Exercises — Plane change maneuvers — Δv = 2v·sin(Δi - 2)

• $\Delta v$ = required burn ki magnitude (velocity vector mein change, m/s ya km/s mein measured).
• $v$ = orbital speed, jo pehle aur baad mein same hoti hai (ek pure plane change speed fixed rakhta hai).
• $\Delta i$ = woh angle jitna aap velocity vector ko rotate karte ho = inclination mein change.
• Half geometry se aata hai: triangle ke apex (angle $\Delta i$) se base $\Delta \vec{v}$ par ek perpendicular daalo. Yeh apex ko $\Delta i/2$ mein bisect karta hai aur base ko half mein kaat deta hai. Har half-base $v\sin (\Delta i/2)$ hai, isliye poori base $2v\sin (\Delta i/2)$ hai.

Kyunki $\Delta v=2v\sin (\Delta i/2)$ aur $\Delta i$ dono ke liye same hai, $\Delta v$ directly $v$ ke proportional hai. Isliye satellite A (faster) zyada pay karta hai. Ratio exactly speed ratio hai: $\frac{\Delta v_A}{\Delta v_B}=\frac{v_A}{v_B}=\frac{7.7}{3.1}\approx \mathrm{2.48.}$ A lagbhag 2.5× zyada pay karta hai. (Yahi wajah hai ki plane changes wahin karte hain jahan aap slow ho.)

(a) $\Delta i=0°$: $\Delta v=2v\sin (0°)=0$. Koi rotation nahi, koi burn nahi — obvious hai. (b) $\Delta i=180°$: $\Delta v=2v\sin (90°)=2v$. Aapko direction poori tarah reverse karni hai: apni speed $v$ ko khatam karo, phir opposite taraf $v$ build karo, total $2v$. Formula ise exactly capture karta hai.

Half-angle: $\Delta i/2=5°$. Phir $\sin 5°\approx 0.08716$. $\Delta v=2(7.7)(0.08716)\approx 1.342\text{ km/s}=1342\text{ m/s}.$ Sirf $10°$ mein $1.3$ km/s se zyada cost hoti hai — orbital speed ka lagbhag paanchwa hissa.

$\Delta v=2v\sin (\Delta i/2)$ ko $\Delta i$ ke liye rearrange karo. Pehle sine ko isolate karo: $\sin \left(\frac{\Delta i}{2}\right)=\frac{\Delta v}{2v}=\frac{0.200}{2(7.7)}=\frac{0.200}{15.4}=\mathrm{0.012987.}$ $\arcsin$ kyun? Hum half-angle ka sine jaante hain aur angle wapas chahiye — $\arcsin$ woh operation hai jo sine ko undo karta hai. Kyunki $0.012987$ bahut chhota hai, angle bhi chhota hai (aur positive), isliye koi quadrant ambiguity nahi: $\frac{\Delta i}{2}=\arcsin (0.012987)=0.012987\text{ rad}\Rightarrow \Delta i=0.025974\text{ rad}.$ Degrees mein convert karo: $0.025974\times \frac{180}{\pi }\approx 1.488°$. Isliye sabse bada change lagbhag $1.49°$ hai — yeh dikhata hai ki chhote budget mein plane changes kitne "kanjoosi" hote hain.

Pehle, units unify karo. Formula mein $\Delta v$ aur $v$ dono hain, isliye dono same unit mein hone chahiye. Budget $\Delta v=400$ m/s $=0.400$ km/s convert karo ($1000$ se divide karo), aur $v$ km/s mein nikalega. $v$ ke liye rearrange karo: $v=\frac{\Delta v}{2\sin (\Delta i/2)}$. Half-angle $\Delta i/2=10°$, $\sin 10°\approx 0.17365$. $v=\frac{0.400\text{ km/s}}{2(0.17365)}=\frac{0.400}{0.34730}\approx 1.152\text{ km/s}.$ Aapko lagbhag $1.15$ km/s par rengna padhega — yeh woh speed hai jo sirf ek high ellipse ke apoapsis par milti hai. Woh jagah dhundhne ke liye vis-viva use karo.

