3.3.2 · D1 · Physics › Rocket Propulsion › Δv = v_e · ln(m₀ - m_f) — understanding each term
Ek rocket apne bahar kisi cheez ko push nahi kar sakta, isliye woh apni khud ki mass ko peeche fenk ke aage ki taraf kick leta hai. Poora topic isi idea ko ek single equation mein convert karne ke baare mein hai jo kehta hai: aap kitni extra speed kamate ho yeh depend karta hai kitni tez aap exhaust fenkte ho aur apni starting mass aur ending mass ke ratio par.
Isse pehle ki aap Δ v = v e ln ( m 0 / m f ) padh sako, us line par har ek mark ka aapko kuch matlab hona chahiye. Yeh page har ek cheez ko kuch nahi se banata hai, us order mein jo har ek idea ko pehle wale ke upar rest karne deta hai. Yahan kuch bhi assumed nahi hai — agar parent note ne likha hai, toh hum usse define karte hain.
m
m rocket mein "stuff" ki quantity hai, kilograms (kg) mein measured. Weight nahi (uske liye gravity chahiye) — bas kitna material hai wahan jo push kiya ja sake.
Picture yeh hai: ek frictionless ice rink par ek box. Box jitna heavy hoga, uski motion change karna utna hi mushkil hoga. Ek rocket ek aisa box hai jiska mass continuously shrink karta rehta hai jaise woh fuel jalata hai — yahi poori story hai.
Humein teen alag mass-labels chahiye kyunki ek rocket burn ke start aur end mein alag weight ka hota hai:
m 0 = initial (wet) mass — ignition par FULL rocket: structure + payload + engines + saara propellant . "Wet" = tanks bhar ke.
m f = final (dry) mass — burn ke BAAD jo kuch bachi: structure + payload + engines, lekin koi usable fuel nahi . "Dry" = tanks khaali.
m (plain) = burn ke kisi instant par mass — m 0 aur m f ke beech kahin.
m f woh fuel hai jo use hua."
Kyun sahi lagta hai: "final" sun ke lagta hai "woh stuff jo jal gaya."
Fix: m f woh mass hai jo bachi rehti hai . Jo propellant jala woh difference m 0 − m f hai. Picture: ek fuel tank — khaali hua hissa m 0 − m f hai, bacha hua rocket-body m f hai.
v
v yeh hai ki rocket kitni tez, aur kis direction mein move kar raha hai (m/s). Hamare ice-rink picture mein yeh ek arrow hai: length = speed, pointing = direction of travel.
Ab do tareekon se velocity badalta hai:
d v vs Δ v
d v = speed change ka chhota sa sliver , woh chhoti si kick jo aapko ek chhoti puff exhaust fenk ke milti hai. "d " ka matlab hai "infinitesimally chhota bit of."
Δ v = kul speed change jo poore burn mein add up hoti hai. "Δ " (Greek capital delta ) ka matlab hai "mein change." Yeh saare d v 's ka sum hai.
Δ v rocket ki final speed hai."
Kyun sahi lagta hai: iske units m/s hain aur burn se nikalta hai.
Fix: Δ v ek budget hai — ek change jo aap apni starting speed mein ADD karte ho. Agar aap 3000 m/s par start karte ho aur Δ v = 4000 m/s hai, toh aap 7000 m/s tak pahunch sakte ho (gravity aur drag ke khaane se pehle). Picture: do arrows tip-to-tail rakhe: purani velocity plus kamaaya hua Δ v .
Definition Effective exhaust velocity
v e
v e yeh hai ki exhaust gas nozzle se kitni tez nikalti hai, rocket ke relative measure ki gayi (m/s). Yeh woh "throwing speed" hai jis mass ko aap eject karte ho.
Picture: aap ek skateboard par khade ho aur baseball peeche fenk rahe ho. v e yeh hai ki aap har ball kitni zor se fenkte ho. Zyada zor se fenka → aap har ball ke liye zyada tez aage khisakte ho. Yahi wajah hai ki engineers high v e ke liye bhaagte hain: yeh fuel ke har kilogram par zyada speed khareedta hai.
Intuition "Rocket ke relative" kyun?
Agar rocket already v par move kar raha hai aur gas v e speed par rocket ke relative peeche nikalti hai, toh ice-rink ke fixed frame mein gas v − v e par move karti hai. Is bookkeeping ko straight rakhna derivation ka sabse tricky part hai — hum throw ko thrower se measure karte hain, lekin momentum ko fixed frame mein count karna hoga.
v e ek aur number se bhi connect hota hai jo rocket engineers hamesha quote karte hain, Specific Impulse (Isp) , v e = g 0 I s p ke through, jahan g 0 = 9.81 m/s 2 ek fixed reference number hai. Poori story ke liye woh note dekho.
p
p = m v hai mass times velocity : "kitni motion" ek object carry karta hai uska measure. Bada aur slow, chhote aur fast ke barabar ho sakta hai.
Humein isko kyun chahiye: koi road nahi, koi air nahi, aur koi gravity nahi, toh rocket-plus-exhaust system par koi outside force nahi hai. Jab koi outside force act nahi karta, total momentum change nahi ho sakta — yeh conserved hai. Yahi woh law hai jo hamein "before" aur "after" throw ko equate karne deta hai. Poori derivation Conservation of Momentum par built hai, jo khud Newton's Third Law ka disguise hai: gas rocket ko exactly utna hi push karta hai jitna rocket gas ko push karta hai.
