Equations of motion — 3DOF point mass (trajectory analysis): Worked examples

3.4.4 · D3 · Physics › Rocket Flight Mechanics › Equations of motion — 3DOF point mass (trajectory analysis)

Shuru karne se pehle, ek reminder notation ki jo hum use karenge (har symbol parent mein explain ho chuka hai):

Definition Chaar forces (aur do engine numbers) jo hum plug in karenge

$T$ = thrust = engine ka aage ki taraf push, newtons (N) ya kilonewtons mein measure hota hai ( $1 kN = 1000 N$ ). Yeh velocity arrow ke saath act karta hai (zero angle of attack).
$D$ = drag = air ka vehicle par peeche ki taraf friction. Yeh hamesha velocity arrow ko oppose karta hai, isliye minus sign ke saath aata hai.
$L$ = lift = velocity ke perpendicular ek force (arrow ke sideways). Ek plain ballistic rocket mein $L = 0$ ; wings ya flared body ise non-zero banate hain.
$g$ = local gravitational acceleration $= 9.81 m/s^{2}$ ground ke paas — weight har kilogram ko straight neeche kitni tezi se kheenchta hai.
$I_{s p}$ = specific impulse (seconds) = engine ke liye ek fuel-efficiency score; zyada matlab har kilogram propellant mein zyada push.
$g_{0}$ = standard gravity $= 9.80665 \approx 9.81 m/s^{2}$ = ek fixed reference constant jo sirf $I_{s p}$ ko exhaust speed mein convert karne ke liye use hota hai. Yeh rocket ko kheenchne wala local $g$ NAHIN hai — yeh altitude ke saath kabhi nahi badlata; yeh sirf woh bookkeeping number hai jo $I_{s p}$ ki definition mein baka hua hai.

Scenario matrix

Har 3DOF problem in hi cells mein se ek hoti hai. Neeche ke examples [C1], [C2], … se label hain taaki dekh sako ki har cell cover ho rahi hai.

Cell	Case class	Kya special hai	Key term jo sign badlta / vanish hota hai
C1	Steep climb, $0 < γ < 9 0^{\circ}$	thrust on, gravity slows & turns	$sin γ > 0, cos γ > 0$
C2	Vertical, $γ = 9 0^{\circ}$	local accel, drag climb oppose karta hai	$cos γ = 0 \Rightarrow \overset{γ}{˙} = 0$
C3	Level flight, $γ = 0^{\circ}$	gravity speed nahi ghataati	$sin γ = 0, cos γ = 1$ (fastest turn)
C4	Descending / diving, $γ < 0^{\circ}$	gravity ab speed badhaata hai	$sin γ < 0 \Rightarrow - g sin γ > 0$
C5	Coast / ballistic, $T = 0$	engine off; drag $- D / m$ abhi bhi slow karta hai, ya $D = 0$ ⇒ pure projectile	thrust term vanish; drag term survive karta hai jab tak $D = 0$ na ho
C6	Mass changing, $\overset{m}{˙} \neq = 0$	same thrust, rising acceleration	$m$ burn ke dauran ghatta hai
C7	Degenerate speed, $V \to 0$	turning rate blow up karta hai	$\overset{γ}{˙} = - g cos γ / V \to \infty$
C8	Word problem (gravity turn)	initial pitch-over	small $γ$ , full set use karta hai
C9	Exam twist (lift added)	$L \neq = 0$ path ko upar curve karta hai	normal equation full form

Ab hum har cell hit karte hain.

C1 — Steep climb (sabhi signs positive)

Worked example Example 1 —

$γ = 4 0^{\circ}$ par climbing instantaneous acceleration $T = 180 kN$ , $D = 25 kN$ , $m = 9000 kg$ , $γ = 4 0^{\circ}$ , $g = 9.81$ . $\dot{V}$ nikalo.

Forecast: guess karo — kya $\dot{V}$ level-flight value se bada hoga ya chhota? (Gravity kuch speed chura rahi hai, isliye yeh $γ$ ke $0$ hone se chhota hona chahiye.)

