Terminal landing — propulsive descent, suicide burn: Question bank

3.4.26 · D5 · Physics › Rocket Flight Mechanics › Terminal landing — propulsive descent, suicide burn

True or false — justify

Har claim ek line hai. True/false decide karo, phir mechanism do — sirf label nahi.

Ek rocket suicide burn mein apna zyaadatar descent engine off karke guzarta hai.

True — poora point hi ek single continuous burn end mein hai; ignition se pehle wo free-fall karta hai, isliye fall ke bulk mein engine off rehta hai.

Pehle fire karna hamesha gentle aur safer landing deta hai.

False — bahut pehle fire karo toh

a_{T} > g

tumhe zyaada upar rokta hai, aur har second fuel jalata hai sirf gravity se ladne ke liye jab tak tank khaali na ho jaaye aur tum giro.

Braking distance linearly depend karti hai tumhari fall speed pe.

False — ye

v^{2}

ke saath scale hoti hai, isliye speed double karne se altitude chahiye wo chaar guna ho jaati hai; speed ko linearly se kahin zyaada saza milti hai.

Agar engine ka thrust exactly vehicle ke weight ke barabar ho, toh suicide burn tab bhi barely kaam kar sakta hai.

False — tab

a_{T} = g

hoga isliye

a_{n e t} = 0

; rocket hamesha ke liye constant speed se girta rahega aur kabhi zero velocity nahi aa sakti.

Suicide burn fuel-optimal hai kyunki ye sabse chhoti possible thrust use karta hai.

False — ye aksar full thrust use karta hai; ye optimal isliye hai kyunki ye sabse kam possible time ke liye thrust use karta hai, gravity se ladne mein waste hone wala impulse minimize karta hai.

Moon pe wahi lander, wahi speed pe, Earth se zyaada ignition altitude chahega.

False — kam

g

matlab bada

a_{n e t} = a_{T} - g

, isliye tum kam distance mein ruk jaate ho; Moon pe tum neecha ignite kar sakte ho.

Mass ko constant treat karna formula ko burn perfectly predict karne deta hai.

False — jaise propellant nikalta hai,

m

girta hai isliye

a_{T} = F / m

badhta hai aur deceleration burn ke through badhti hai; constant-mass answer sirf ek pehla estimate hai.

Ek baar ignition altitude par bina fire kiye nikal jao, thoda aur wait karna theek hai.

False —

h_{b u r n}

se aage koi altitude bacha nahi rukne ke liye, isliye zero velocity se pehle ground aa jaati hai; ye "no recovery" property exactly isliye hai ise suicide burn kehte hain.

Spot the error

Har line mein ek flawed statement hai. Correction reveal karo.

a_{T}

ko deceleration use karo kyunki engine hi tumhe slow karta hai."

Gravity burn ke dauran downward speed add karna band nahi karti, isliye real braking

a_{T} - g

hai;

a_{T}

use karna altitude ko underestimate karta hai jo tumhe actually chahiye.

0 = v^{2} + 2 a_{n e t} h_{b u r n}

set karo aur solve karo."

Braking acceleration motion ke opposite hai, isliye minus ke saath enter hoti hai:

0 = v^{2} - 2 a_{n e t} h_{b u r n}

, jo

h_{b u r n} = v^{2} / [2 a_{n e t}]

deta hai — ek positive distance.

"Fire karo jab tumhari altitude

v^{2} / (2 a_{T})

ke barabar ho."

Ye denominator mein gravity bhool jaata hai; sahi trigger hai

h = v^{2} / [2 (a_{T} - g)]

, jo zyaada hai, isliye naive rule bahut neecha fire karta hai aur crash hota hai.

"Kyunki burn short hai, thrust aur acceleration dono constant hain."

Thrust roughly constant ho sakti hai, lekin mass girta hai, isliye

a_{T} = F / m

badhta hai — acceleration constant nahi hai chahe force constant bhi ho.

"Agar

a_{n e t}

negative hai toh tum bas dheere decelerate karoge."

Negative

a_{n e t}

matlab net force neeche ki taraf hai; tum speed up karte ho, slow down nahi — deceleration impossible hai aur tum kabhi us regime mein nahi jaana chahte.

v = 2 g H

tumhe fall speed batata hai chahe engine fire ho raha ho."

Wo free-fall result sirf ignition se pehle valid hai; ek baar engine on hai, acceleration

a_{n e t}

hai (upward),

g

nahi, isliye relation ab apply nahi hota.

Why questions

Underlying cause ke saath jawab do, ek ya do sentences.

Required altitude speed ke square ke saath kyun badhti hai, sirf proportionally nahi?

Kinetic energy

v^{2}

ke saath scale hoti hai aur constant braking force energy per unit distance remove karti hai, isliye distance ko

v^{2}

ke saath scale karna padta hai saari energy absorb karne ke liye.

