Plane change maneuvers — Δv = 2v·sin(Δi - 2): Question bank

3.2.22 · D5 · Physics › Orbital Mechanics & Astrodynamics › Plane change maneuvers — Δv = 2v·sin(Δi - 2)

Lagbhag har trap ke peeche ek hi governing fact hai: $Δ v = 2 v sin (\frac{Δ i}{2})$ Yahan $v$ orbital speed hai (ek magnitude, km/s mein), $Δ i$ woh angle hai jitna orbital plane rotate hota hai, aur $Δ v$ ==velocity-vector change ki magnitude== hai jo burn supply karta hai.

True or false — justify

A satellite ki speed ek pure plane change se pehle aur baad mein same rehti hai, isliye koi fuel nahi chahiye.

False. Speed (ek magnitude) unchanged rehti hai, lekin

v

ek vector hai aur uski direction rotate hoti hai;

Δ v = ∣ v_{2} - v_{1} ∣ = 2 v sin (Δ i /2) \neq = 0

. Kisi bhi vector ko rotate karne mein thrust lagta hai.

Orbital speed

v

ko double karne se same

Δ i

ke liye plane-change cost

Δ v

bhi double ho jaati hai.

True.

Δ v = 2 v sin (Δ i /2)

directly proportional hai

v

se; sine factor sirf

Δ i

par depend karta hai, isliye double speed par double

Δ v

pay karna padta hai.

Angle

Δ i

ko double karne se

Δ v

double ho jaati hai.

False.

Δ v

linearly depend karta hai

v

par lekin

Δ i

par

sin (Δ i /2)

ke through, jo linear nahi hai. Jaise

60°

120°

jaane par

sin

0.5

\approx 0.866

ho jaata hai, jo

1.73

ka factor hai,

2

nahi.

Chhote

Δ i

ke liye, cost roughly

Δ i

ke proportional badhti hai.

True. Tiny angles ke liye

sin (Δ i /2) \approx Δ i /2

(radians mein), isliye

Δ v \approx v Δ i

— nearly linear. Isliye parent note mein small-angle GPS nudge cleanly scale hua tha.

180°

plane change mein

Δ v = 0

hota hai kyunki

sin 180° = 0

False. Half angle plug karna hota hai:

Δ v = 2 v sin (90°) = 2 v

180°

reversal sabse mehnga case hai — tum

v

khatam karte ho aur doosri taraf se wapas banate ho.

Speed change aur plane change ko ek burn mein combine karna hamesha alag-alag karne se sasta hota hai.

True. Triangle inequality se, velocity triangle ka ek vector side kabhi bhi do sequential sides se lamba nahi hota; ek combined burn

v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos Δ i

do burns ke sum se sasta rehta hai.

Plane change orbit ki size ya shape kabhi nahi badalta.

True (by definition). Ek pure plane change

∣ v ∣

aur burn point par radius ko unchanged rakhta hai, isliye Orbital velocity — vis-viva equation ke zariye semi-major axis unchanged rehta hai; sirf plane ka tilt badalta hai.

Plane changes jaldi khatam karne ke liye periapsis par karne chahiye.

False. Kyunki

Δ v \propto v

aur speed periapsis par maximum hoti hai, yeh sabse mehnga spot hai. Inhe apoapsis par karo, jahan

v

sabse chhota hota hai.

Spot the error

"Kyunki

∣ v_{1} ∣ = ∣ v_{2} ∣ = v

, isliye

Δ v = ∣ v_{2} ∣ - ∣ v_{1} ∣ = 0

Error yeh hai ki magnitudes subtract ki ja rahi hain vectors ki jagah. Sahi hai

Δ v = ∣ v_{2} - v_{1} ∣

: pehle vectors subtract karo, tab difference ki magnitude lo.

Δ i = 40°

ke liye,

Δ v = 2 v sin (40°)

Sine ke andar angle half hona chahiye:

Δ v = 2 v sin (20°)

. Half-angle isosceles-triangle geometry se aata hai (perpendicular bisector

Δ i

ko do equal halves mein split karta hai).

"Velocity triangle ki sides

v_{1}

v_{2}

hain aur included angle

90°

minus

Δ i

hai."

v_{1}

aur

v_{2}

ke beech included angle exactly

Δ i

hi hai, uska complement nahi — tumne

Δ i

se rotate kiya, isliye woh angle hai jitna dono vectors span karte hain.

10°

change sasta hai, isliye cost per degree har angle par chhota hota hai."

Cost per degree angle badhne ke saath badh jaata hai, kyunki

sin (Δ i /2)

Δ i = 180°

ki taraf upar curve karta hai. Near-linear sasta behavior sirf chhote angles ke liye hota hai.

"Plane-change

Δ v

chhota hai, isliye propellant mass bhi chhota hai."

Tsiolkovsky mass ratio

m_{0} / m_{f} = e^{Δ v / v_{e}}

Δ v

ke saath exponentially badhata hai. Ek modest-looking

Δ v

exhaust speed

v_{e}

ke paas bhi payload se zyada propellant maang sakta hai.

"Orbit direction reverse karne ke liye (

180°

), bas ek baar

Δ v = v

se fire karo."

