Lambert's problem — connecting two positions in given time

3.2.28 · Physics › Orbital Mechanics & Astrodynamics

WHY hum yeh chahte hi kyun hain?

WHAT actually given hai aur kya unknown hai?

Woh magic fact jo ise solvable banata hai:

HOW time-of-flight equation derive karein

Hum classic Lagrange / universal-variable form geometry se build karenge.

Step 1 — Geometry: chord aur space triangle

Dono radii aur chord place karo:

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

Yeh step kyun? $Δ θ$ dono positions ke beech ka transfer angle hai; chord sirf triangle Earth– $r_{1}$ – $r_{2}$ par law of cosines hai. Short way $0 < Δ θ < π$ use karta hai, long way $π < Δ θ < 2 π$ use karta hai.

Uss space triangle ka semiperimeter define karo: $s = \frac{1}{2} (r_{1} + r_{2} + c)$

Kyun? Kyunki Lambert's theorem kehta hai time $r_{1} + r_{2}$ aur $c$ par depend karta hai — aur woh $s$ ke roop mein sabse naturally combine hote hain.

Step 2 — Auxiliary angles introduce karo

Ek elliptical transfer ( $a > 0$ ) ke liye hum do angles $α, β$ se parametrize karte hain: $sin^{2} \frac{α}{2} = \frac{s}{2 a}, sin^{2} \frac{β}{2} = \frac{s - c}{2 a}$

Yeh step kyun? Yeh arc ke across eccentric-anomaly difference ko geometry ke terms mein likhne se aata hai. $α, β$ exactly woh combinations hain eccentric anomalies ke dono ends par jo geometry fold hone ke baad bachte hain — yeh $a, c, s$ ko pure angles mein package karte hain.

Step 3 — Arc ke across Kepler's time law

Kepler's equation kehti hai ki do eccentric anomalies $E_{1}, E_{2}$ ke beech sweept "mean anomaly" hai: $\frac{μ}{a ^{3}} Δ t = (E_{2} - e sin E_{2}) - (E_{1} - e sin E_{1}) .$

Kyun? Yeh literally Kepler's equation $M = E - e sin E$ hai jo dono endpoints par apply ki gayi aur subtract ki gayi — mean anomaly time ke saath uniformly advance karta hai, toh uska change $n Δ t$ equal hota hai jahan $n = μ / a^{3}$ .

Half-angle identities substitute karke aur trig collapse karke (eccentricity $e$ $α, β$ combination mein drop ho jaata hai — yeh hi miracle hai), hume Lambert's time-of-flight equation milti hai:

Yeh kyun poora game hai: right side ek jaani hui function hai single unknown $a$ ki. Baaki sab ( $s, c, μ$ ) geometry hai. Toh Lambert reduce ho jaata hai: $a$ dhundho jis par yeh diya hua $Δ t$ equal kare, phir velocities reconstruct karo.

Step 4 — Minimum-energy orbit (ek key landmark)

Ek sabse choti ellipse hoti hai jo dono points ko bilkul connect kar sakti hai: $a_{m i n} = \frac{s}{2}$ $a_{m i n}$ par, $α = π$ , aur time of flight minimum-energy time hai — koi bhi faster ya slower transfer ek bade $a$ ki zaroorat hai (zyada energy). Yeh tumhe ek sanity check aur starting guess deta hai.

$a = s /2$ kyun? $sin (α /2) = s /2 a$ se, argument 1 se zyada nahi ho sakta, toh $2 a \geq s$ , yani $a \geq s /2$ . Equality sabse extreme (patli) allowed ellipse hai.

Step 5 — Velocities recover karo (Lagrange coefficients)

Jab $a$ mil jaaye, f aur g functions use karo velocity paane ke liye: $r_{2} = f r_{1} + g v_{1} \Rightarrow v_{1} = \frac{r _{2} - f r _{1}}{g}, v_{2} = \frac{g ˙ r _{2} - r _{1}}{g}$

Kyun? Lagrange coefficients future state ko initial position aur velocity ke linear combination ke roop mein express karte hain. Kyunki ab hum $f, g, \overset{g}{˙}$ jaante hain (jo $a, Δ θ, \dots$ ke functions hain) aur dono positions bhi, hum algebraically $v_{1}$ , phir $v_{2}$ solve karte hain. Yeh velocities woh answer hain jo mission planner chahta hai.

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

Worked Example 1 — Short-way elliptic transfer

Given (canonical units, $μ = 1$ ): $r_{1} = 1$ , $r_{2} = 1.5$ , $Δ θ = 9 0^{\circ}$ , $Δ t = 1.2$ .

Step A — chord & semiperimeter. $c = 1^{2} + 1. 5^{2} - 2 (1) (1.5) cos 9 0^{\circ} = 1 + 2.25 - 0 = 3.25 = 1.803$ . $s = \frac{1}{2} (1 + 1.5 + 1.803) = 2.152$ . Kyun: pehle geometry — baaki sab $s, c$ par depend karta hai.

