acceleration $a_{n e t} = a_{T} - g$ hai (up positive minus downward pull), jo positive honi chahiye taaki fall actually slow ho sake.

Common mistake Yeh formula kab use karna allowed hai? (domain of validity)

Denominator $a_{T} - g$ hai, isliye teen regimes hain:

$a_{T} > g$ → denominator positive → $h_{b u r n}$ finite aur positive. Yeh ek hi landable case hai, aur neeche ke saare problems yahan rehte hain.
$a_{T} = g$ → denominator $= 0$ → $h_{b u r n} = v^{2} /0 \to \infty$ . Engine gravity ko exactly cancel karta hai, net braking zero hai, to chahe kitna bhi altitude ho aap kabhi slow nahi hote. Formula "infinite runway" return karta hai — yeh red flag hai, landing nahi.
$a_{T} < g$ → denominator negative → $h_{b u r n}$ negative nikalta hai, jo physically meaningless hai (aap ground ke neeche ignite nahi kar sakte). Yeh algebra ka cheekh ke kehna hai "impossible": gravity se kamzor thrust ke saath aap aur tezi se girte rehte ho. Koi bhi number trust karne se pehle hamesha $a_{T} > g$ check karo.

Formula

$v^{2} = 2 a d$ kahan se aata hai (poore steps, taaki yeh page apne aap mein complete ho) Hum speed aur distance ke beech ek aisa relation chahte hain jo time kabhi mention na kare — kyunki burn problems mein hum "kitna altitude" care karte hain, "kitne seconds" nahi. Constant-acceleration ke do facts se shuru karo (derive kiye hain Kinematics — Equations of Motion mein): $v_{f} = v_{i} + a t (1)$ $d = v_{i} t + \frac{1}{2} a t^{2} (2)$ $t$ eliminate kyun karein? Kyunki $t$ ek aisi quantity hai jo hum jaante nahi aur care bhi nahi karte; hamare paas sirf speeds aur distance hai. To hum ise hata dete hain. Step A — (1) se time solve karo: $t = \frac{v _{f} - v _{i}}{a}$ . Step B — yeh $t$ (2) mein substitute karo: $d = v_{i} (\frac{v _{f} - v _{i}}{a}) + \frac{1}{2} a (\frac{v _{f} - v _{i}}{a})^{2} .$ Step C — common denominator $a$ par har piece expand karo: $d = \frac{v _{i} ( v _{f} - v _{i} )}{a} + \frac{( v _{f} - v _{i} ) ^{2}}{2 a} = \frac{2 v _{i} v _{f} - 2 v _{i}^{2} + v _{f}^{2} - 2 v _{i} v _{f} + v _{i}^{2}}{2 a} .$ Step D — $2 v_{i} v_{f}$ terms cancel ho jaate hain aur $- 2 v_{i}^{2} + v_{i}^{2} = - v_{i}^{2}$ , bacha: $d = \frac{v _{f}^{2} - v _{i}^{2}}{2 a} ⟹ v_{f}^{2} = v_{i}^{2} + 2 a d .$ Iska matlab: distance speed ke square ke saath badhti hai — yeh seed hai har " $v^{2}$ " ka jo tumhe neeche dikhega.

Level 1 — Recognition

Problem 1.1

Ek lander gir raha hai. Uska engine, akele, use upar $a_{T} = 25 m/s^{2}$ par push karega. Local gravity $g = 10 m/s^{2}$ hai. Net braking acceleration kya hai jo vehicle actually feel karti hai jab engine lit ho jaata hai?

Recall Solution

Jab engine fire hota hai tab gravity off nahi hoti — woh poore time neeche pull karti rehti hai. To useful deceleration sirf woh bacha hua push hai jo gravity ne jo churaya usse upar hai: $a_{n e t} = a_{T} - g = 25 - 10 = 15 m/s^{2} .$ Answer: $15 m/s^{2}$ .

Problem 1.2

Do landers, A aur B, dono Mars par approach kar rahe hain ( $g = 3.7 m/s^{2}$ ). A ka $a_{T} = 5 m/s^{2}$ hai, B ka $a_{T} = 3 m/s^{2}$ . Kaun physically suicide burn perform kar sakta hai, aur kaun bilkul ruk nahi sakta?

