Converting between orbital elements and state vectors (r, v): Worked examples

3.2.17 · D3 · Physics › Orbital Mechanics & Astrodynamics › Converting between orbital elements and state vectors (r, v)

Pehle, har symbol ko kamao

Inhe hum scenario matrix ki pehli line pe hi use karenge, isliye inhe sab abhi build karte hain — koi bhi symbol is list se pehle nahi aayega.

Definition Woh symbols jo hum use karenge

$r$ — position vector: Earth ke centre se spacecraft tak ka ek arrow, km mein measure kiya hua.
$v$ — velocity vector: ek arrow jiska length speed hai aur jiska direction motion hai, km/s mein.
$r = ∣ r ∣$ , $v = ∣ v ∣$ — un arrows ki seedhi lengths (koi direction nahi, sirf size).
$μ = GM$ — central body ki "gravitational strength." Earth ke liye $μ = 398600 km^{3} / s^{2}$ .
$h = r \times v$ — cross product: ek arrow jo $r$ aur $v$ dono ke perpendicular hai, isliye woh orbit plane se seedha bahar nikalti hai. Iska length $h$ batata hai ki area kitni tezi se sweep ho raha hai. Iska teesra component $h_{z}$ woh "kitna north ki taraf jhuka hai" wala part hai.
$ε = \frac{v ^{2}}{2} - \frac{μ}{r}$ — energy per kilogram. Negative = bound (ellipse), zero = parabola, positive = escaping (hyperbola).
$\overset{z}{^} = (0, 0, 1)$ — fixed frame ka "north" reference direction; reference plane flat $x y$ -plane hai (Earth ka equator).
$i$ — inclination: $h$ aur $\overset{z}{^}$ ke beech ka tilt angle, yaani orbit plane equator se kitna steep away lean karta hai. Yeh $0^{\circ}$ (flat, equator mein) se $18 0^{\circ}$ (flat lekin "ulti" taraf ja raha, retrograde) tak chalta hai. Milta hai $cos i = h_{z} / h$ se.
$n = \overset{z}{^} \times h$ — node vector: ek arrow jo equator mein lie karta hai, ascending node ki taraf point karta hai (woh spot jahan orbit equator ko north ki taraf cross karta hai). Iska length $n = ∣ n ∣$ hai, aur iske components $n_{x}, n_{y}$ hain. Agar orbit equator mein flat ho, toh $h ∥ \overset{z}{^}$ aur $n$ zero pe collapse kar jaata hai.
$e$ — eccentricity vector: ek arrow jo Earth se periapsis ki taraf point karta hai (orbit ka sabse kareeb wala point). Iska length $e = ∣ e ∣$ eccentricity hai (shape). Iska teesra component $e_{z}$ hai. Formula: $e = \frac{1}{μ} [(v^{2} - μ / r) r - (r \cdot v) v]$ .
$Ω$ — RAAN: equator mein $x$ -axis se node vector $n$ tak ka angle.
$ω$ — argument of periapsis: node $n$ se periapsis $e$ tak ka angle, orbit plane mein measure kiya hua.
$ν$ — true anomaly: periapsis $e$ se current position $r$ tak ka angle.
$r \cdot v$ — dot product: ek single number jo positive hota hai jab $r$ aur $v$ ek-doosre ke "saath" point karte hain (satellite door ja raha, $r$ badh raha hai) aur negative jab "against" ( $r$ chhota ho raha, andar gir raha hai). Yeh $ν$ ka tie-breaker hai.
$p$ — semi-latus rectum: orbit ka "width scale," $p = a (1 - e^{2}) = h^{2} / μ$ . Yeh radius $r$ hai jab $ν = 9 0^{\circ}$ , aur $a$ ke unlike yeh parabola ke liye bhi finite rehta hai.
$u = ω + ν$ — argument of latitude: ascending node se seedha satellite tak ka angle. Tab use hota hai jab $e = 0$ se $ω$ mar jaata hai.
$ϖ = Ω + ω$ — true longitude of periapsis: fixed $x$ -axis se measure kiya hua periapsis direction. Tab use hota hai jab $i = 0$ se $Ω$ mar jaata hai.
$ℓ$ — true longitude: $x$ -axis se $r$ ka angle. Tab use hota hai jab orbit dono circular aur equatorial ho.

