Combined maneuvers — optimal split between plane change and velocity change: Worked examples

3.2.23 · D3 · Physics › Orbital Mechanics & Astrodynamics › Combined maneuvers — optimal split between plane change and

Yeh page parent topic ko itna drill karti hai ki koi bhi case aapko surprise na kar sake. Shuru karne se pehle ek vaada: neeche har symbol pehle se parent note mein earn kiya ja chuka hai, lekin har ek ko jab woh aaye tab re-anchor karte hain taaki aap line one se padh sako.

Ek master formula jo parent note ne build kiya:

Is page par sab kuch bas yahi formula hai (aur iska optimization) jo har tarah ke input se milta hai jo aap ise de sakte ho.

The scenario matrix

Ek "scenario" ko inputs ka ek choice samjho. Inputs hain: do speeds ( $v_{1}$ vs $v_{2}$ ), aur angle $θ$ . Yeh har distinct class of case hai, har ek ke saath woh example jo ise pakka karta hai.

Cell	Inputs mein kya khaas hai	Kyun yeh aapko trip kar sakta hai	Example
A. $θ = 0$ (degenerate angle)	bilkul bhi turn nahi	formula ko plain speed change mein collapse karna chahiye	Ex 1
B. $v_{1} = v_{2}$ (equal speeds, pure tilt)	sirf direction badlti hai	$2 v sin (θ /2)$ reproduce karna chahiye	Ex 2
C. $θ = 18 0^{\circ}$ (reverse direction)	bilkul opposite arrows	$cos$ hits $- 1$ ; speeds add hoti hain	Ex 3
D. General $v_{1} \neq = v_{2}$ , $0 < θ < 9 0^{\circ}$	roz ka combined burn	magnitude + direction mix karna	Ex 4
E. Optimal split, two burns (real mission)	LEO→GEO with inclination	equal-marginal-cost, not equal-angle	Ex 5
F. Limiting case $s \to i$ vs $s \to 0$	saari tilt ek end par rakhna	kaun sa endpoint jeetega, aur kyun	Ex 6
G. Combined vs separate (word problem)	kya ek throw sach mein sasta hai?	triangle inequality ko numeric banana	Ex 7
H. Exam twist: obtuse $θ$ , unequal $v$	$cos θ < 0$	cosine ka sign middle term ko flip kar deta hai	Ex 8

In sab mein, cells A–H cover karte hain: zero input, degenerate equal input, extreme $18 0^{\circ}$ , generic acute, generic obtuse, full optimization, uske do limiting endpoints, aur ek real design decision. Kuch bhi nahi chhoota.

Cell A — zero-angle degenerate case

Forecast: Bina kisi turn ke, dono arrows ek hi direction mein point karte hain. Guess: answer sirf unki lengths ka difference hona chahiye. Aage padhne se pehle apna guess likhو.

$θ = 0$ substitute karo, to $cos θ = 1$ . Yeh step kyun? $cos 0 = 1$ sabse bada value hai jo cosine le sakta hai, jo subtracted middle term ko as big as possible banata hai — yeh "koi sideways effort waste nahi" aisa dikhta hai. $Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2}$
Perfect square pehchano. Yeh step kyun? $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ . Root ke neeche pura $(v_{2} - v_{1})^{2}$ hai. $Δ v = (v_{2} - v_{1})^{2} = ∣ v_{2} - v_{1} ∣$
Numbers lagao. $Δ v = ∣3.07 - 1.61∣ = 1.46 km/s$

Verify: Units km/s ✅. Answer exactly dono speeds ka gap hai — aapka forecast. Koi bhi turn ( $θ > 0$ ) $cos θ < 1$ bana deta, subtracted term ko shrink karta aur $Δ v$ raise karta; yahan woh penalty zero hai.

Cell B — equal speeds, pure tilt

Forecast: Same length arrows, $3 0^{\circ}$ se fele hue. Unki tips ke beech ka gap thoda-sa hai lekin zero nahi. Kya yeh full $v$ se zyada hoga ya kam? Guess karo.

Figure — Combined maneuvers — optimal split between plane change and velocity change

Figure s01 — Equal-speed pure tilt. Violet arrow $v_{1}$ hai aur orange arrow $v_{2}$ hai; dono ki same length $v$ hai, plane-change angle $θ = 3 0^{\circ}$ se alag. Magenta arrow unki tips ko jodta hai $Δ v = v_{2} - v_{1}$ . Kyunki dono lambi sides equal hain, triangle isosceles hai, aur magenta side ki length exactly $2 v sin (θ /2)$ nikalti hai — magenta dot uska midpoint mark karta hai, "half-base" jo half-angle $θ /2$ control karta hai.

