Payload fraction as function of Δv and Isp: Question bank

3.3.47 · D5 · Physics › Rocket Propulsion › Payload fraction as function of Δv and Isp

Yeh ek rapid-fire deck hai conceptual questions ka, jo yeh batata hai ki payload fraction $λ$ kaise respond karta hai $Δ v$ , specific impulse $I_{s p}$ , aur structural coefficient $ϵ$ ke saath. Yahan heavy arithmetic nahi hai — har item ek misconception ya ek boundary case ko target karta hai. Answer cover karo, reason karo, phir reveal karo.

Shuru karne se pehle, parent topic se meanings refresh karo:

Recall Ek hi saans mein charon symbols

== $λ$ == ::: payload fraction, $m_{payload} / m_{0}$ — launch mass ka woh hissa jo actual cargo hai.
== $ϵ$ == ::: structural coefficient, $m_{structure} / m_{propellant}$ — fuel ke har kg par lagne wala "tank tax".
== $R$ == ::: characteristic mass ratio $e^{Δ v / (I_{s p} g_{0})}$ — loaded rocket, empty rocket se kitne guna bhaari hai.
== $g_{0}$ == ::: standard gravity, ek fixed number $9.81 m/s^{2}$ . Yeh sirf ek conversion factor ke roop mein aata hai jo $I_{s p}$ (seconds mein measured) ko m/s mein exhaust velocity mein convert karta hai, $v_{e} = I_{s p} g_{0}$ ke zariye. Yeh kisi bhi planet ki local gravity nahi hai — sirf ek bookkeeping constant hai jo specific impulse ki definition mein baki hai. Dekho Specific Impulse.

Mass budget ko picture karna

Neeche har question actually ek sawaal hai ki launch mass kaise slice hota hai. Bar chart dekho: poora launch mass $m_{0}$ (poori bar) teen pieces mein bant jaata hai — cargo, tanks/engines, aur fuel. Payload fraction $λ$ bas blue slice ki length ko poori bar se divide karna hai.

Figure — Payload fraction as function of Δv and Isp

Figure se notice karo kyun master formula ki woh shape hai jo hai. Picture padhte hue:

Blue slice woh hai jo hum rakhte hain — woh ratio $λ$ hai.
Propellant jalana (orange) woh hai jo $Δ v$ khareedta hai; jitna zyada $Δ v$ hum demand karte hain, orange slice utna hi lamba hona chahiye, aur Tsiolkovsky kehta hai ki woh height $R = e^{Δ v / (I_{s p} g_{0})}$ ki tarah badhti hai.
Gray structure slice orange ke saath chained hai: orange ke har kg ke liye $ϵ$ kg gray add karna padta hai. Toh structure free real estate nahi hai — yeh fuel ke saath lockstep mein badhta hai.

Wahi chaining exactly woh jagah hai jahan $ϵ (1 - R)$ term janam leti hai. Ideal slice $1/ R$ (blue agar gray na hota) se shuru karke, gray tax subtract karna padta hai, jo orangey kitna burn kiya uske saath scale karta hai — aur woh amount $R$ ke saath badhta hai. Ise algebraically work karo (poori tarah parent note mein kiya gaya hai) toh collapse hota hai:

Exponential tyranny ko dekhna

Agla figure $λ$ ko $Δ v$ ke against plot karta hai. Curve ek gentle ramp nahi hai — yeh plunge karta hai aur zero ke neeche bhi jaata hai. Woh crossing hai "single-stage wall": uske baad koi cargo fit hi nahi hota.

Deck work karte waqt dono pictures dhyan mein rakho.

True ya false — justify karo

Ek perfectly efficient rocket (

ϵ = 0

) ka hamesha

λ = 1/ R

hota hai.

True —

ϵ = 0

set karo aur gray tax term gayab ho jaata hai, bacha rehta hai

λ = 1/ R = e^{- Δ v / (I_{s p} g_{0})}

, pure Tsiolkovsky ceiling (blue slice bina gray ke).

Ek fixed mission ke liye,

I_{s p}

double karne se payload fraction exactly double ho jaata hai.

False —

I_{s p}

double karne se exponent half ho jaata hai, toh

R \to R^{1/2} = R

. Concretely, agar

R = 16

toh

λ_{ϵ = 0}

1/16 = 0.0625

1/4 = 0.25

ho jaata hai — 4× jump, 2× nahi. Kyunki

I_{s p}

exponent ke andar hai, iska effect non-linear hai.

Payload fraction kabhi bhi us value se zyada nahi ho sakta jo

ϵ = 0

par hota hai.

True —

ϵ (1 - R)

term negative hota hai jab bhi

R > 1

(hamesha, real missions ke liye), toh koi bhi positive structure

λ

1/ R

ideal se neeche hi kheenchta hai.