Recall Solution

Suicide burn tabhi kaam karta hai jab net acceleration upar ki taraf (positive) point kare, yani $a_{T} > g$ ho.

Lander A: $a_{T} - g = 5 - 3.7 = + 1.3 m/s^{2} > 0$ → brake kar sakta hai.
Lander B: $a_{T} - g = 3 - 3.7 = - 0.7 m/s^{2} < 0$ → net acceleration abhi bhi neeche hai; engine use hold bhi nahi kar sakta, slow karna to door ki baat. Ruk nahi sakta. Answer: A kar sakta hai, B nahi kar sakta.

Level 2 — Application

Problem 2.1

Ek booster seedha neeche $v = 200 m/s$ par gir raha hai. Engine $a_{T} = 30 m/s^{2}$ deta hai; $g = 10 m/s^{2}$ . Kis altitude par use ignite karna chahiye taaki velocity aur altitude saath mein zero ho jaayein?

Recall Solution

Net braking: $a_{n e t} = 30 - 10 = 20 m/s^{2}$ . Burn altitude: $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} = \frac{20 0 ^{2}}{2 \cdot 20} = \frac{40000}{40} = 1000 m .$ Answer: $1000 m$ par ignite karo.

Problem 2.2

Wahi booster, wahi engine, lekin ab woh do guna tez gir raha hai, $v = 400 m/s$ . Pehle predict karo ki required altitude double hogi ya nahi, phir compute karo.

Recall Solution

Forecast: height $v^{2}$ ke saath badhti hai, to $v$ double karne par height chaar guni honi chahiye → $4000 m$ expect hai, $2000$ nahi. Verify: $h_{b u r n} = \frac{40 0 ^{2}}{2 \cdot 20} = \frac{160000}{40} = 4000 m . ✓$ Answer: $4000 m$ — chaar guna, do guna nahi.

Problem 2.3

Ek lander ko $600 m$ se zyada upar ignite nahi karna (zyada upar gaya to ek ridge se takraayega). Uske engine ka $a_{n e t} = 12 m/s^{2}$ hai. Ignition par maximum falling speed kya ho sakti hai?

Recall Solution

Problem dhyan se padho: ridge sabse bada allowed burn altitude set karti hai, to ceiling $600 m$ fastest permitted speed par burn altitude hai hi. Isliye hum $h_{b u r n} = 600 m$ set karte hain aur $v$ solve karte hain. Burn formula ko $v$ ke liye rearrange karo: $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} ⟹ v = 2 a_{n e t} h_{b u r n} .$ $h_{b u r n} = 600 m$ aur $a_{n e t} = 12 m/s^{2}$ substitute karo: $v = 2 \cdot 12 \cdot 600 = 14400 = 120 m/s .$ Answer: $120 m/s$ — isse tez ho to $600 m$ ki runway waqt par rokne ke liye kaafi nahi hai.

Level 3 — Analysis

Problem 3.1

Ek rocket rest se total height $H_{t o t} = 2000 m$ se girta hai. Engine $a_{T} = 30 m/s^{2}$ deta hai, $g = 10$ . Woh altitude $h^{*}$ dhundho ground ke upar jahan suicide burn ignite hoti hai.

Neeche ki figure is problem ka map hai — jaate jaate isko refer karo. Vertical axis (white arrow, "altitude h" label) ground ke upar ki height hai, up positive. Upar amber ▼ start point ko mark karta hai ( $h = 2000 m$ , rest mein). Lambi cyan arrow Phase 1 hai, free fall jahan speed badhti hai. Amber dashed line ignition altitude $h^{*}$ hai jo hum solve kar rahe hain, amber ● se mark ki hui. Iske neeche amber arrow Phase 2 hai, braking burn $h = 0$ , $v = 0$ par cyan ground line tak.

Figure — Terminal landing — propulsive descent, suicide burn

Recall Solution

Figure dekho: fall do arrows mein split hoti hai, aur har ek ek equation deta hai.