Master formula mein $v_{1} = v_{2} = v$ set karo. $Δ v = v^{2} + v^{2} - 2 v^{2} cos θ = v 2 - 2 cos θ$ Yeh step kyun? $v^{2}$ factor out karne se pure angle penalty $2 - 2 cos θ$ isolate ho jaati hai, aap kitne tez ja rahe ho usse independent.
Half-angle identity $2 - 2 cos θ = 4 sin^{2} (θ /2)$ use karo. Yeh step kyun? Identity $1 - cos θ = 2 sin^{2} (θ /2)$ ugly root ko ek clean sine mein badal deti hai — aur geometrically $sin (θ /2)$ dono equal arrows jo triangle banate hain uski half the base hai (figure mein magenta segment dekho). $Δ v = v 4 sin^{2} (θ /2) = 2 v sin (θ /2)$
Numbers lagao. $θ /2 = 1 5^{\circ}$ , $sin 1 5^{\circ} = 0.2588$ . $Δ v = 2 (7.35) (0.2588) = 3.80 km/s$

Verify: $3.80$ km/s aapki orbital speed ka aadha se zyada hai sirf $3 0^{\circ}$ tilt karne ke liye — exactly isliye parent note chillata hai "turn where slow." Units km/s ✅. Master formula se cross-check: $7.3 5^{2} + 7.3 5^{2} - 2 (7.35)^{2} cos 3 0^{\circ} = 3.80$ ✅.

Cell C — $18 0^{\circ}$ extreme

Forecast: Arrows bilkul ek dusre ke against point kar rahe hain. Reverse karne ke liye aur speed up karne ke liye, aapko apna $2.0$ cancel karna hai AND doosri taraf $3.0$ build karna hai. Guess: dono speeds add hongi.

$cos 18 0^{\circ} = - 1$ substitute karo. Yeh step kyun? $- 1$ sabse chhota cosine ho sakta hai, to " $- 2 v_{1} v_{2} cos θ$ " term $+ 2 v_{1} v_{2}$ ban jaata hai — middle term subtract karne se add karne mein flip ho jaata hai. Yeh worst-case, maximum-cost geometry hai. $Δ v = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2}$
Doosra perfect square pehchano. $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ . $Δ v = (v_{1} + v_{2})^{2} = v_{1} + v_{2} = 2.0 + 3.0 = 5.0 km/s$

Verify: Speeds add ho gayi, forecast se match. Note karo Ex 1 ( $θ = 0$ ) ne $∣ v_{2} - v_{1} ∣ = 1.0$ diya aur Ex 3 ( $θ = 18 0^{\circ}$ ) ne $v_{1} + v_{2} = 5.0$ diya — master formula ke do extremes, in speeds ke liye possible sabse chhota aur sabse bada $Δ v$ . Har doosra angle inke beech land karta hai.

Cell D — roz ka combined burn (acute angle)

Forecast: Pure-speed change ( $2.0$ ) aur tilt se kuch bade ke beech kaheen. km/s mein ek number guess karo.

Combined burn — directly master formula mein plug karo. $Δ v = 6. 0^{2} + 8. 0^{2} - 2 (6.0) (8.0) cos 4 0^{\circ}$ Yeh step kyun? Yeh direct third-side length hai; $cos 4 0^{\circ} = 0.7660$ . $= 36 + 64 - 96 (0.7660) = 100 - 73.54 = 26.46 = 5.14 km/s$
Separate burns — speed change phir plane change. Yeh step kyun? Alag karne ki penalty measure karne ke liye. Pehle $∣8 - 6∣ = 2.0$ speed change, phir $v_{2} = 8.0$ par pure tilt. Tilt pure plane change hai speed $8.0$ par, to Ex 2 ka equal-speed formula use hoga $2 v sin (θ /2)$ — aur kyunki poora tilt $θ = 4 0^{\circ}$ is ek leg mein ho raha hai, half-angle $θ /2 = 2 0^{\circ}$ hai. Isliye $2 (8.0) sin 2 0^{\circ} = 16 (0.3420) = 5.47$ . $Δ v_{sep} = 2.0 + 5.47 = 7.47 km/s$
Compare karo. $5.14 < 7.47$ : ek single throw $2.33$ km/s bachata hai. Yeh step kyun? Kyunki propellant Delta-v Budget ke saath exponentially badhta hai (upar definition callout dekho), $2.33$ km/s bachana koi chhoti correction nahi — iska matlab bahut zyada payload ho sakta hai.

