Systems with variable mass — rocket equation derivation preview: Question bank

1.4.12 · D5 · Physics › Momentum & Collisions › Systems with variable mass — rocket equation derivation prev

Poore note mein, in do headline results ko yaad rakho jin par hum depend karte hain:

General variable-mass law: $m \frac{d v}{d t} = F_{ext} - u \frac{d m}{d t}$ .
Ideal (force-free) rocket equation: $Δ v = u ln \frac{m _{0}}{m _{f}}$ .

Yahan $m$ rocket ki current mass hai, $\overset{m}{˙} = \frac{d m}{d t} < 0$ (mass ja rahi hai), aur $u$ exhaust speed hai rocket ke relative. Is sab ki machinery ke liye Conservation of Linear Momentum, Newton's Third Law, aur Impulse–Momentum Theorem dekho.

True or false — justify

A rocket in deep space needs something (air, ground) to push against to accelerate.

False — wo apni khud ki ejected gas ke against push karta hai Newton's Third Law ke through; gas peeche jaati hai, rocket aage jaata hai. Koi external medium involve nahi hota, isliye exactly rockets vacuum mein kaam karte hain.

Agar ek rocket apna fuel zero speed par apne relative se eject kare (

u = 0

), to bhi speed up hoga kyunki mass lose ho rahi hai.

False —

u = 0

ke saath thrust term

- u \overset{m}{˙}

vanish ho jaata hai, to

m \overset{v}{˙} = F_{ext}

; sirf halka hona koi push nahi deta. Tumhe fuel ki zaroorat hai ki wo actually nonzero relative speed par tumse peeche chhoot kar jaaye.

Exhaust speed

u

ko double karne se

Δ v

bhi double ho jaata hai (baaki sab equal ho to).

True —

Δ v = u ln (m_{0} / m_{f})

mein factor

u

aage hai aur linear hai, to

u

double karne se

Δ v

double hota hai. Isko mass ratio se compare karo, jo sirf ek stingy logarithm ke through help karta hai.

Mass ratio

m_{0} / m_{f}

ko double karne se

Δ v

double ho jaata hai.

False —

Δ v

depend karta hai

ln (m_{0} / m_{f})

par, ratio par nahi. Ratio double karne se sirf

u ln 2

add hota hai (roughly

0.69 u

), answer double nahi hota.

Force-free rocket ke liye,

Δ v

depend karta hai is baat par ki tum fuel kitni tezi se jalaate ho.

False — burn rate

\overset{m}{˙}

cancel ho jaata hai jab tum

d v = - u d m / m

integrate karte ho; sirf start aur end masses bachti hain. Fast ho ya slow, same fuel same

Δ v

deta hai (ideal, gravity-free case mein).

Derivation mein velocity terms

v d m

aur

- v d m

ka cancel hona algebra ki coincidence hai.

False — yeh physically meaningful hai: jis ground-frame speed par mass leave karta hai wo cancel ho jaati hai, sirf relative speed

u

bachti hai thrust produce karne ke liye. Isliye

u

, ground speed nahi, woh hai jo matter karta hai.

Mist mein girta hua aur barhta hua raindrop variable-mass problem nahi hai kyunki kuch bahar nahi jaaya.

False — variable mass dono mass loss aur gain ko cover karta hai; ek accreting drop same

F_{ext} = d p / d t

bookkeeping use karta hai, bas

\overset{m}{˙} > 0

aur incoming relative velocity ke saath.

Spot the error

F = \frac{d ( m v )}{d t} = m \overset{v}{˙} + v \overset{m}{˙}

rocket ki motion directly deta hai."

v \overset{m}{˙}

term secretly assume karta hai ki lost mass rocket ki apni velocity

v

par depart karti hai (zero relative speed), to yeh no thrust predict karta hai.

F = \overset{p}{˙}

sirf closed system ke liye valid hai, to tumhe rocket plus ejected gas dono track karne chahiye.

"Thrust

= u \frac{d m}{d t}

Minus sign missing hai: kyunki

\overset{m}{˙} < 0

hai, yeh thrust negative bana dega. Sahi form hai

F_{thrust} = - u \frac{d m}{d t} = u ∣ \overset{m}{˙} ∣ > 0

— minus rakho aur

\overset{m}{˙}

ki sign ko apna kaam karne do.

u

exhaust ki speed hai jo ground se measure ki gayi hai."

u

exhaust speed hai rocket ke relative (ek engine property, ~2000–4500 m/s). Ground-frame exhaust velocity

v - u

hai, jo sign bhi change kar leti hai jab rocket

u

se tez ho jaata hai. Dekho Relative Velocity.

"Kyunki

F_{thrust} = u ∣ \overset{m}{˙} ∣

constant hai, acceleration bhi constant hai."

Acceleration hai

a = F_{thrust} / m

jisme current

m

hai, jo continuously shrink karta rehta hai. To constant thrust par bhi,

a

badhta hai jaise rocket halka hota jaata hai.

"Launch pad par, agar thrust weight ke equal ho to rocket slowly utha."