Drop ka function banate hue fall speed (cyan Phase-1 arrow). Rest se $(H_{t o t} - h^{*})$ ki drop se freely girne par — cyan arrow ki length — amber ignition line par pahunchi speed hai (from $v^{2} = 2 a d$ with $a = g$ , $v_{i} = 0$ ): $v_{i g n} = 2 g (H_{t o t} - h^{*}) .$
Burn requirement (amber Phase-2 arrow). $v_{i g n}$ se baaki bacha $h^{*}$ height par rokne ke liye — amber arrow ki length ground line tak: $h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )} .$

(1) ko (2) mein substitute karo. $v_{i g n}^{2} = 2 g (H_{t o t} - h^{*})$ term upar jaata hai: $h^{*} = \frac{2 g ( H _{t o t} - h ^{*} )}{2 ( a _{T} - g )} = \frac{g ( H _{t o t} - h ^{*} )}{a _{T} - g} .$ Numbers daalo: $\frac{g}{a _{T} - g} = \frac{10}{20} = 0.5$ . $h^{*} = 0.5 (2000 - h^{*}) = 1000 - 0.5 h^{*} ⟹ 1.5 h^{*} = 1000 ⟹ h^{*} = 666.7 m .$ Exactly wahan amber dashed line figure mein baithti hai. Ignition speed check karo (amber ● dot ke paas ki value): $v_{i g n} = 2 \cdot 10 \cdot (2000 - 666.7) = 26667 \approx 163.3 m/s$ . Answer: $h^{*} \approx 667 m$ par ignite karo, $\approx 163 m/s$ se girte hue.

Problem 3.2

Rest se fall ke liye, dikhao ki ignition altitude total drop height ka ek fixed fraction hai, aur woh fraction $a_{T}$ aur $g$ ke terms mein dhundho.

Recall Solution

Problem 3.1 ke substituted equation se: $h^{*} = \frac{g}{a _{T} - g} (H_{t o t} - h^{*}) .$ Maano $r = \frac{g}{a _{T} - g}$ . To $h^{*} = r H_{t o t} - r h^{*} ⟹ h^{*} (1 + r) = r H_{t o t}$ , isliye $\frac{h ^{*}}{H _{t o t}} = \frac{r}{1 + r} = \frac{g}{a _{T}} .$ Ignition fraction simply $g / a_{T}$ hai. Zyada thrust → chhota fraction → aap zyada der ruko (neeche fire karo). Check 3.1: $g / a_{T} = 10/30 = 1/3$ , aur $2000/3 = 666.7 m$ ✓. Answer: $h^{*} / H_{t o t} = g / a_{T}$ .

Level 4 — Synthesis

Problem 4.1

Ek lander ka dry-plus-payload mass aisa hai ki ignition par $a_{T} = 24 m/s^{2}$ milta hai ( $g = 10$ ). Woh $v = 180 m/s$ se gir raha hai. Engine ka exhaust velocity ise specific-impulse-based mass-flow deta hai jisse woh full thrust ke har second mein apni $8%$ mass burn karta hai (from Thrust and Specific Impulse). (a) Constant-mass formula se burn altitude dhundho. (b) Constant- $a_{n e t}$ approximation use karke burn duration estimate karo. (c) Consumed propellant fraction estimate karo.

Recall Solution

(a) $a_{n e t} = 24 - 10 = 14 m/s^{2}$ . $h_{b u r n} = \frac{18 0 ^{2}}{2 \cdot 14} = \frac{32400}{28} \approx 1157.1 m .$ (b) Constant deceleration ke saath, speed $v$ khatam karne ka time $t = v / a_{n e t}$ hai: $t = \frac{180}{14} \approx 12.86 s .$ (c) $8% = 0.08$ per second at $\approx 12.86 s$ (linear first estimate): $Δ m / m \approx 0.08 \times 12.86 \approx 1.03.$ Yeh $100%$ se zyada hai — red flag. Matlab constant-mass model break down ho jaata hai: is flow par itni der burn nahi kar sakte bina dry hue. Reality mein mass girta hai, $a_{T} = F / m$ badhta hai, to $a_{n e t}$ badhti hai, burn 12.86 s se chhota hota hai, aur kam propellant use hota hai. Sahi exact treatment ke liye Problem 5.1 dekho. Answers: (a) $\approx 1157 m$ , (b) $\approx 12.9 s$ (over-estimate), (c) linear estimate $\approx 103%$ — signal karta hai ki constant-mass model yahan invalid hai.