Definition Rotation matrix

$\tilde{R}$ (perifocal → inertial) Ek perifocal vector ko inertial mein badalne ke liye hum ek combined matrix lagate hain. Hum compact shorthand $c θ = cos θ$ aur $s θ = sin θ$ use karte hain; jab $θ$ hamare elements mein se ek ho toh yeh padhta hai $c O = cos Ω$ , $s O = sin Ω$ , $c w = cos ω$ , $s w = sin ω$ , aur $c i = cos i$ , $s i = sin i$ — yaani $c i$ aur $s i$ wahi shorthand hai inclination $i$ pe laagoo ki hui.

c\Omega\,c\omega - s\Omega\,s\omega\,ci & -c\Omega\,s\omega - s\Omega\,c\omega\,ci & s\Omega\,si\\ s\Omega\,c\omega + c\Omega\,s\omega\,ci & -s\Omega\,s\omega + c\Omega\,c\omega\,ci & -c\Omega\,si\\ s\omega\,si & c\omega\,si & ci \end{bmatrix}$$ Phir $\vec r=\tilde R\,\vec r_{pqw}$ aur $\vec v=\tilde R\,\vec v_{pqw}$. Rotation hone ki wajah se $\tilde R$ kabhi bhi kisi vector ki length nahi badalta — ek fact jo hum Ex 6 mein sanity-check ke liye use karte hain.

Scenario matrix

Ab jab har symbol define ho gaya hai, hum list karte hain kya-kya vary ho sakta hai taaki kuch bhi skip na ho. Har row ek "cell" of behaviour hai; worked examples un cells ke saath tagged hain jo woh hit karte hain.

#	Cell (jo cheez test ho rahi hai)	Kyun dangerous hai	Covered by
C1	True anomaly $ν$ doosri half mein ( $r \cdot v < 0$ )	$cos^{- 1}$ chhupaata hai; $r \cdot v$ ke sign se flip karna padta hai	Ex 1
C2	RAAN $Ω$ jab $n_{y} < 0$ (node past $18 0^{\circ}$ )	$Ω$ par wahi $cos^{- 1}$ trap	Ex 2
C3	Argument of periapsis $ω$ jab $e_{z} < 0$	$ω$ par wahi trap	Ex 2
C4	Degenerate: circular non-equatorial orbit ( $e = 0$ , $i \neq = 0$ ) → $ω$ undefined	$e = 0$ se division; argument of latitude $u$ use karo	Ex 3
C5	Degenerate: equatorial orbit ( $i = 0$ ) → $Ω$ undefined	node vector $n \to 0$	Ex 4
C6	Limiting: hyperbola ( $e > 1$ , $ε > 0$ , $a < 0$ )	$a$ ka sign flip ho jaata hai; panic mat karo	Ex 5
C7	Elements → State with $ν$ off periapsis (full rotation)	$\tilde{R}$ matrix sahi se lagaana zaroori	Ex 6
C8	Real-world word problem (ISS-jaisi circular orbit)	English ko vectors mein translate karna	Ex 7
C9	Exam twist: parabola limit ( $e \to 1$ , $ε = 0$ )	$a \to \infty$ ; $p$ use karo, $a$ nahi	Ex 8

Ex 1 — Second-half true anomaly (cell C1)

Forecast: pehle guess karo — kya satellite periapsis ki taraf ja raha hai ya door ja raha hai? Agar $r$ aur $v$ ek-doosre ke "saath" point karein (positive dot product), toh woh door ja raha hai; agar "against" toh andar aa raha hai. Apna guess likh lo.

$r$ aur $v$ compute karo. $r = 604 5^{2} + 349 0^{2} + 250 0^{2} = 7414.0$ km, $v = 3.45 7^{2} + 6.61 8^{2} + 2.53 3^{2} = 7.884$ km/s. Ye step kyun? Kisi bhi ratio se pehle hamein plain lengths chahiye; $e$ aur $ν$ ke formulas mein $r$ aur $v$ directly use hote hain.
Angular momentum. $h = r \times v = (- 25384.4, 6669.5, - 52070.7)$ , $h = 58311.7$ . Ye step kyun? $h$ plane ko fix karta hai aur aage $e$ banane mein help karta hai.
Eccentricity vector. $e = \frac{1}{μ} [(v^{2} - μ / r) r - (r \cdot v) v]$ use karte hue, pehle dot product carefully compute karo: $r \cdot v = (- 6045) (- 3.457) + (- 3490) (6.618) + (2500) (2.533) = 20897.9 - 23096.8 + 6332.5 = + 4133.6$ . Phir $e = (- 0.1090, - 0.1244, 0.02386)$ , toh $e = 0.1655 \approx 0.166$ . Ye step kyun? $e$ periapsis ki taraf point karta hai; $ν$ woh angle hai $e$ se $r$ tak. Note karo dot product positive aaya — step 5 ke liye yaad rakho.
True anomaly (raw). $cos ν = \frac{e \cdot r}{e r} = 0.6295 \Rightarrow ν = 51. 0^{\circ}$ ya $309. 0^{\circ}$ . Ye step kyun? $cos^{- 1}$ $51. 0^{\circ}$ aur $309. 0^{\circ}$ mein fark nahi bata sakta — dono ka cosine same hai. Tie todna padega.
Quadrant fix. $r \cdot v = + 4133.6 > 0$ , matlab satellite periapsis se door ja raha hai, isliye $ν < 18 0^{\circ}$ . Lo $ν = 51. 0^{\circ}$ . Ye step kyun? Positive $r \cdot v$ = $r$ badh raha hai = hum outbound half pe hain. Yeh exactly woh sign rule hai jo parent recipe mein hai.

