Combined maneuvers — optimal split between plane change and velocity change: Exercises

3.2.23 · D4 · Physics › Orbital Mechanics & Astrodynamics › Combined maneuvers — optimal split between plane change and

Yeh page parent topic ko test karti hai. Har problem ka ek collapsible worked solution hai — pehle khud try karo, phir reveal karo. Hum formula pehchanne se shuru karke poore mission plan banana tak jaate hain.

Yahan sab kuch parent note ke ek hi formula par tikaa hai, toh use ek baar apne words mein restate kar lete hain.

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

Us triangle ki amber side dekho — woh amber length hi woh $Δ v$ hai jo tumhare engine ko supply karni hai. Do cyan sides $v_{1}$ aur $v_{2}$ hain; white arc $θ$ hai. Jab $θ = 0$ ho toh do sides flat hokar ek doosre par fold ho jaati hain aur amber side sirf simple difference $∣ v_{2} - v_{1} ∣$ tak sir jaati hai; jab sides equal hon toh yeh isoceles-triangle base $2 v sin (θ /2)$ ban jaata hai; jab $θ = 18 0^{\circ}$ ho toh sides opposite directions mein point karti hain aur amber side $v_{1} + v_{2}$ tak stretch ho jaati hai. Apne dimaag mein yeh ek picture rakkho — neeche har problem bas yahi triangle hai alag side lengths aur alag arc ke saath.

Level 1 — Recognition

L1.1

Problem. Ek spacecraft ek circular orbit mein $v = 7.8$ km/s speed par hai. Use orbit plane ko $θ = 1 5^{\circ}$ tilt karna hai bina speed change ke. Ek single-burn $Δ v$ kitni cost karega?

Recall Solution

Equal speeds matlab $v_{1} = v_{2} = v$ — triangle figure mein dono cyan sides ki same length hai. Master formula mein plug karo: $Δ v = v^{2} + v^{2} - 2 v^{2} cos θ = v 2 - 2 cos θ = 2 v sin (θ /2)$ Beech ka step identity $1 - cos θ = 2 sin^{2} (θ /2)$ use karta hai — kyun? Kyunki yeh ugly $2 - 2 cos θ$ ko clean $2 sin (θ /2)$ mein convert kar deta hai, jo exactly isoceles velocity triangle ka base hai (dono equal sides $v$ , apex angle $θ$ ) — apex se perpendicular girao aur har half-base $v sin (θ /2)$ hai. $Δ v = 2 (7.8) sin (7. 5^{\circ}) = 15.6 \times 0.13053 = 2.036 km/s$ Answer: $\approx 2.04$ km/s. Yeh sirf $1 5^{\circ}$ tilt ke liye orbital speed ka chautha hissa se zyada hai — plane changes bahut mehenga hota hai.

L1.2

Problem. Usi triangle mein, $Δ v$ kya hogi agar $θ = 0^{\circ}$ , $v_{1} = 3.07$ km/s, $v_{2} = 1.61$ km/s?

Recall Solution

$θ = 0$ ke saath do velocity arrows same direction mein point karte hain, toh triangle flat collapse ho jaata hai aur burn pure speed change hai (seedha subtraction): $Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos 0 = (v_{1} - v_{2})^{2} = ∣ v_{1} - v_{2} ∣$ $= ∣3.07 - 1.61∣ = 1.46 km/s$ Answer: $1.46$ km/s. $cos 0 = 1$ triangle ko flat line mein collapse kar deta hai — koi turning nahi, sirf braking.

Level 2 — Application

L2.1

Problem. Ek angled burn speed ko $v_{1} = 10.15$ km/s se $v_{2} = 7.73$ km/s karta hai aur $θ = 1 0^{\circ}$ se turn karta hai. $Δ v$ nikalo.

Recall Solution

Seedha master formula mein — do cyan sides ki ab alag lengths hain aur arc non-zero hai, toh hume full cosine chahiye (koi shortcut nahi chalega): $Δ v = 10.1 5^{2} + 7.7 3^{2} - 2 (10.15) (7.73) cos 1 0^{\circ}$ Piece by piece compute karo (KYA: har term banao; KYUN: do squared terms do sides ki lengths hain, aur cross term dikhata hai ki chota angle tips ko kitna "pull" karta hai saath mein — ek almost-flat triangle):

$10.1 5^{2} = 103.0225$ (side one squared)
$7.7 3^{2} = 59.7529$ (side two squared)
$2 (10.15) (7.73) = 156.919$ , aur $cos 1 0^{\circ} = 0.98481$ (1 ke kareeb kyunki turn chota hai), toh cross term $= 154.536$ $Δ v = 103.0225 + 59.7529 - 154.536 = 8.239 = 2.87 km/s$ Answer: $\approx 2.87$ km/s. Note karo yeh pure speed change $∣10.15 - 7.73∣ = 2.42$ ke kareeb hai — kyunki $1 0^{\circ}$ triangle ko barely tilt karta hai.

