Lambert's problem — connecting two positions in given time: Question bank

3.2.28 · D5 · Physics › Orbital Mechanics & Astrodynamics › Lambert's problem — do positions ko given time mein connect karna

Traps se pehle, hum neeche use hone wale har symbol ko earn karte hain — inme se koi bhi kisi question mein tab tak nahi aata jab tak yahan define na ho.

Definition Is page ke symbols

$r_{1}, r_{2}$ = central body se start aur end points tak ke do position vectors; $r_{1}, r_{2}$ unki lengths hain.
$c$ = chord = do points ke beech ki straight-line distance $∣ r_{2} - r_{1} ∣$ (unke beech rakha ek ruler, flight path nahi).
$r_{1} + r_{2}$ = do radial distances ka sum.
$s = \frac{1}{2} (r_{1} + r_{2} + c)$ = us triangle ki semiperimeter jiske corners central body, point 1, aur point 2 hain.
$a$ = transfer orbit ka semi-major axis (uska "size"); $a_{m i n} = s /2$ sabse chhota ellipse hai jo dono points tak pahunch sakta hai.
$e$ = orbit ki eccentricity — ek pure number jo batata hai ellipse kitna stretched hai ( $e = 0$ circle hai, $e \to 1$ parabola hai).
$Δ θ$ = transfer angle, $r_{1}$ se $r_{2}$ tak motion ki direction mein swept angle (neeche sign convention dekho).
$Δ t$ = time of flight; $N$ = arrival se pehle complete revolutions ki sankhya.

Definition Mean anomaly

$M$ , eccentric anomaly $E$ , aur auxiliary angles $α, β$ Mean anomaly $M$ ek kalpnik angle hai jo time ke saath uniformly badhta hai (yeh woh hai jo orbit par ek clock read karta). Eccentric anomaly $E$ ek geometric angle hai jo ek body ko uske ellipse par ellipse ke centre se measure karta hai (focus se nahi). Kepler's equation inhe link karta hai: $M = E - e sin E$ , jahan $e$ upar define ki gayi eccentricity hai. Yahan $E_{1}, E_{2}$ start aur end points par eccentric anomaly ki values hain. Jab Kepler's law dono ends par likha jaata hai aur subtract kiya jaata hai, to do eccentric anomalies do auxiliary angles $α, β$ mein fold ho jaate hain jo in equations se define hote hain: $sin \frac{α}{2} = \frac{s}{2 a}, sin \frac{β}{2} = \frac{s - c}{2 a} .$ Geometrically, $α$ "full-triangle" size $s$ ko package karta hai aur $β$ "chord-shortened" size $s - c$ ko, dono orbit size $a$ se divided. Ye pure-angle handles hain jo Lambert ko ek-unknown ( $a$ ) equation mein badal dete hain.

Figure — Lambert's problem — connecting two positions in given time

Figure 1 — space triangle. Do teal radius arrows

r_{1}, r_{2}

central body (focus par black dot) se do mission points tak pahunchte hain. Unke beech ki seedhi burnt-orange line chord

c

hai; bahar bulging plum dashed curve woh actual flight arc hai jo spacecraft fly karta hai. Focus ke paas ka chhota black arc transfer angle

Δ θ

mark karta hai. Isse chord (ek ruler) aur arc (trajectory) ke beech ka farq feel karo, jo TF4, SE1, aur EC-items mein referenced hai.

Figure 2 — kyun $a \geq s /2$ . Horizontal axis orbit size

a

hai; curves hain

sin (α /2) = s /2 a

(orange) aur

sin (β /2) = (s - c) /2 a

(teal). Height 1 par dotted horizontal line woh ceiling hai jo har sine ko maanni hi hoti hai. Plum dashed vertical line

a_{m i n} = s /2

mark karti hai, jahan orange curve 1 ko touch karti hai (

α = π

); uske left mein shaded plum band woh forbidden region hai jahan formula ek sine 1 se bada maangta. Yeh picture SE2, WHY3, aur EC7 ke peeche hai.

