Mass ratio m₀ - m_f — why it's so critical: Question bank

3.3.3 · D5 · Physics › Rocket Propulsion › Mass ratio m₀ - m_f — why it's so critical

True or false — justify

The mass ratio

R

can equal 1.

False.

R = 1

ka matlab hai

m_{0} = m_{f}

, yaani zero propellant (

m_{p} = m_{0} - m_{f} = 0

); tab

Δ v = u ln 1 = 0

. Jo rocket koi fuel carry nahi karta woh apni velocity change nahi kar sakta, isliye physically

R > 1

hamesha hota hai.

R

can be less than 1.

False. Iske liye

m_{f} > m_{0}

chahiye hoga — yaani rocket shuru se zyaada bhaari hokar khatam ho, jo is fact ko violate karta hai ki sirf mass bahar jaata hai (exhaust ke roop mein).

R < 1

ek burning rocket ke liye unphysical hai.

Doubling the propellant mass doubles

Δ v

False.

Δ v = u ln R

log of

R

ke saath badhta hai,

R

ke saath nahi. Roughly double fuel roughly double

R

karta hai, jo sirf

u ln 2 \approx 0.69 u

ka ek fixed chhota bonus deta hai — doubling nahi.

If two rockets have the same

R

but different exhaust speeds

u

, they get the same

Δ v

False.

Δ v = u ln R

$u$ mein linear hai. Same

R

(same fraction burnt) hone par bhi jis rocket ka

u

bada hai woh proportionally zyaada speed gain karta hai, kyunki har kilogram zyaada tez bahar pheka jaata hai.

The propellant fraction

ζ

can reach 1.

False.

ζ = 1 - 1/ R

sirf approach karta hai 1 ko jab

R \to \infty

; yeh kabhi equal nahi hota, kyunki structure ka nonzero mass

m_{f} > 0

hota hai, isliye

1/ R > 0

hamesha hota hai aur

ζ < 1

strictly rehta hai. "Asymptotically close" ka matlab "equal" nahi hota.

A rocket with

ζ = 0.9

(90% fuel) has

R = 10

True.

ζ = 1 - 1/ R = 0.9

deta hai

1/ R = 0.1

, toh

R = 10

. Yeh single stage ke liye practical structural ceiling ke paas hai.

Increasing exhaust speed

u

always lowers the required mass ratio for a fixed mission

Δ v

True. Invert karo:

R = e^{Δ v / u}

. Bada

u

exponent

Δ v / u

ko chhota karta hai, isliye

R

(aur woh fuel jo carry karna padta hai) kam ho jaata hai — yahi reason hai ki high-specific-impulse engines itne valued hain.

Why questions

Why does the mass ratio appear inside a logarithm and not, say, as a plain multiplier?

Kyunki har infinitesimal step deta hai

d v = - u d m / m

: same fractional mass loss

d m / m

hamesha same

d v

khareedta hai. Sum karne par (integrate karne par) equal fractional cuts

ln

produce karte hain, jo count karta hai ki tumne mass ke kitne "

e

-foldings" kharche.

Why must we track the exhaust's momentum, not just the rocket's?

Rocket ko push karne ke liye koi external force nahi hai; woh sirf mass peeche phenkkar accelerate karta hai. Newton ka law actually hai "force = rate of change of momentum," isliye rocket ka forward momentum gain exhaust ke backward momentum ke equal hona chahiye — ek ko doosre ke bina nahi dekh sakte.

Why does a 47% increase in

u

nearly triple the useful (dry-mass) fraction?

Kyunki

R = e^{Δ v / u}

exponent mein

1/ u

ke saath exponential hai.

u

badhane se exponent chhota hota hai, aur exponential us shrink ko amplify karta hai: exponent mein ek modest change

R

mein ek bada change ban jaata hai, aur isliye

1/ R

mein bhi, jo surviving fraction hai.

Why do we build Multistage Rockets instead of one giant tank?

