Terminal landing — propulsive descent, suicide burn: Visual walkthrough

3.4.26 · D2 · Physics › Rocket Flight Mechanics › Terminal landing — propulsive descent, suicide burn

Step 1 — Situation draw karo: kya gir raha hai aur ground kahan hai?

KYA. Ek rocket (ise lander kaho) seedha neeche flat ground ki taraf gir raha hai. Hum height ko ground se lander tak ki doori ke roop mein measure karte hain. Is height ko hum $h$ kehte hain — metres (m) mein ek akela number. Ground hai $h = 0$ .

KYUN. Koi bhi maths karne se pehle, humein agree karna hoga ki kaunsa direction positive hai. Hum choose karte hain upar = positive. Yeh ek choice baaki sab cheezein ka sign fix kar deti hai, toh baad mein kabhi sign guess nahi karna padega.

PICTURE. Blue arrow lander ki velocity hai — kitni tez aur kis direction mein move kar raha hai. Kyunki woh gir raha hai, woh arrow neeche point karta hai, isliye uski value negative hai. Girti speed (ek positive number) ko hum $v$ likhte hain, toh velocity khud $- v$ hai.

Figure — Terminal landing — propulsive descent, suicide burn

Step 2 — Doono accelerations ko naam do: gravity neeche, thrust upar

KYA. Do cheezein lander ki speed change karti hain:

Gravity use neeche kheenchti hai, ek steady rate par downward speed add karti hai. Woh rate hai $g$ (Earth par, lagbhag $9.8 m/s^{2}$ ; hum aksar $10$ round karte hain).
Engine use upar push karta hai. Woh push ek upward change-of-speed rate produce karta hai jise hum $a_{T}$ kehte hain ("acceleration from thrust").

YEH symbols kyun, force nahi? Tum force $F$ newtons mein baat kar sakte the, lekin jo cheez actually motion ko mod deti hai woh hai acceleration — har second mein kitni m/s speed gain hoti hai. Kyunki thrust deta hai $a_{T} = F / m$ (force divided by mass), seedha $a_{T}$ mein kaam karne se humein $m$ carry karna nahi padta. Track karne ke liye ek kam symbol.

PICTURE. Ek hi lander par do vertical arrows: ek red gravity arrow neeche point karta hua (length $g$ ), ek orange thrust arrow upar point karta hua (length $a_{T}$ ). Gaur karo orange arrow lamba drawn hai — yeh Step 4 mein matter karega.

Step 3 — Arrows add karo: burn ke dauran net acceleration

KYA. Jab engine on hai, doono arrows ek saath kaam karte hain. Upar positive hai, isliye hum add karte hain upward thrust ( $+ a_{T}$ ) aur downward gravity ( $- g$ ). Jo bachta hai woh hai net acceleration: $a_{n e t} = engine pushes up, + a_{T} - gravity pulls down, - g$

KYUN. Gravity kabhi band nahi hoti. Engine ko gravity se ladna aur uske baad bhi kuch bachana hoga taaki fall actually slow ho sake. Sirf woh bachta hua hissa, $a_{T} - g$ , braking karta hai. Agar tum $- g$ bhool gaye toh hamesha bahut jaldi rokne ka plan banaaoge aur crash karoge (parent ke mistake box dekho).

PICTURE. Red gravity arrow ko orange thrust arrow se subtract kiya gaya hai, ek chhota green "net" arrow upar pointing chhodke. Woh green arrow hi sirf fall ko decelerate kar raha hai.

Step 4 — Zindagi-maut ki condition: $a_{T}$ ko $g$ se jeetna hoga

KYA. Green arrow ka sign dekho. Teen cases hain:

$a_{T} > g$ → $a_{n e t} > 0$ → green arrow upar point karta hai → tum slow ho sakte ho. ✅
$a_{T} = g$ → $a_{n e t} = 0$ → koi green arrow nahi → speed kabhi nahi badlti; tum hamesha constant speed se girte raho. ❌
$a_{T} < g$ → $a_{n e t} < 0$ → green arrow neeche point karta hai → engine gravity ko bhi rok nahi sakta; tum aur tez hote rehte ho. ❌

KYUN teeno dikhao? Yeh woh degenerate case hai jo reader ko kabhi bina jaane hit nahi karna chahiye. Suicide burn tabhi possible hai jab engine gravity se zyada accelerate kare. Kisi heavy-gravity world par ya weak engine ke saath, koi bhi ek ignition altitude nahi hai jo kaam kare — tum is tarah simply land nahi kar sakte.