Agar

Δ v = 0

hai, toh

λ = 1

hoga chahe

ϵ

kuch bhi ho.

True — koi burn nahi,

R = e^{0} = 1

, toh

λ = (1 + ϵ \cdot 0) /1 = 1

: tum sirf payload carry kar rahe ho kyunki tumhe koi propellant nahi chahiye aur isliye koi tank bhi nahi.

Ek negative computed

λ

ka matlab sirf bahut chhota par real payload hai.

False — negative

λ

physically impossible hai; yeh signal karta hai ki structure plus propellant pehle se hi poore launch mass se zyada hai, toh yeh rocket single stage mein ban hi nahi sakta. Wahi SSTO wall hai jo doosri figure mein curve cross karta dikha.

ϵ

badhane se woh

Δ v

shift ho jaata hai jahan

λ

zero hita hai, chhoti value par.

True — zero-crossing

R = 1 + 1/ ϵ

par hai; bada

ϵ

means

1/ ϵ

chhota, toh wall chhote

R

par aati hai, yani chhote

Δ v

par. Algebra:

ϵ = 0.05 \Rightarrow R = 21

ϵ = 0.20 \Rightarrow R = 6

Do rockets jinke paas same

R

hai par alag

Δ v

aur

I_{s p}

hain, unka payload fraction same hoga (same

ϵ

ke saath).

True —

λ

sirf

R = e^{Δ v / (I_{s p} g_{0})}

ke combination ke zariye

Δ v

aur

I_{s p}

par depend karta hai, toh equal

R

ka matlab equal

λ

hai.

Formula predict karta hai ki kuch inputs ke liye

λ

1 se bada ho sakta hai.

Real missions ke liye False — uske liye

R < 1

chahiye, yaani

Δ v < 0

, jo meaningless hai. Sab

Δ v > 0

ke liye

R > 1

aur

λ < 1

hai.

Error pakdo

"Kyunki Tsiolkovsky

m_{0} / m_{f}

deta hai, final mass

m_{f}

sirf payload hai."

Galat —

m_{f}

hai payload plus structure (figure 1 mein blue aur gray slices). Structure

ϵ (m_{0} - m_{f})

dead weight hai jo burn ke baad bachi rehti hai, toh woh

m_{f}

ke andar hai aur jo payload ho sakta tha usse churaati hai.

"10 ka mass ratio hit karne ke liye, rocket ko 90% fuel aur 10% payload banao."

Galat — woh 10% jo propellant nahi hai woh payload aur tanks/engines hai.

ϵ = 0.1

ke saath woh structure akele poore 10% consume kar sakta hai, zero ya negative payload chhodke.

"1 km/s

Δ v

add karne ka har baar same payload cost hai, toh ise linearly budget karo."

Galat —

λ

mein

e^{- Δ v / (I_{s p} g_{0})}

hai, ek exponential. Har extra km/s

R

ko ek fixed factor se multiply karta hai, toh baad ke kilometres pehle waalon se kaafi zyada hurt karte hain — bilkul figure 2 mein woh plunge.

"Ek better engine (

ϵ = 0

assumed) structural mass ko irrelevant kar deta hai, toh

ϵ

ko har jagah ignore karo."

Galat —

ϵ

tanks aur plumbing measure karta hai, engine efficiency nahi. High-

I_{s p}

engine

R

ko shrink karta hai par tank tax par kuch nahi karta;

ϵ

phir bhi cap karta hai ki tum kitna payload rakh sakte ho.

"Agar

λ = 0

hai, toh mission barely achievable hai — bas thoda aur fuel add karo."

Galat —

λ = 0

par poora rocket structure aur propellant hai, koi cargo nahi. Zyada fuel add karne se zyada structure add hota hai (

ϵ

per kg),

λ

ko aur zyada negative push karta hai, positive nahi. Tumhe stage karna hoga ya

I_{s p} / ϵ

improve karna hoga.

"Structural coefficient aur structural mass ek hi cheez hai."

Galat —

ϵ

ek ratio hai (structure per unit propellant), dimensionless; structural mass ek absolute quantity hai kg mein. Ek bada rocket aur ek chhota rocket same

ϵ

share kar sakte hain.

"Payload fraction aur propellant mass fraction hamesha 1 tak add ho jaate hain."

Galat — teen fractions payload + structure + propellant 1 tak add hote hain (figure 1 ke teen slices). Dekho Propellant Mass Fraction aur Structural Coefficient; structural slice woh missing teesra piece hai.

Why questions

λ

Δ v

aur

I_{s p}

par sirf ek exponential ke andar unke ratio ke zariye kyun depend karta hai, kabhi alag-alag nahi?