L2.2

Problem. $28. 5^{\circ}$ plane change (speed change $1.61 \to 3.07$ km/s ke saath) ko (a) do alag burns ya (b) ek combined burn ke roop mein karne ki comparison karo. Kaun jeetta hai aur kitne se?

Recall Solution

(a) Separate. Pehle speed change karo ( $θ = 0$ ), phir nai speed $v_{2} = 3.07$ par turn karo — yeh triangle ki do sides par chalna hai seedhe teesre ki bajay: $Δ v_{sep} = ∣3.07 - 1.61∣ + 2 (3.07) sin (14.2 5^{\circ}) = 1.46 + 6.14 (0.24615) = 1.46 + 1.511 = 2.97 km/s$ (b) Combined. Ek angled burn — triangle ki direct amber side: $Δ v_{comb} = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 28. 5^{\circ}$ $= 2.5921 + 9.4249 - 9.8854 (0.87882) = 12.017 - 8.687 = 3.330 = 1.825 km/s$ Answer: combined $= 1.83$ km/s, separate $= 2.97$ km/s se better hai — lagbhag 1.14 km/s ki saving. Yeh Law of Cosines triangle inequality concrete form mein hai: seedha teesra side doosre do par chalne se chota hota hai.

Level 3 — Analysis

L3.1

Problem. LEO→GEO transfer ke liye, total plane-change budget $i = 28. 5^{\circ}$ hai (yaad karo: $i$ woh poora turn hai jo mission ko do burns mein share karna hai). Perigee burn: $v_{1, i} = 7.73$ , $v_{1, f} = 10.15$ km/s. Apogee burn: $v_{2, i} = 1.61$ , $v_{2, f} = 3.07$ km/s. Saara $28. 5^{\circ}$ perigee par karte hue, perigee $Δ v$ nikalo, aur compare karo saara apogee par karne se.

Recall Solution

"All at perigee" matlab perigee share $s = 28. 5^{\circ}$ hai (toh uska turn ka slice poora $i$ hai), aur apogee share $i - s = 0^{\circ}$ hai. Har burn par master formula apply karo apni speeds ke saath: $Δ v_{1} = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos 28. 5^{\circ}$ $= 59.7529 + 103.0225 - 156.919 (0.87882) = 162.775 - 137.905 = 24.871 = 4.987 km/s$ Apogee burn phir pure speed change hai ( $i - s = 0$ ): $∣3.07 - 1.61∣ = 1.46$ . Total $\approx 4.99 + 1.46 = 6.45$ km/s.

All at apogee ( $s = 0$ , apogee slice $= i$ ): perigee pure speed change hai $∣10.15 - 7.73∣ = 2.42$ ; apogee combined $= 1.825$ (L2.2 se). Total $\approx 2.42 + 1.825 = 4.25$ km/s.

Answer: perigee-turn total $\approx 6.45$ km/s vs apogee-turn total $\approx 4.25$ km/s. Fast perigee par turning karna lagbhag 2.2 km/s zyada cost karta hai — kyunki cross term mein $v \cdot v$ products wahan bahut bade hain. Dekho Delta-v Budget.

L3.2

Problem. Verify karo ki marginal-cost (optimal-split) condition satisfy nahi hoti jab saara plane change apogee par ho. Dono sides compute karo $s = 0$ par (perigee share $s = 0$ , toh apogee share $i - s = 28. 5^{\circ}$ ).

Recall Solution

Condition (parent note se) yeh hai $\frac{v _{1, i} v _{1, f} sin s}{Δ v _{1}} = \frac{v _{2, i} v _{2, f} sin ( i - s )}{Δ v _{2}}$ , jahan left side perigee par ek aur degree turn ki marginal cost hai aur right side apogee par wahi.

Left (perigee) at $s = 0$ : $sin 0 = 0$ , toh LHS $= 0$ .

Right (apogee) at $s = 0$ : numerator $= (1.61) (3.07) sin 28. 5^{\circ} = 4.9427 (0.47716) = 2.358$ ; denominator $Δ v_{2} = 1.825$ (L2.2 se). RHS $= 2.358/1.825 = 1.292$ .