PICTURE. Teen landers side by side, har ek ke thrust/gravity/net arrows ke saath. Sirf pehle ka upward green arrow hai.

Step 5 — Tool: kaunsa equation of motion speed ko distance mein convert karta hai?

KYA. Hum chahte hain distance jaanna — steady braking $a_{n e t}$ use karke falling speed $v$ ko $0$ tak laane ke liye. Hamare paas teen quantities hain — starting speed, acceleration, distance — aur koi time nahi. Exactly ek equation of motion hai jo teeno ko time ke bina connect karta hai: $v_{f}^{2} = v_{i}^{2} + 2 a d$

YEH tool kyun, koi aur kyun nahi? Baaki kinematic equations ( $v_{f} = v_{i} + a t$ , $d = v_{i} t + \frac{1}{2} a t^{2}$ ) sab mein time $t$ hai. Humne kabhi "kitna time?" nahi pucha — humne "kitna door?" pucha. Time-free equation choose karne ka matlab hai pehle $t$ solve nahi karna padega. Kam steps, slip karne ke kam jagah.

PICTURE. Ek speed-vs-distance graph. $v^{2}$ ki curve distance ke against plot karne par straight line hoti hai — woh straightness hi kyun hai ki $v^{2} = 2 a d$ yahan clean tool hai.

Step 6 — Plug in karo aur burn distance ke liye solve karo

KYA. Braking burn ke dauran:

Start speed $v_{i} = v$ (ignition par falling speed, positive magnitude ke roop mein).
End speed $v_{f} = 0$ (hum chahte hain rukke hue pahunche).
Acceleration = braking magnitude $a_{n e t}$ (yeh fall ka virodh karta hai, isliye $v^{2}$ kam karta hai).
Distance = $h_{b u r n}$ , woh height jo braking ke dauran humein chhorni hogi.

Substitute karo: $0 = start energy-per-mass v^{2} - taken away by braking 2 a_{n e t} h_{b u r n}$

Ab rearrange karo — dono sides mein $2 a_{n e t} h_{b u r n}$ add karo, phir $2 a_{n e t}$ se divide karo: $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} = \frac{v ^{2}}{2 ( a _{T} - g )}$

KYUN. Yeh ek line poora result hai. Iska matlab hai: speed $v$ ka ek fall mitaane ke liye, tumhe height ka ek runway chahiye jitna $v^{2}$ ko braking rate ke double se share kiya jaaye. Speed double karo → height chaar guna (square ki wajah se). Woh $v^{2}$ hi reason hai ki landings extra speed ke saath itni unforgiving hoti hain.

PICTURE. Height axis par braking stretch drawn hai: lander $- v$ par enter karta hai, green net arrow speed ko khaata jaata hai, aur exactly $h = 0$ par $0$ reach karta hai. Us stretch ki length hai $h_{b u r n}$ .

Step 7 — Trigger: tab fire karo jab altitude required burn height tak gir jaaye

KYA. Jab lander girta hai, uski speed $v$ badhti hai, isliye required burn height $h_{b u r n} (v)$ bhi badhti hai. Saath hi uski actual altitude $h$ simaatti hai. Ek instant aisa hota hai jahan doono milte hain: $h = h_{b u r n} (v) = \frac{v ^{2}}{2 ( a _{T} - g )} ⟹ abhi ignite karo.$

KYUN. Milne se pehle fire karo ( $h > h_{b u r n}$ ) toh tum upar rok ke hover karoge aur fuel waste karoge — "bahut jaldi" wala crash. Baad mein fire karo ( $h < h_{b u r n}$ ) toh runway khatam ho jaayegi — "bahut deri" wala crash. Crossing woh aakhri safe instant hai: woh "suicide" moment.