Kyunki Tsiolkovsky unhe

Δ v = I_{s p} g_{0} ln R

ke roop mein link karta hai;

R

ke liye solve karne se dono ek exponent mein bundle ho jaate hain, aur

λ

sirf

R

ka function hai. Dekho Tsiolkovsky Rocket Equation.

Upper stages mein hydrogen (high-

I_{s p}

) engines kyun dominate karte hain haandling mein awkward hone ke bawajood?

Upper-stage

Δ v

tab spend hota hai jab thodi mass bachi hoti hai, aur

R

mein exponential high

I_{s p}

ko itni strongly reward karta hai ki modest

I_{s p}

gain ek bade

λ

gain mein compound ho jaata hai. Dekho Specific Impulse.

LEO ke liye single-stage-to-orbit chemical rocket negative

λ

kyun deta hai?

~9.4 km/s LEO

Δ v

aur

I_{s p} \sim 350

s ke saath

R \approx 15

force hota hai; structure tax

ϵ (R - 1)

tab 1 se zyada ho jaata hai, toh poora launch mass spend ho jaata hai kisi cargo se pehle. Dekho Staging.

Staging woh payload fraction rescue kyun karta hai jo single stage achieve nahi kar sakta?

Har stage apne empty tanks discard kar deta hai, toh agla stage us dead structure ko accelerate nahi karna padta. Yeh mass ratio problem ko aadhe raste mein reset karta hai, exponential wall se side-stepping karte hue. Dekho Optimal Staging.

Exponential relationship ko "tyranny" kyun kaha jaata hai na ki sirf "steep"?

Kyunki yeh cost multiplicatively grow karti hai: mission

Δ v

mein modest increases disproportionately bade rockets demand karte hain, aur yahi wajah hai ki interplanetary travel orbit se kaafi zyada mushkil hai.

Oberth effect clever trajectory design ko

λ

par

Δ v

burden kyun ease karne deta hai?

Gravity well mein deep burn karna propellant energy ko orbital energy mein zyada efficiently convert karta hai, toh ek given goal ke liye required

Δ v

kam ho jaata hai — aur kyunki

λ

exponentially

Δ v

ke saath girta hai, koi bhi reduction richly pay off karta hai. Dekho Oberth Effect.

Do engineers ek hi hardware ke saath rocket ke "payload fraction" par disagree kyun kar sakte hain?

Kyunki

λ

assigned mission

Δ v

par depend karta hai; same rocket ka ek short hop ke liye high

λ

hoga aur ek ambitious burn ke liye low (ya negative) wala.

Edge cases

Δ v \to \infty

ki limit mein

λ

kya hoga (jab

ϵ > 0

R \to \infty

, toh

λ \to \frac{- ϵ R}{R} = - ϵ

. Yeh negative ho jaata hai aur

- ϵ

par settle karta hai: impossible regime gehra hota hai par

- \infty

par nahi bhagta.

Jab

ϵ = 0

aur

Δ v \to \infty

ho toh

λ

kya hoga?

λ = 1/ R \to 0^{+}

. Ek structureless ideal rocket zero payload ke paas jaata hai par kabhi negative nahi hota — negative branch entirely structure tax hai.

Kis

R

par

λ

zero cross karta hai, aur us value ka kya matlab hai?

1 + ϵ (1 - R) = 0

set karne se

R = 1 + 1/ ϵ

milta hai. Is mass ratio se aage koi payload fit nahi hota; yeh woh hard ceiling hai jo single stage attempt kar sakta hai.

ϵ \to 0

hone par zero-crossing

R = 1 + 1/ ϵ

ka kya hota hai?

Yeh infinity tak jaata hai, yani ek perfectly light structure ka koi finite

Δ v

wall nahi hai — principle mein tum kisi bhi

Δ v

tak positive (chahe tiny) payload ke saath pahunch sakte ho.

Kya

λ = 1

physically reachable hai, aur kaunse inputs par?

Sirf degenerate case

Δ v = 0

mein (toh

R = 1

), ek rocket jo kabhi fire nahi karta. Koi bhi real burn

R > 1

aur

λ < 1

deta hai.

ϵ = 1

kya represent karta hai, aur kya yeh survivable hai?

Fuel ke har kg ko structure ka ek kg chahiye — ek 50/50 tank. Zero-crossing

R = 2

par hai, toh bahut chhote

Δ v

par bhi almost koi payload nahi bachta; aisa design real missions ke liye essentially unusable hai.

Agar koi rocket

λ = 0.04

report karta hai (jaise Saturn V), toh kaunsa fraction payload nahi hai, aur woh kahan jaata hai?

96% non-payload hai: overwhelmingly propellant, baaki tanks, engines, aur discarded stages hain — orbital

Δ v

ke liye exponential

R

climb karne ki direct cost.