Ek single stage ka

R

structural mass se cap hota hai (

m_{f}

kuch percent se neeche nahi ja sakta

m_{0}

ka). Staging mid-flight mein dead structure drop karta hai, toh baad ke stages chhote

m_{f}

se shuru karte hain aur fresh high

R

tak pahunch sakte hain — total

Δ v

har stage ke

u ln R

ka sum ban jaata hai, single-stage wall se bachte hue.

Why do the two

v d m

terms cancel in the derivation, and why does that matter?

Rocket ka momentum change mass khone se (

v d m

) bilkul exhaust ke saath cancel hota hai jo usi mass ko ground-frame reference par carry karta hai (

- v d m

); jo bachta hai woh hai

u d m

, relative ejection se momentum. Yeh matter karta hai kyunki yeh prove karta hai ki sirf relative exhaust speed

u

thrust drive karta hai — ground frame drop out ho jaata hai.

Edge cases

A rocket at rest ejects its very first tiny bit of fuel. What is

v - u

(ground-frame exhaust speed) at that instant?

v = 0

par, exhaust

v - u = - u

par move karta hai: bilkul peeche full relative speed

u

par. Jaise rocket speed up karta hai, ground-frame exhaust speed zero ki taraf shrink karta hai aur sign bhi reverse ho sakta hai, lekin relative speed

u

fixed rehti hai.

What happens to

Δ v

R \to 1^{+}

(almost no fuel)?

ln R \to 0

, toh

Δ v \to 0

smoothly. Jo rocket essentially saari structure hai aur koi propellant nahi, woh barely move karta hai — log mein

R = 1

par koi discontinuity nahi, woh bas vanish ho jaata hai.

Take the limit

R \to \infty

: does

Δ v

blow up quickly?

Nahi — yeh sirf logarithmically diverge karta hai. Equal chunks of

Δ v

add karte rehne ke liye

R

ko multiply karte rehna padta hai (har extra

u

of speed fuel ko

e \approx 2.72

se multiply karta hai). Infinite speed ke liye infinite fuel chahiye, aur yeh approach agonizingly dheemi hai.

Set

u = 0

(no exhaust speed relative to rocket). What is

Δ v

Δ v = 0 \cdot ln R = 0

kisi bhi

R

ke liye. Zero relative speed par mass eject karna koi momentum transfer nahi karta — fuel bas tumhare paas float karta rahega, kuch gain nahi hoga. Yeh confirm karta hai ki

u

, fuel ki matra nahi, engine ka asli lever hai.

A rocket with

R = e^{2.3} \approx 10

using

u = 3

km/s: does it clear the single-stage-to-orbit bar of

\approx 9.4

km/s?

Nahi.

Δ v = 3 \times 2.3 \approx 6.9

km/s, jo

9.4

km/s se kaafi kam hai. Structural ceiling

R \approx 10

ek stage ko near

2.3 u

par cap karta hai — bilkul yahi reason hai ki LEO (dekho Delta-v Budget) staging ya bahut bada

u

demand karta hai.

m_{f}

(dry mass) is doubled while

m_{0}

is held fixed, does

Δ v

rise or fall?

Girta hai.

R = m_{0} / m_{f}

half ho jaata hai, toh

Δ v = u ln R

mein

u ln 2 \approx 0.69 u

ki kami aa jaati hai. Structure mein extra dead weight doubly costly hai — woh

m_{f}

bhi badhata hai aur poora safar drag bhi kiya jaata hai.

Connections

Tsiolkovsky Rocket Equation — woh formula jisko yahan har trap probe karta hai.
Specific Impulse Isp — kyun $u$ badhana fuel pile karne se behtar hai.
Multistage Rockets — exponential wall se engineered escape.
Conservation of Momentum — $d v = - u d m / m$ ke peeche ka first principle.
Delta-v Budget — jahan se "9.4 km/s to LEO" bar aata hai.
Exhaust Velocity and Thrust — $u$ ka physical source.

Question bank — Mass ratio m₀ - m_f — why it's so critical

True or false — justify

Spot the error

Why questions

Edge cases

Connections