PICTURE. Ek altitude-vs-speed chart par do curves: simaatti actual altitude (blue) aur badhti required burn height (orange). Unka crossing point red circle mein hai — woh hai ignition instant.

Step 8 — Girne ki speed kahan se aati hai? (rest se girna)

KYA. Agar lander rest par shuru hua aur ignite karne se pehle freely distance $H$ gira, toh gravity akele ne uski speed build ki. Wahi time-free tool use karke $v_{i} = 0$ , $a = g$ , $d = H$ ke saath: $v^{2} = 0 + 2 g H ⟹ v = 2 g H$

Fall distance ko $H_{t o t} - h^{*}$ set karo (total drop minus ground ke upar ignition height $h^{*}$ ) aur demand karo ki fall speed burn requirement se equal ho: $v_{i g n} = 2 g (H_{t o t} - h^{*}), h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )}$

KYUN. Yeh loop close karta hai: yeh tumhe ignition altitude $h^{*}$ purely drop geometry se bata deta hai, speedometer dekhe bina bhi. Do equations, do unknowns ( $v_{i g n}$ , $h^{*}$ ).

PICTURE. Ek single vertical drop do coloured segments mein divided: upar gray free-fall of height $H_{t o t} - h^{*}$ (speed ek curve ke saath badhti hui), phir neeche green braking segment of height $h^{*}$ (speed wapas zero tak girti hui).

Ek-picture summary

Upar ki sari cheez ek diagram mein collapse ho jaati hai: rest se puri descent, upar gray free-fall, ignition crossing marked, neeche green braking wedge, aur boxed formula jagah par annotated. Agar tum sirf ek image yaad rakh sako, toh yahi yaad rakho.

Recall Feynman retelling — plain words mein poora walkthrough

Tum ek rocket drop karte ho. Neeche "minus" hai, upar "plus" hai — hum ne yeh ek baar choose kiya taaki signs ke baare mein kabhi argue na karen (Step 1). Do cheezein ise kheenchti hain: gravity hamesha $g$ rate par neeche kheenchti hai, engine $a_{T}$ rate par upar push karta hai (Step 2). Unhe add karo aur sirf bachta hua hissa, $a_{T} - g$ , actually tumhe slow karta hai (Step 3) — aur agar woh bachta hua hissa positive nahi hai, bhool jaao, tum kabhi ruk nahi sakte (Step 4). Rokne ke liye kitni jagah chahiye yeh jaanne ke liye, hum woh ek motion equation lete hain jismein koi time nahi, $v^{2} = 2 a d$ , kyunki humne "kitna door?" pucha tha, "kitna time?" nahi (Step 5). "Speed $v$ se shuru, zero par khatam, $a_{T} - g$ par brake" feed karo, aur required height $v^{2} / [2 (a_{T} - g)]$ nikalta hai — square notice karo, isliye double speed ko chaar guna jagah chahiye (Step 6). Tum engine tab fire karte ho jab tumhari real altitude us required height tak gir jaaye — pehle nahi (tum hover karke starve ho jaate), baad mein nahi (tum smash ho jaate) (Step 7). Aur agar tum way up high ek still hover se shuru kiye the, toh gravity ne khud tumhari speed $v = 2 g H$ se set ki, jo exactly woh altitude pin down karti hai jahan candle jalaani hai (Step 8).

Active recall

Connections

Parent topic — poori theory aur worked examples
Kinematics — Equations of Motion — $v^{2} = 2 a d$ kahan se aata hai (Steps 5, 8)
Thrust and Specific Impulse — $a_{T} = F / m$ ka source (Step 2)
Tsiolkovsky Rocket Equation — real burn ke liye constant-mass shortcut ko correct karta hai
Powered Descent Guidance — is idea ko real 3-D flight mein steer karna