True or false — justify

TF1. "Do points aur ek fixed flight time hamesha exactly ek orbit pin down karte hain."

False — har

a > a_{m i n}

ke liye generally do ellipses hote hain (focus chord ke either side par), plus short-way/long-way choice aur multi-rev branches, isliye kaafi saare orbits same

Δ t

share kar sakte hain.

TF2. "Time of flight transfer ki eccentricity

e

par depend karta hai."

False us sense mein jo matter karta hai — Lambert's theorem kehta hai

Δ t

sirf

a

c

, aur

r_{1} + r_{2}

par depend karta hai; eccentricity

e

kabhi independent input ke roop mein appear nahi hoti, woh trig collapse mein drop ho jaati hai.

TF3. "Transfer orbit ko bada karna (larger

a

) hamesha trip ko zyada lamba banata hai."

False —

a_{m i n}

ke paas time,

a

mein monotonic nahi hai; do solution branches ek minimum-time point bracket karte hain, isliye

a

badhana trip ko chhota ya lamba dono kar sakta hai, branch ke hisaab se.

TF4. "Chord

c

woh distance hai jo spacecraft actually travel karta hai."

False — craft curved arc fly karta hai;

c

sirf straight-line gap

∣ r_{2} - r_{1} ∣

hai. Geometry Lambert mein purely is straight chord ke through enter karti hai, jo Lambert's theorem ka exactly woh surprising content hai.

TF5. "Agar main short way ko long way se swap karun lekin

r_{1}

r_{2}

same rakhun, toh chord

c

badal jaata hai."

False —

c = ∣ r_{2} - r_{1} ∣

sirf do endpoints se set hota hai, isliye yeh short aur long way dono ke liye identical hai; jo badalta hai woh swept arc hai, jo auxiliary angle

β

ka sign flip karke handle hota hai (upar ke sign convention ke anusaar).

TF6. "

a_{m i n} = s /2

least time of flight ka orbit hai."

False —

a_{m i n} = s /2

minimum-energy orbit hai (sabse chhota possible

a

), minimum-time orbit nahi; ye

Δ t

versus

a

curve par alag special points hain.

TF7. "Kepler's equation jaisa ek propagator Lambert's problem ko directly solve kar sakta hai."

False — ek propagator initial value problem solve karta hai (position aur velocity chahiye); Lambert ek boundary value problem hai (do positions aur ek time, velocity unknown), isliye ise invert karna hoga, sirf propagate nahi.

TF8. "Lambert's problem orbit ki shape output karta hai; velocities ek alag mission-design step hai."

False — Lagrange

f, g

functions ke zariye

v_{1}, v_{2}

recover karna Lambert solve karne ka hi hissa hai; woh velocities (aur isliye

Δ v

) hi ise solve karne ka poora point hain.

TF9. "Ek fixed geometry ke liye ek shortest possible flight time hota hai, jiske neeche koi ellipse kaam nahi karta."

True — parabolic escape limit ek floor set karti hai; parabola ke

Δ t

se chhota maango aur tumhe hyperbolic transfer (

a < 0

) use karna hoga, ellipse nahi.

Spot the error

SE1. "

c = r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos Δ θ

, aur kyunki

cos

negative ho sakta hai,

c

r_{1} + r_{2}

se bada ho sakta hai."

Error —

c

kabhi

r_{1} + r_{2}

se zyada nahi ho sakta; yeh triangle inequality hai. Maximum

c = r_{1} + r_{2}

sirf us degenerate case mein reach hota hai jab

r_{1}

aur

r_{2}

exactly opposite directions mein point karein.

SE2. "Mujhe

sin (α /2) = s /2 a = 1.2

mila, toh

α /2 \approx 7 4^{\circ}

Error — ek sine kabhi 1 se zyada nahi ho sakta, isliye

s /2 a = 1.2

impossible hai; iska matlab hai chosen

a

a_{m i n} = s /2

se neeche hai, yaani ek nonphysical ellipse.

a

badha do.

