Plane change maneuvers — Δv = 2v·sin(Δi - 2): Worked examples

3.2.22 · D3 · Physics › Orbital Mechanics & Astrodynamics › Plane change maneuvers — Δv = 2v·sin(Δi - 2)

Intuition Ye page kya hai

Parent note ne formula $Δ v = 2 v sin (Δ i /2)$ ek isosceles velocity triangle se build kiya tha. Yahan hum ise har us case ke against stress-test karte hain jo universe throw kar sakti hai: tiny angles, giant angles, exact $180°$ reversal, degenerate $0°$ "no-op", different-speed combined burns, ek slow-point strategy, aur ek full exam twist. Har number page ke bottom par machine-check hai.

Koi bhi symbol se pehle: $v$ aapki speed hai (kitni tez, ek plain number km/s mein). $v$ (upar arrow) aapki velocity hai — speed plus aap kis direction mein point kar rahe ho. $Δ i$ woh angle hai jitna aap us direction ko rotate karte ho. $Δ v$ us dhakke ki length hai (bhi km/s mein) jo rotation karne ke liye chahiye.

Figures ke naam kaise hain. Har figure filename ka end -s01, -s02, wagera hota hai — ye sirf reading order mein figure numbers hain (s01 = "step figure 1", s02 = "step figure 2", …). Jab text kehta hai "figure s03 dekho", teen numbers wali embedded picture par scroll karo. Kuch cryptic nahi: s0N = is page par $N$ -vi figure.

Scenario matrix

Har plane-change problem in boxes mein se ek mein hota hai (ye ek table hai, figure nahi). Neeche ke examples us box ke saath labelled hain jo wo hit karta hai — milaakar ye poori grid fill karte hain.

Cell	Case class	Kya special hai	Example
A	Degenerate: $Δ i = 0°$	koi turn nahi, dhakka zero hona chahiye	Ex 1
B	Small angle $Δ i \leq 5°$	$sin$ almost linear; nudge	Ex 2
C	Mid angle ( $30°$ – $60°$ )	"expensive turn" regime	Ex 3
D	Exactly $Δ i = 60°$	famous $Δ v = v$	Ex 4
E	Boundary $Δ i = 90°$	$Δ v / v = 2$ exactly	Ex 5
F	Large angle ( $90°$ – $180°$ )	orbit speed se zyada cost	Ex 6a
G	Limiting: $Δ i = 180°$	full reversal, $Δ v = 2 v$	Ex 6b
H	Slow-point strategy	kahan burn karein (min $v$ )	Ex 7
I	Combined burn, $v_{1} \neq = v_{2}$	general law of cosines	Ex 8
J	Real-world word problem	Tsiolkovsky se fuel mass	Ex 9
K	Exam twist: $Δ i$ solve karo	formula invert karo (arcsin)	Ex 10

Jinpar hum rely karte hain: Vector addition and law of cosines (triangle), Orbital velocity — vis-viva equation ( $v$ deta hai), Apoapsis and periapsis (slow point), Tsiolkovsky rocket equation ( $Δ v$ ko fuel mein convert karta hai). Parent: Plane change maneuvers ($\Delta v = 2v\sin(\Delta i/2)$).

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

Figure s01 (pehli embedded picture upar) $Δ v / v = 2 sin (Δ i /2)$ ko $Δ i$ ke against $0°$ se $180°$ tak plot karta hai. Ise menu ki tarah padho: neeche ke har worked example ka ek labelled dot is curve par hai (letters B, C, D, E, F, G cell labels se match karte hain). Notice karo ye zero par start hota hai (Ex 1), chhote angles ke liye almost ek straight line hai (dashed blue line, Ex 2), $60°$ par $1$ se guzarta hai (Ex 4), $90°$ par $2$ cross karta hai (Ex 5), aur $180°$ par $2$ tak climb karta hai (Ex 6b). Har example ke baad figure s01 par wapas aao aur uska dot dhundho.

