Δv = v_e · ln(m₀ - m_f) — understanding each term

3.3.2 · Physics › Rocket Propulsion

Tsiolkovsky rocket equation. Yeh batata hai ki kitna speed change (Δv) ek rocket ko milta hai jab woh ek given amount of fuel jalata hai.

Scratch se derive karna (log KYUN aata hai)

Setup. Ek inertial frame mein kaam karo, koi gravity nahi, koi drag nahi (ideal case). Kisi instant par rocket ki mass $m$ hai aur velocity $v$ hai. Time $d t$ mein woh thoda exhaust mass eject karta hai aur $d v$ se speed up hota hai.

Momentum conserve karo. Total momentum constant hai kyunki koi external force nahi lag raha.

Pehle: $p_{before} = m v$ .

Baad mein: rocket ki mass $∣ d m ∣$ se kam hoti hai (note karo $d m < 0$ kyunki mass decrease ho rahi hai). Rocket ki mass ab $m + d m$ hai aur velocity $v + d v$ hai. Ejected chunk ki mass $- d m$ hai aur velocity $v - v_{e}$ hai (uski speed rocket ke relative $v_{e}$ backward hai).

$p_{after} = (m + d m) (v + d v) + (- d m) (v - v_{e})$

Yeh step kyun? Hum system ko "rocket abhi" + "gas abhi throw ki" mein split karte hain, jisme har ek apna momentum carry karta hai, taki hum before-state se equate kar sakein.

Expand karo. $p_{after} = m v + m d v + v d m + d m d v - v d m + v_{e} d m$

Yeh step kyun? Multiply out karo aur $v d m$ terms cancel karo. Product $d m d v$ ek second-order infinitesimal hai → isse drop karo.

$p_{after} = m v + m d v + v_{e} d m$

Before = after equate karo: $m v = m v + m d v + v_{e} d m$ , toh

$m d v + v_{e} d m = 0 \Rightarrow d v = - v_{e} \frac{d m}{m}$

Yeh step kyun? Yahi toh core hai: velocity gain fractional mass loss $d m / m$ ke proportional hai. Yeh $1/ m$ hi logarithm ko paida karta hai.

Integrate karo start ( $v = v_{i}, m = m_{0}$ ) se end ( $v = v_{f}, m = m_{f}$ ) tak:

$\int_{v_{i}}^{v_{f}} d v = - v_{e} \int_{m_{0}}^{m_{f}} \frac{d m}{m}$

$v_{f} - v_{i} = - v_{e} [ln m]_{m_{0}}^{m_{f}} = - v_{e} (ln m_{f} - ln m_{0}) = v_{e} ln \frac{m _{0}}{m _{f}}$

Figure — Δv = v_e · ln(m₀ - m_f) — understanding each term

Equation padhna (physically KYA kehti hai)

Agar $m_{0} = m_{f}$ (koi fuel nahi jala): $ln 1 = 0 \Rightarrow Δ v = 0$ . ✔ samajh aata hai.
Fixed $v_{e}$ par Δv double karne ke liye, mass ratio ko square karna padta hai (kyunki $ln R^{2} = 2 ln R$ ). Fuel ki demand blast ho jaati hai.
Zyada $v_{e}$ → har kg fuel par zyada Δv. Isliye engineers high exhaust speed ke peeche bhagte hain (ion engines, hot gases).

Worked examples

Common mistakes (steel-manned)

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum ek frozen lake par skateboard par ho aur tumhare haath mein ek backpack hai jisme baseballs bhari hain. Har baar jab tum ek ball peeche throw karte ho, tum thoda aage slide karte ho. Jaise backpack khaali hoti hai, tum halke ho jaate ho, toh har throw tumhe thoda aur push karta hai. Rocket equation kehti hai: tum kitne tez ho jaate ho yeh depend karta hai kitni tez tum balls throw karte ho ( $v_{e}$ ) aur kitne halke tum ho gaye — ek ratio ke roop mein measure kiya gaya (heavy-you divided by light-you), ek "shrinking machine" jise $ln$ kehte hain se pass kiya gaya. Balls harder throw karna bahut help karta hai; bahut zyada balls carry karna sirf dheere dheere help karta hai, kyunki balls khud bhi carry karne mein heavy hoti hain.

Connections

Conservation of Momentum — derivation ki foundation.
Newton's Third Law — thrust mechanism.
Specific Impulse (Isp) — $v_{e} = g_{0} I_{s p}$ link karta hai.
Multistage Rockets — staging log ki tyranny ko kaise beat karta hai.
Thrust and Mass Flow Rate — $F = v_{e} \overset{m}{˙}$ , differential cousin.
Natural Logarithm and Integration of 1/x — log kyun aata hai.

State the Tsiolkovsky rocket equation.

Δ v = v_{e} ln (m_{0} / m_{f})

Δv physically kya represent karta hai?

Rocket ki velocity mein change (ek speed budget), uski final speed nahi.

m_{0}

(wet mass) define karo.

Total initial mass: structure + payload + saara propellant.

m_{f}

(dry mass) define karo.

Burn ke baad bachi mass: structure + payload, koi usable propellant nahi.

v_{e}

kya hai?

Effective exhaust velocity — rocket ke relative gas ki speed;

v_{e} = g_{0} I_{s p}

Derivation mein logarithm kyun aata hai?

Kyunki velocity gain

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

obey karta hai;

1/ m

integrate karne par

ln

milta hai.

Rocket motion ki starting differential equation (no gravity)?

m d v + v_{e} d m = 0

Agar

m_{0} = m_{f}

ho, toh Δv kya hai aur kyun?

0, kyunki

ln (1) = 0

— koi fuel nahi jala.

Fixed

v_{e}

par Δv double karne ke liye mass ratio ko kya karna hoga?

Square karna hoga, kyunki

ln R^{2} = 2 ln R

Equation ke symbols mein jale hue propellant ki mass?

m_{0} - m_{f}

Kaun si ek change har kg fuel par zyada Δv deti hai?

Exhaust velocity

v_{e}

badhao.

Rocket:

v_{e} = 3000

m/s,

R = 5

. Δv?

3000 ln 5 \approx 4828

m/s.

Ideal rocket equation ke peeche assumptions?

Koi gravity nahi, koi drag nahi, constant

v_{e}

Δv = v_e · ln(m₀ - m_f) — understanding each term

The Big Picture

Har term ka matlab