SE3. "Long way (

Δ θ = 27 0^{\circ}

) ke liye maine same

+ β

use kiya jaise short way ke liye."

Error — long way ko

β \to - β

chahiye (sign convention ke anusaar);

β

ka sign short-vs-long encode karta hai, aur ise positive rakhna same

a

ke liye badha hua galat

Δ t

deta hai.

SE4. "Newton's method converge ho gayi, toh mere paas the answer hai."

Error — Lambert ke multiple branches hain; Newton bina bracketed guess ke (start

a_{m i n}

-based bounds se aur

N

specify karke) galat ellipse ya galat revolution number par land kar sakta hai.

SE5. "Transfer angle

Δ θ = arccos (\frac{r _{1} \cdot r _{2}}{r _{1} r _{2}})

hai, done."

Error —

arccos

sirf

0

π

return karta hai, isliye short way aur long way mein fark nahi bata sakta; tumhe direction orbit normal / prograde-retrograde choice se resolve karni hogi taaki pata chale ki

Δ θ

2 π - Δ θ

apply hota hai.

SE6. "Kyunki

α, β

(α - sin α) - (β - sin β)

mein symmetrically appear karte hain, unhe swap karna sirf ek sign flip karta hai, koi badi baat nahi."

Error —

α

s

se tied hai aur

β

s - c

se, alag-alag arguments ke zariye; ye interchangeable nahi hain, aur inhe swap karna ek alag (galat) geometry describe karta hai.

SE7. "

a_{m i n} = s /2

jahan

s = \frac{1}{2} (r_{1} + r_{2} + c)

, toh

a_{m i n} = r_{1} + r_{2} + c

Error — arithmetic slip:

a_{m i n} = s /2 = \frac{1}{4} (r_{1} + r_{2} + c)

r_{1} + r_{2} + c

nahi.

Why questions

WHY1. Time of flight sirf

a

c

, aur

r_{1} + r_{2}

par depend karne tak kyun collapse hota hai?

Kyunki jab Kepler's equation dono endpoints par likhi jaati hai aur subtract ki jaati hai, to eccentricity

e

aur individual eccentric anomalies

E_{1}, E_{2}

(upar define) do auxiliary angles

α, β

mein combine ho jaate hain, jo khud sirf

s

aur

s - c

over

a

par depend karte hain — yahi Lambert's theorem ka algebraic miracle hai.

WHY2.

E_{1}, E_{2}

mein directly solve karne ki jagah auxiliary angles

α, β

kyun introduce karte hain?

Kyunki

α, β

exactly woh geometry-folded combinations of

E_{1}, E_{2}

hain jo subtraction ke baad bachte hain; ye do-endpoint problem ko ek clean function of ek unknown

a

mein badal dete hain, jo root-find ko one-dimensional banata hai.

WHY3.

a \geq s /2

ek hard lower bound kyun hai?

Kyunki

sin (α /2) = s /2 a

aur ek sine 1 se zyada nahi ho sakta, jo

2 a \geq s

force karta hai; equality sabse patla allowed ellipse, minimum-energy orbit deta hai.

WHY4. Solve karne se pehle prograde vs retrograde (short vs long way) kyun choose karna zaroori hai?

Kyunki transfer angle

Δ θ

— aur isliye

β

ka sign — sirf do positions se fix nahi hota; same endpoints do arcs admit karte hain, aur har ek same

a

ke liye alag time deta hai.

WHY5. Lagrange

f, g

functions sirf do positions se velocity recover kyun karne dete hain?

Kyunki

r_{2} = f r_{1} + g v_{1}

(upar definition dekho) future state ko initial position aur velocity ke linear combination ke roop mein express karta hai; ek baar

a

(hence

f, g, \overset{g}{˙}

) known ho aur dono positions known hon, hum algebraically

v_{1}

phir

v_{2}

solve kar lete hain.