Cell A — degenerate no-turn

Worked example Ex 1 ·

$Δ i = 0°$ (figure s01 ke origin par dot) Ek satellite pehle se usi plane mein hai jahan wo chahta hai. Koi phir bhi formula run karta hai $Δ i = 0°$ , $v = 7.7$ km/s ke saath. Kya $Δ v$ ?

Forecast: kya na morne ke liye kuch cost aana chahiye? Padhne se pehle guess karo — aur figure s01 par $Δ i = 0°$ kahan baithta hai dhundho (bilkul left edge par).

Figure s02 chaar turn angles ke liye velocity triangle dikhata hai. Sabse left wale mein, $v_{1}$ aur $v_{2}$ ek doosre ke upar lie karte hain — koi triangle nahi hai, isliye uski teesri side (pink $Δ v$ ) ki length zero hai.

Plug in: $Δ v = 2 \cdot 7.7 \cdot sin (0°/2) = 2 \cdot 7.7 \cdot sin (0°)$ . Ye step kyun? Direct substitution taaki check ho sake ki formula apne floor par behave karta hai.
$sin (0°) = 0$ , isliye $Δ v = 15.4 \cdot 0 = 0$ km/s. Ye step kyun? Koi rotation nahi matlab $v_{2} = v_{1}$ , toh triangle ek single arrow mein collapse hota hai aur uski "teesri side" ki length zero hoti hai.

Verify: velocity vector move nahi karta, isliye $Δ v = v_{2} - v_{1} = 0$ . Zero turn ke liye zero dhakka — formula apna sanity floor pass karta hai. ✔

Cell B — small-angle nudge

Worked example Ex 2 ·

$Δ i = 2°$ , GPS orbit (figure s01 par dot B) $v = 3.9$ km/s. Inclination sirf $2°$ badlao. $Δ v$ kya hai? Kya $2°$ "negligible" hai?

Forecast: $2°$ trivial lagta hai — guess karo ki $Δ v$ $50$ m/s se kam hai ya nahi, phir check karo dot B figure s01 par kitna low baithta hai.

Pehle half-angle: $Δ i /2 = 1°$ . Radians mein $1° = π /180 \approx 0.01745$ rad. Ye step kyun? Chhote angle ka $sin$ radians mein angle ke almost equal hota hai ( $sin θ \approx θ$ ), isliye curve origin ke paas straight lagta hai — figure s01 par dashed blue line dekho.
$sin (1°) \approx 0.017452$ . Ye step kyun? Hume actual sine chahiye, lekin small-angle picture kehti hai ki roughly $0.01745$ expect karo.
$Δ v = 2 \cdot 3.9 \cdot 0.017452 = 7.8 \cdot 0.017452 \approx 0.1361$ km/s $= 136.1$ m/s. Ye step kyun? Multiply karo; sirf $2°$ tilt bhi $136$ m/s cost karta hai — bilkul negligible nahi.

Verify: small-angle approximation predict karti hai $Δ v \approx 2 v \cdot (Δ i /2 in rad) = v \cdot Δ i_{rad} = 3.9 \cdot 0.034907 = 0.1361$ km/s. Digit tak match. Units: (km/s) $\times$ (dimensionless) $=$ km/s. ✔

Cell C — mid-angle expensive turn

Worked example Ex 3 ·

$Δ i = 45°$ , LEO (figure s01 par dot C) $v = 7.7$ km/s. Plane ko $45°$ rotate karo. Orbital speed ka kaunsa fraction ye cost karta hai?

Forecast: ratio $Δ v / v$ guess karo — $0.5$ se bada ya chhota? Phir figure s01 par dot C ki height eyeball karo.

Figure s03 isosceles velocity triangle draw karta hai: do equal sides of length $v$ unke beech apex angle $Δ i = 45°$ ke saath, aur pink base $Δ v$ use close karti hai. Dashed line woh perpendicular bisector hai jo apex ko do $22.5°$ halves mein split karta hai — yahi se half-angle aata hai.

