Trajectory optimization — minimum gravity loss, minimum drag loss

3.4.15 · Physics › Rocket Flight Mechanics

1. "Losses" hain kya? (WHY hum inki baat karte hain)

Ek rocket ko apne fuel se "milni chahiye" thi jo ideal velocity, wo Tsiolkovsky ideal hai: $Δ v_{ideal} = v_{e} ln \frac{m _{0}}{m _{f}}$

Lekin orbit mein actually jo velocity milti hai, wo kam hoti hai. Poora hisaab kuch aisa hai: $Δ v_{ideal} = Δ v_{orbit} + Δ v_{grav} + Δ v_{drag} + Δ v_{steer}$

Har ek "loss" ek aisa $Δ v$ hai jiske liye propellant chuka diya, lekin useful orbital speed ke roop mein nahi mila.

2. Ye terms KAISE aate hain — Newton se derive karo (Scratch se derivation)

Equation of motion velocity vector ke saath (tangential direction) lao. Variable-mass rocket ke liye Newton's 2nd law, flight ki direction par project karo:

$m \frac{d v}{d t} = thrust T - drag D - gravity component along path m g sin γ$

$m g sin γ$ kyun? Gravity seedhi neeche point karti hai. Velocity horizontal se $γ$ angle banati hai, isliye gravity ka wo component jo velocity ko oppose karta hai, wo $g sin γ$ hai. Seedha upar: $γ = 9 0^{\circ}$ , poora $g$ tumhare against. Horizontal: $γ = 0^{\circ}$ , gravity path ke saath koi kaam nahi karti (sirf usse curve karti hai).

Ab $T = \overset{m}{˙}_{p} v_{e}$ aur $m \frac{d v}{d t}$ ke saath, $m$ se divide karo aur burn par integrate karo:

$\int_{0}^{t_{b}} \frac{d v}{d t} d t = \int_{0}^{t_{b}} \frac{T}{m} d t - \int_{0}^{t_{b}} \frac{D}{m} d t - \int_{0}^{t_{b}} g sin γ d t$

$\int \frac{T}{m} d t = v_{e} \int \frac{m ˙ _{p}}{m} d t = v_{e} ln \frac{m _{0}}{m _{f}} = Δ v_{ideal}$ ✔ (yahi se Tsiolkovsky aata hai)
Left side $Δ v_{achieved}$ hai.

Toh: $Δ v_{achieved} = Δ v_{ideal} - Δ v_{drag} - Δ v_{grav}$

Poora bookkeeping equation bas yahi hai, derive ho gaya. Do integrals do loss terms hain.

3. Trade-off: WHY dono ko minimize nahi kar sakte

Dono integrands dekho aur $γ$ (pitch) unhe kaise control karta hai.

Zyada vertical steer karo ( $γ \to 9 0^{\circ}$ )	Zyada horizontal steer karo ( $γ \to 0^{\circ}$ )
$sin γ \to 1$ ⇒ gravity loss BADA	$sin γ \to 0$ ⇒ gravity loss chhota
Dense atmosphere jaldi chhodte hain, neeche $v$ kam ⇒ drag loss chhota	Neeche low & fast rehte hain, $ρ v^{2}$ bahut bada ⇒ drag loss BADA

Figure — Trajectory optimization — minimum gravity loss, minimum drag loss

4. Gravity turn (practical optimum)

Turning rate derive karo. Normal (velocity se perpendicular) equation of motion, jab thrust $v$ ke saath hai (no lift, no side thrust):

$m v \frac{d γ}{d t} = - m g cos γ$

$\frac{d γ}{d t} = - \frac{g cos γ}{v}$

Minus sign kyun / yeh efficient kyun hai: sirf gravity ka perpendicular component $g cos γ$ path ko curve karta hai. Koi propellant turning mein nahi lagta — yeh free hai. Zyada jaldi pitch over karne ke liye thrust ko off-velocity point karna padta (steering loss). Zyada dheere pitch over karne se zyada vertical rehte (gravity loss). Gravity turn nature ka sasta compromise hai.

Gaur karo: zyada $v$ par turn rate chhoti hoti hai ⇒ initial pitch kick pehle lagani padti hai (jab $v$ kam ho) warna fuel khatam hone se pehle horizontal nahi ho paoge.

5. Worked examples

6. Common mistakes (Steel-man → fix)

7. Recall & Feynman

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho tum apne dost ko door ek ball phenk rahe ho. Agar seedha upar phenko, wapas aa jaati hai — gravity se ladne mein saari energy waste ho gayi, koi sideways dost ki taraf nahi gayi. Agar bilkul flat aur hard phenko, hawa (drag) isse bahut slow kar deti hai aur pahunchne se pehle gir jaati hai. Sabse achha throw ek curve hai: pehle thoda upar taaki thick hawa se niklo, phir use jhuka do. Rocket bhi yahi "curve" karta hai — pehle upar jaata hai dense air se niklne ke liye, phir dheere se tip over karta hai taaki gravity khud turning free mein kar de. Zyada seedha = gravity se bahut zyada laro. Zyada flat = hawa kha jaaye. Perfect lean beech mein hai.

Recall Active self-test

Scratch se do loss integrals likho. 2. Dono zero kyun nahi ho sakte? 3. Gravity-turn pitch rate kya set karta hai? 4. Real launch mein kaun sa loss dominate karta hai?

Gravity loss formula

Δ v_{g r a v} = \int_{0}^{t_{b}} g sin γ d t

, jahan

γ

flight-path angle hai.

Drag loss formula

Δ v_{d r a g} = \int_{0}^{t_{b}} \frac{D}{m} d t = \int \frac{\frac{1}{2} ρ v ^{2} C _{D} A}{m} d t

Full delta-v bookkeeping

Δ v_{i d e a l} = Δ v_{or bi t} + Δ v_{g r a v} + Δ v_{d r a g} + Δ v_{s t eer}

Gravity loss

γ = 9 0^{\circ}

par sabse zyada kyun hoti hai

sin 9 0^{\circ} = 1

, toh poora

g

motion ko oppose karta hai; vertical flight gravity loss ke liye worst hai.

Dono losses ko zero kyun nahi kar sakte

Pitch

γ

dono ko opposite control karta hai — chhota

γ

gravity loss ghataata hai lekin drag loss badhata hai (dense air mein fast flight), aur vice versa.

Gravity-turn pitch rate

\frac{d γ}{d t} = - \frac{g cos γ}{v}

— sirf gravity ka normal component path ghoomata hai, wo bhi free mein.

Gravity turn efficient kyun hai

Thrust velocity ke saath rehti hai (

α = 0

) ⇒ zero steering loss; gravity turning karta hai bina kisi propellant cost ke.

Steering loss expression

\int \frac{T}{m} (1 - cos α) d t

, jab thrust velocity se

α

angle par misaligned ho.

Real launch mein kaun sa loss dominate karta hai

Gravity loss (≈1–2 km/s) drag loss (≈30–150 m/s) se kahin zyada hoti hai.

Gravity loss aur altitude ke baare mein common misconception

Yeh height se scale nahi karta; yeh

\int g sin γ d t

hai — pitch angle aur burn duration se tay hota hai.

Connections

Tsiolkovsky Rocket Equation — $Δ v_{ideal}$ ka source
Flight-path angle and equations of motion
Gravity Turn Ascent
Aerodynamic Drag and Max-Q
Thrust-to-weight ratio and burn time
Steering losses and thrust vectoring
Orbital insertion and required orbital velocity