Verify: $5.14 \leq 7.47$ triangle inequality obey karta hai (direct side kabhi bhi do-leg path se lambi nahi hoti). Units km/s ✅. Yeh "combine karo, alag mat karo" ka numeric heart hai.

Cell E — full optimal split (real mission)

Intuition "Equal marginal cost" optimum kyun hai — scratch se banaya

Socho aapne already kuch split choose ki hai: fast burn par $s$ degrees of tilt, slow burn par $(i - s)$ . Ab poochho: agar main ek tiny extra degree burn 2 se burn 1 par move karun toh total mein kya hoga? Burn 1 ka cost kuch amount se upar jayega — ise uska price per degree, $p_{1}$ kahte hain. Burn 2 ka cost apni price per degree, $p_{2}$ se neeche jayega. To total mein net change $p_{1} - p_{2}$ per degree moved hoga.

Agar $p_{1} < p_{2}$ : ek degree burn 1 par move karna total kam karta hai — to karo.
Agar $p_{1} > p_{2}$ : ek degree waapis burn 2 par move karna total kam karta hai — woh karo.
Aap tabhi improve karna band kar sakte ho jab $p_{1} = p_{2}$ . Woh equality hi minimum HAI.

"Price per degree" exactly woh hai jo ek derivative measure karta hai: koi quantity kitni tez badlti hai jab aap uska input thoda nudge karte ho. Isliye yahan tool $\frac{d Δ v}{d s}$ hai aur kuch nahi — yeh precise sawaal ka jawab deta hai "tilt ka ek degree move karne par cost kitni badlti hai?" Parent note ne yeh derivatives compute kiye: $\frac{d Δ v _{1}}{d s} = = p_{1} \frac{v _{1, i} v _{1, f} sin s}{Δ v _{1}}, - \frac{d Δ v _{2}}{d s} = = p_{2} \frac{v _{2, i} v _{2, f} sin ( i - s )}{Δ v _{2}}$ Total ka derivative zero set karna, $\frac{d Δ v _{1}}{d s} + \frac{d Δ v _{2}}{d s} = 0$ , exactly $p_{1} = p_{2}$ likhne jaisa hai — equal-marginal-cost condition. $Δ v_{2}$ par minus sign sirf bookkeeping hai: $s$ badhana burn 2 ka angle kam karta hai, to uska cost girta hai.

Forecast: Kya tilt 50/50 split hogi? Mostly-perigee? Mostly-apogee? Compute karne se pehle guess karo.

Figure s02 — Total cost jab split $s$ vary karta hai. Horizontal axis $s$ hai, $28. 5^{\circ}$ tilt ke degrees ki sankhya jo fast perigee burn ko assign ki gayi hai (to $28. 5^{\circ} - s$ slow apogee burn ko jaati hai). Violet dashed curve burn 1 ka cost $Δ v_{1} (s)$ hai — jab aap fast burn par tilt load karte ho toh yeh steeply climb karta hai. Orange dashed curve burn 2 ka cost $Δ v_{2} (s)$ hai — jab tilt slow burn se jaati hai toh yeh gently girta hai. Magenta curve unka sum $Δ v_{t o t} (s)$ hai; navy dot uska minimum mark karta hai, left edge ke bilkul paas $s \approx 2. 2^{\circ}$ par. Picture visually dikhata hai kyun almost sari tilt apogee par belong karti hai.

Har burn ko ek combined maneuver ki tarah likho. Yeh step kyun? Har burn khud apna "third side" problem hai, master formula ki apni copy: $Δ v_{1} (s) = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos s$ $Δ v_{2} (s) = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos (i - s)$
"Sab apogee par" test karo ( $s = 0$ ). Yeh step kyun? Apogee mein sabse chhoti speeds hain, to parent note predict karta hai yeh sasta hona chahiye. $Δ v_{1} (0) = ∣10.15 - 7.73∣ = 2.42, Δ v_{2} (0) = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 28. 5^{\circ} = 1.83$ $Δ v_{tot} (0) = 2.42 + 1.83 = 4.25 km/s$
"Sab perigee par" test karo ( $s = 28. 5^{\circ}$ ). Yeh step kyun? High speed par tilt karne ki disaster expose karne ke liye. $Δ v_{1} = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos 28. 5^{\circ} = 5.00, Δ v_{2} = ∣3.07 - 1.61∣ = 1.46$ $Δ v_{tot} = 5.00 + 1.46 = 6.46 km/s$
True optimum nikalne ke liye equal-marginal-cost condition apply karo (upar intuition callout mein build kiya): $\frac{v _{1, i} v _{1, f} s i n s}{Δ v _{1} ( s )} = \frac{v _{2, i} v _{2, f} s i n ( i - s )}{Δ v _{2} ( s )}$ Yeh step kyun? Minimum par dono prices per degree equal hain. Numerically yeh $s \approx 2. 2^{\circ}$ perigee par, $26. 3^{\circ}$ apogee par deta hai, $Δ v_{tot} \approx 4.24$ km/s ke saath.

