Terminal landing — propulsive descent, suicide burn

3.4.26 · Physics › Rocket Flight Mechanics

Suicide burn KYA hota hai?

NAAM kyun yeh hai? Agar ignition thodi bhi der se ho, toh recovery ka koi chance nahi — ground pehle aa jaata hai, rukne se pehle. Yeh "all or nothing" hai, isliye suicide.

Yeh karte kyun hain (80/20 reason)? Rockets ka mass fraction budget low hota hai. Har second ka thrust propellant $\propto$ thrust jalata hai. Koi bhi hovering ya early braking fuel waste karta hai kyunki gravity se zyada der tak ladna padta hai. End mein single-continuous-burn karna total impulse against gravity ko minimize karta hai → maximum payload / minimum landing fuel.

Ignition altitude derive kaise karein (first principles se)

1-D vertical model set up karo. Upar ko positive lo. Vehicle mass $m$ (short burn ke dauran ~constant maano), gravitational acceleration $g$ , engine thrust se upward acceleration $a_{T} = F / m$ milta hai.

Phase 1 — free fall: rest (ya given speed) se altitude ke through, speed gain karte hue. Phase 2 — braking burn: net upward acceleration $a_{n e t} = a_{T} - g$ (decelerate karne ke liye yeh positive hona chahiye!).

Step 1: Burn ke dauran net deceleration

Thrusting ke waqt net force: $F - m g$ upward. $a_{n e t} = \frac{F}{m} - g = a_{T} - g$ Yeh step kyun? Gravity kabhi switch off nahi hoti — engine ko downward momentum aur gravity dono se ladna hota hai, isliye useful braking sirf weight se zyada thrust hai.

Step 2: Velocity $v$ ko khatam karne ke liye kitni distance chahiye

Burn mein downward speed $v$ se enter karo (matlab velocity $= - v$ ). Hum chahte hain yeh $0$ ho jaaye. $v_{f}^{2} = v_{i}^{2} + 2 a d$ use karo, braking acceleration magnitude $a_{n e t}$ braking distance $h_{b u r n}$ pe act karta hai: $0 = v^{2} - 2 a_{n e t} h_{b u r n}$ $h_{b u r n} = \frac{v ^{2}}{2 a _{n e t}} = \frac{v ^{2}}{2 ( a _{T} - g )}$ Yeh step kyun? Yeh pure kinematics hai (equations of motion ki energy form). Yeh kehta hai: jitni tez giroge, zaroorat ka runway utna zyada — distance $v^{2}$ ke saath badhta hai.

Step 3: Trigger condition

Agar vehicle altitude $h$ pe speed $v$ se gir raha hai, toh tumhe us moment ignite karna hai jab $h \leq h_{b u r n} (v) = \frac{v ^{2}}{2 ( a _{T} - g )}$ Kyunki $h$ aur $v$ dono girate waqt change hote rehte hain, tum continuously $h_{b u r n} (v)$ compute karte raho aur jab current altitude us tak pahunche tab fire karo.

Step 4: Girne ki speed ka altitude se rishta (agar rest se start karo)

Agar tum rest se height $H$ ke through freely gire burn se pehle, toh energy deta hai $v = 2 g H$ . Substitute karo aur dekho ignition altitude kahan trigger hogi. Ignition altitude $h^{*}$ (ground se measure ki gayi), total drop height $H_{t o t}$ ke saath fall speed aur burn requirement ko equate karo: $v_{i g n} = 2 g (H_{t o t} - h^{*}), h^{*} = \frac{v _{i g n}^{2}}{2 ( a _{T} - g )}$

Figure — Terminal landing — propulsive descent, suicide burn

Steel-manned mistakes

Worked examples

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum ek toy floor pe gentle land karna chahte ho drop karke, uske neeche ek tiny upward jet use karke. Agar jet jaldi on karo, toy mid-air mein ruk jaata hai aur phir, jab jet ki hawa khatam ho, woh gir ke floor pe thappad lagaata hai. Agar bahut der se on karo, tum use time pe nahi rok sakte — phir thappad. Ek perfect moment hota hai, neeche ke paas, jab jet on karo taaki woh bilkul zero speed pe touch down kare. Tez girna = jet upar se on karna padta hai, aur bahut zyada upar agar speed bahut zyada ho (double speed ko chaar guna room chahiye). Aur yaad raho: gravity poore time kheenchti rehti hai, isliye jet itna strong hona chahiye ki gravity aur fall dono ko beat kare.

Active recall

Suicide burn (hoverslam) kya hota hai?

Braking engine ko latest possible instant pe fire karna taaki velocity aur altitude simultaneously zero ho jaayein — fuel-optimal single continuous burn.

Fall speed v diye hue burn (ignition) altitude ka formula?

h_{b u r n} = v^{2} / [2 (a_{T} - g)]

Thrust acceleration se g kyun subtract karte hain?

Gravity burn ke dauran neeche kheenchti rehti hai, isliye sirf excess thrust

a_{T} - g

actually decelerate karta hai.

Required braking distance velocity ke saath kaise scale karta hai?

v^{2}

ke saath — double falling speed ko chaar guna altitude chahiye.

Bahut jaldi ignite karna crash risk kyun hai?

Tum upar hi stop/hover ho jaate ho, gravity se ladne mein fuel waste hota hai, khatam ho jaata hai, phir gir jaate ho.

Bahut der se ignite karna crash risk kyun hai?

Zero velocity tak pahunchne ke liye enough distance nahi, toh ground pe still fast speed se lagte ho.

Suicide burn kaam karne ke liye

a_{T} > g

kyun hona chahiye?

Agar thrust acceleration gravity se zyada nahi, toh net acceleration neeche ki taraf hai aur tum kabhi decelerate nahi kar sakte.

Fuel jalte waqt (constant thrust)

a_{n e t}

ka kya hota hai?

Mass girta hai, isliye

a_{T} = F / m

badhta hai aur deceleration burn ke through increase hoti hai — real burn nonlinear hai.

Rest se height H ke through fall speed?

v = 2 g H

Connections

Tsiolkovsky Rocket Equation — burn ke dauran exact mass loss
Kinematics — Equations of Motion — $v^{2} = 2 a d$ ka source
Thrust and Specific Impulse — jahan se $a_{T} = F / m$ aata hai
Gravity Turn Ascent — trajectory optimization ka launch analogue
Powered Descent Guidance — real-world closed-loop version (G-FOLD)
Terminal Velocity (atmospheric) — jab drag, thrust nahi, descent limit karta hai