Orbital elements (Keplerian) — semi-major axis a, eccentricity e, inclination i, RAAN Ω, argument of perigee ω, true ano: Worked examples

3.2.8 · D3 · Physics › Orbital Mechanics & Astrodynamics › Orbital elements (Keplerian) — semi-major axis a, eccentrici

Yeh parent topic ka drill page hai. Parent ne tumhe chhe numbers aur formulas sikhaye. Yahaan hum un formulas par har tarah ka case daalte hain — har sign, har degenerate shape, angle ka har quadrant, plus ek real-world problem aur ek exam trap — aur har ek ko haath se grind karte hain.

Koi bhi symbol aane se pehle, woh toolkit yaad karo jo parent ne banayi thi (sab wahan define hai):

Recall Woh formulas jo hum baar baar use karenge (sab parent se hain)

Endpoints ke radii: perigee $r_{p} = a (1 - e)$ , apogee $r_{a} = a (1 + e)$ . Yahaan $a$ semi-major axis hai (lambe diameter ka aadha) aur $e$ eccentricity hai (kitna dabba hua hai, $0$ = circle).
$a, e$ recover karo: $a = \frac{r _{p} + r _{a}}{2}$ , $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}}$ .
Vis-viva: $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ , jahan $μ$ gravitational parameter hai (Earth ka "gravity strength number").
Orbit shape: $r = \frac{p}{1 + e cos ν}$ , semi-latus rectum $p = a (1 - e^{2})$ .
Angular momentum magnitude $h$ (area-sweeping rate $\times 2$ ) $p$ se $h = μ p$ ke zariye juda hai, isliye $p = h^{2} / μ$ .
True anomaly $ν$ = perigee se satellite tak ka angle, Earth ke centre par measure kiya hua.
Energy: $ε = - \frac{μ}{2 a}$ (negative = bound, zero = escape, positive = flyby).

Is poore page par $μ_{E} = 3.986 \times 1 0^{5} km^{3} / s^{2}$ aur Earth radius $R_{E} = 6378 km$ hai.

The scenario matrix

Is topic ka har problem in cells mein se ek hogi. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Kya khaas hai	Example
A. Standard ellipse	$0 < e < 1$ , seedhe numbers	Ex 1
B. Circle (degenerate $e = 0$ )	perigee = apogee, $ν$ meaningless	Ex 2
C. Escape (limiting $e \to 1$ )	$ε = 0$ , $a \to \infty$	Ex 3
D. Hyperbolic flyby ( $e > 1$ )	$a < 0$ , energy positive	Ex 4
E. Angle quadrant trap	$ν$ recover karo jahan sirf $cos$ ambiguous ho	Ex 5
F. Orientation angles: signs & quadrants	$i$ , $Ω$ ek vector se, saare sign cases	Ex 6
G. Real-world word problem	GEO / mission framing	Ex 7
H. Exam twist	do formulas milaata hai, ek hidden step	Ex 8

A. Standard ellipse

Worked example Ex 1 —

$a, e$ banao, phir har jagah speed (Cell A) Ek satellite ka perigee altitude $500$ km aur apogee altitude $2500$ km hai. $a$ , $e$ , aur dono apsides par speed nikalo.

Forecast: abhi andaaza lagao — perigee speed apogee speed se zyada hogi ya kam? Circular LEO ( $\approx 7.6$ km/s) se zyada ya kam?

Altitudes → radii. $r_{p} = 6378 + 500 = 6878$ km, $r_{a} = 6378 + 2500 = 8878$ km. Yeh step kyun? Elements distance measure karte hain focus (Earth ke centre) se, surface se nahi.
Semi-major axis. $a = \frac{r _{p} + r _{a}}{2} = \frac{6878 + 8878}{2} = 7878$ km. Yeh step kyun? Poora lamba axis $r_{p} + r_{a} = 2 a$ hota hai.
Eccentricity. $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{2000}{15756} = 0.1269$ . Yeh step kyun? Kyunki $r_{a} - r_{p} = 2 a e$ aur $r_{a} + r_{p} = 2 a$ hai, ratio mein $a$ cancel ho jaata hai aur $e$ reh jaata hai.
Perigee speed vis-viva se $r = r_{p}$ par: $v_{p}^{2} = 3.986 \times 1 0^{5} (\frac{2}{6878} - \frac{1}{7878})$ , jisse $v_{p} = 8.075$ km/s milti hai. Yeh step kyun? Vis-viva speed ko ek single position se sirf $a$ use karke jodhta hai; perigee sabse nazdik point hai isliye yahaan sabse zyada speed expect karte hain.
Apogee speed $r = r_{a}$ par: $v_{a}^{2} = 3.986 \times 1 0^{5} (\frac{2}{8878} - \frac{1}{7878})$ , jisse $v_{a} = 6.256$ km/s milti hai. Yeh step kyun? Same universal formula, ab sabse door point par — energy fixed hai, isliye bada $r$ matlab chota $v$ hona chahiye.