Verify: All-apogee ( $4.25$ ) all-perigee ( $6.46$ ) ko $2.21$ km/s se pehle se hi crush karta hai. Optimal split ( $4.24$ ) sirf aur $\sim 0.01$ shave karta hai — ek baal jaisi baat. Conclusion: almost sari tilt apogee par belong karti hai; perigee ka hissa tiny hai. Units km/s ✅.

Cell F — do limiting endpoints

Forecast: Hum pehle se dekh chuke hain ki all-apogee ( $s = 0$ ) bahut sasta hai. Guess karo kaun se endpoint ki marginal price chhoti hai — degrees hamesha saste price ki taraf migrate karte hain.

Har burn ki marginal price likho. Yeh step kyun? Intuition callout se, ek burn par tilt load karne ki price per radian us burn ke derivative ka size hai: $p_{1} (s) = \frac{v _{1, i} v _{1, f} s i n s}{Δ v _{1} ( s )}, p_{2} (s) = \frac{v _{2, i} v _{2, f} s i n ( i - s )}{Δ v _{2} ( s )} (arguments in radians)$
Speed products note karo jo har price ko scale karte hain. Yeh step kyun? Aage ka factor poori scale set karta hai. Perigee par $v_{1, i} v_{1, f} = 7.73 \times 10.15 = 78.46$ ; apogee par $v_{2, i} v_{2, f} = 1.61 \times 3.07 = 4.943$ . Perigee factor lagbhag $16 \times$ bada hai — yahi akela reason hai ki fast burn par tilt expensive hai.
$s = 0$ par apogee price evaluate karo (to $i - s = 28. 5^{\circ} = 0.4974$ rad). Yeh step kyun? Yeh apogee par baithe last radian of tilt ki price hai, woh tilt jo aap perigee par move karne ka soch rahe ho. $p_{2} (0) = \frac{( 1.61 ) ( 3.07 ) s i n ( 0.4974 )}{Δ v _{2} ( 0 )} = \frac{4.943 \times 0.4772}{1.83} = \frac{2.359}{1.83} = 1.289 km/s per radian$
Woh jagah dhundho jahan perigee price pakad leti hai — root-finding dikhate hue. $s = 0$ par, $sin s = 0$ , isliye $p_{1} (0) = 0$ : perigee par tilt ka pehla sliver almost free hai. Lekin $p_{1}$ tezi se climb karta hai $78.46$ factor ki wajah se. Hum woh $s$ (radians mein) chahte hain jahan $p_{1} (s)$ badhke $1.289$ se mil jaye. Kyunki perigee par tilt move karna $p_{2}$ ko bhi thoda change karta hai, exact balance wahan hai jahan $p_{1} (s) = p_{2} (s)$ ; values try karke numerically solve karte hain:

$s$ (deg) $s$ (rad) $Δ v_{1} (s)$ $p_{1} (s) = \frac{78.46 sin s}{Δ v _{1}}$

$1. 0^{\circ}$ $0.0175$ $2.421$ $0.566$

$2. 0^{\circ}$ $0.0349$ $2.427$ $1.128$

$2. 2^{\circ}$ $0.0384$ $2.429$ $1.240$

$2. 4^{\circ}$ $0.0419$ $2.431$ $1.352$

Yeh step kyun? Table answer bracket karti hai: $p_{1}$ apogee price ( $\approx 1.28$ – $1.29$ , jo khud thoda dip karti hai jab uski apni tilt shrink hoti hai) ko $2. 2^{\circ}$ aur $2. 4^{\circ}$ ke beech cross karta hai. Full equal-price solution ( $p_{2}$ ko bhi move hote dekhte hue) $s \approx 2. 2^{\circ}$ par land karta hai — Ex 5 ke numeric minimization se match karta hai.