Answer: LHS $= 0 \neq = 1.29 =$ RHS. Turning ki marginal cost perigee par zero hai lekin apogee par badi — toh turn ka ek sliver perigee ko shift karna total cost ghataata hai. Optimum $s = 0$ se thoda off hai (lagbhag $s \approx 2. 2^{\circ}$ perigee par), exactly wahan jahan do sides balance hoti hain.

Level 4 — Synthesis

L4.1

Problem. L3.1 ke numbers ke liye, numerically confirm karo ki optimal split $s \approx 2. 2^{\circ}$ (perigee) / $i - s \approx 26. 3^{\circ}$ (apogee) all-at-apogee plan se better hai, aur optimum par total $Δ v$ report karo.

Recall Solution

Hum har burn ki cost evaluate karte hain turn ke apne slice ke saath: perigee slice $s = 2. 2^{\circ}$ hai, apogee slice $i - s = 26. 3^{\circ}$ hai.

Perigee ( $θ = s = 2. 2^{\circ}$ ). Itna chota angle yahan kyun? Perigee fast hai, isliye hum use sirf ek whisker turn carry karne dete hain: $Δ v_{1} = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos 2. 2^{\circ}$ Yahan $cos 2. 2^{\circ} = 0.99926$ — almost 1, kyunki angle tiny hai, toh cross term $= 156.919 (0.99926) = 156.803$ apni flat-triangle value se barely shrink hota hai. Triangle almost collapsed hai, isliye cost pure speed change ke kareeb hai: $= 162.775 - 156.803 = 5.972 = 2.444 km/s$ Apogee ( $θ = i - s = 26. 3^{\circ}$ ). Bada slice yahan kyun? Apogee slow hai, isliye mehenga turning yahan sabse sasta hai: $Δ v_{2} = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 26. 3^{\circ}$ $cos 26. 3^{\circ} = 0.89661$ (1 se kaafi neeche — ek real tilt), cross term $= 9.8854 (0.89661) = 8.863$ , jo flat value se chhota hai, toh triangle khulta hai aur amber side badi hoti hai: $= 12.017 - 8.863 = 3.154 = 1.776 km/s$ Total $= 2.444 + 1.776 = 4.220$ km/s.

All-at-apogee total $4.245$ km/s (L3.1) se compare karo. Answer: optimum $\approx 4.22$ km/s lagbhag 0.025 km/s (25 m/s) bachata hai — chota lekin free hai, aur confirm karta hai ki split real hai. Hohmann Transfer Orbit budgeting se relate karo.

L4.2

Problem. Ek satellite circular orbit mein ( $v = 7.8$ km/s) hai aur use $4 0^{\circ}$ plane change chahiye. Option A: single burn. Option B (bi-elliptic-style): prograde burn karke apoapsis raise karo jahan speed $v_{high} = 1.5$ km/s tak drop ho jaati hai, wahan poora plane change karo, phir wapas aao. Raise/lower cost ko comparison of the turn itself ke liye ignore karo — sirf turning costs compare karo.

Recall Solution

Dono options pure equal-speed turns hain, toh dono $2 v sin (θ /2)$ use karte hain (isoceles-triangle base) — sirf speed $v$ alag hai.

Option A turn cost. $7.8$ kyun plug in karte hain? Turn full orbital speed par hota hai, toh dono cyan sides lambi hain: $2 v sin (2 0^{\circ}) = 2 (7.8) (0.34202) = 5.336 km/s$ Option B turn cost. $1.5$ kyun plug in karte hain? Humne deliberately turn ko apoapsis tak move kiya jahan sides $1.5$ tak shrink ho jaati hain, toh wahi $4 0^{\circ}$ apex angle bahut chota base span karta hai: $2 v_{high} sin (2 0^{\circ}) = 2 (1.5) (0.34202) = 1.026 km/s$ Answer: turn akela $5.34$ se $1.03$ km/s tak drop karta hai — $5.2$ guna sasta. Exactly $5.2$ kyun? Kyunki ek isoceles triangle ka base side length ke saath linearly scale karta hai, toh cost ratio sirf speed ratio $v / v_{high} = 7.8/1.5 = 5.2$ hai. Isliye Bi-elliptic Transfer bade plane changes ke liye shine karta hai: jahan crawl karo wahan turning dramatically sasta hai. Principle: jahan slow ho wahan turn karo.

Level 5 — Mastery

L5.1

Problem. Prove karo ki combined burn hamesha utna hi ya usse kam expensive hota hai jitna koi bhi two-step "speed change karo, phir turn karo" plan usi $v_{1} \to v_{2}$ aur turn $θ$ ke liye. Phir woh single condition batao jab dono exactly equal hote hain.