Agar

u ∣ \overset{m}{˙} ∣ = m g

exactly ho to

a = 0

: rocket hover karta hai, utha nahi. Lift off ke liye thrust-to-weight strictly greater than 1 chahiye (

u ∣ \overset{m}{˙} ∣ > m g

Δ v = u ln (m_{0} / m_{f})

mein hum

m_{f} = 0

plug karte hain fully burnt rocket ke liye max speed paane ke liye."

m_{f}

burnout (empty structure + payload) mass hai, kabhi zero nahi — engines, tanks, aur payload bacha rehta hai. Jaise

m_{f} \to 0

hota hai, log infinity tak jaata hai, jo unphysical hai: tum apna sab kuch kabhi nahi jala sakte.

ln

isliye aata hai kyunki humne integrate karna choose kiya; ek alag method isse avoid kar leta."

ln

physics ki wajah se forced hai: har kg jalane se ever-smaller remaining mass push hoti hai, to

d v = - u d m / m

mein

m

denominator mein hai, aur

\int d m / m = ln m

inevitable hai. Dekho Logarithms and Exponential Growth.

Why questions

Hum box kyun rocket ke plus us fuel ke around draw karte hain jo abhi eject honi hai, sirf rocket ke around kyun nahi?

Kyunki Newton's law

F_{ext} = d p / d t

form mein sirf us system ke liye valid hai jiske members $d t$ ke upar fixed hain; rocket akela members lose karta hai, to uski momentum ek aisi wajah se change hoti hai jo external force nahi hai.

Second-order term

d m d v

ko kyun drop kiya jaata hai?

Yeh do infinitesimally small quantities ka product hai, to yeh first-order terms

m d v

aur

u d m

ke comparison mein vanishingly tiny hai —

d t \to 0

limit mein yeh kuch contribute nahi karta.

High

Δ v

achieve karna fuel mein itna brutally expensive kyun hai?

Kyunki mass ratio ek logarithm ke andar baithe hai, to

Δ v

linearly badhane ke liye

m_{0} / m_{f}

ko exponentially badhana padta hai — "tyranny of the rocket equation."

Same rocket deep space mein burnout ke paas launch ke waqt se zyada tezi se accelerate kyun karta hai (constant thrust)?

Thrust

u ∣ \overset{m}{˙} ∣

fixed rehta hai lekin

a = F_{thrust} / m

mein

m

shrink ho gayi hai, to same push kam cheez ko tezi se move karti hai.

Burn rate

\overset{m}{˙}

final Tsiolkovsky formula mein kyun nahi dikhta, jabki yeh clearly thrust set karta hai?

Thrust aur time dono

\overset{m}{˙}

par depend karte hain, lekin jab total velocity change paane ke liye integrate karte ho to ye cancel ho jaate hain;

Δ v

sirf total mass expelled ki care karta hai, yeh nahi ki wo kitni jaldi gayi.

Thrust determine karne mein relative speed

u

kyun important hai, rocket ki speed nahi?

Kyunki exhaust ko jo momentum tum dete ho wo set hota hai is baat se ki wo tumse kitni tezi se jaata hai; ground-frame velocity terms bookkeeping mein cancel ho jaate hain, sirf relative kick

u

thrust ka source bachta hai.

Edge cases

Rocket equation ka kya hoga agar

F_{ext} \neq = 0

ho (maano gravity)?

Tum cleanly variables separate nahi kar sakte

d v = - u d m / m

mein; term

F_{ext}

m \overset{v}{˙} = F_{ext} - u \overset{m}{˙}

mein rehta hai aur tumhe poori equation integrate karni padti hai, jo Tsiolkovsky se chhota

Δ v

deta hai.

Δ v

kya hoga agar rocket koi fuel nahi jalaata, to

m_{f} = m_{0}

Δ v = u ln (m_{0} / m_{0}) = u ln 1 = 0

— koi mass eject nahi matlab koi thrust nahi aur koi speed change nahi, exactly jaise intuition demand karta hai.

Kya ek variable-mass body kabhi mass change ki wajah se akele slow down ho sakti hai?

Haan — ek accreting body (

\overset{m}{˙} > 0

) jo ground ke relative rest par mass pick up karti hai, incoming mass usse drag karti hai; relative velocity ki sign flip hoti hai aur thrust ek retarding force ban jaata hai.

Agar exhaust somehow aage eject ho (motion ki same direction mein,

u < 0

To thrust

- u \overset{m}{˙}

ka sign reverse ho jaata hai aur rocket ko peeche dhakelta hai (decelerate karta hai) — exactly yehi hai retro-rockets aur reverse thrusters ek craft ko kaise brake karte hain.

Ground frame mein, jab rocket apni exhaust speed se zyada ho jaaye

v > u

, to exhaust ki velocity kya hoti hai?

Exhaust velocity

v - u

positive ho jaati hai — gas ground ke relative aage move kar rahi hoti hai, chahe usne rocket ko uske relative peeche jaate hue chhoda ho. Thrust unaffected rehta hai kyunki sirf relative speed matter karta hai. Dekho Relative Velocity.

Agar tum mass ratio arbitrarily large bana sako to

Δ v

ko kya limit karta hai?

Ideal equation mein mathematically kuch cap nahi karta (

ln \to \infty

), lekin physically tum infinite fuel aur zero structure nahi rakh sakte — real tanks, engines, aur payload ek finite

m_{f}

fix karte hain, to

Δ v

mathematical limit se kaafi neeche bounded hai.

Recall One-line self-test

Agar tum "why $u$ and not $v$ ?", "why $ln$ and not the raw ratio?", aur "why the closed box?" har ek ko ek sentence mein answer kar sako, to yeh topic tumhara hai.

Parent: topic note · builds on Conservation of Linear Momentum, Impulse–Momentum Theorem, Newton's Third Law, Logarithms and Exponential Growth.