Tumhe

Δ v = 2 v sin (90°) = 2 v

chahiye,

v

nahi. Tumhe apni poori velocity

v

hatani padti hai aur phir

v

doosri direction mein banani padti hai — do full speeds ka kaam.

Why questions

Half-angle kyun aata hai poore

Δ i

ki jagah?

Isosceles velocity triangle ke apex se base

Δ v

par perpendicular daalne se apex angle

Δ i

aur base dono bisect ho jaate hain; har half-base

v sin (Δ i /2)

hoti hai, isliye poori base

2 v sin (Δ i /2)

hoti hai.

Velocity triangle isosceles kyun hota hai general triangle ki jagah?

Kyunki pure plane change speed ko unchanged rakhta hai,

∣ v_{1} ∣ = ∣ v_{2} ∣ = v

— do equal sides. Agar speeds alag hote (combined burn), triangle scalene hota aur tum poora law of cosines use karte.

Yahan simple subtraction ki jagah law of cosines kyun use karte hain?

Humein do side lengths (

v

v

) aur unke beech ka angle (

Δ i

) pata hai, aur teesri side (

Δ v

) chahiye. Law of cosines exactly woh tool hai jo "do sides + included angle" ko opposite side mein convert karta hai.

Δ v \propto v

apoapsis ko smart burn point kyun banata hai?

Cost directly tumhari current speed ke saath scale hoti hai, aur energy conservation se speed apoapsis par sabse kam hoti hai; slow vector rotate karna sasta hai, fast vector rotate karna mehnga hai.

60°

plane change ko "orbit tak pahunchne jitna mehnga" kyun kaha jaata hai?

Δ v = 2 v sin (30°) = v

, isliye burn poori orbital speed ke barabar hota hai (~

7.7

km/s LEO mein) — orbit tak pahunchne ki poori cost ke comparable.

Engineers plane changes ko Hohmann apogee burn mein kyun fold karte hain?

High apogee par speed

v

chhoti hoti hai, isliye

\propto v

cost chhoti hoti hai; aur plane rotation ko circularization burn ke saath combine karna (ek vector step) triangle inequality se do alag burns se sasta hota hai.

Derivation mein

1 - cos Δ i

2 sin^{2} (Δ i /2)

mein kyun rewrite kiya jaata hai?

Half-angle identity awkward

1 - cos

ko perfect square mein convert karti hai, jisse tum clean square root le sakte ho aur

2 v sin (Δ i /2)

par land karte ho ek messy radical ki jagah.

Edge cases

Δ i = 0

par

Δ v

kya hoga?

Zero:

2 v sin (0) = 0

. Koi rotation nahi matlab koi vector change nahi, isliye koi burn nahi — trivially correct base case.

Maximum possible plane-change

Δ v

kya hai, aur kis angle par?

Maximum

2 v

hai,

Δ i = 180°

par jahan

sin (90°) = 1

; tum completely direction reverse karte ho. Koi bhi plane change tumhari speed ke double se zyada cost nahi kar sakta.

Agar orbital speed

v = 0

ho (hypothetically rest par), toh

Δ v

kya hoga?

Zero, kyunki

Δ v = 2 v sin (Δ i /2)

zero ho jaata hai jab

v = 0

. Koi velocity vector nahi rotate karna hai, toh "plane change" meaningless hai — koi motion hi nahi hai.

Combined-burn formula mein kya hoga jab

Δ i \to 0

aur

v_{1} \neq = v_{2}

cos 0 = 1

deta hai

Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} = ∣ v_{2} - v_{1} ∣

— koi rotation nahi, pure speed change, exactly ek scalar difference.

Combined-burn formula mein kya hoga jab

v_{1} = v_{2} = v

Yeh collapse ho jaata hai

2 v^{2} - 2 v^{2} cos Δ i = 2 v sin (Δ i /2)

par, pure plane-change formula recover karta hai — ek accha consistency check.

Kya plane change orbital period badalta hai?

Nahi. Period semi-major axis par depend karta hai, jo burn point par speed aur radius par depend karta hai (vis-viva se); pure plane change inhe nahi chhuta, isliye period preserved rehta hai.

Agar tum ek

Δ i

ko do equal plane changes of

Δ i /2

mein split karo same speed par, kya fuel bachta hai?

Nahi — tum zyada kharchte ho. Do burns

2 v sin (Δ i /4)

each ka sum

4 v sin (Δ i /4)

hota hai, jo kisi bhi positive angle ke liye single

2 v sin (Δ i /2)

se zyada hai (sine curve concave hai). Split tabhi help karta hai jab koi slower point available ho.

Connections

Parent topic
Vector addition and law of cosines (kyun magnitudes nahi, vectors subtract karte hain)
Orbital velocity — vis-viva equation (woh $v$ jo plug karte hain)
Inclination and orbital elements ( $i$ aur $Δ i$ define karta hai)
Apoapsis and periapsis (slow-point strategy)
Hohmann transfer orbit (jahan burns combine hote hain)
Tsiolkovsky rocket equation ( $Δ v$ ko fuel mein convert karta hai)