Lambert's problem — connecting two positions in given time: Visual walkthrough

3.2.28 · D2 · Physics › Orbital Mechanics & Astrodynamics › Lambert's problem — do positions ko given time mein connect karna

Step 1 — Do points draw karo aur wo jo triangle banate hain

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

Figure dekho. Teen labelled cheezein:

$r_{1}$ — Sun $F$ se start point $P_{1}$ tak ka arrow. Iski length $r_{1} = ∣ r_{1} ∣$ hai.
$r_{2}$ — $F$ se arrival point $P_{2}$ tak ka arrow. Iski length $r_{2} = ∣ r_{2} ∣$ hai.
$Δ θ$ — un do arrows ke beech Sun par ka angle. Yeh transfer angle hai: spacecraft sky mein kitna far around travel karta hai.

$P_{1}$ se $P_{2}$ tak ki straight line chord hai — shortcut across, flown path nahi:

$c = ∣ r_{2} - r_{1} ∣ = triangle par law of cosines r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos Δ θ$

Step 2 — Teen lengths ko semiperimeter mein bundle karo

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

Step 3 — Do angles $α$ aur $β$ se miliye

$sin \frac{α}{2} = \frac{s}{2 a}, sin \frac{β}{2} = \frac{s - c}{2 a}$

Intuition Yeh do equalities kahan se aati hain (right-triangle picture)

Yeh guess nahi ki gayi — yeh ellipse ki apni geometry se nikli hain. Kisi bhi ellipse par, focus se ek point ki distance $r = a (1 - e cos E)$ hai, jahan $E$ us point ki eccentric anomaly (uski "clock angle") hai. Ise dono ends par likho aur chord $c$ bhi do points ke coordinates $(a cos E, b sin E)$ use karke likho. Algebra grind karo (ek standard half-angle regrouping) aur wo ban jaata hai $r_{1} + r_{2} = 2 a (1 - cos \frac{α}{2} cos \frac{β}{2}), c = 2 a sin \frac{α}{2} cos \frac{β}{2} (short-way sign)$ jahan $α = E_{2} + E_{1}$ aur $β = E_{2} - E_{1}$ . Inn dono ko add aur subtract karo: $s = \frac{1}{2} (r_{1} + r_{2} + c) = 2 a sin^{2} \frac{α}{2}, s - c = 2 a sin^{2} \frac{β}{2},$ $1 - cos x = 2 sin^{2} \frac{x}{2}$ use karke. Square roots lo aur tum exactly do boxed equalities par pahunch jaate ho. Figure har ek ko ek right triangle ki tarah draw karta hai: vertical leg $s$ (ya $s - c$ ), hypotenuse $2 a$ , isliye enclosed angle hai $α /2$ (ya $β /2$ ) aur uska sine hai $s /2 a$ (ya $(s - c) /2 a$ ).

Step 4 — Kepler's clock, dono ends par apply karo aur subtract karo

$= n \frac{μ}{a ^{3}} Δ t = (E_{2} - e sin E_{2}) - (E_{1} - e sin E_{1})$

Step 5 — Miracle: eccentricity cancel ho jaati hai

Intuition Eccentricity kyun cancel hoti hai (crucial algebra, shown)

Step 4 ka right side lo, $(E_{2} - E_{1}) - e (sin E_{2} - sin E_{1})$ . Pehla bracket sirf $β$ hai. Doosre ke liye, sum-to-product identity use karo $sin E_{2} - sin E_{1} = 2 cos \frac{E _{2} + E _{1}}{2} sin \frac{E _{2} - E _{1}}{2} = 2 cos \frac{α}{2} sin \frac{β}{2}$ . Ab $e cos \frac{α}{2}$ ki value chahiye — aur Step 3 ki geometry humein deti hai. $r_{1} + r_{2} = 2 a (1 - cos \frac{α}{2} cos \frac{β}{2})$ ko $r_{1} r_{2}$ relations ke saath combine karne par, ellipse identity reduce hoti hai $e cos \frac{α}{2} = cos \frac{β}{2}$ — $e cos \frac{α}{2}$ ki ek clean, geometry-only value; eccentricity pinned hai, free nahi. Ise substitute karo: $e (sin E_{2} - sin E_{1}) = 2 (e cos \frac{α}{2}) sin \frac{β}{2} = 2 cos \frac{β}{2} sin \frac{β}{2} = sin β .$ $e$ gayab ho gaya, $sin β$ mein swallow ho gaya. $α$ piece par same bookkeeping karne se $sin α$ milta hai, aur poora right side collapse ho jaata hai $(α - sin α) - (β - sin β)$ mein.