Verify: perigee sabse nazdik → sabse fast, apogee sabse door → sabse slow: $8.08 > 6.26$ . ✓ Aur $r_{p} v_{p} = 6878 \cdot 8.075 = 55, 540$ vs $r_{a} v_{a} = 8878 \cdot 6.256 = 55, 540$ — barabar (angular momentum $h = r v$ apsides par conserved). ✓

B. Circle — ek degenerate case

Jab $e = 0$ hota hai toh ellipse collapse hokar circle ban jaata hai. Do cheezein quietly toot jaati hain: perigee aur apogee ek hi point ban jaate hain, aur argument of perigee $ω$ aur true anomaly $ν$ apna matlab kho dete hain (koi "closest point" nahi hai jisse measure karein). Neeche ki figure mein yeh collapse dikhti hai — woh do kaale dots jo perigee aur apogee the ab focus se same distance par hain, aur laal loop poore chakkar mein same radius ka hai.

Figure — Orbital elements (Keplerian) — semi-major axis a, eccentricity e, inclination i, RAAN Ω, argument of perigee ω, true ano

Worked example Ex 2 — circular orbit

(Cell B) Ek satellite $780$ km altitude par perfect circle mein orbit karta hai. Uski speed nikalo aur check karo ki "perigee = apogee".

Forecast: circle mein, kya speed ghoomte waqt change hoti hai? $e$ kya hoga?

Radius. $r = 6378 + 780 = 7158$ km. Circle ke liye, $a = r$ hai ("lamba diameter" aur "chhota diameter" barabar hain). Yeh step kyun? Circle woh special ellipse hai jisme dono axes barabar hain, isliye $a = r_{p} = r_{a} = r$ .
Eccentricity. $r_{a} = r_{p}$ ⇒ $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = 0$ . Yeh step kyun? Dono ends mein zero difference matlab zero squash.
Speed vis-viva se $a = r$ ke saath: $v^{2} = μ (\frac{2}{r} - \frac{1}{r}) = \frac{μ}{r}$ . Toh $v = 3.986 \times 1 0^{5} /7158 = 7.462$ km/s. Yeh step kyun? Jab $a = r$ hota hai toh vis-viva simplify hokar famous circular-orbit speed $v = μ / r$ ban jaata hai.

Verify: kyunki $r$ kabhi nahi badlta, vis-viva har point par same $v$ deta hai — constant speed, jaisi circle demand karta hai. ✓ Units: $km^{3} s^{- 2} / km = km^{2} / s^{2} =$ km/s. ✓

C. Escape limit — $e \to 1$

Jaise $e$ $1$ ki taraf badhta hai orbit ek closed loop rehna band kar deta hai. Dekho formulas exactly $e = 1$ par kya karte hain: energy zero ho jaati hai aur $a$ infinity ki taraf bhaag jaata hai.

Worked example Ex 3 — parabolic escape speed

(Cell C, limiting case) $r = 7000$ km radius ke circular LEO se, escape speed kya hai (us point se guzarne wale parabolic $e = 1$ orbit ki speed)?

Forecast: escape speed vs same radius par circular speed — exactly kis factor ka?

Escape ki energy. Parabola ki total energy $ε = 0$ hoti hai (bas barely unbound). Yeh step kyun? Escape ka matlab hai "infinity tak pahuncho aur zero speed baaki bache": $ε = v^{2} /2 - μ / r \to 0 - 0 = 0$ .
$ε = 0$ set karo. $\frac{v _{esc}^{2}}{2} - \frac{μ}{r} = 0 \Rightarrow v_{esc} = \frac{2 μ}{r}$ . Yeh step kyun? $ε = 0$ hone par energy equation ko $v$ ke liye solve karna.
Number. $v_{esc} = 2 \cdot 3.986 \times 1 0^{5} /7000 = 10.673$ km/s.