Tangential rule likho: $\dot{V} = \frac{T - D}{m} - g sin γ$ . Yeh step kyun? Sirf along-velocity forces velocity arrow ki length badlte hain. Thrust ise lamba karta hai, drag aur gravity ka along-track slice ise ghataate hain.
Net propulsive term: $\frac{180000 - 25000}{9000} = \frac{155000}{9000} = 17.22 m/s^{2}$ . Yeh step kyun? Thrust minus drag woh force hai jo gravity se pehle survive karta hai; $m$ se divide karne par force acceleration ban jaata hai (Newton).
Gravity along-track slice: $g sin 4 0^{\circ} = 9.81 \times 0.6428 = 6.306 m/s^{2}$ . Yeh step kyun? $γ = 4 0^{\circ}$ par weight ka sirf $sin 4 0^{\circ}$ fraction motion ke saath peeche point karta hai — wahi piece speed arrow ko slow karta hai.
Combine karo: $\dot{V} = 17.22 - 6.31 = 10.92 m/s^{2}$ . Yeh step kyun? Tangential rule arrow ke saath accelerations ka sum hai: propulsive push (step 2) minus gravity slice (step 3). Inhe subtract karne par iss instant mein speed ke change ka single net rate milta hai.

Verify: units $\frac{N}{kg} = m/s^{2}$ ✓. Yeh positive hai (abhi bhi accelerating) aur pure $17.22$ se chhota hai, forecast se match ✓.

C2 — Vertical flight ( $γ = 9 0^{\circ}$ )

Worked example Example 2 — Seedha upar, aur kya yeh turn karta hai?

$T = 180 kN$ , $D = 25 kN$ , $m = 9000 kg$ , $γ = 9 0^{\circ}$ , $V = 120 m/s$ , $L = 0$ . $\dot{V}$ aur $\overset{γ}{˙}$ nikalo.

Forecast: $γ = 9 0^{\circ}$ par, gravity sab climb ko oppose karta hai — isliye $\dot{V}$ poore $g$ se girna chahiye. Aur ek vertical rocket vertical hi rehna chahiye, isliye $\overset{γ}{˙} = 0$ .

$sin 9 0^{\circ} = 1$ , isliye $\dot{V} = \frac{155000}{9000} - 9.81 (1) = 17.22 - 9.81 = 7.41 m/s^{2}$ . Yeh step kyun? Seedha upar, "velocity ke saath peeche" matlab seedha neeche hai — poora weight motion se ladhta hai. Yeh vertical-climb equation ki local acceleration form hai; ise drag-free vertical burn ke saath integrate karne par Tsiolkovsky Rocket Equation $Δ V$ formula milta hai — lekin woh integral $Δ V$ result aur yeh instantaneous $\dot{V}$ do alag quantities hain, same expression nahin.
$cos 9 0^{\circ} = 0$ , isliye $\overset{γ}{˙} = - \frac{g cos γ}{V} = - \frac{9.81 \times 0}{120} = 0 rad/s$ . Yeh step kyun? Ek vertical velocity ke perpendicular, gravity ka koi sideways slice nahin hai — kuch bhi arrow ko turn nahi karta.

Verify: $γ = 9 0^{\circ}$ par along-track gravity maximal hai aur turning gravity zero hai — "S-along, C-across" rule ke do extreme ✓. Vertical climb vertical rehta hai ✓.

Figure s01 kaise padhen. Horizontal axis flight-path angle $γ$ hai jo level ( $0^{\circ}$ ) se vertical ( $9 0^{\circ}$ ) tak sweep karta hai; vertical axis $m/s^{2}$ mein gravity acceleration hai. Amber curve $g sin γ$ hai — gravity ka woh slice jo velocity ke saath act karta hai aur speed equation $\dot{V}$ mein aata hai. Cyan curve $g cos γ$ hai — woh slice jo velocity ke across act karta hai aur turning equation $\overset{γ}{˙}$ ko drive karta hai. Notice karo ki woh $4 5^{\circ}$ par cross karte hain aur places trade karte hain: left edge par (level, C3) turning maximal hai, right edge par (vertical, C2) slowing maximal hai. Har curve literally woh number hai jise tum corresponding worked example mein multiply karte ho.

Figure — Equations of motion — 3DOF point mass (trajectory analysis)

C3 — Level flight ( $γ = 0^{\circ}$ ): fastest turn

Worked example Example 3 — Flat fly karna: koi speed loss nahi, sabse zyada bend

$V = 250 m/s$ , $γ = 0^{\circ}$ , $L = 0$ , $g = 9.81$ . Gravity ka $\dot{V}$ contribution aur $\overset{γ}{˙}$ nikalo.