$\frac{μ}{a ^{3}} Δ t = β (E_{2} - E_{1}) - = s i n β via e c o s \frac{α}{2} = c o s \frac{β}{2} e (sin E_{2} - sin E_{1}) ⟶ (α - sin α) - (β - sin β)$

Step 6 — Sabse chhoti possible orbit (degenerate edge)

Sine kabhi $1$ se zyada nahi ho sakti. Isliye $sin (α /2) = s /2 a$ mein humein chahiye $\frac{s}{2 a} \leq 1 ⟹ a \geq \frac{s}{2} .$

Sabse chhoti allowed value minimum-energy semi-major axis hai: $a_{m i n} = \frac{s}{2}, jahan sin \frac{α}{2} = 1 \Rightarrow α = π .$

Step 7 — Which way around? Branches: $β$ sign, $α$ branch, aur multi-rev

Definition Exact boundary

$Δ θ = π$ (transfer angle of 180°) Right at $Δ θ = 18 0^{\circ}$ do radii opposite directions mein point karte hain, isliye $P_{1}$ , $F$ , $P_{2}$ seedha line up ho jaate hain aur chord focus se guzarti hai. Yahan:

$c = r_{1} + r_{2}$ (chord poora straight span hai), isliye $s - c = \frac{1}{2} (r_{1} + r_{2} + c) - c = \frac{1}{2} (r_{1} + r_{2} - c) = 0$ .
Phir $sin (β /2) = (s - c) /2 a = 0 \Rightarrow β = 0$ , isliye $β$ -sign ambiguity gayab ho jaati hai — short aur long way yahaan milte hain, dono $β = 0$ dete hain.
Transfer plane sirf do radii se undefined hai (wo collinear hain, isliye infinitely many planes unhe contain karti hain). Tumhe plane (ya angular-momentum direction) alag se supply karna hoga. Yeh classic $18 0^{\circ}$ singularity hai — solvers $∣ s - c ∣ \to 0$ detect karte hain aur user-chosen plane par switch karte hain.

Definition

$α$ -branch rule (ek $a$ ke liye do ellipses) Ek given $a > a_{m i n}$ ke liye, equation $sin \frac{α}{2} = s /2 a$ ke $[0, 2 π]$ mein do solutions hain: principal wala $α_{0} = 2 arcsin s /2 a$ (jo $[0, π]$ mein hai), aur uska supplement $α = 2 π - α_{0}$ .

$α = α_{0}$ use karo jab transfer ka swept angle arc ke far point (focus geometry ke relative) ko near side par rakhe — lower-time solution.
$α = 2 π - α_{0}$ use karo other ellipse ke liye (focus chord ke opposite side par) — higher-time solution.

Switch kyun hota hai: dono ellipses ka same $a$ hai aur dono points se guzarti hain, lekin unka empty focus chord ke opposite sides par hai. $arcsin$ hamesha sirf $[0, π]$ mein principal value return karta hai, isliye woh dono mein se sirf ek ko name kar sakta hai — doosra $2 π - α_{0}$ ke roop mein recover hota hai. Tum choose karte ho is basis par ki do branches kaunsa $Δ t$ dete hain: wo minimum-time transfer ko $a_{m i n}$ par bracket karte hain.

Figure dono arcs ko ek green chord share karte dikhata hai, long-way arc opposite side bulge karta hua aur $β$ negative marked hai.

Worked check — short-way numbers line up karte hain

Canonical units, $μ = 1$ : $r_{1} = 1$ , $r_{2} = 1.5$ , $Δ θ = 9 0^{\circ}$ .

Chord: $c = 1 + 2.25 - 0 = 3.25 \approx 1.8028$ .
Semiperimeter: $s = \frac{1}{2} (1 + 1.5 + 1.8028) = 2.1514$ .
Minimum-energy: $a_{m i n} = s /2 = 1.0757$ .
$a = 1.30$ par (principal $α$ -branch): $sin (α /2) = 2.1514/2.6 = 0.9096 \Rightarrow α = 2.2810$ rad; $sin (β /2) = 0.3486/2.6 = 0.3662 \Rightarrow β = 0.7503$ rad. Phir $Δ t = a^{3/2} [(α - sin α) - (β - sin β)] \approx 1.06.$

Isliye direct short-way, principal-branch transfer at $a = 1.30$ deta hai $Δ t \approx 1.06$ . Parent note ka target $Δ t = 1.2$ hit karne ke liye tum $a$ nudge karte ho (Step 5 ka root-find); value near $a \approx 1.25$ land karti hai. Point yeh hai: ek scalar $a$ time tune karta hai. (Recovered $v_{1}, v_{2}$ phir Lagrange Coefficients (f and g functions) se follow karte hain, Orbital Rendezvous and Targeting aur Porkchop Plots and Launch Windows ko feed karte hain.)

