Tsiolkovsky rocket equation — full first-principles derivation from momentum

3.3.1 · Physics › Rocket Propulsion

WHAT are we deriving?

WHY it must be true (before the algebra)

Ek hi baahri tool hai: Newton ka law momentum form mein: $F_{e x t} = \frac{d p}{d t}$ . Free space mein $F_{e x t} = 0$ , isliye (rocket + already-ejected gas) ka total momentum conserved hai. Rocket jo bhi speed gain karta hai, uski cost exhaust mein le jaaye gaye momentum se pay karni padti hai. Woh ek akela bookkeeping fact log ko force karta hai.

HOW — the derivation, step by step

Time $t$ par ek snapshot consider karo: rocket ka mass $m$ aur velocity $v$ hai (sab ek dimension mein, ground frame).

Step 1 — Abhi momentum. $p (t) = m v$ Yeh step kyun? Hum poore system ko track karte hain. Abhi woh fuel jo jalne wala hai woh andar hai, $v$ par move kar raha hai, isliye woh $m v$ ka part hai.

Step 2 — Thode time $d t$ baad. Rocket ek chota sa mass $d m_{g a s} > 0$ peeche eject karta hai. Rocket ka mass utna kam ho jaata hai, isliye agar hum rocket mass ko $m$ likhein, toh $d m$ (rocket mass mein change) negative hai, aur $d m_{g a s} = - d m$ .

Rocket ab: mass $m + d m$ , velocity $v + d v$ .
Ejected gas: mass $- d m$ , ground frame mein velocity $(v - v_{e})$ par move kar raha hai.

$v - v_{e}$ kyun? Gas $v_{e}$ speed par rocket ke relative, peeche ki taraf nikalti hai. Ground frame mein yeh rocket ki speed minus $v_{e}$ hai.

Step 3 — Tab momentum. $p (t + d t) = rocket (m + d m) (v + d v) + gas (- d m) (v - v_{e})$

Yeh step kyun? Total momentum = rocket ka naya momentum + gas jo momentum le gaya.

Step 4 — Expand aur cancel karo. $p (t + d t) = m v + m d v + v d m + d m d v - v d m + v_{e} d m$ $+ v d m$ aur $- v d m$ cancel ho jaate hain. Second-order term $d m d v$ drop karo (do tiny quantities ka product → negligible). $p (t + d t) = m v + m d v + v_{e} d m$

Step 5 — Conservation apply karo. Free space ⇒ $p (t + d t) = p (t) = m v$ . Subtract karo: $0 = m d v + v_{e} d m$

Yeh step kyun? Koi external force nahi matlab $d t$ mein total momentum ka change zero hai. Yahi poori physics hai — iske baad sab calculus hai.

Step 6 — Variables separate karo. $m d v = - v_{e} d m \Rightarrow d v = - v_{e} \frac{d m}{m}$

Step 7 — Integrate karo start se ( $v = 0$ ... ya $v_{i}$ , $m = m_{0}$ ) finish tak ( $v = v_{f}$ , $m = m_{f}$ ): $\int_{v_{i}}^{v_{f}} d v = - v_{e} \int_{m_{0}}^{m_{f}} \frac{d m}{m}$ $v_{f} - v_{i} = - v_{e} [ln m]_{m_{0}}^{m_{f}} = - v_{e} (ln m_{f} - ln m_{0})$

Figure — Tsiolkovsky rocket equation — full first-principles derivation from momentum

The thrust connection (bonus HOW)

Step 5 se, $d t$ se divide karo: $m \frac{d v}{d t} = - v_{e} \frac{d m}{d t}$ . Kyunki $\frac{d m}{d t} < 0$ , mass flow rate define karo $\overset{m}{˙} = - \frac{d m}{d t} > 0$ . Toh rocket par force: $F_{t h r u s t} = v_{e} \overset{m}{˙}$ Yeh kyun matter karta hai: thrust is baat par depend karta hai ki aap mass kitni tez throw karte ho ( $v_{e}$ ) times kitna per second ( $\overset{m}{˙}$ ). Rocket equation iska time-integrated version hai.