Half-angle: $45°/2 = 22.5°$ . Ye step kyun? Isosceles-triangle geometry hamesha half angle use karti hai (figure s03 mein apex ka perpendicular-bisector split, dashed line).
$sin (22.5°) \approx 0.38268$ . Ye step kyun? Velocity triangle mein ye ratio (half-base)/(side) hai.
$Δ v = 2 \cdot 7.7 \cdot 0.38268 \approx 5.893$ km/s. Ratio $Δ v / v = 2 sin (22.5°) \approx 0.7654$ . Ye step kyun? Multiply karo; $45°$ pehle se 76.5% poori orbital speed cost karta hai.

Verify: figure s01 par $Δ i = 45°$ halfway height se kaafi upar baithta hai — $0.765$ se match karta hai. Bhi $Δ v < 2 v = 15.4$ (reversal cost kabhi exceed nahi karta). ✔

Cell D — famous 60°

Worked example Ex 4 ·

$Δ i = 60°$ , LEO (figure s01 par dot D) $v = 7.7$ km/s. Textbook ka "shock" case. Dikhao ki $Δ v = v$ exactly.

Forecast: parent claim karta hai $Δ v = v$ exactly — matlab dot D figure s01 ki $Δ v / v = 1$ line par exactly land karta hai. Check karne se pehle believe karo.

Half-angle: $60°/2 = 30°$ . Ye step kyun? Rotation ka half, jaise hamesha.
$sin (30°) = \frac{1}{2}$ exactly (ek $30$ – $60$ – $90$ triangle). Ye step kyun? Ye woh ek special value hai jo algebra ko collapse karta hai.
$Δ v = 2 v \cdot \frac{1}{2} = v = 7.7$ km/s. Ye step kyun? $2$ aur $\frac{1}{2}$ cancel ho jaate hain — $60°$ turn karna poori extra orbit ki speed cost karta hai.

Verify: Ex 3 ( $45°$ ) mein humne $0.765 v$ paya; bada turn ( $60°$ ) zyada cost karna chahiye, aur $1.0 v > 0.765 v$ . Monotone increasing — rising s01 curve ke saath consistent. Figure s01 par dot D pink $Δ v / v = 1$ line par exactly baithta hai. ✔

Cell E — 90° par boundary

Worked example Ex 5 ·

$Δ i = 90°$ , LEO (figure s01 par dot E) $v = 7.7$ km/s. Right-angle plane change — "mid" aur "large" ke beech exact border. Dikhao $Δ v / v = 2$ .

Forecast: poora right angle turn — guess karo ki ye $1.4 v$ se zyada cost karta hai ya kam, phir figure s01 par dot E locate karo.

Figure s04 $Δ i = 90°$ ke liye velocity triangle dikhata hai: kyunki apex right angle hai, $v_{1} ⊥ v_{2}$ , isliye $Δ v$ ek right triangle ki hypotenuse hai jiske do legs length $v$ ke hain. Pythagoras akela answer de deta hai.

Right-angle apex matlab hum Pythagoras use kar sakte hain: $Δ v = v^{2} + v^{2} = v 2$ . Ye step kyun? Jab do velocity arrows perpendicular hain to law of cosines ka $cos (90°) = 0$ term vanish ho jaata hai, plain Pythagoras reh jaata hai — formula par ek khoobsurat cross-check.
Formula route: $Δ v = 2 v sin (45°) = 2 v \cdot \frac{2}{2} = v 2$ . Ye step kyun? $sin (45°) = \frac{2}{2}$ , aur do halves combine ho jaate hain — formula Pythagoras se bilkul agree karta hai.
Numbers: $Δ v = 7.7 2 \approx 10.89$ km/s, ratio $2 \approx 1.4142$ . Ye step kyun? $90°$ turn pehle se orbit speed ka $1.41 \times$ cost karta hai.

