Terminal landing — propulsive descent, suicide burn: Worked examples

3.4.26 · D3 · Physics › Rocket Flight Mechanics › Terminal landing — propulsive descent, suicide burn

Yeh page tumhare saamne suicide burn ke har tarah ke problems rakhti hai: normal numbers, moving start, "engine bahut weak hai" wala degenerate case, limiting behaviour, ek real-world word problem, aur ek exam twist jo mass loss add karta hai. Har ek ko poori tarah se chew kiya gaya hai. Agar koi bhi symbol unfamiliar lage toh parent note se shuru karo — lekin main har cheez neeche re-define kar raha hoon.

Scenario matrix

Is topic mein jo bhi problem aa sakti hai, woh in cells mein se kisi ek mein fit hoti hai. Last column mein us cell ko cover karne wala Example likha hai.

Cell	Kya special hai	$a_{T} - g$ ka sign / regime	Covered by
A. Plain	Diya $v, a_{T}, g$ → nikalo $h_{b u r n}$	positive, comfortable margin	Ex 1
B. Weak margin	$a_{T}$ barely $g$ se upar	positive but chhota → bahut badi height	Ex 2
C. Degenerate	$a_{T} \leq g$	zero ya negative → impossible	Ex 3
D. Zero input	$v = 0$ (pehle se ruk gaya)	kuch bhi	Ex 4
E. Free-fall start	Rest se $H$ tak giraya, ignition altitude $h^{*}$ nikalo	positive	Ex 5
F. Different world	$g$ badlo (Moon/Mars)	positive, large margin	Ex 6
G. Word problem	Real lander, sensor trigger	positive	Ex 7
H. Limiting behaviour	Kya hota hai jab $a_{T} \to g^{+}$ ya $a_{T} \to \infty$	limit	Ex 8
I. Exam twist	Burn ke dauran mass girti hai (Tsiolkovsky-flavoured)	positive, badhta hua	Ex 9

Worked examples

Cell A — plain case

Worked example Example 1 — seedha substitution

Ek lander $v = 80 m/s$ ki speed se gir raha hai. Engine $a_{T} = 25 m/s^{2}$ deta hai, aur $g = 10 m/s^{2}$ . Ignition altitude $h_{b u r n}$ nikalo.

Forecast: Bacha hua push hai $25 - 10 = 15$ . Guess: kaafi sau metres? Aage padhne se pehle apna number likho.

Net braking acceleration: $a_{n e t} = a_{T} - g = 25 - 10 = 15 m/s^{2}$ . Yeh step kyun? Gravity burn ke dauran bhi kheenchti rehti hai, isliye sirf extra thrust hi actually tumhe slow karta hai. $25$ use karna tumhari braking ko over-promise karna hoga.
Formula lagao: $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} = \frac{8 0 ^{2}}{2 \cdot 15} = \frac{6400}{30}$ . Yeh step kyun? Yeh kinematic relation $v_{f}^{2} = v_{i}^{2} + 2 a d$ hai jisme $v_{f} = 0$ hai, distance $d$ ke liye solve kiya gaya.
Divide karo: $\frac{6400}{30} \approx 213.3 m$ .

Verify: Units — $(m/s)^{2} / (m/s^{2}) = m$ . ✓ Achha, yeh ek length hai. Sanity: $80$ ki speed se girte hue strong engine ke saath, ~ $213$ m ka runway reasonable hai.

Cell B — weak thrust margin

Worked example Example 2 — same speed, weak engine

Same $v = 80 m/s$ , $g = 10$ , lekin weak engine: $a_{T} = 11 m/s^{2}$ . $h_{b u r n}$ nikalo.

Forecast: Margin sirf $1$ hai, Ex 1 ke $15$ ke mukable — pandrah guna chhota. Kya tumhe lagta hai $h_{b u r n}$ $\times 15$ badhta hai?

Net braking: $a_{n e t} = 11 - 10 = 1 m/s^{2}$ . Yeh step kyun? Engine barely gravity ko beat karta hai — brake karne ke liye almost kuch nahi baca.
Formula: $h_{b u r n} = \frac{8 0 ^{2}}{2 \cdot 1} = \frac{6400}{2} = 3200 m$ . Yeh step kyun? Same kinematics; chhota denominator → bada answer.