Worked Examples

Common Mistakes (Steel-manned)

Assumptions baked in (the fine print)

Koi external forces nahi (koi gravity loss nahi, koi drag nahi). Real launches mein ek $- \int g d t$ gravity loss term add hota hai.
Constant $v_{e}$ .
Ejection continuous hai (calculus limit), discrete lumps nahi.

Recall Feynman: explain to a 12-year-old

Imagine karo tum space mein ek skateboard par float kar rahe ho, haath mein baseballs ka bada bag hai. Space slippery hai — push karne ke liye kuch nahi. Toh tum ek baseball peeche throw karte ho. Tum thoda sa aage slide karte ho. Ek aur throw karo — thoda aur. Really fast jaane ke liye tum SAARI balls throw karte ho. Yahan twist hai: jaise jaise tumhara bag halka hota jaata hai, har throw tumhe (ab bhi halka) tezi se aage dhakelta hai. Lekin balls khatam ho jaati hain, isliye speed dheere dheere aati hai. Math kehta hai: do guni speed ke liye, do guni balls nahi chahiye — balls ki sankhya multiply hoti rehni chahiye. Woh "speed add karne ke liye multiply karna" exactly wahi hai jo logarithm hai.

Active Recall Flashcards

Tsiolkovsky rocket equation state karo.

Δ v = v_{e} ln (m_{0} / m_{f})

, jahan

v_{e}

= exhaust speed relative to rocket,

m_{0}

initial mass,

m_{f}

final mass.

Poori derivation kis conservation law par built hai?

Conservation of momentum (free space mein koi external force nahi ⇒ rocket+ejected gas ka total momentum constant hai).

Ground frame mein ejected gas ki velocity kya hai?

v - v_{e}

(rocket ki velocity minus relative exhaust speed).

Hum

d m d v

term kyun drop karte hain?

Yeh second-order hai (do infinitesimals ka product) aur limit mein vanish ho jaata hai.

d m

(rocket mass mein change) ka sign kya hai aur kyun?

Negative — rocket mass lose karta hai jab propellant eject karta hai.

Mass ratio kya hai aur uska symbol kya hai?

R = m_{0} / m_{f}

, initial se final mass ka ratio.

Δ v

mass mein linear kyun nahi hai, logarithmic kyun hai?

Kyunki

d v = - v_{e} d m / m

fractional mass change par depend karta hai;

d m / m

integrate karne se

ln

milta hai.

Rocket apni khud ki exhaust speed kab exceed karta hai?

Jab

R > e \approx 2.718

, matlab

ln R > 1

aur

Δ v > v_{e}

Step 5 se thrust derive karo.

m d v / d t = - v_{e} d m / d t

;

\overset{m}{˙} = - d m / d t > 0

ke saath, thrust

F = v_{e} \overset{m}{˙}

Ek rocket ko

Δ v = v_{e}

chahiye. Kitna mass ratio required hai?

R = e \approx 2.72

(roughly 63% mass jaalaana padega).

Do effects batao jo ideal equation ignore karta hai.

Gravity loss aur atmospheric drag (aur yeh constant

v_{e}

assume karta hai).

Connections

Conservation of Momentum — ek hi physical principle use hua.
Newton's Second Law (momentum form) — $F = d p / d t$ , Step 5 ka source.
Thrust and Mass Flow Rate — differential ( $F = v_{e} \overset{m}{˙}$ ) form.
Multistage Rockets — staging, log ke diminishing returns ko kyun beat karta hai.
Specific Impulse — $I_{s p} = v_{e} / g_{0}$ , $v_{e}$ ka engineering measure.
Gravity Loss and Drag Loss — ideal $Δ v$ ke corrections.
Variable Mass Systems — general framework jiska yeh ek special case hai.