Verify: sign positive tha, isliye humne first-half value rakhi — physical meaning se match karta hai "periapsis se door ja raha hai." Agar negative hota toh $309. 0^{\circ}$ le lete. ✓ (units: $cos ν$ dimensionless hai, $km^{2} \cdot$ km $^{- 1} \cdot$ km $^{- 1}$ ka ratio — clean.)

Ex 2 — RAAN aur $ω$ door ke quadrants mein (cells C2, C3)

Forecast: $h_{z} < 0$ ke saath (Ex 1 se, $h_{z} = - 52070.7$ ), orbit ka "north pole" equator ke neeche point karta hai. Guess karo: kya $i$ , $9 0^{\circ}$ se kam hai ya zyada?

Inclination. $cos i = \frac{h _{z}}{h} = \frac{- 52070.7}{58311.7} = - 0.8930 \Rightarrow i = 153. 2^{\circ}$ . Ye step kyun? $i$ (upar define hua) $h$ ka $\overset{z}{^}$ se tilt hai; negative $h_{z}$ matlab $h$ equator ke neeche point karta hai → retrograde ( $i > 9 0^{\circ}$ ). Koi quadrant fix nahi chahiye: $i$ sirf $[0, 18 0^{\circ}]$ mein rehta hai.
Node vector. $n = \overset{z}{^} \times h = (- h_{y}, h_{x}, 0) = (- 6669.5, - 25384.4, 0)$ , $n = 26247.9$ . Ye step kyun? $n$ (upar define hua) us line ki taraf point karta hai jahan orbit reference plane ko upar cross karta hai — ascending node, $Ω$ ka anchor.
RAAN (raw). $cos Ω = \frac{n _{x}}{n} = \frac{- 6669.5}{26247.9} = - 0.2541 \Rightarrow Ω = 104. 7^{\circ}$ ya $255. 3^{\circ}$ . Ye step kyun? Phir se $cos^{- 1}$ do-valued hai.
RAAN fix. $n_{y} = - 25384.4 < 0 \Rightarrow Ω = 36 0^{\circ} - 104. 7^{\circ} = 255. 3^{\circ}$ . Ye step kyun? Parent rule: agar $n_{y} < 0$ toh node lower half-plane mein hai, isliye $Ω$ past $18 0^{\circ}$ hai.
Argument of periapsis (raw + fix). $cos ω = \frac{n \cdot e}{n e}$ . $e = (- 0.1090, - 0.1244, 0.02386)$ (Ex 1 se) aur $n \cdot e = (- 6669.5) (- 0.1090) + (- 25384.4) (- 0.1244) = 727.0 + 3157.8 = 3884.8$ ke saath, $cos ω = \frac{3884.8}{26247.9 \times 0.1655} = 0.8944 \Rightarrow ω = 26. 6^{\circ}$ ya $333. 4^{\circ}$ . Kyunki $e_{z} = + 0.02386 > 0$ , $ω = 26. 6^{\circ}$ rakho. Ye step kyun? $e_{z} > 0$ matlab periapsis equator ke upar hai, isliye $ω < 18 0^{\circ}$ .

Verify: teeno angles apni proper ranges mein hain ( $i \in [0, 180]$ , $Ω \in [0, 360)$ , $ω \in [0, 360)$ ). Retrograde read ( $i = 153. 2^{\circ}$ ) $h_{z} < 0$ ke consistent hai. ✓

Ex 3 — Circular non-equatorial orbit: $ω$ undefined, $u$ use karo (cell C4)

Forecast: agar orbit ek perfect circle hai lekin phir bhi tilted, periapsis kahan hai — aur kya hum abhi bhi measure kar sakte hain "orbit pe kitna aage" satellite hai? Padhne se pehle guess karo.