Forecast: level fly karte waqt, gravity seedha neeche hai, ek horizontal velocity ke exactly perpendicular — isliye yeh koi speed nahi churaati lekin path ko sabse zyada bend karti hai.

Gravity along-track: $- g sin 0^{\circ} = - 9.81 \times 0 = 0$ . Yeh step kyun? Horizontal motion matlab "neeche" velocity ke saath $9 0^{\circ}$ angle banata hai — koi along-track component nahin.
Turning rate: $\overset{γ}{˙} = - \frac{g cos 0 ^{\circ}}{V} = - \frac{9.81 \times 1}{250} = - 0.03924 rad/s$ . Yeh step kyun? $cos 0^{\circ} = 1$ : pura weight ab motion ke sideways act karta hai, isliye iss speed ke liye path apni maximum rate par bend karta hai.
Degrees mein: $- 0.03924 \times \frac{180}{π} = - 2.24 8^{\circ} / s$ . Yeh step kyun? Formula radians per second produce karta hai ( $\overset{γ}{˙}$ ki natural unit), lekin reader degrees mein sochta hai. $180/ π$ se multiply karne par consistently convert hota hai taaki dono forms ek hi physical turn rate describe karen — rad aur deg mix karna classic wrong answers ka source hai.

Verify: parent ke level-vs-vertical statement se compare karo — turning $γ = 0$ par fastest hai, $γ = 9 0^{\circ}$ par zero ✓ (Example 2 ne $\overset{γ}{˙} = 0$ diya).

C4 — Descending / diving ( $γ < 0^{\circ}$ ): gravity speed badhata hai

Worked example Example 4 —

$γ = - 3 0^{\circ}$ par coasting downhill Coast phase: $T = 0$ , $D = 12 kN$ , $m = 9000 kg$ , $γ = - 3 0^{\circ}$ , $g = 9.81$ . $\dot{V}$ nikalo.

Forecast: neeche point karte waqt, gravity ab motion ke saath aage kheeenchti hai — isse rocket speed up hona chahiye, partly drag se cancel.

$\dot{V} = \frac{0 - 12000}{9000} - g sin (- 3 0^{\circ})$ . Yeh step kyun? Thrust off hai (cell C5 yahan overlap karta hai); hum full formula rakhte hain aur $γ$ ka sign kaam karne dete hain.
$sin (- 3 0^{\circ}) = - 0.5$ , isliye $- g sin (- 3 0^{\circ}) = - 9.81 (- 0.5) = + 4.905 m/s^{2}$ . Yeh step kyun? Ek negative $γ$ gravity term ko positive kar deta hai — yeh gravity ka helping karna hai, mathematical reason ki kyon ek diving vehicle accelerate karta hai.
Drag term: $\frac{- 12000}{9000} = - 1.333 m/s^{2}$ . Yeh step kyun? Engine off hone ke saath, drag hi woh akela along-track force hai jo arrow ko slow karta hai; yeh $- D / m$ ke roop mein aata hai kyunki drag hamesha velocity oppose karta hai. Ise alag compute karne par helping (gravity) aur hurting (drag) pieces add karne se pehle visible rehte hain.
Combine karo: $\dot{V} = - 1.333 + 4.905 = + 3.572 m/s^{2}$ . Yeh step kyun? Tangential rule along-track accelerations ko sum karta hai: drag slice (step 3) plus ab-positive gravity slice (step 2). Unka sum speed change ka single net rate hai.

Verify: engine off, phir bhi $\dot{V} > 0$ — sirf isliye possible hai kyunki $γ < 0$ ne gravity ko energy add karne diya; drag akela $- 1.333$ deta ✓. Units m/s² ✓.

C5 — Coast / ballistic ( $T = 0$ ): parabola recover karo

Worked example Example 5 — Engine-off, drag-free: kya yeh parabola hai?

$T = 0$ , $D = 0$ , $γ = 4 5^{\circ}$ , $V = 200 m/s$ , $g = 9.81$ . $\dot{V}$ , $\overset{γ}{˙}$ , $\overset{x}{˙}$ , $\dot{h}$ nikalo.

Forecast: koi thrust nahin aur koi drag nahin, toh yeh sirf ek throw ki hui ball hai — classic projectile. (Example 4 se compare karo, jo bhi thrust-off tha lekin drag $- D / m$ abhi bhi vehicle ko slow kar raha tha; yahan $D = 0$ set karne par woh term bhi hat jaata hai.)