Verify: Ex 1 se ratio: $3200/213.3 = 15.0$ . ✓ Exactly $15 \times$ , $15 \times$ chhote margin se match karta hua — $h_{b u r n}$ inversely proportional hai $a_{n e t}$ ke. Lesson: ek kamzor engine tumhe kilometres upar se ignite karne pe majboor karta hai, jo dangerous hai. Isliye real landers $a_{T}$ ko $g$ se comfortably upar rakhte hain.

Cell C — degenerate case (thrust brake karne ke liye bahut weak hai)

Worked example Example 3 — jab maths mana kar deta hai

Same $v = 80$ , $g = 10$ , lekin $a_{T} = 8 m/s^{2}$ . $h_{b u r n}$ nikalo.

Forecast: Engine $8$ par upar push karta hai, gravity $10$ par neeche kheenchti hai. Kaun jeetta hai? Formula ka kya hoga?

Net acceleration: $a_{n e t} = 8 - 10 = - 2 m/s^{2}$ . Yeh step kyun? Negative matlab net acceleration ab bhi neeche ki taraf hai — full throttle par bhi tum zameen ki taraf aur tezi se jaate rehte ho.
Formula try karo: $h_{b u r n} = \frac{6400}{2 \cdot ( - 2 )} = - 1600 m$ . Yeh step kyun? Negative height physically meaningless hai — yeh maths chilla raha hai "koi bhi altitude tumhe nahi bachaa sakti."

Verify: Jab $a_{T} \leq g$ ho toh koi suicide burn exist nahi karta. Vehicle decelerate nahi kar sakta; woh crash kar jaayega ignition timing chahe kuch bhi ho. Requirement ek hard inequality hai: $a_{T} > g$ . Neeche diya figure sign map dikhata hai.

Figure dekho: sirf vertical line ke daayein ( $a_{T} > g$ ) wala region real, positive braking distance deta hai. Line par ( $a_{T} = g$ ) height infinity tak blow up ho jaati hai; uske baayein, koi solution nahi.

Cell D — zero input

Cell E — free-fall start, ignition altitude nikalo

Worked example Example 5 — rest se ek known height se giraya gaya

Ek vehicle rest se release hota hai aur free-fall karta hai. Zameen tak available total drop $H_{t o t} = 2000 m$ hai. Engine: $a_{T} = 30$ , $g = 10$ . Zameen se kitni altitude $h^{*}$ par ignite karna chahiye?

Forecast: Girne se speed badhti hai aur zaroori burn height bhi badhti hai. Dono kahin milenge. Halfway mark $1000$ m se upar ya neeche?

Ignition par speed, $(H_{t o t} - h^{*})$ gir ke: rest se free-fall deta hai $v_{i g n} = 2 g (H_{t o t} - h^{*})$ , toh $v_{i g n}^{2} = 2 g (H_{t o t} - h^{*})$ . Yeh step kyun? Energy / kinematics: rest se distance $d$ gir ke speed build hoti hai $v = 2 g d$ . Yahan $d = H_{t o t} - h^{*}$ (pehle se gira hua hissa).
Ignition altitude = required burn height: $h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )}$ . Yeh step kyun? Hum exactly tab ignite karte hain jab current altitude $h^{*}$ us distance ke barabar ho jo rukne ke liye chahiye.
Step 1 ko step 2 mein substitute karo: $h^{*} = \frac{2 g ( H _{t o t} - h ^{*} )}{2 ( a _{T} - g )} = \frac{g ( H _{t o t} - h ^{*} )}{a _{T} - g}$ . Yeh step kyun? $v$ eliminate ho jaata hai; ab ek unknown $h^{*}$ .
Algebraically solve karo: multiply out karo: $h^{*} (a_{T} - g) = g H_{t o t} - g h^{*}$ , toh $h^{*} (a_{T} - g + g) = g H_{t o t}$ , milta hai $h^{*} = \frac{g H _{t o t}}{a _{T}}$ . Yeh step kyun? $g$ 's neatly recombine ho jaate hain — ignition altitude bas total drop ka $g / a_{T}$ hissa hai.
Numbers lagao: $h^{*} = \frac{10 \cdot 2000}{30} = \frac{20000}{30} \approx 666.7 m$ .