Verify: do independent methods (Pythagoras aur formula) dono $v 2$ dete hain. Figure s01 par dot E, dot D ( $1.0$ ) aur $180°$ ceiling ( $2.0$ ) ke beech, height $1.414$ par baithta hai. ✔

Cells F & G — large angle aur 180° limit (do alag cases)

Worked example Ex 6a ·

$Δ i = 120°$ (cell F, figure s01 par dot F) $v = 5$ km/s. Bade $120°$ turn ki cost.

Forecast: kya $Δ v$ kabhi orbit speed $v$ se zyada ho sakta hai? Guess karo, phir figure s01 par dot F dhundho (kya $Δ v / v = 1$ line se upar hai?).

Figure s05 dikhata hai ki triangle flatten hota hai jab $Δ i$ $120°$ se $180°$ ki taraf badhta hai: sabse left wala triangle yahan $120°$ case hai (cell F), sabse right wala Ex 6b ka $180°$ case hai (cell G).

Half-angle $= 120°/2 = 60°$ , aur $sin (60°) = \frac{3}{2} \approx 0.86603$ . Ye step kyun? Ek $60$ – $30$ – $90$ triangle, is baar long leg lete hain.
$Δ v = 2 \cdot 5 \cdot 0.86603 \approx 8.660$ km/s. Ye step kyun? Pehle se $1.73 v$ — orbit speed se bhi zyada.

Verify: $8.660 < 2 v = 10$ , isliye hum ceiling ke neeche rehte hain. Figure s01 par dot F, dot E ( $2 \approx 1.41$ ) aur top ( $2$ ) ke beech, height $1.73$ par baithta hai. ✔

Worked example Ex 6b ·

$Δ i = 180°$ (cell G, figure s01 par dot G) $v = 5$ km/s. Direction ka poora reversal — limiting case.

Forecast: apni velocity ko end-for-end flip karne ki exact cost guess karo, phir check karo ki dot G figure s01 par sabse upar wala point hai.

Half-angle $= 180°/2 = 90°$ , aur $sin (90°) = 1$ . Ye step kyun? Direction reverse karna extreme hai: triangle ek straight line mein flatten ho jaata hai (figure s05 mein sabse right), base $= 2 v$ .
$Δ v = 2 \cdot 5 \cdot 1 = 10$ km/s $= 2 v$ . Ye step kyun? Tumhe apni saari speed khatam karni padti hai ( $- 5$ ) aur doosri taraf rebuild karni padti hai ( $+ 5$ ): total $10$ .

Verify: $[0°, 180°]$ par $2 sin (Δ i /2)$ ka maximum $180°$ par hai jahan ye $2$ equal karta hai. Isliye $Δ v \leq 2 v$ hamesha — Ex 6a ( $8.66 < 10$ ) ye ceiling maanta hai. Ye common mistake bhi kill karta hai $sin (Δ i)$ use karne ki: woh $sin (180°) = 0$ dega, foolishly claim karta hua ki reversal free hai. ✔

Cell H — slow-point strategy

Worked example Ex 7 · Same

$30°$ turn, ellipse mein do jagah Ek elliptical orbit mein periapsis speed $v_{p} = 10$ km/s aur apoapsis speed $v_{a} = 1.6$ km/s hai (dekho Apoapsis and periapsis). Tumhe $30°$ plane change chahiye. Har end par cost compare karo.

Forecast: same angle, same formula — guess karo apoapsis kitne times sasta hai. (Hint: figure s01 par angle-dot same hai, isliye sirf $v$ differ kar sakta hai.)

Figure s06 ellipse sketch karta hai: periapsis par ek bada velocity arrow (fast, Earth ke paas) aur apoapsis par ek chhota arrow (slow, door). Bade arrow ko $30°$ se turn karna bada dhakka maangta hai; chhote arrow ko same $30°$ se turn karna chhota dhakka maangta hai — kyunki $Δ v \propto v$ .