$\Delta v=v$ set karo: $2v\sin (\Delta i/2)=v\Rightarrow \sin (\Delta i/2)=0.5\Rightarrow \Delta i/2=30°\Rightarrow \Delta i=60°$. $\Delta v=2v$ set karo: $\sin (\Delta i/2)=1\Rightarrow \Delta i/2=90°\Rightarrow \Delta i=180°$. Isliye burn exactly $\Delta i=60°$ par poori orbital speed ke barabar hoti hai, aur apna maximum $2v$ sirf full reversal $\Delta i=180°$ par pahunchti hai. Kyunki $\sin (\Delta i/2)$ kabhi $0°$–$180°$ par $1$ se zyada nahi hota, $\Delta v=2v\sin (\Delta i/2)$ kabhi $2v$ se zyada nahi hoti — yeh $2v$ par capped hai. Function $0°$ se $180°$ tak monotonically increasing hai (half-angle ka sine steadily $0$ se $1$ tak badhta hai), isliye bada tilt hamesha zyada cost karta hai, us hard ceiling $2v$ tak. Lesson yeh hai: ek baar $\Delta i>60°$ ho jaaye toh single plane change poori orbit ki speed se zyada cost karti hai, isliye engineers bade changes ko slow orbit points par split karte hain.

Derivative kyun? Hum rate $d(\Delta v)/d(\Delta i)$ chahte hain — "cost priti unit angle kitni change hoti hai" — aur derivative exactly woh tool hai jo "ek quantity ka dusri ke saath change ki rate" ka jawab deta hai. $\frac{d}{d(\Delta i)}[2v\sin (\Delta i/2)]=2v\cdot \frac{1}{2}\cos (\Delta i/2)=v\cos (\Delta i/2).$ Chhote $\Delta i$ par, $\cos (\Delta i/2)\approx 1$, isliye slope $\approx v=7.7$ km/s per radian. Per degree mein convert karo $\pi /180\approx 0.01745$ se multiply karke: $7.7\times 0.01745\approx 0.1344\text{ km/s per degree}=134\text{ m/s per degree}.$ Zero ke paas har extra degree LEO mein lagbhag 134 m/s cost karta hai. (Parent note ke $2°$-GPS example se speed ke anusaar match karta hai.)

Figure dobara dekho: red arrow $\Delta \vec{v}$ ek isosceles triangle ki base hai. ${\vec{v}}_1$ aur $\Delta \vec{v}$ ke beech ka angle us triangle par $90°+\Delta i/2$ ke barabar hai — yaani burn almost exactly perpendicular velocity ke saath point karta hai (yeh $\Delta i/2$ se thoda peeche jhukta hai). Infinitesimally chhote change ke liye ($\Delta i\to 0$) burn purely sideways hai ($\vec{v}$ ke normal), jo sense deta hai: ek sideways nudge first order par length change kiye bina vector ko rotate karta hai. Sign (left vs right / north vs south) decide karta hai ki inclination badhti hai ya ghatti hai.

Combined (law of cosines, do alag sides angle $\Delta i$ par): $\Delta v_{\text{comb}}=\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \Delta i}.$ $\cos 28°\approx 0.88295$ ke saath: $\Delta v_{\text{comb}}=\sqrt{1.6^2+3.1^2-2(1.6)(3.1)(0.88295)}$ $=\sqrt{2.56+9.61-8.7589}=\sqrt{3.4111}\approx 1.847\text{ km/s}.$ Alag alag = pehle speed change, phir nayi speed $v_2=3.1$ par pure plane change: $\Delta v_{\text{sep}}=\underset{\text{speed}}{\underbrace{(3.1-1.6)}}+\underset{\text{plane}}{\underbrace{2(3.1)\sin 14°}}=1.5+6.2(0.24192)=1.5+1.500=3.000\text{ km/s}.$ Combined ($1.85$ km/s) alag alag ($3.00$ km/s) se ek km/s se zyada behtar hai — triangle inequality action mein: ek seedha step do bent path ke legs se chhota hota hai. Isliye Hohmann transfers plane change ko apogee burn mein fold karte hain.

Ek shot: $\Delta v=2(7.7)\sin 20°=15.4(0.34202)\approx 5.267$ km/s. Do $20°$ burns mein split karo: har ek costs $2(7.7)\sin 10°=15.4(0.17365)\approx 2.674$ km/s, total $2\times 2.674=5.348$ km/s. Same speed par splitting thoda worse hai ($5.35>5.27$): har chhota chord chhota hai, lekin do chords ek bade chord se zyada sum karte hain (phir triangle inequality, ab dusri taraf). Splitting tabhi help karta hai jab aap sub-burns wahin karo jahan $v$ chhoti ho — asli saving lower speed se aati hai, angle ko subdivide karne se nahi.