Verify: yahaan circular speed hai $μ / r = 7.548$ km/s, aur $10.673/7.548 = 1.4142 = 2$ . ✓ Escape speed hamesha circular speed ka $2$ guna hoti hai — ek saaf check. Vis-viva terms mein $e = 1 \Rightarrow a = \infty \Rightarrow 1/ a = 0 \Rightarrow v^{2} = 2 μ / r$ , same result. ✓

D. Hyperbolic flyby — $e > 1$ aur negative $a$

Ab escape se aage jao. $e > 1$ ke liye "ellipse" hyperbola ban jaata hai, energy positive hoti hai, aur semi-major axis $a$ negative nikalta hai. Phir bhi kaam karta hai — bas sign rakhna hoga. Figure mein laal curve open hyperbolic path hai: yeh perigee (kaala dot) ke paas swing karke aata hai aur wapas infinity ki taraf nikal jaata hai, kabhi close nahi hota — yeh $e > 1$ unbound orbit ka visual signature hai.

Worked example Ex 4 — hyperbolic flyby speed

(Cell D, e>1, sign case) Ek probe $e = 1.5$ aur perigee radius $r_{p} = 8000$ km ke saath Earth ke paas se flyby karta hai. $a$ aur perigee par speed nikalo.

Forecast: kya $a$ positive hoga ya negative? Kya $v_{p}$ us radius par escape speed se zyada hogi?

Perigee abhi bhi $r_{p} = a (1 - e)$ follow karta hai. Toh $a = \frac{r _{p}}{1 - e} = \frac{8000}{1 - 1.5} = \frac{8000}{- 0.5} = - 16000$ km. Yeh step kyun? Perigee formula geometric hai aur kisi bhi conic ke liye hold karta hai; $e > 1$ ke saath factor $(1 - e)$ negative hota hai, $a < 0$ force karta hai. Negative $a$ unbound orbit ka flag hai.
Speed vis-viva se, $a$ ka sign rakhte hue. $v_{p}^{2} = μ (\frac{2}{r _{p}} - \frac{1}{a}) = 3.986 \times 1 0^{5} (\frac{2}{8000} - \frac{1}{- 16000})$ . Yeh step kyun? Vis-viva universal hai; $- 1/ a$ $+ 1/16000$ ban jaata hai, energy add hoti hai — yehi reason hai ki hyperbolic orbits zyada fast hote hain.
Number. $v_{p}^{2} = 3.986 \times 1 0^{5} (2.5 \times 1 0^{- 4} + 6.25 \times 1 0^{- 5}) = 3.986 \times 1 0^{5} (3.125 \times 1 0^{- 4})$ , toh $v_{p} = 11.163$ km/s.

Verify: $8000$ km par escape speed hai $2 μ /8000 = 9.983$ km/s. Hamara $v_{p} = 11.16 > 9.98$ ✓ — hyperbola ka escape speed se zyada hona zaroori hai. Energy $ε = - μ / (2 a) = - 3.986 \times 1 0^{5} / (2 \cdot (- 16000)) = + 12.46$ km²/s² positive hai ✓, unbound confirm karta hai. ✓

E. Angle-quadrant trap — $ν$ recover karna

Yeh classic pitfall hai. $r$ diya ho toh orbit equation tumhe $cos ν$ deta hai — lekin $cos$ nahi bata sakta "perigee se door ja raha hai" ( $0 < ν < 18 0^{\circ}$ ) ya "wapas aa raha hai" ( $18 0^{\circ} < ν < 36 0^{\circ}$ ), kyunki $cos ν = cos (- ν)$ . Quadrant pick karne ke liye tumhe ek aur piece of information chahiye. Figure mein, ek laal radius $ν = 10 4^{\circ}$ (outbound, axis ke upar) ki taraf point karta hai aur doosra $ν = 25 6^{\circ}$ (inbound, axis ke neeche) — dono same radius tak pahunchte hain, aur yahi woh ambiguity hai jo hume resolve karni hai.

Worked example Ex 5 — konsa

$ν$ ? (Cell E, quadrant ambiguity) Ex 1 se $a = 7878$ km, $e = 0.1269$ use karte hue, satellite $r = 8000$ km par hai. $ν$ nikalo — aur note karo ki do answers hain.

Forecast: ek radius par, orbit ke kitne points share karte hain? (Ek horizontal chord draw karo.)