Problem 4.2

Same $v = 150 m/s$ , $g = 10$ descent ke liye do candidate engines: Engine X $a_{T} = 20 m/s^{2}$ deta hai, Engine Y $a_{T} = 40 m/s^{2}$ deta hai. Kaun neeche ignite karne deta hai, aur Y ka burn altitude X se kitne factor se chhota hai?

Recall Solution

$h_{b u r n} \propto 1/ (a_{T} - g)$ .

X: $a_{n e t} = 10$ , $h_{X} = \frac{15 0 ^{2}}{2 \cdot 10} = \frac{22500}{20} = 1125 m$ .
Y: $a_{n e t} = 30$ , $h_{Y} = \frac{22500}{2 \cdot 30} = \frac{22500}{60} = 375 m$ . Ratio $h_{Y} / h_{X} = 375/1125 = 1/3$ . Engine Y neeche ignite karta hai (safer margin) aur sirf ek-tehai altitude chahiye. Dhyan do $a_{T}$ 20→40 double karne se height aadhi nahi hui — net braking teen guni ho gayi (10→30), to height teen gune se giri. Answer: Y neeche ignite karta hai; $h_{Y} = h_{X} /3$ .

Level 5 — Mastery

Problem 5.1

Mass-varying burn par exact-ish check. Ek lander apna burn $v_{0} = 100 m/s$ speed par enter karta hai, thrust se initial $a_{T} = 20 m/s^{2}$ ; $g = 10$ . Kyunki propellant nikal raha hai, true deceleration burn ke through badhti jaati hai. Constant-mass burn altitude ko two-segment estimate ke saath compare karo jo speed loss ki pehli aadhi ke liye $a_{T} = 20$ aur doosri aadhi ke liye $a_{T} = 25$ use karta hai (mass girne se thrust-accel badhne ka mimicry). Kaun sa kam altitude chahta hai, aur kya woh physics se match karta hai?

Recall Solution

Constant-mass baseline: $a_{n e t} = 20 - 10 = 10$ , $h_{co n s t} = \frac{10 0 ^{2}}{2 \cdot 10} = \frac{10000}{20} = 500 m .$ Two-segment estimate. Speed drop $100 \to 0$ ko $v = 50$ par do halves mein split karo. Har segment ke liye $v_{f}^{2} = v_{i}^{2} + 2 a d$ ko $d = (v_{i}^{2} - v_{f}^{2}) / (2 a_{n e t})$ rearrange karo (yahan $v_{i}$ = segment enter karte waqt tez speed, $v_{f}$ = chhodni waqt slow speed).

Segment 1: $100$ se $50 m/s$ slow karo, mass abhi bhi zyada hai to $a_{T} = 20$ , $a_{n e t} = 20 - 10 = 10$ : $h_{1} = \frac{10 0 ^{2} - 5 0 ^{2}}{2 \cdot 10} = \frac{10000 - 2500}{20} = \frac{7500}{20} = 375 m .$
Segment 2: $50$ se $0$ slow karo, mass gir gaya to $a_{T} = 25$ , $a_{n e t} = 25 - 10 = 15$ : $h_{2} = \frac{5 0 ^{2} - 0 ^{2}}{2 \cdot 15} = \frac{2500}{30} \approx 83.3 m .$
Total burn altitude: dono segment heights add karo: $h_{2 se g} = h_{1} + h_{2} = 375 + 83.3 \approx 458.3 m .$

Conclusion: mass-varying estimate ko kam altitude chahiye ( $\approx 458 m$ vs $500 m$ constant-mass). Yeh physics se match karta hai: jaise fuel burn hota hai, $a_{T} = F / m$ badhta hai, net braking strengthen hoti hai, aur vehicle kam distance mein rukta hai. To constant-mass formula tumhare chahiye runway ka conservative over-estimate hai — jo achha hai, kyunki braking room over-estimate karna safe side par error karta hai. Answer: mass-varying $\approx 458 m < 500 m$ constant-mass; badhti thrust-acceleration runway ko shorten karti hai.