$s$ (deg)	$s$ (rad)	$Δ v_{1} (s)$	$p_{1} (s) = \frac{78.46 sin s}{Δ v _{1}}$
$1. 0^{\circ}$	$0.0175$	$2.421$	$0.566$
$2. 0^{\circ}$	$0.0349$	$2.427$	$1.128$
$2. 2^{\circ}$	$0.0384$	$2.429$	$1.240$
$2. 4^{\circ}$	$0.0419$	$2.431$	$1.352$

Verify: Perigee par huge speed product ( $78.46$ vs $4.94$ , ek $16 \times$ ratio) matlab hai ki wahan sirf kuch degrees bhi utne hi expensive hain jitne apogee par kaafi zyada degrees — isliye optimum $s = 0$ end ke bilkul paas baithta hai, $s \approx 2. 2^{\circ}$ par. $2. 2^{\circ}$ aur $2. 4^{\circ}$ ke beech table ka crossing Ex 5 ki split confirm karta hai ✅.

Cell G — combined vs separate, ek word problem ki tarah

Forecast: Parent note ka triangle-inequality argument keh ta hai combined hamesha jeetega. Dono numbers guess karo.

Alag plan. Speed change $∣3.07 - 1.61∣ = 1.46$ , phir $v_{2} = 3.07$ par pure tilt poore $θ = 28. 5^{\circ}$ ke saath equal-speed formula use karte hue (half-angle $14.2 5^{\circ}$ ): $2 (3.07) sin 14.2 5^{\circ} = 6.14 (0.2462) = 1.512$ . $Δ v_{sep} = 1.46 + 1.512 = 2.972 km/s$ Yeh step kyun? Yeh intern ka total hai — do saaf legs.
Combined plan. Ek angled throw: $Δ v = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 28. 5^{\circ} = 1.83 km/s$ Yeh step kyun? Velocity triangle ki direct third side.
Difference. $2.972 - 1.83 = 1.14$ km/s combine karne se bacha. Yeh step kyun? Delta-v Budget callout mein exponential rocket-equation logic se, intern ke budget ka ek tihaai se zyada gaayab ho jaata hai — ek bada propellant saving.

Verify: $1.83 \leq 2.972$ — triangle inequality holds ✅. Combining $1.14$ km/s bachata hai. Units km/s ✅. (Yeh exactly Ex 5 ke step 2 ka all-apogee burn hai, "kya combine karna chahiye?" angle se dekha gaya.)

Cell H — exam twist: obtuse angle, unequal speeds

Forecast: Kyunki $12 0^{\circ} > 9 0^{\circ}$ , arrows ek dusre se door lean karte hain. Guess: acute cases se bada — middle term ab add karta hai.

$cos 12 0^{\circ} = - \frac{1}{2}$ evaluate karo. Yeh step kyun? Obtuse angles ke liye cosine negative hota hai, isliye $- 2 v_{1} v_{2} cos θ$ ban jaata hai $+ 2 v_{1} v_{2} \cdot \frac{1}{2} = + v_{1} v_{2}$ . Classic exam trap yeh sign flip bhoolna aur subtract karna hai. $Δ v = 4. 0^{2} + 5. 0^{2} - 2 (4.0) (5.0) (- 0.5)$
Simplify karo. $= 16 + 25 + 20 = 61 = 7.81 km/s$

Verify: $Δ v = 7.81$ ki value $θ = 0$ value $∣5 - 4∣ = 1$ aur $θ = 18 0^{\circ}$ value $5 + 4 = 9$ ke beech hai — jaise har combined-maneuver answer hona chahiye. Agar koi student galat $+ 2 v_{1} v_{2} cos θ$ use karta to $21 = 4.58$ aata, ek obtuse turn ke liye impossibly small answer — sanity band error expose kar deta. Units km/s ✅.

Recall Matrix par quick self-test

Fixed speeds ke liye kaun se cell mein sabse bada possible $Δ v$ hota hai? ::: Cell C, $θ = 18 0^{\circ}$ , jo $v_{1} + v_{2}$ deta hai. Kaun sa cell sabse chhota deta hai? ::: Cell A, $θ = 0$ , jo $∣ v_{2} - v_{1} ∣$ deta hai. Cell E mein almost saara plane change kahan jaata hai? ::: Apogee (slow burn) par; sirf ~2° perigee par rehta hai. Optimum "equal marginal cost" par kyun hota hai? ::: Agar ek burn ka price per degree sasta hota, aap ek degree wahan move karte aur total kam hota; aap tabhi rukoge jab dono prices match karein. Cell H mein trap kya hai? ::: Obtuse $θ$ ke liye $cos θ < 0$ bhool jaana, jo middle term ko add kara deta hai.