Semi-latus rectum. $p = a (1 - e^{2}) = 7878 (1 - 0.01610) = 7751$ km. Yeh step kyun? $p$ orbit equation ka numerator hai.
Orbit equation ko $cos ν$ ke liye invert karo. $r = \frac{p}{1 + e cos ν}$ se: $cos ν = \frac{1}{e} (\frac{p}{r} - 1) = \frac{1}{0.1269} (\frac{7751}{8000} - 1) = \frac{- 0.03113}{0.1269} = - 0.2453$ . Yeh step kyun? Algebraically $cos ν$ ke liye solve karo; bracket ka negative sign batata hai ki hum $9 0^{\circ}$ se aage hain.
DONO angle solutions list karo. Equation $cos ν = - 0.2453$ ke $[0^{\circ}, 36 0^{\circ})$ par do solutions hain: ek principal $ν_{1} = arccos (- 0.2453) = 104.2 0^{\circ}$ , aur iska mirror major axis ke neeche $ν_{2} = 36 0^{\circ} - 104.2 0^{\circ} = 255.8 0^{\circ}$ . Yeh step kyun? $cos$ even hai, isliye koi bhi value axis ke upar ek angle aur neeche ek angle deta hai — kabhi mat maano ki akela $arccos$ answer hai.
Radial velocity sign se quadrant resolve karo. $r$ ke change ki rate $\overset{r}{˙} = \frac{μ}{h} e sin ν$ hai (positive = bahar ja raha hai, negative = andar aa raha hai), jahan $h = μ p$ upar recall box se angular-momentum magnitude hai — is orbit ke liye ek fixed positive constant. Toh $\overset{r}{˙}$ ka sign sirf $sin ν$ ka sign hai. Poori tarah unambiguous recipe hai $ν = atan2 (sin ν, cos ν)$ (yaad karo: numerator = $sin ν$ , denominator = $cos ν$ ), jahan $sin ν = + 1 - cos^{2} ν$ agar outbound ( $\overset{r}{˙} > 0$ ) aur $- 1 - cos^{2} ν$ agar inbound ( $\overset{r}{˙} < 0$ ) ho. Yeh step kyun? Kyunki $h > 0$ aur $e > 0$ , sirf $sin ν$ $\overset{r}{˙}$ ka sign carry karta hai. Dono $sin ν$ aur $cos ν$ ko $atan2$ mein feed karne par seedha correct quadrant milta hai — outbound $+ 104.2 0^{\circ}$ deta hai, inbound $- 104.2 0^{\circ} \equiv 255.8 0^{\circ}$ deta hai. Manual case-splitting ki zaroorat nahi.

Verify: dono ko $r = p / (1 + e cos ν)$ mein plug karo. $104.2 0^{\circ}$ ke liye: $cos = - 0.2453$ , $r = 7751/ (1 - 0.03113) = 8000$ km ✓. $255.8 0^{\circ}$ ke liye: $cos (255.8 0^{\circ}) = - 0.2453$ bhi hai, same $r = 8000$ km ✓. Dono radii match karte hain — prove karta hai ki ambiguity real hai, koi mistake nahi, aur $atan2$ / $\overset{r}{˙}$ sign hi tie breaker hai.

F. Orientation angles — $i$ aur $Ω$ ke saare signs

Tilt $i$ (0°–180°) aur swing $Ω$ (0°–360°) angular-momentum vector $h$ se aate hain. Inhe sahi karne ka matlab hai saare quadrants aur dono hemispheres handle karna. 3D figure geometry fix karta hai: kaala arrow Earth ki spin axis $Z$ (North) hai, faint plane equator hai, aur laal arrow $h$ hai — inke beech angle $i$ yehi inclination hai.

Worked example Ex 6 —

$h$ se inclination & RAAN, saare cases (Cell F, signs/quadrants) Specific angular-momentum vector $h = (h_{x}, h_{y}, h_{z})$ diya ho, $i$ aur $Ω$ nikalo. Phir teen vectors evaluate karo: (a) $h_{a} = (0, 0, 60000)$ , (b) $h_{b} = (30000, - 30000, 42426)$ , (c) $h_{c} = (0, 0, - 60000)$ . Units km²/s.

Forecast: inme se kaun equatorial hai, kaun retrograde, kaun prograde-inclined?