$\dot{V} = 0 - 0 - g sin 4 5^{\circ} = - 9.81 \times 0.7071 = - 6.936 m/s^{2}$ . Yeh step kyun? Dono $T = 0$ aur $D = 0$ ke saath, sirf gravity ka along-track slice bachta hai; climbing, yeh speed slow karta hai. (Example 4 mein extra $- D / m$ term survive kiya kyunki wahan $D \neq = 0$ tha.)
$\overset{γ}{˙} = - \frac{g cos 4 5 ^{\circ}}{V} = - \frac{9.81 \times 0.7071}{200} = - 0.03468 rad/s$ . Yeh step kyun? Gravity ka sideways slice arrow ko neeche tip karta hai — ball arc hoti hai.
$\overset{x}{˙} = V cos 4 5^{\circ} = 200 \times 0.7071 = 141.4 m/s$ , $\dot{h} = V sin 4 5^{\circ} = 141.4 m/s$ . Yeh step kyun? Speed $γ$ ke $cos$ / $sin$ se horizontal aur vertical mein split hoti hai — woh numbers jo tum arc plot karne ke liye integrate karte ho.

Verify: drag-free projectile ke liye horizontal acceleration zero hona chahiye. Check karo $\overset{x}{¨} = \frac{d}{d t} (V cos γ) = \dot{V} cos γ - V sin γ \overset{γ}{˙} = (- 6.936) (0.7071) - (141.4) (- 0.03468) = - 4.905 + 4.905 = 0$ ✓. Woh exact cancellation hi parabola hai — horizontal speed constant hai. Beautiful.

Figure s02 kaise padhen. Axes ground frame hain: horizontal downrange $x$ (m) hai, vertical altitude $h$ (m) hai. Cyan curve coast trajectory hai — ek perfect parabola. Amber arrows iske saath teen instants par draw hain velocity arrows hain; dekho unka horizontal reach same length rehta hai jabki unka vertical reach ghatta hai, tip karta hai, phir neeche badhta hai. Woh constant horizontal component exactly woh $\overset{x}{¨} = 0$ result hai jo humne abhi verify kiya — picture aur algebra ek hi baat kehte hain.

C6 — Mass changing ( $\overset{m}{˙} \neq = 0$ ): acceleration badhti hai

Worked example Example 6 — Same thrust, burn mein baad mein

Engine $T = 180 kN$ rakhta hai, $D = 25 kN$ , $γ = 4 0^{\circ}$ (Example 1 jaisa hi) lekin tank khali ho gaya hai isliye $m = 5000 kg$ . Agar $I_{s p} = 280 s$ , $g_{0} = 9.81$ ho toh $\overset{m}{˙}$ bhi nikalo. $\dot{V}$ aur $\overset{m}{˙}$ nikalo.

Forecast: kam mass, same push → Example 1 ke $10.92$ se zyada acceleration.

$\dot{V} = \frac{180000 - 25000}{5000} - 9.81 sin 4 0^{\circ} = 31.0 - 6.306 = 24.69 m/s^{2}$ . Yeh step kyun? Newton ka $a = F / m$ : $m$ ghataao aur same net force zyada $a$ dega. Isliye burnout acceleration brutal hoti hai.
Mass rate: $\overset{m}{˙} = - \frac{T}{I _{s p} g _{0}} = - \frac{180000}{280 \times 9.81} = - 65.53 kg/s$ . Yeh step kyun? Yahan $g_{0}$ woh fixed reference constant hai jo $I_{s p}$ ki definition mein baka hua hai (woh local gravity nahin jo rocket ko kheenchti hai). Product $I_{s p} g_{0}$ m/s mein exhaust speed hai; thrust ko exhaust speed se divide karne par propellant ke kilograms per second milte hain jo nikal rahe hain — negative isliye kyunki mass sirf kabhi girti hai.

Verify: Example 1 mein $m = 9000$ tha aur $10.92$ mila; yahan $m = 5000$ se $24.69 > 10.92$ milta hai, forecast se match ✓. $\overset{m}{˙} < 0$ (mass sirf kabhi ghatti hai) ✓.

C7 — Degenerate speed ( $V \to 0$ ): turning rate blow up karta hai

Worked example Example 7 — Liftoff par hi,

$V$ almost zero Vehicle barely moving, $V = 5 m/s$ , $γ = 1 0^{\circ}$ , $L = 0$ , $g = 9.81$ . $\overset{γ}{˙}$ nikalo. Phir $V \to 0$ discuss karo.