Verify: Halfway mark $1000$ m se neeche, jaise forecast tha — tum zyaadatar raste girte ho, phir neeche hard brake karte ho. Cross-check speed: $2000 - 666.7 = 1333.3$ m gire, toh $v_{i g n} = 2 \cdot 10 \cdot 1333.3 = 26666.7 \approx 163.3 m/s$ ; phir $h_{b u r n} = 163. 3^{2} / (2 \cdot 20) = 26666.7/40 = 666.7$ m. ✓ Dono taraf same number.

Cell F — ek alag duniya

Cell G — real-world word problem

Worked example Example 7 — radar-triggered landing

Ek Mars lander (treat karo $g = 3.6 m/s^{2}$ ) ek aisi parachute ke neeche $v = 60 m/s$ ki steady speed se descend karta hai jo usne abhi kaat di. Uska engine $a_{T} = 15 m/s^{2}$ deliver kar sakta hai. Uska landing radar seedha altitude read karta hai. (a) Flight computer ko engine kis altitude par light karna chahiye? (b) Agar radar mein $\pm 5 m$ error ho, toh kya yeh yahan dangerous hai?

Forecast: Mars gravity Moon aur Earth ke beech hai. Nearest 50 m tak ignition altitude guess karo.

Net braking: $a_{n e t} = 15 - 3.6 = 11.4 m/s^{2}$ . Yeh step kyun? Mars gravity subtract karo — woh burn ke dauran bhi kheenchti rehti hai.
Ignition altitude: $h_{b u r n} = \frac{6 0 ^{2}}{2 \cdot 11.4} = \frac{3600}{22.8} \approx 157.9 m$ . Yeh step kyun? Standard formula; yeh woh trigger hai jiska computer wait karta hai.
(b) Radar error impact: $\approx 158 m$ mein se $5 m$ matlab lagbhag $3.2%$ . $5$ m baad trigger hone se stopping distance khaati hai; yeh fatal hai ya nahi yeh thrust reserve par depend karta hai — yahan margin comfortable hai, isliye ek chhota percentage error survivable hai lekin pad kiya jaana chahiye. Yeh step kyun? Hamesha ek absolute error ko us quantity ke fraction mein translate karo jo matter karti hai.

Verify: $5/157.9 = 0.0317 = 3.17%$ . ✓ $h_{b u r n}$ par units: $(m/s)^{2} / (m/s^{2}) = m$ . ✓ Answer Moon-like aur Earth-like values ke beech baitha hai, jaise Mars ke liye expect kiya tha.

Cell H — limiting behaviour

Worked example Example 8 — knobs ko extremes par push karna

$v = 100$ , $g = 10$ fixed rakho. $h_{b u r n}$ describe karo jab (a) $a_{T} \to g^{+} = 1 0^{+}$ (engine barely gravity ko beat karta hai) aur (b) $a_{T} \to \infty$ (infinitely strong engine). (a) ke liye $a_{T} = 10.1$ par numeric value do.

Forecast: Ek limit blow up hogi, ek kuch nahi tak simt jaayegi. Kaun kaun sa hai?

Limit (a), $a_{T} \to g^{+}$ : $a_{n e t} = a_{T} - g \to 0^{+}$ , toh $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} \to + \infty$ . Yeh step kyun? Ek fixed number ko zero ki taraf simt rahe kisi cheez se divide karna → unbounded. Tumhe infinite altitude chahiye hogi — Cell C ki boundary se match karta hai.
$a_{T} = 10.1$ par numeric taste: $a_{n e t} = 0.1$ , $h_{b u r n} = \frac{10000}{0.2} = 50000 m = 50 km$ . Yeh step kyun? Dikhata hai ki height boundary ke paas kitni brutally explode hoti hai.
Limit (b), $a_{T} \to \infty$ : $a_{n e t} \to \infty$ , toh $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} \to 0$ . Yeh step kyun? Ek arbitrarily strong engine essentially koi distance nahi mein tumhe rok deta hai — last metre par ignite karo.