Picture: baseballs ka backpack peeche fenk do, aur aap aage recoil karte ho — aapko jo forward momentum milti hai woh exactly baseballs ke backward momentum se match karti hai. Total: unchanged.
Parent note likhta hai d v = − v e m d m aur phir integrate karta hai. Wahan do symbols ko zero se build karna padega.
m d m — ek fractional change
d m ek tiny bit of mass change hai. Kyunki burn ke dauran mass girti hai , d m ek negative number hai. m se divide karne par fraction milta hai jitna current mass ka us instant mein lose hua — e.g. "abhi jo hai uska 0.1% lose hua."
Intuition Tool fraction kyun hona chahiye, raw amount kyun nahi?
Har kick d v = − v e m d m follow karta hai: speed gain jo mass lost par depend karta hai woh jo bacha hai uska fraction hai, na ki raw kilograms. 1 kg fenk na zyada matter karta hai jab sirf 2 kg bache hon, jab 500 kg bache hon usse zyada. Koi bhi quantity jo khud ke fraction se change hoti hai woh logarithm ka fingerprint hai.
∫
Stretched-S symbol ∫ ka matlab hai "infinitely many tiny slivers add karo ." Hum poore burn mein har chhoti d v ko add karte hain total Δ v paane ke liye. Yeh sum ka continuous version hai.
ln — "shrinking machine"
ln ( x ) answer karta hai: ==e ≈ 2.718 special number ko kis power tak raise karna hoga x paane ke liye?== Yeh bade multiplicative jumps leke unhe chhote additive steps mein turn karta hai.
Key behaviour, aur parent note log ko "villain" kyun kehta hai:
Intuition Andar multiply karna bahar add karne banta hai
ln ( a ⋅ b ) = ln a + ln b , toh ln ( R 2 ) = 2 ln R . Apna Δ v double karne ke liye mass ratio R ko square karna padega. Thodi si extra speed double karna kai zyada fuel demand kar sakta hai. Log bahut dheere grow karta hai.
e x — undo button
e x (exponential) ln ko reverse karta hai: agar ln R = 2.765 toh R = e 2.765 = 15.9 . Hamein iska zarurat tab padti hai jab hum backward solve karte hain — "mujhe jo Δ v chahiye woh pata hai, uske liye mass ratio kya chahiye?"
R
R = m f m 0 — starting mass divided by ending mass , ek pure number bina units ke (kg cancels kg). R = 5 ka matlab hai rocket start mein end se 5× heavy tha. Yeh woh single quantity hai jo ln actually "khata" hai.
Worked example Mass ratio padhna
m 0 = 500 , 000 kg, m f = 100 , 000 kg se R = 5 milta hai. Phir Δ v = v e ln 5 . Agar v e = 3000 m/s: Δ v = 3000 × 1.609 = 4828 m/s. ln ne 5× mass jump ko sirf 1.6 ke factor tak shrink kar diya.
Delta v total speed change
Momentum p equals m times v
dv equals minus ve times dm over m
Tsiolkovsky Delta v equals ve times ln R
Mass ratio R equals m0 over mf
Upar se neeche padho: masses aur velocities raw materials hain; momentum conservation (Newton's third law se) unhe ek differential kick d v = − v e d m / m mein convert karta hai; d m / m ka integral logarithm paida karta hai; aur sab kuch Tsiolkovsky equation mein land karta hai.
Right side dhako aur khud test karo. Agar koi bhi answer surprise kare, aage badhne se pehle us section ko dobara padho.
m 0 (wet mass) mein kya include hai?Structure + payload + engines + SAARA propellant — ignition par full rocket.
m f (dry mass) mein kya include hai?Burn ke baad jo kuch bacha — structure + payload + engines, koi usable fuel nahi.
Propellant burned kaise compute karte hain? m 0 − m f (difference, m f khud nahi).
Δ v precisely kya hai?Speed mein kul change — ek budget jo aap apni starting velocity mein add karte ho, final speed nahi.
v e kya hai aur ise kiske relative measure kiya jaata hai?Exhaust gas speed, rocket ke relative measured.
Yahan momentum conserved kyun hai? Koi external force (ideal case: koi gravity nahi, koi drag nahi) rocket-plus-exhaust system par act nahi karta.
m d m kya represent karta hai?Ek tiny instant mein current mass ka fraction jo lost hua (aur d m negative hai).
Subtraction ki jagah logarithm kyun aata hai? Kyunki ∫ m d m = ln m ; velocity gain fractional mass loss par depend karta hai.
Mass ratio R kya hai aur iske units kya hain? R = m 0 / m f , ek dimensionless pure number.
e x aapko kya karne deta hai?ln ko undo karna — required Δ v se required mass ratio tak backward solve karna.
ln ( R 2 ) = 2 ln R physically kya batata hai?Δ v double karne ke liye mass ratio square karna padega — fuel demand explode karti hai.
Yeh note Hinglish mein padho →
Conservation of Momentum — woh law jo hamein before aur after equate karne deta hai.
Newton's Third Law — exhaust aur rocket equally kyun push karte hain.
Specific Impulse (Isp) — v e = g 0 I s p ka source.
Natural Logarithm and Integration of 1/x — ∫ d m / m se ln kyun milta hai.
Thrust and Mass Flow Rate — instantaneous cousin F = v e m ˙ .
Multistage Rockets — staging log ki dheemi growth se kaise ladhta hai.