Common factor: $2 sin (30°/2) = 2 sin (15°) \approx 2 \cdot 0.258819 = 0.517638$ . Ye step kyun? Angle-part dono ends par identical hai; sirf $v$ differ karta hai, kyunki $Δ v \propto v$ .
Periapsis par: $Δ v_{p} = 0.517638 \cdot 10 \approx 5.176$ km/s. Ye step kyun? Fast point → expensive rotation.
Apoapsis par: $Δ v_{a} = 0.517638 \cdot 1.6 \approx 0.828$ km/s. Ye step kyun? Slow point → sasti rotation.
Ratio $= v_{p} / v_{a} = 10/1.6 = 6.25 \times$ sasta apoapsis par. Ye step kyun? Kyunki $Δ v$ $v$ mein linear hai, saving exactly speed ratio hai.

Verify: $Δ v_{p} /Δ v_{a} = 5.176/0.828 = 6.25 = v_{p} / v_{a}$ . Mantra "slow par turn karo, kam kharo" literally speeds ka ratio hai. ✔

Cell I — different speeds ke saath combined burn

Worked example Ex 8 · GTO apogee: speed change

aur plane change ek saath Apogee par tumhare paas $v_{1} = 1.6$ km/s hai (transfer ellipse par) aur circular geostationary speed $v_{2} = 3.07$ km/s chahiye, saath mein plane bhi $Δ i = 28.5°$ se turn karna hai (Cape Canaveral launch). Combined burn ke liye $Δ v$ nikalo, aur alag alag karne se compare karo.

Forecast: guess karo ki combine karna better hai ya nahi — aur kitna. Phir figure s07 dekho aur predict karo kaunsa pink path chhota hai.

Figure s07 same start se do paths contrast karta hai: combined triangle (ek chhota pink diagonal, sides $v_{1}$ aur $v_{2}$ ke saath $28.5°$ angle unke beech) versus separate L-shaped path (pehle speed change, phir ek plane-change leg). Diagonal visibly chhota hai — ye triangle inequality hai jo tumhara fuel budgeting kar rahi hai.

Combined (general law of cosines, unequal sides): $Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos Δ i$ Ye step kyun? Jab do velocity arrows ki different lengths hain, isosceles trick fail hoti hai; hume full law of cosines chahiye.
Numbers: $v_{1}^{2} = 2.56$ , $v_{2}^{2} = 9.4249$ , $2 v_{1} v_{2} = 9.824$ , $cos (28.5°) \approx 0.87882$ . Ye step kyun? Triangle ke teen pieces assemble karo.
$Δ v = 2.56 + 9.4249 - 9.824 \cdot 0.87882 = 11.9849 - 8.6357 = 3.3492 \approx 1.830$ km/s. Ye step kyun? Triangle ka single diagonal dhakka.
Separate: pehle speed change $∣ v_{2} - v_{1} ∣ = 1.47$ km/s, phir $v_{2}$ par pure plane change: $2 \cdot 3.07 \cdot sin (14.25°) = 6.14 \cdot 0.246154 \approx 1.5114$ km/s. Sum $= 1.47 + 1.5114 = 2.981$ km/s. Ye step kyun? Do sequential arrows end to end rakh do.

Verify: combined $1.830 <$ separate $2.981$ . Triangle inequality guarantee karta hai ki direct side do-leg path se chhoti hogi — combining yahan $\approx 1.15$ km/s bachata hai. ✔

Cell J — real-world fuel word problem

Worked example Ex 9 ·

$10°$ change ke liye kitna propellant chahiye? Ek $2000$ kg satellite $v = 7.5$ km/s par inclination $10°$ se change karna chahta hai. Uske thruster ki exhaust velocity $v_{e} = 3.0$ km/s hai. Ye burn kitna fuel mass use karta hai?