Forecast: chhote $V$ se divide karne par $\overset{γ}{˙}$ bahut bada ho jaata hai — jab rocket almost stationary ho toh path direction whip around karta hai.

$\overset{γ}{˙} = - \frac{g cos 1 0 ^{\circ}}{V} = - \frac{9.81 \times 0.9848}{5} = - \frac{9.663}{5} = - 1.933 rad/s$ ( $\approx - 110. 7^{\circ} / s$ ). Yeh step kyun? Normal equation mein denominator mein $V$ hai; ek non-zero $cos γ$ numerator ke saath chhoti speed ratio ko explode kar deti hai — almost-stationary arrow ki direction wildly swing ho sakti hai.
Limit: jab $V \to 0^{+}$ , $\overset{γ}{˙} \to - \infty$ ( $cos γ \neq = 0$ ke liye). Yeh step kyun? Physically model toot jaata hai — ek truly motionless point mass ka koi well-defined "direction of velocity" nahin hota jise turn kiya ja sake. Real launches near-vertical ( $cos γ \approx 0$ ) shuru hote hain is term ko chhota rakhne ke liye, aur exactly isliye rocket pehle upar jaata hai pitching se pehle.

Verify: danger sirf off-vertical hai: $γ = 9 0^{\circ}$ par, $cos γ = 0$ isliye $V = 0$ par bhi $\overset{γ}{˙} = 0$ — koi blow-up nahi ✓ (Example 2 se consistent). Yeh Gravity Turn Trajectory ke vertical se shuru hone ka mathematical justification hai.

C8 — Word problem: gravity-turn pitch-over

Worked example Example 8 —

$8 9^{\circ}$ se $8 8^{\circ}$ tak tip hone mein kitna time? Liftoff ke baad ek rocket $γ = 8 9^{\circ}$ par hai, ek steady $V = 100 m/s$ par climb kar raha hai (iss brief instant par roughly constant maano), $L = 0$ . Sirf gravity se $1^{\circ}$ flight-path angle lose karne ka time estimate karo.

Forecast: near vertical $cos γ$ tiny hai, isliye turn slow hai — yeh gentle start hi gravity turn ka poora point hai.

$\overset{γ}{˙} = - \frac{g cos 8 9 ^{\circ}}{V} = - \frac{9.81 \times 0.017452}{100} = - 1.712 \times 1 0^{- 3} rad/s$ . Yeh step kyun? Gravity hi akela bender hai; near vertical uska sideways slice $cos 8 9^{\circ}$ minute hai.
$1^{\circ}$ ko radians mein convert karo: $1^{\circ} = 0.017453 rad$ . Yeh step kyun? $\overset{γ}{˙}$ rad/s mein hai, isliye target change bhi radians mein hona chahiye — units match karne chahiye.
Time $= \frac{∣Δ γ ∣}{∣ γ ˙ ∣} = \frac{0.017453}{1.712 \times 1 0 ^{- 3}} = 10.19 s$ . Yeh step kyun? Yahan $\overset{γ}{˙}$ negative hai (angle ghatt raha hai) aur poocha gaya hai "ek degree lose hone mein kitna time" — ek duration, jo positive hona chahiye. Dono change aur rate ke absolute value lene par direction sign hat jaata hai aur elapsed time milta hai, kyunki time = size-of-change ÷ size-of-rate.

Verify: sirf ek degree shed karne mein karib das second — pitch-over sach mein gentle hai, exactly isliye koi control effort waste nahi hota ✓. flight-path-angle geometry use karta hai.

C9 — Exam twist: lift path ko upar curve karta hai

Worked example Example 9 — Normal equation mein lift add karo

$m = 7000 kg$ , $V = 220 m/s$ , $γ = 3 0^{\circ}$ , $g = 9.81$ , aur ab ek lift $L = 90 kN$ velocity ke perpendicular (upar) act karta hai. $\overset{γ}{˙}$ nikalo. $L = 0$ se compare karo.

Forecast: lift arrow ki tip ko upar push karta hai, gravity ke downward bend se ladhta hai — $\overset{γ}{˙}$ lift-free case se kam negative (shayad positive bhi) hona chahiye.