Verify: $a_{T} = 10.1$ par: $10000/ (2 \cdot 0.1) = 10000/0.2 = 50000$ . ✓ $g$ ke paas blow-up confirm hoti hai; aur thrust badhne par $h_{b u r n} \to 0$ , physical intuition se match karta hua (bada deceleration ⇒ tiny stopping distance).

Cell I — exam twist: burn ke dauran mass girti hai

Worked example Example 9 — constant-mass estimate vs. mass loss

Ek lander burn mein $v = 200 m/s$ , mass $m_{0} = 1000 kg$ , constant thrust $F = 30000 N$ , $g = 10$ ke saath enter karta hai. (a) Initial mass use karke constant-mass burn-height estimate do. (b) Qualitatively explain karo ki true height kam kyun hai, Tsiolkovsky Rocket Equation invoke karke.

Forecast: Jaise fuel nikalti hai, mass girti hai, toh $a_{T} = F / m$ badhta hai. Kya yeh real stopping distance ko constant-mass guess se lamba ya chhota banata hai?

Initial thrust acceleration: $a_{T} = F / m_{0} = 30000/1000 = 30 m/s^{2}$ . Yeh step kyun? Thrust and Specific Impulse se $a_{T} = F / m$ — woh acceleration jo thrust akele us instant par deta hai.
Constant-mass net braking: $a_{n e t} = 30 - 10 = 20 m/s^{2}$ . Yeh step kyun? Initial thrust acceleration se gravity subtract karo — sirf bacha hua push brake karta hai, exactly jaise har pehle example mein.
Constant-mass height: $h_{b u r n} = \frac{20 0 ^{2}}{2 \cdot 20} = \frac{40000}{40} = 1000 m$ . Yeh step kyun? Yeh sabse kamzor $a_{T}$ (sabse badi mass) use karta hai, isliye required height ka over-estimate hai.
(b) True height kam kyun hai: jaise propellant jalta hai, $m$ shrink hoti hai, toh $a_{T} = F / m$ burn ke dauran badhta jaata hai; jitna zyaada brake karo deceleration utna strong hoti jaati hai. Constant- $m$ model se zyaada strong braking ⇒ kam distance mein rukna. Exact answer $a_{T} (t) = F / (m_{0} - \overset{m}{˙} t)$ integrate karke aata hai, wahi mass-varying idea jo Tsiolkovsky Rocket Equation ke peeche hai. Yeh step kyun? Exam chahta hai ki tum jaano ki constant-mass formula ek safe upper bound hai, aur exactness ke liye sahi tool ka naam lo.

Verify: $20 0^{2} / (2 \cdot 20) = 40000/40 = 1000$ . ✓ Reasoning " $m ↓\Rightarrow a_{T} ↑\Rightarrow h_{b u r n} ↓$ " monotonic hai aur parent note ke "mass constant 80/20 estimate hai" se consistent hai.

Recall Tumhare liye sabse mushkil cell kaun si thi?

Table cover karo aur Ex 5 ka clean result $h^{*} = g H_{t o t} / a_{T}$ scratch se re-derive karo. Free-fall start ke liye $h^{*}$ ki derivation ::: $h^{*} = 2 g (H_{t o t} - h^{*}) / [2 (a_{T} - g)]$ set karo, cross-multiply karo, $h^{*}$ terms collect karo → $h^{*} = g H_{t o t} / a_{T}$ . Kisi bhi real solution ke liye sign condition ::: $a_{T} > g$ (strictly), warna $h_{b u r n} \leq 0$ ya infinite — koi burn nahi bachaa sakta. $a_{T} \to g^{+}$ hone par $h_{b u r n}$ ka behaviour ::: Yeh $+ \infty$ tak diverge ho jaata hai — tumhe unlimited altitude chahiye hogi.

Connections

Terminal landing — propulsive descent, suicide burn (index 3.4.26) — parent note & core derivation
Kinematics — Equations of Motion — $v^{2} = 2 a d$ jo har cell mein use hua
Thrust and Specific Impulse — jahan se $a_{T} = F / m$ (Ex 9) aata hai
Tsiolkovsky Rocket Equation — exact mass-loss treatment (Cell I)
Powered Descent Guidance — kaise ek real computer trigger schedule karta hai (Ex 7)
Gravity Turn Ascent — is vertical model ka ascent counterpart