Forecast: $10°$ chhota hai — guess karo ki fuel $100$ kg se kam hai ya nahi. (Figure s01 par $Δ v / v$ dot low hai; lekin Tsiolkovsky ise exponentially fuel mein bend karta hai.)

$Δ v = 2 \cdot 7.5 \cdot sin (5°) = 15 \cdot 0.0871557 \approx 1.3073$ km/s. Ye step kyun? Half-angle $= 5°$ ; sirf $10°$ bhi $1.3$ km/s maangta hai.
Mass ratio via Tsiolkovsky rocket equation: $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{1.3073/3.0} = e^{0.43578} \approx 1.5463$ . Ye step kyun? Fuel $Δ v$ ke saath exponentially badhta hai, linearly nahi — isliye plane changes itne painful hain.
Final (dry-side) mass: $m_{f} = m_{0} /1.5463 = 2000/1.5463 \approx 1293.4$ kg. Ye step kyun? $m_{0} = 2000$ kg burn se pehle poora wet mass hai.
Fuel burned: $m_{0} - m_{f} = 2000 - 1293.4 \approx 706.6$ kg. Ye step kyun? Difference woh propellant hai jo expel hua.

Verify: $706.6$ kg poore satellite ka 35% hai sirf $10°$ nudge ke liye. Exponential mapping matlab chhota $Δ v$ bhi bada fuel fraction kha jaata hai. ✔

Cell K — exam twist: $Δ i$ ke liye invert karo

Worked example Ex 10 · "Tumhare paas

$Δ v = 2$ km/s hai $v = 7.7$ km/s par. Sabse bada plane change kya hai?" $Δ v$ nahi, $Δ i$ solve karo.

Forecast: guess karo ki answer $15°$ se bada hai ya chhota. (Figure s01 par height $Δ v / v = 2/15.4 \approx 0.13$ dhundho aur angle tak left padho — ye inverse move hai.)

$Δ v = 2 v sin (Δ i /2)$ ko sine ke liye rearrange karo: $sin (\frac{Δ i}{2}) = \frac{Δ v}{2 v} = \frac{2}{2 \cdot 7.7} = \frac{2}{15.4} \approx 0.12987$ Ye step kyun? Unknown ko sine ke andar isolate karo — algebra, koi geometry nahi gayi.
Sine ko $arcsin$ se undo karo ("is sine ka kaunsa angle hai?" question): $\frac{Δ i}{2} = arcsin (0.12987) \approx 7.462°$ Ye step kyun? $arcsin$ , $[0°, 90°]$ par $sin$ ka exact inverse hai, aur half-angles yahan kabhi us range se nahi nikalte (kyunki $Δ i \leq 180°$ ), isliye ye unambiguous hai.
Double karo: $Δ i = 2 \cdot 7.462° \approx 14.92°$ . Ye step kyun? Half se poori rotation recover karo.

Verify: plug back karo — $2 \cdot 7.7 \cdot sin (14.92°/2) = 15.4 sin (7.46°) = 15.4 \cdot 0.12984 \approx 2.00$ km/s. Diye gaye budget par round-trip karta hai. Aur $14.92° < 15°$ , ek $Δ v$ ko match karta hai jo sirf $2 v$ ka $13%$ hai. ✔

Active recall

Recall Kya tumne har cell cover kiya?

Kaunse example ne kya handle kiya: degenerate $0°$ ? $90°$ boundary? exact $180°$ ceiling? unequal-speed combined burn? $Δ i$ ke liye inverting? Degenerate 0° ::: Ex 1 (Δv = 0) 90° boundary (Δv/v = √2) ::: Ex 5 (Pythagoras cross-check) 180° reversal (ceiling Δv = 2v) ::: Ex 6b Combined burn with v₁ ≠ v₂ ::: Ex 8 (general law of cosines) Solving for Δi from a Δv budget ::: Ex 10 (arcsin, then double) Apoapsis sasta kyun hai? ::: Δv ∝ v, isliye minimum speed → minimum cost (Ex 7)

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