Ek-picture summary

Recall Feynman retelling — plain words mein kaho

Main Sun ko ek point $F$ par rakhta hoon aur do arrows draw karta hoon jahan main start karta hoon ( $P_{1}$ ) aur jahan mujhe end karna hai ( $P_{2}$ ). Wo do arrow-tips plus Sun ek flat triangle banate hain. Main lambi sides measure karta hoon ( $r_{1}, r_{2}$ ) aur shortcut across ( $c$ ), aur main inhe sab add karta hoon aur aadha karta hoon ek tidy number $s$ paane ke liye. Ab main ellipse ka ek size guess karta hoon jisme main fly karunga — iski half-width $a$ kaho. Ellipse ki apni geometry se (focus tak distance $a (1 - e cos E)$ hai) main do angles build karta hoon: $α$ , do endpoint clock-angles $E_{1} + E_{2}$ ka sum, aur $β$ , unka difference $E_{2} - E_{1}$ , aur main paata hoon $sin \frac{α}{2} = s /2 a$ aur $sin \frac{β}{2} = (s - c) /2 a$ . Phir main trip ke dono ends par Kepler's steady clock use karta hoon aur subtract karta hoon, isliye sirf time difference bachta hai — actual flight time, $n Δ t$ ke barabar. Jab main sum aur difference angles substitute karta hoon aur $sin E_{2} - sin E_{1} = 2 cos \frac{α}{2} sin \frac{β}{2}$ ko geometry fact $e cos \frac{α}{2} = cos \frac{β}{2}$ ke saath use karta hoon, eccentricity cancel ho jaati hai aur main $μ Δ t = a^{3/2} [(α - sin α) - (β - sin β)]$ par pahunch jaata hoon. Right side sirf ek number hai jo meri guess $a$ par depend karta hai, isliye main $a$ adjust karta rehta hoon jab tak time target na aa jaaye. Agar main sabse sasti trip chahta hoon to main thinnest legal ellipse use karta hoon, $a = s /2$ . Agar main long way round fly karta hoon to main $β$ ka sign flip karta hoon; agar transfer angle exactly $18 0^{\circ}$ hai to $β = 0$ hai aur mujhe plane hand-pick karna hoga; agar main same size ki doosri ellipse chahta hoon to main $α$ ko $2 π - α$ se swap karta hoon; aur agar mere paas bahut time hai to main $2 π N$ ke saath whole loops add kar sakta hoon. Yeh sab ek bound elliptic orbit ke liye hai — parabola ya hyperbola in sines ko unke straight-line ya hyperbolic cousins se swap kar deta. Orbit nikalti hai — aur us se velocities aur fuel.

Recall Test yourself

Lambert's time of flight kin teen scalars par depend karta hai? ::: $a$ , chord $c$ , aur $r_{1} + r_{2}$ (equivalently $s$ ) — eccentricity par kabhi alag se nahi. $α, β$ eccentric anomalies se kaise relate karte hain? ::: $α = E_{2} + E_{1}$ (sum) aur $β = E_{2} - E_{1}$ (difference). Kaunsa geometry fact $e$ cancel karta hai? ::: $e cos \frac{α}{2} = cos \frac{β}{2}$ , jo $e (sin E_{2} - sin E_{1})$ ko $sin β$ mein turn karta hai. $a \geq s /2$ kyun? ::: Kyunki $sin (α /2) = s /2 a$ aur sine 1 se exceed nahi kar sakti, isliye $s /2 a \leq 1$ forced hai. Formula mein short-way ko long-way se kya differentiate karta hai? ::: $β$ ka sign: $Δ θ < π$ ke liye positive, $Δ θ > π$ ke liye negative flip kiya. $Δ θ = 18 0^{\circ}$ par kya hota hai? ::: $s - c = 0$ , isliye $β = 0$ aur transfer plane undefined ho jaata hai — tumhe ise supply karna hoga. Multi-revolution solutions kaise enter karte hain? ::: $N$ extra full loops ke liye bracket ke andar $2 π N$ add karo; har $N$ ke do $α$ -branches hain.