Full normal equation: $\overset{γ}{˙} = \frac{L - m g cos γ}{mV}$ . Yeh step kyun? Perpendicular forces turning rate set karte hain; lift aur sideways gravity slice dono yahan hain. Yeh $\overset{γ}{˙}$ rule ki general form hai pehle ki hum har pehle example mein $L = 0$ assume karen.
Gravity slice: $m g cos 3 0^{\circ} = 7000 \times 9.81 \times 0.8660 = 59470 N$ . Yeh step kyun? $cos 3 0^{\circ}$ weight ka woh fraction pick karta hai jo velocity ke across act karta hai — downward-bending force.
Numerator: $L - m g cos γ = 90000 - 59470 = 30530 N$ (net upward). Yeh step kyun? Lift (upar) minus gravity ka cross-slice (neeche) net perpendicular force hai. Yeh positive nikla, isliye surviving push upar hai.
$\overset{γ}{˙} = \frac{30530}{7000 \times 220} = \frac{30530}{1 , 540 , 000} = + 0.01982 rad/s$ ( $\approx + 1.13 6^{\circ} / s$ ). Yeh step kyun? Net perpendicular force ko $mV$ se divide karne par turning rate milta hai (normal-equation form). Ab positive — lift gravity ko overpower karta hai, isliye path upar curve karta hai instead of neeche.
Lift-free comparison: $\overset{γ}{˙} = - \frac{g cos 3 0 ^{\circ}}{V} = - \frac{9.81 \times 0.8660}{220} = - 0.03862 rad/s$ . Yeh step kyun? $L = 0$ set karne par har pehle example mein use ki gayi equation recover hoti hai, isliye hum exactly dekh sakte hain ki lift ne kya badla.

Verify: sign negative se (neeche curve, lift-free) positive (upar curve) ho gaya jab lift ne gravity ka slice beat kiya — forecast se match ✓. Numerically $+ 0.01982$ vs $- 0.03862 rad/s$ : lift add karne ne sirf turn soft nahi kiya, usne direction reverse kar di. Yeh woh term hai jo ek ballistic rocket mein drop ho jaata hai lekin winged vehicles aur 6DOF Rigid-Body Dynamics mein central hai, jahan woh attitude jo $L$ generate karta hai apne equations ka set ban jaata hai.

Recall

Recall Rocket dive karne par kaunsa term sign badlta hai?

Speed equation mein gravity term, $- g sin γ$ ::: $γ < 0$ ke liye, $sin γ < 0$ isliye $- g sin γ > 0$ — gravity dive mein speed add karta hai.

Recall

$V = 0$ par bhi vertical rocket ke liye $\overset{γ}{˙} = 0$ kyun? Kyunki $\overset{γ}{˙} = - g cos γ / V$ aur $cos 9 0^{\circ} = 0$ ::: numerator zero hai, blow-up khatam; koi sideways gravity nahi matlab kuch bhi arrow ko turn nahi karta.

Recall Same thrust, half the mass —

$\dot{V}$ ka kya hota hai? Yeh (roughly) double ho jaata hai ::: $a = F / m$ ; $m$ ghataane par acceleration badhti hai, isliye burnout violent hoti hai.

Recall

$g_{0}$ kya hai aur yeh $g$ se kaise alag hai? $g_{0} = 9.80665 m/s^{2}$ ek fixed reference constant hai jo sirf $I_{s p} g_{0}$ ke andar exhaust speed nikalne ke liye use hota hai ::: $g$ woh local gravity hai jo actually rocket ko kheechti hai aur altitude ke saath vary kar sakti hai; $g_{0}$ kabhi nahi badlata.

Recall Thrust-off coast mein,

$- D / m$ term kab gayab hota hai? Sirf jab $D = 0$ ho (drag-free) ::: $T = 0$ ke saath drag term abhi bhi vehicle ko slow karta hai jab tak drag khud zero na ho, jo pure-projectile subcase hai jo parabola recover karta hai.

Connections

Equations of motion — 3DOF point mass (trajectory analysis) — parent rules jo yeh examples drill karte hain
Tsiolkovsky Rocket Equation — vertical, drag-free limit (C2)
Gravity Turn Trajectory — C7 aur C8 ka pitch-over
Drag and Atmospheric Models — $D$ supply karta hai jo throughout use hota hai
Flight-Path Angle and Velocity Frame — har $sin γ$ / $cos γ$ split ki geometry
6DOF Rigid-Body Dynamics — jahan lift (C9) aur rotation wapas aate hain
Numerical Integration of Trajectories (RK4) — in instantaneous rates ko ek full path mein chain karna