Engine ko gravity se kyun ladna padta hai chahe usne fall "cancel" kar diya ho?

Gravity kabhi switch off nahi hoti; jis second thrust rukti hai vehicle phir girne lagti hai, isliye engine ko continuously

g

worth of acceleration dena padta hai sirf rokne ke liye, plus aur zyaada decelerate karne ke liye.

Bahut pehle ignite karna fuel problem kyun hai, timing problem nahi?

Tum zero velocity ground se upar pahuncho aur phir hover karo, har second propellant kharch karo sirf gravity counter karne ke liye, tank touchdown se pehle khaali ho jaata hai. ::: Phir tum us height se bina fuel ke girate ho.

Kam thrust margin (

a_{T}

barely

g

se upar) tumhe dangerously zyaada upar ignite kyun karta hai?

Chota

a_{n e t}

h_{b u r n} = v^{2} / (2 a_{n e t})

ko huge banata hai, isliye tumhe ground se bahut dur braking start karni padti hai jahan errors aur terrain judge karna mushkil hai.

Suicide burn "all or nothing" kyun hai late start ka koi margin nahi?

Trigger altitude exactly minimum stopping distance hai, isliye koi bhi delay physics se kam runway chodti hai — koi reserve nahi hai wapas karne ka.

Slowly hover down kyun nahi kar sakte safe rehne ke liye?

Hovering matlab

a_{T} = g

lamba time sustain karna, jo ek brief maximal burn se kahin zyaada propellant consume karta hai; ek real lander ke paas slow descent ke liye fuel nahi hota.

Parent note

m

ko constant kyun treat karta hai jabki maan leta hai ki ye hai nahi?

Ye ek 80/20 first estimate hai — constant-mass formula scaling aur trigger logic clearly capture karta hai, aur exact mass loss (Tsiolkovsky Rocket Equation) tabhi layer hoti hai jab precision chahiye.

Velocity ka sign (

- v

, downward)

v^{2} = 2 a d

likhte waqt kyun matter karta hai?

Sirf magnitude

v^{2}

mein enter hoti hai, lekin direction batata hai ki braking acceleration motion ke opposite point karni chahiye (upward), jo energy equation mein minus sign fix karta hai.

Edge cases

Boundary aur degenerate scenarios — jahan naive picture toot jaati hai.

Exactly

v = 0

par kya hota hai (kisi height pe already rest mein ho)?

h_{b u r n} = 0

: tumhe koi braking distance nahi chahiye, isliye "ignition altitude" ground hi hai — tum sirf free-fall karoge aur zero speed pe tab hi touch karoge jab tum already ground pe ho. ::: Practice mein tum hover karoge ya drop karoge; suicide-burn logic degenerate ho jaata hai.

Jab

a_{T} \to g

upar se aata hai tab kya hota hai?

a_{n e t} \to 0

isliye

h_{b u r n} \to \infty

; required ignition altitude diverge ho jaati hai, matlab koi finite starting height tumhe rok nahi sakti.

Agar

a_{T} < g

ho (thrust weight se weaker)?

a_{n e t} < 0

: vehicle full thrust pe bhi neeche accelerate karta hai, isliye suicide burn impossible hai — tum sirf crash slow kar sakte ho, rok nahi sakte.

Agar tum burn already upward move karte hue shuru karo?

Suicide-burn model downward entry assume karta hai; agar tum rise kar rahe ho, pehle coast up karo, gravity wapas laati hai, aur braking analysis sirf subsequent fall pe apply hoti hai.

h_{b u r n}

ka limiting behaviour kya hai jab thrust enormous ho jaaye (

a_{T} \to \infty

h_{b u r n} \to 0

: infinitely strong engine tumhe essentially koi distance mein rok leta hai, isliye tum right at the ground ignite kar sakte — ideal, unreachable extreme.

Ek airless world versus atmosphere wali world pe, kya same

v

ko same

h_{b u r n}

chahiye?

Exactly nahi — atmospheric drag fall aur burn ke dauran extra upward force add karta hai, effectively braking increase karta hai, isliye real atmosphere (Terminal Velocity (atmospheric)) needed altitude vacuum formula se kam kar deti hai.

Agar do landers ka identical

a_{n e t}

ho lekin different masses hoon, kya unhe same

v

pe same ignition altitude chahiye?

Haan —

h_{b u r n}

sirf

v

aur

a_{n e t}

pe depend karta hai, directly mass pe nahi; mass matter karta hai sirf isliye kyunki ye

a_{T} = F / m

set karta hai shuruaat mein.

Connections

Parent: Terminal landing
Kinematics — Equations of Motion — yahan har trap ke peeche $v^{2} = 2 a d$ hai
Thrust and Specific Impulse — isliye $a_{T} = F / m$ aur margin kyun matter karta hai
Tsiolkovsky Rocket Equation — constant-mass traps ka mass-loss correction
Terminal Velocity (atmospheric) — drag edge case