Tsiolkovsky kyun? Yeh woh tool hai jo required $\Delta v$ ko mass ratio mein convert karta hai. Dekho Tsiolkovsky rocket equation. $\frac{m_0}{m_f}=e^{\Delta v/v_e}=e^{1.342/3.0}=e^{0.44733}\approx \mathrm{1.5641.}$ Propellant fraction $=1-\frac{m_f}{m_0}=1-\frac{1}{1.5641}=1-0.6393=0.3607$. Lagbhag 36% spacecraft ka mass fuel ke roop mein jalta hai — sirf ek $10°$ tilt ke liye. Plane changes bahut mahenge hote hain.

(a) Combined burn. Do sides $v_a=1.60$, $v_{\text{GEO}}=3.07$, included angle $27°$, $\cos 27°\approx 0.89101$: $\Delta v=\sqrt{1.60^2+3.07^2-2(1.60)(3.07)(0.89101)}$ $=\sqrt{2.5600+9.4249-8.7526}=\sqrt{3.2323}\approx 1.798\text{ km/s}.$

(b) Alag alag burns. Circularize: $|3.07-1.60|=1.47$ km/s. Phir GEO speed $3.07$ par pure plane change: $2(3.07)\sin 13.5°=6.14(0.23345)=1.433$ km/s. Total: $\Delta v_{\text{sep}}=1.47+1.433=2.903\text{ km/s}.$ Combine karne se $2.903-1.798=1.105$ km/s bacha — ek bada margin.

(c) Propellant fraction (combined). Tsiolkovsky use karo $\Delta v=1.798$ km/s aur $v_e=3.1$ km/s ke saath: $\frac{m_0}{m_f}=e^{1.798/3.1}=e^{0.58000}\approx 1.7860,\text{fraction}=1-\frac{1}{1.7860}=1-0.5599=\mathrm{0.4401.}$ Combined burn ke saath lagbhag 44% mass propellant hai. Comparison ke liye, alag alag strategy ($\Delta v=2.903$ km/s) deti hai $\frac{m_0}{m_f}=e^{2.903/3.1}=e^{0.93645}\approx 2.551$, fraction $1-1/2.551=0.6081$, yaani lagbhag 61% — kaafi bura.

(d) Design lesson. Plane change (i) sabse slow point par karo — apogee, jahan $v$ sabse chhoti hoti hai — aur (ii) circularization burn ke saath combined karo, law-of-cosines triangle ka faida uthao. Dono tricks stack hokar propellant bill ko ~61% se ~44% tak slash karti hain. Isliye real GEO missions inclination removal ko apogee kick mein fold karte hain.

Separate cost, $\Delta i$ ke function ke roop mein: $\Delta v_{\text{sep}}(\Delta i)=|v_{\text{GEO}}-v_a|+2v_{\text{GEO}}\sin (\Delta i/2)=1.47+6.14\sin (\Delta i/2).$ Combined cost: $\Delta v_{\text{comb}}(\Delta i)=\sqrt{1.60^2+3.07^2-2(1.60)(3.07)\cos \Delta i}=\sqrt{11.9849-9.8248\cos \Delta i}.$ $\Delta i=0$ par: combined $=|3.07-1.60|=1.47$, separate $=1.47$ — equal (jaisa hona chahiye, koi plane change nahi). Sab $\Delta i>0$ ke liye combined value strictly separate se neeche hai (triangle inequality), isliye combine karna hamesha kam se kam utna hi sasta aur $\Delta i>0$ par strictly sasta hai.

$\Delta i=27°$ par numerical check: combined $=\sqrt{11.9849-9.8248(0.89101)}=\sqrt{11.9849-8.7538}=\sqrt{3.2311}\approx 1.797$ km/s vs separate $1.47+6.14(0.23345)=2.903$ km/s. Isliye koi break-even nahi hai jahan separate jeet jaaye — combining har $\Delta i>0$ ke liye dominate karta hai, aur dono sirf trivial $\Delta i=0$ par equal hain.

$7.7$ km/s par $2°$ tilt se speed change? ::: $\approx 269$ m/s ($2\cdot 7.7\cdot \sin 1°$ use karke) $7.7$ km/s par $200$ m/s budget ke liye sabse bada $\Delta i$? ::: $\approx 1.49°$ $7.7$ km/s par zero ke paas $\Delta v$ per degree? ::: $\approx 134$ m/s per degree Combined burn formula? ::: $\Delta v=\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \Delta i}$ Combining $\Delta i>0$ ke liye hamesha kyun jeet jaata hai? ::: triangle inequality — ek vector leg do sequential legs se chhota hota hai