WHY6. Kisi bhi mission ke liye hum

a = a_{m i n}

pick karke kyun nahi khatam kar sakte?

Kyunki

a_{m i n}

ek specific flight time fix karta hai (minimum-energy time); alag required

Δ t

wale mission ko ek bada

a

chahiye, jo is tarah choose kiya jaaye ki Lambert's equation us

Δ t

ke barabar ho, minimum-energy wale ke nahi.

WHY7. Lambert's problem launch-window aur rendezvous planning ki backbone kyun hai?

Kyunki har aisa task "abhi yahan raho, ek set baad ke time par wahan jao" hota hai; Lambert us boundary condition ko required departure velocity aur hence

Δ v

mein convert karta hai, jo exactly woh hai jo fuel budgets aur porkchop plots ko chahiye.

Edge cases

EC1. Jab

Δ θ \to 0

ho (do points direction mein coincide karein, same radius) toh

c

aur

s

ka kya hoga?

Chord

c \to 0

aur

s \to r_{1}

; "space triangle" ek line mein degenerate ho jaata hai, aur Lambert ill-posed ban jaata hai — kisi bhi finite time ke liye ek full revolution (

N \geq 1

) chahiye.

EC2.

Δ θ = π

par (points diametrically opposite) kya hoga?

Agar

r_{1}

aur

r_{2}

exactly antiparallel hain toh transfer plane undefined hai, isliye short/long way ambiguous hai aur orbit normal sirf positions se fix nahi ho sakta; chord apne maximum,

c = r_{1} + r_{2}

tak pahunch jaata hai.

EC3. Agar requested

Δ t

kisi bhi elliptic transfer se chhota ho toh?

Koi ellipse kaam nahi karega; tum hyperbolic transfer mein chale jaate ho jahan

a < 0

hai, aur elliptic formula (real

α, β

ke saath) ko

sinh

use karte hue uske hyperbolic analog se replace karna hoga.

EC4. Parabolic case kya hai, aur

a < 0

a > s /2

classification mein yeh kahan fit hota hai?

Parabola exact boundary

a \to \infty

(zero orbital energy) hai; yeh geometry ke liye single fastest bound-adjacent transfer hai. Parabola se zyada lamba time elliptic hai (

s /2 \leq a < \infty

); chhota time hyperbolic hai (

a < 0

). Toh parabola do families ko alag karne wala hinge hai.

EC5. Agar

r_{1} = r_{2}

exactly ho toh?

Bilkul valid — geometry symmetric hai,

α, β

ab bhi

s

aur

s - c

se follow karte hain; yeh same-altitude transfers correspond karta hai (Hohmann-like family yahan ke paas rehti hai jab

Δ θ = π

EC6.

N \geq 1

(multi-revolution) maangne se solutions ki sankhya par kya asar padta hai?

Yeh branches multiply karta hai — har revolution count ke liye do aur solutions ho sakte hain (ek "low" aur ek "high"

a

), isliye long-duration transfers mein kaafi saare valid orbits ho sakte hain, aur tumhe

N

aur branch specify karna hoga.

EC7.

α = π

wale solution ka kya matlab hai?

Yeh minimum-energy orbit

a = s /2

hai; auxiliary angle

α

us geometry ke liye apne maximum par pahunch gaya hai, jo ek fixed short/long way ke liye do

a

-branches ke beech ki boundary mark karta hai.

Recall Ek-sentence summary

Lambert traps almost always ek branch, ek sign, ya ek lower bound bhoolne se aate hain — ek geometry ke liye multiple orbits, direction ke liye $β$ ka sign, aur $a \geq s /2$ — plus chord ko arc se confuse karna aur time-with- $a$ monotonicity ko.

See also: Kepler's Equation and Time of Flight · Lagrange Coefficients (f and g functions) · Universal Variable Formulation · Hohmann Transfer · Porkchop Plots and Launch Windows · Orbital Rendezvous and Targeting