Question bank — Combined maneuvers — optimal split between plane change and velocity change

True. Triangle inequality se teesri side $\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \theta }$ kabhi bhi two-leg path $|v_2-v_1|+2v_2\sin (\theta /2)$ se zyada nahi ho sakti; yeh sirf degenerate cases mein equal hoti hain (jaise $\theta =0$).

False. Equal-speed cost hai $2v\sin (\theta /2)=2v\sin 30^{\circ }=v$ exactly — sirf turn karne ke liye tu apni poori orbital speed phir se kharch karta hai.

False — chote angles ke liye bhi aur generally bhi — cost $\sin (\theta /2)$ ke saath jaati hai, $\theta$ ke saath nahi. Bahut chote angles ke liye $\sin (\theta /2)\approx \theta /2$ toh roughly linear hai, lekin bade angles mein sine flat ho jaata hai aur eventually cost saturate ho jaati hai.

True. $2v\sin (90^{\circ })=2v$ — tujhe apni poori velocity khatam karni hai aur ulti direction mein dobara banana hai, yeh sabse expensive turn hai.

False. Jo equal hona chahiye woh degrees nahi balki marginal cost per extra degree hai. Kyunki burns bahut alag speeds par chalte hain, $i$ ka optimal split lopsided hota hai — zyaatar $i$ slow burn par jaata hai.

True. Jab $v_{1,i}{v}_{1,f}=v_{2,i}{v}_{2,f}=k$ ho, toh balance condition ban jaata hai $\frac{k\sin s}{\Delta v_1}=\frac{k\sin (i-s)}{\Delta v_2}$; equal speed products ke saath dono $\Delta v$ functions ka identical shape hota hai, toh equation tab satisfy hoti hai jab $\sin s=\sin (i-s)$, yaani $s=i-s\Rightarrow s=i/2$.

False. Apoapsis raise karne aur waapis laane ke liye extra pay karna padta hai. Yeh tabhi jeet ta hai jab plane-change angle itna bada ho ki $2v\sin (\theta /2)$ par savings raise/lower cost ko beat kar den — yahi Bi-elliptic Transfer ki logic hai.

True. $v_1=v_2=v$ set karne par $\sqrt{2v^2-2v^2\cos \theta }=v\sqrt{2-2\cos \theta }=2v\sin (\theta /2)$ milta hai half-angle identity se.

True. Woh speeds $v_{1,i},v_{1,f},v_{2,i},v_{2,f}$ jo condition mein enter karti hain, woh actual transfer ellipse par Vis-viva Equation evaluate karne se aati hain, toh alag transfer (jaise Bi-elliptic Transfer) shift kar deta hai ki plane change kahan rehna chahta hai.

Periapsis woh jagah hai jahan $v$ sabse zyada hota hai, aur turning cost $v$ ke saath scale hoti hai. Wahan sab kuch turn karna ($2v_p\sin (\theta /2)$) sabse expensive choice hai; turn apoapsis par hona chahiye jahan $v$ sabse chhota hota hai.

Woh formula do alag burns add karta hai (velocity triangle ki do sides par chalna) aur overcount karta hai. Single combined burn triangle ki teesri side $\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \theta }$ hai, jo triangle inequality se $\le$ sum hai.

Nahi — yeh dono derivatives $\frac{d\Delta v}{ds}$ ko magnitude mein equal set karti hai (marginal costs), yaani $\frac{v_{1,i}{v}_{1,f}\sin s}{\Delta v_1}=\frac{v_{2,i}{v}_{2,f}\sin (i-s)}{\Delta v_2}$. Total burns khud usually kaafi unequal hoti hain.

$\theta =0$ par, $\cos \theta =1$, jo deta hai $\sqrt{v_1^2-2v_1{v}_2+v_2^2}=|v_2-v_1|$, na ki $\sqrt{v_1^2+v_2^2}$. Cross term vanish nahi hoti — yeh maximal hoti hai.

Apoapsis par speed sabse chhoti hoti hai, zero nahi. Cost $2v_a\sin (\theta /2)$ toh $v_a$ ke saath chhoti hoti hai lekin real orbit ke liye kabhi zero nahi hoti ($v_a>0$ hamesha).

$\cos \theta$ har angle ke liye defined hai; formula $0$ se $180^{\circ }$ tak sabhi $\theta$ ke liye kaam karta hai. $90^{\circ }$ ke baad, $\cos \theta$ simply negative ho jaata hai, jisse $-2v_1{v}_2\cos \theta$ term cost mein add ho jaata hai.