Recall Solution

Pehle notation par ek note. Ab tak $v_{1}, v_{2}$ plain speeds (cyan sides ki lengths) rahe hain. Velocities subtract karne ki baat karne ke liye unke arrow form ki zaroorat hai — velocity mein size aur direction dono hote hain. Hum upar arrow likhte hain, $v_{1}$ aur $v_{2}$ , matlab "poori velocity, size aur direction"; unke plain-letter versions $v_{1} = ∣ v_{1} ∣$ aur $v_{2} = ∣ v_{2} ∣$ sirf lengths hain. Burn literally woh arrow hai jo tumhe $v_{1}$ ki tip se $v_{2}$ ki tip tak le jaata hai.

Setup. Combined burn vector $Δ v = v_{2} - v_{1}$ hai, jiska length master formula hai (figure mein triangle ki amber side). Ek two-step plan triangle ki do sides traverse karta hai $v_{1}$ ki tip se $v_{2}$ ki tip tak jaane ke liye: ek leg speed change ke liye, ek leg turn ke liye. Theorem. Triangle inequality se (seedha raasta kabhi detour se lamba nahi hota): $∣ v_{2} - v_{1} ∣ \leq ∣ leg_{1} ∣ + ∣ leg_{2} ∣ = Δ v_{sep}$ Concretely, speed $v_{2}$ par turn ke saath: $v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos θ \leq ∣ v_{2} - v_{1} ∣ + 2 v_{2} sin (θ /2)$ . Equality condition. Ek triangle ek straight line mein degenerate hota hai (equality) sirf tab jab do legs collinear hon — yaani jab koi turn hi na ho, $θ = 0$ . Tab dono sides $∣ v_{2} - v_{1} ∣$ ke equal hain. Answer: combined $\leq$ separate hamesha; equal iff $θ = 0$ . $v_{1}, v_{2}$ kahan se aate hain, uske liye Law of Cosines aur Vis-viva Equation dekho.

L5.2

Problem. Ek mission ko $v_{1} = 7.73 \to v_{2} = 10.15$ km/s ke saath perigee par $28. 5^{\circ}$ turn chahiye aur apogee par $v_{1} = 1.61 \to v_{2} = 3.07$ km/s, lekin tum $28. 5^{\circ}$ ka koi bhi split $s$ perigee par rakh sakte ho. Marginal condition use karke show karo kyun optimum $s$ small aur positive hai (zero nahi, bada nahi). Sirf qualitative sign argument do.

Recall Solution

Define karo $g (s) = \frac{d Δ v _{tot}}{d s} = \frac{v _{1, i} v _{1, f} sin s}{Δ v _{1} ( s )} - \frac{v _{2, i} v _{2, f} sin ( i - s )}{Δ v _{2} ( s )}$ . Yahan $s$ total turn $i = 28. 5^{\circ}$ ka perigee share hai, aur $i - s$ apogee share. Hum $g (s) = 0$ chahte hain.

$s = 0$ par: pehla term $= 0$ (kyunki $sin 0 = 0$ ); doosra term $= 2.358/1.825 = 1.292 > 0$ , toh $g (0) = - 1.292 < 0$ . Slope negative hai — $s$ badhana ab bhi total cost ghataata hai.
$s = i = 28. 5^{\circ}$ par: ab doosra term vanish ho jaata hai ( $sin (i - s) = sin 0 = 0$ ), aur pehla term bahut bada aur positive hai (bade perigee $v \cdot v$ products). Toh $g (i) > 0$ — saara turn perigee par karna optimum se bahut aage hai.
$g$ continuous hai aur negative se positive jaata hai, toh ek root beech mein hai, aur kyunki perigee products $v_{1, i} v_{1, f} = 78.4$ apogee products $v_{2, i} v_{2, f} = 4.94$ se kaafi bade hain (ratio $\approx 16$ ), balance ek chote $s$ par aata hai: perigee par ek tiny $sin s$ bhi woh marginal cost generate karta hai jo apogee par bahut bade $sin (i - s)$ se produce hoti hai. Answer: optimum chota aur positive hai — numbers dete hain $s \approx 2. 2^{\circ}$ . Heavy perigee $v \cdot v$ term turn ko almost entirely apogee par force karta hai, lekin poora nahi.

Recall One-line self-test

Zyaadatar plane change apogee par kyun end up hota hai? ::: Kyunki slow apogee par $v \cdot v$ cross-term products tiny hote hain, toh turn ka har degree wahan sasta hota hai — marginal cost ek chote perigee share par balance hoti hai.