Problem 5.2

Ek booster Earth par rest se girta hai ( $g = 9.81$ ) aur ground par zero speed tak pahunchna chahiye. Uska engine $a_{T} = 29.43 m/s^{2}$ deta hai (exactly $3 g$ ). Total drop $H_{t o t} = 3000 m$ . (a) Ignition altitude fraction aur altitude dhundho. (b) Ignition speed dhundho. (c) Ignition speed ko woh ground-impact speed jo vehicle ki burn ke bina hoti se compare karo, aur comment karo.

Recall Solution

(a) Problem 3.2 se, ignition fraction $= g / a_{T} = 1/3$ . To $h^{*} = \frac{1}{3} \times 3000 = 1000 m .$ (b) $H_{t o t} - h^{*} = 2000 m$ se girne ke baad ignition speed: $v_{i g n} = 2 g (2000) = 2 \cdot 9.81 \cdot 2000 = 39240 \approx 198.09 m/s .$ Cross-check burn formula se: $h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )} = \frac{39240}{2 \cdot 19.62} = \frac{39240}{39.24} = 1000 m$ ✓. (c) Burn ke bina poore $3000 m$ ke baad impact speed hogi $v_{g r o u n d} = 2 \cdot 9.81 \cdot 3000 = 58860 \approx 242.6 m/s .$ To vehicle $\approx 198 m/s$ par ignite karta hai (already bahut tez) aur, last kilometre mein, woh sab plus jo extra gain hota — burn aakhir mein sabse zyada kaam karta hai. Woh "almost poora rasta girne do, phir ek dum brakes maro" exactly suicide burn ki fuel-optimal logic hai. Answers: (a) fraction $1/3$ , $h^{*} = 1000 m$ ; (b) $v_{i g n} \approx 198.1 m/s$ ; (c) no-burn impact $\approx 242.6 m/s$ .

Common mistake L5 trap — yeh assume karna ki constant-mass altitude

unsafe hai kyunki woh rising thrust ignore karta hai. Kyun sahi lagta hai: "mera model approximate hai, to shayad mujhe bahut neeche fire karne ko bol de aur main crash ho jaoon." Kyun galat hai: mass loss ignore karna teri braking under-estimate karta hai (real $a_{n e t}$ zyada hoti hai), isliye constant-mass formula reality se zyada required altitude deta hai — tum thoda pehle fire karoge aur end mein ek tiny hover margin hogi, crash nahi. Dangerous direction braking over-estimate karna hai (e.g. $g$ bhool jaana), jo tumhe bahut neeche fire karwa deta hai. Fix: jano ki har approximation kis taraf error karti hai. $g$ bhoolna → bahut neeche fire karo (dangerous). Mass loss ignore karna → thoda upar fire karo (conservative).

Recall Poore page ka one-line summary

$h_{b u r n} = v^{2} / [2 (a_{T} - g)]$ (sirf $a_{T} > g$ ke liye valid); rest se, $g / a_{T}$ fraction par ignite karo drop ka; height $v^{2}$ se scale hoti hai; $g$ bhoolna dangerous hai, mass loss ignore karna conservative hai.

Connections

3.4.26 Terminal landing — propulsive descent, suicide burn (Hinglish) — parent note jise yeh exercises drill karte hain
Kinematics — Equations of Motion — $v^{2} = 2 a d$ engine har problem ke peeche
Tsiolkovsky Rocket Equation — exact mass-loss treatment jo L4/L5 mein invoke hota hai
Thrust and Specific Impulse — $a_{T} = F / m$ ka source (jahan $F$ = thrust force hai) aur Problem 4.1 mein mass-flow
Powered Descent Guidance — flight mein $h_{b u r n} (v)$ ka real-time recomputation