Total cost $\Delta v_1(s)+\Delta v_2(s)$ wahan minimize hoti hai jahan uski derivative zero ho: $\frac{d\Delta v_1}{ds}+\frac{d\Delta v_2}{ds}=0$. Kyunki $s$ badhane se burn 1 badhta hai lekin burn 2 ghatta hai, yeh kehta hai ki ek aur degree turn ki marginal cost dono burns par equal honi chahiye — agar nahi hoti, toh ek degree saste burn par shift karte aur save karte, toh already minimum par nahi the.

Ek burn ki cost differentiate karo: sirf $\cos s$ mein $s$ hai, toh $\frac{d\Delta v_1}{ds}=\frac{v_{1,i}{v}_{1,f}\sin s}{\Delta v_1}$. Ek degree turn ki marginal cost local speed products $v_i{v}_f$ ke proportional hai. Slow (apoapsis) burn ke products tiny hote hain, toh uski marginal cost tab tak low rehti hai jab tak usne zyaatar turn absorb nahi kar liya.

Pythagoras ko $v_1$ aur $v_2$ ke beech right angle chahiye; yahan included angle $\theta$ arbitrary hai (upar velocity triangle dekho). Law of Cosines Pythagoras ko kisi bhi included angle tak generalize karta hai, jo exactly velocity-triangle situation hai.

Equal-speed velocity triangle isosceles hota hai; apex se perpendicular girane par $\theta$ aur base $\Delta v$ dono bisect ho jaate hain. Resulting right half-triangle mein $\sin (\theta /2)=(\Delta v/2)/v$, jo deta hai $\Delta v=2v\sin (\theta /2)$ — half-angle identity ka geometric roop.

Speed change sirf $|v_2-v_1|$ kharcha karta hai — speeds ka difference. Plane change kharcha karta hai $2v\sin (\theta /2)$, jo poori speed $v$ se scale hota hai; tujhe poori velocity vector ko sideways redirect karna padta hai.

Delta-v vector magnitude measure karta hai, tasks nahi. Do alag burns velocity triangle ki do sides trace karte hain; combined burn seedha teesri side hai, aur seedhi raah kabhi detour se lambi nahi hoti (triangle inequality).

Dono orderings same do legs trace karti hain ulte order mein, aur addition commutative hai — toh total identical hai. $2v_2$ factor sirf plane change ko us speed par fix karta hai jab tu turn karta hai; $v_2$ tak speed up karne ke baad karna isliye doosri leg $v_2$ use karta hai, $v_1$ nahi.

Balance condition har burn ko uske speed products $v_i{v}_f$ se weight karta hai; woh transfer orbit par Vis-viva Equation se aate hain, toh faster/slower endpoints reshufle karte hain ki turn kahan sasta hai.

Zero — speed ya direction mein koi change nahi toh koi burn nahi. Formula deta hai $\sqrt{v^2+v^2-2v^2}=0$, sanity floor.

Dono burns par speed products equality ki taraf approach karti hain, toh equal-marginal-cost condition split ko symmetric ki taraf push karti hai, $s\to i/2$.

$v_a\to 0$, toh turn cost $2v_a\sin (\theta /2)\to 0$ — wahan asymptotically free. Lekin total free nahi hai: pehle ek raise burn $\Delta v_{\text{up}}=|v_{\text{trans, peri}}-v_p|$ pay karna padta hai apoapsis ko bahar fenkne ke liye aur ek return burn $\Delta v_{\text{down}}$ waapis aane ke liye, aur yeh dono burns finite rehte hain (aur zyada jaane par badhte hain), isliye poora Bi-elliptic Transfer delta-v kabhi actually zero nahi pahunchta.

Us point par combine karne ke liye kuch hai hi nahi, toh tu pure $2v\sin (\theta /2)$ pay karta hai. Lesson location par shift ho jaata hai: jahan $v$ sabse chhota ho wahan perform karo, yaani apoapsis par.

$\Delta v=\sqrt{0+v_2^2-0}=v_2$ — tujhe simply scratch se poori final velocity banani hai; plane-change angle ab matter nahi karta kyunki turn karne ke liye koi initial direction hi nahi hai.

Tie hoti hai. Combined deta hai $2v$; separating deta hai $|v-v|+2v\sin 90^{\circ }=2v$. Yeh woh degenerate case hai jahan "triangle" ek straight line mein collapse ho jaata hai aur inequality equality ban jaati hai.