$z$ -component se inclination. $cos i = \frac{h _{z}}{∣ h ∣}$ , toh $i = arccos (h_{z} /∣ h ∣)$ , range $0$ – $18 0^{\circ}$ . Yeh step kyun? $i$ $h$ aur $Z$ -axis (North) ke beech ka angle hai. $arccos$ over $[0, 18 0^{\circ}]$ ko koi quadrant fix nahi chahiye — range already full inclination range hai.
Node vector. $n = \hat{Z} \times h = (- h_{y}, h_{x}, 0)$ equatorial plane mein ascending node ki taraf point karta hai. ( $\hat{Z}$ notation legend se $Z$ ke along unit vector hai.) Yeh step kyun? RAAN us plane mein measure hota hai; $n$ line of nodes hai.
Quadrant fix ke saath RAAN. $cos Ω = n_{x} /∣ n ∣$ ; phir agar $n_{y} < 0$ ho, $Ω = 36 0^{\circ} - arccos (\cdot)$ . Yeh step kyun? Bilkul parent ke style mein $arctan$ jaisa: sirf $cos Ω$ east swing aur west swing mein fark nahi kar sakta. Node ka $n_{y}$ sign (node ka $Y$ -component) correct half-circle pick karta hai.

Ab teen cases:

(a) $h_{a} = (0, 0, 60000)$ : $cos i = 60000/60000 = 1 \Rightarrow i = 0^{\circ}$ — equatorial, prograde. Node $n = (- h_{y}, h_{x}, 0) = (0, 0, 0)$ : node undefined hai (plane kabhi equator cross nahi karta), isliye $Ω$ meaningless hai — ek degenerate case (bilkul waise jaisa circle ke liye $ν$ mar jaata hai).
(b) $h_{b} = (30000, - 30000, 42426)$ : $∣ h ∣ = 3000 0^{2} + 3000 0^{2} + 4242 6^{2} = 60000$ , toh $cos i = 42426/60000 = 0.7071 \Rightarrow i = 45.0 0^{\circ}$ — inclined, prograde. Node $n = (- h_{y}, h_{x}, 0) = (30000, 30000, 0)$ , $∣ n ∣ = 42426$ , $cos Ω = 30000/42426 = 0.7071 \Rightarrow arccos = 4 5^{\circ}$ ; yahaan $n_{y} = 30000 > 0$ hai isliye koi flip nahi: $Ω = 45.0 0^{\circ}$ .
(c) $h_{c} = (0, 0, - 60000)$ : $cos i = - 60000/60000 = - 1 \Rightarrow i = 18 0^{\circ}$ — equatorial, retrograde ("galat" direction mein ja raha hai, $h$ South ki taraf point karta hai). Node $n = (0, 0, 0)$ phir se undefined, isliye $Ω$ yahan bhi meaningless hai.

Verify: (a) $i = 0$ ✓ equatorial; (b) $i = 4 5^{\circ}$ , $Ω = 4 5^{\circ}$ ✓; (c) $i = 18 0^{\circ}$ ✓ retrograde. Teeno cases $i = 0$ , $0 < i < 9 0^{\circ}$ , aur $i = 18 0^{\circ}$ cover karte hain — tilt ki poori sign story, plus dono degenerate ( $Ω$ undefined) endpoints. ✓

G. Real-world word problem

Worked example Ex 7 — geostationary orbit design karna

(Cell G, real world) Ek comms satellite ko equator par ek jagah hover karna hai, isliye uska orbital period ek sidereal day $T = 86164$ s ke barabar hona chahiye. Required semi-major axis $a$ nikalo, phir uski altitude aur (circular) speed. "Ek jagah hover karna" $e$ aur $i$ ko kya force karta hai?

Forecast: roughly kitna zyada high — LEO jaisa hundreds of km, ya tens of thousands?

Kepler's third law. $T = 2 π a^{3} / μ \Rightarrow a = (\frac{μ T ^{2}}{4 π ^{2}})^{1/3}$ . Yeh step kyun? Period sirf size se fix hota hai two-body problem mein — yeh woh tool hai jo "ek din" ko distance mein convert karta hai.
Number. $a = (\frac{3.986 \times 1 0 ^{5} \cdot 8616 4 ^{2}}{4 π ^{2}})^{1/3} = 42241$ km. Yeh step kyun? Direct substitution.
Altitude & speed. Altitude $= a - R_{E} = 42241 - 6378 = 35863$ km. Circular speed $v = μ / a = 3.986 \times 1 0^{5} /42241 = 3.072$ km/s. Yeh step kyun? "Altitude" matlab surface se height, isliye Earth radius ghataate hain; aur kyunki orbit circular hai ( $e = 0$ , agla step) vis-viva collapse hokar $v = μ / a$ har point par ban jaata hai.
Forced elements. Ek fixed point ke upar rehne ke liye north–south drift nahi honi chahiye (⇒ $i = 0^{\circ}$ , equatorial) aur Earth ki spin se match karte hue constant angular rate par chalna chahiye (⇒ $e = 0$ , circular). Yeh step kyun? Koi bhi $e > 0$ perigee ke paas speed up karega aur apogee ke paas slow down (Kepler's 2nd law), isliye ground track east–west swing karega; koi bhi $i > 0$ plane tilt karega aur ground par north–south figure-8 trace karega. Sirf $e = 0, i = 0$ satellite ko ek point ke upar rokta hai.