Step B — minimum-energy check. $a_{m i n} = s /2 = 1.076$ . Uska time compute karo yeh dekhne ke liye ki hamare $Δ t$ ko $a$ upar ya neeche chahiye (yaad raho $a > a_{m i n}$ ke liye do branches hoti hain). Kyun: yeh root search ko frame karta hai aur non-physical $a$ lene se bachata hai.

Step C — Lambert's equation mein $a$ ke liye solve karo (numerically, jaise boxed equation par Newton) taaki RHS $= Δ t = 1.2$ ho. Maano yeh $a \approx 1.30$ deta hai. Kyun: $a$ single unknown hai; ek scalar root-find.

Step D — velocities $f, g$ se is $a$ aur $Δ θ$ ke saath. $v_{1}, v_{2}$ output karo; departure burn hai $Δ v = ∣ v_{1} - v_{park} ∣$ . Kyun: velocities → $Δ v$ → fuel, jo deliver karna hai.

Worked Example 2 — Transfer direction answer kyun badalta hai

Same $r_{1}, r_{2}$ lekin long way choose karo, $Δ θ = 27 0^{\circ}$ . $c = 1 + 2.25 - 2 (1.5) cos 27 0^{\circ} = 3.25 - 0 = 1.803$ (same chord!) — lekin arc ab bada side hai, toh $β$ negative branch leta hai: Lambert's equation mein $β \to - β$ use karo.

Yeh step kyun? Chord symmetric hai, lekin tum kis taraf jaate ho yeh change karta hai kitna area/angle sweep hota hai, isliye same $a$ ke liye alag $Δ t$ . $β$ ka sign short-vs-long way encode karta hai. Ise bhoolne par bilkul galat time aata hai.

Common Mistakes

Active Recall

Recall Feynman: ek 12-saal-ke bacche ko samjhao

Socho tum ek ball apne dost ko phenk rahe ho, aur tum yeh bhi kehte ho "ise exactly 2 seconds mein land karna chahiye." Actually alag alag tarike hain isse phenkhne ke — ek fast flat throw ya ek high loopy throw — aur sirf kuch hi exactly 2 seconds lete hain. Lambert's problem woh math hai jo kehta hai: "Yeh hai exact throw (kitni takkat se aur kis direction mein) taaki ball yahan se wahan exactly uss time mein jaaye." Space mein, gravity "throw" ko ek orbit mein curve karti hai, aur mission planners ise use karte hain ek moving planet ko time par hit karne ke liye.

Flashcards

What does Lambert's problem take as input and produce as output?

Input: do position vectors, time of flight,

μ

, transfer direction. Output: connecting conic orbit aur dono velocities

v_{1}, v_{2}

State Lambert's theorem.

Transfer time sirf

a

, chord

c = ∣ r_{2} - r_{1} ∣

, aur

r_{1} + r_{2}

par depend karta hai (equivalently semiperimeter

s

), arc ki individual shape par nahi.

Write the elliptic Lambert time-of-flight equation.

μ Δ t = a^{3/2} [(α - sin α) - (β - sin β)]

with

sin (α /2) = s /2 a

sin (β /2) = (s - c) /2 a

What is the minimum-energy semi-major axis and why is it a floor?

a_{m i n} = s /2

; kyunki

sin (α /2) = s /2 a \leq 1

force karta hai

2 a \geq s

Why does eccentricity drop out of Lambert's equation?

Jab Kepler's equation dono endpoints par apply hoti hai aur difference liya jaata hai, geometry

e

ko auxiliary angles

α, β

mein fold kar deti hai, time ko sirf

a, c, s

ka function chhod ke.

How are velocities recovered after finding

a

Lagrange coefficients ke zariye:

v_{1} = (r_{2} - f r_{1}) / g

aur

v_{2} = (\overset{g}{˙} r_{2} - r_{1}) / g

What does the sign of

β

select physically?

Transfer direction:

β > 0

short-way ke liye (

Δ θ < π

β < 0

long-way ke liye (

Δ θ > π

Formula for chord length between the two positions.

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

Why can there be multiple Lambert solutions?

Har

a

ke liye do focus placements, short/long-way choice, aur multi-revolution (

N

) solutions sab same endpoints+time satisfy karte hain.

Which classic equation supplies the time law used in the derivation?

Kepler's equation

M = E - e sin E

, jo dono endpoints par apply hoti hai aur difference li jaati hai.

Connections

Kepler's Equation and Time of Flight — woh time law jo Lambert arc ke across difference karta hai.
Lagrange Coefficients (f and g functions) — velocities reconstruct karne ke liye use hota hai.
Hohmann Transfer — Lambert ka ek special optimal case (tangential, $Δ θ = π$ ).
Universal Variable Formulation — Lambert solve karne ka unified (ellipse/parabola/hyperbola) tarika.
Porkchop Plots and Launch Windows — Lambert dates ke grid par solve hota hai.
Orbital Rendezvous and Targeting — Lambert intercept velocity deliver karta hai.