Verify: famous GEO altitude $\approx 35786$ km hai; hamara $35863$ km $0.2%$ se match karta hai (difference exact sidereal-day value aur used $μ$ ka hai). ✓ Speed $\approx 3.07$ km/s textbook GEO speed hai. ✓

H. Exam twist

Worked example Ex 8 — do formulas, ek hidden step

(Cell H, exam trap) Ek orbit mein $a = 10000$ km aur $e = 0.6$ hai. Us point par jahan satellite ki speed same radius ke liye circular speed ke barabar ho, $r$ aur $ν$ nikalo.

Forecast: kya aisa koi point exist karta hai, aur kya yeh apogee se pehle hai ya baad mein? (Trap: yeh apogee ya perigee nahi hai.)

Dono speeds likho. Actual: $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ . Local circular: $v_{c}^{2} = \frac{μ}{r}$ . Yeh step kyun? Condition $v = v_{c}$ vis-viva ko circular formula se jodti hai — woh hidden bridge jo examiner test kar raha hai.
Barabar set karo, $r$ ke liye solve karo. $\frac{2}{r} - \frac{1}{a} = \frac{1}{r} \Rightarrow \frac{1}{r} = \frac{1}{a} \Rightarrow r = a = 10000$ km. Yeh step kyun? $μ$ cancel ho jaata hai; algebra collapse hokar elegant fact tak pahunchta hai ki $v = v_{c}$ exactly jab $r = a$ ho (minor axis ke ends).
Orbit equation se $ν$ nikalo. $cos ν = \frac{1}{e} (\frac{p}{r} - 1)$ jahan $p = a (1 - e^{2}) = 10000 (1 - 0.36) = 6400$ km. Toh $cos ν = \frac{1}{0.6} (\frac{6400}{10000} - 1) = \frac{- 0.36}{0.6} = - 0.6 = - e$ . Yeh step kyun? Orbit equation ko invert karna radius ko angle mein convert karta hai; $cos ν = - e$ minor-axis endpoints ka clean signature hai.
Quadrant. $arccos (- 0.6) = 126.8 7^{\circ}$ , aur symmetry se $233.1 3^{\circ}$ bhi (do minor-axis ends). Yeh step kyun? Ex 5 ki tarah, $cos$ do solutions deta hai — major axis ke ek taraf ek aur doosri taraf ek — aur dono physically valid crossings hain.

Verify: $r = 10000 = a$ par, actual $v^{2} = μ (2/10000 - 1/10000) = μ /10000$ , aur circular $v_{c}^{2} = μ /10000$ — identical ✓. Aur $cos ν = - e = - 0.6$ ko $r = p / (1 + e cos ν) = 6400/ (1 - 0.36) = 10000$ mein plug karo ✓. Trap yeh tha ki log maante hain yeh point apogee hai; asliyat mein yeh minor-axis crossing hai $ν \approx 12 7^{\circ}$ par. ✓

Recall Self-test: cell match karo

Circle ka $e = 0$ hota hai aur speed ::: $v = μ / r$ hoti hai, har jagah constant. Parabolic escape speed circular speed se kitne factor se relate hoti hai ::: $2$ . Hyperbola ke liye semi-major axis $a$ hota hai ::: negative, aur energy $ε$ positive hoti hai. Sirf $cos ν$ diya ho, sahi $ν$ pick karne ke liye tumhe yeh bhi chahiye ::: $sin ν$ ka sign (radial-velocity direction), best done with atan2. Node vector se $Ω$ nikaalte waqt $18 0^{\circ}$ se aage flip karte ho jab ::: node ka $Y$ -component $n_{y} < 0$ ho. Woh point jahan orbital speed local circular speed ke barabar ho ::: $r = a$ par, yaani minor-axis ends, $cos ν = - e$ .

Dekho: Kepler's equation and mean anomaly $ν$ ko time mein convert karne ke liye, State vectors to orbital elements poori $r, v \to$ elements pipeline ke liye, Perifocal coordinate frame aur Angular momentum vector h Ex 6 ke peeche ki geometry ke liye, aur Eccentricity vector $e$ aur $ω$ sign-clean tarike se paane ke liye.