Radius aur angular momentum. $r = 495 0^{2} + 495 0^{2} = 7000$ km. $h = r \times v = (4950 \cdot 5.405 - 0, 0 - 4950 \cdot 5.405, 4950 \cdot 3.822 + 4950 \cdot 3.822) = (26754.8, - 26754.8, 37837.8)$ , $h = 52822$ . Ye step kyun? $h_{z} \neq = 0$ lekin $h_{x}, h_{y} \neq = 0$ bhi, isliye plane tilted hai — yeh equatorial nahi hai. Badhiya, yahi cell ka point hai.
Eccentricity. $v = 3.82 2^{2} + 3.82 2^{2} + 5.40 5^{2} = 7.546$ , $v^{2} - μ / r = 56.94 - 56.94 = 0$ , aur $r \cdot v = 4950 (- 3.822) + 4950 (3.822) + 0 = 0$ . Dono terms zero hain, isliye $e = 0$ , $e = 0$ . Ye step kyun? Zero eccentricity vector ki koi direction nahi — point karne ke liye koi periapsis nahi hai.
$ω$ mein singularity. $cos ω = \frac{n \cdot e}{n e}$ mein denominator mein $e = 0$ hai → $0/0$ , undefined. Lekin node vector bachta hai: $n = \overset{z}{^} \times h = (26754.8, 26754.8, 0)$ , $n = 37837.8 \neq = 0$ . Ye step kyun? Kyunki $i \neq = 0$ hai, node well-defined hai; sirf $ω$ (jo periapsis chahiye) toot ta hai. Yeh pure $e = 0$ case hai, equatorial collapse se unpolluted.
Fix — argument of latitude $u$ . Node $n$ se satellite $r$ tak seedha angle measure karo: $cos u = \frac{n \cdot r}{n r} = \frac{( 26754.8 ) ( 4950 ) + ( 26754.8 ) ( 4950 )}{37837.8 \times 7000} = \frac{264872520}{264864600} = 1.000 \Rightarrow u = 0^{\circ}$ . Ye step kyun? $u = ω + ν$ undefined $ω$ aur position $ν$ ko ek single angle mein fold karta hai jo hum measure kar sakte hain — surviving node se. Yahan $u = 0^{\circ}$ matlab satellite bilkul ascending node pe baitha hai. Dekho Orbital singularities and equinoctial elements.

Verify: $e = 0$ ✓ aur $r = a = 7000$ km (circular). Vis-viva check: $v = μ / r = 398600/7000 = 7.546$ km/s ✓. Kyunki $r = (4950, 4950, 0)$ exactly $n = (26754.8, 26754.8, 0)$ ke saath parallel hai, isliye $u = 0^{\circ}$ ✓.

Ex 4 — Equatorial orbit: $Ω$ undefined (cell C5)

Forecast: agar orbit kabhi equator nahi chhodta, toh "upar cross karne wala point" kahan hai? Ascending node ka kya hoga?

Angular momentum. $h = (0, 0, 8000 \times 6.5) = (0, 0, 52000)$ , purely $+ \overset{z}{^}$ . Ye step kyun? $h ∥ \overset{z}{^}$ matlab $i = 0$ (plane hi reference plane hai).
Inclination. $cos i = h_{z} / h = 52000/52000 = 1 \Rightarrow i = 0^{\circ}$ . Ye step kyun? Equatorial case confirm karta hai.
Node vector vanish ho jaata hai. $n = \overset{z}{^} \times h = \overset{z}{^} \times (52000 \overset{z}{^}) = 0$ , $n = 0$ . Ye step kyun? Node define hota hai jahan orbit plane se upar uthta hai — lekin woh kabhi plane nahi chhodta, isliye aisa koi point hai hi nahi. $Ω = cos^{- 1} (n_{x} / n)$ hai $0/0$ .
Eccentricity bachti hai. $v = 6.5$ , $v^{2} - μ / r = 42.25 - 49.825 = - 7.575$ , $r \cdot v = 0$ , toh $e = \frac{- 7.575}{398600} (8000, 0, 0) = (- 0.1520, 0, 0)$ , $e = 0.1520$ . Periapsis $- x$ direction mein hai, isliye satellite abhi apoapsis pe hai. Ye step kyun? Dikhata hai ki eccentricity theek hai tab bhi jab $Ω$ nahi — sirf node-dependent element toot ta hai.
Fix — true longitude of periapsis. $Ω$ aur $ω$ ko $ϖ = Ω + ω$ se replace karo (upar define hua), yahan bas $e$ ka $x$ -axis se angle: $ϖ = 18 0^{\circ}$ . Ye step kyun? Koi node nahi measure karne ke liye, toh periapsis seedha fixed $x$ -axis se measure karo.

Verify: $i = 0$ ✓, $n = 0$ ✓ (equatorial), $e = 0.152 \neq = 0$ toh shape genuine ellipse hai. Energy $ε = \frac{6. 5 ^{2}}{2} - \frac{398600}{8000} = 21.125 - 49.825 = - 28.700$ , toh $a = - μ /2 ε = 6944$ km. Phir apoapsis distance $a (1 + e) = 6944 (1.152) = 8000$ km $= r$ — satellite sach mein apoapsis pe hai. ✓

Ex 5 — Hyperbolic flyby: $a < 0$ (cell C6)

Forecast: 6000 km pe 12 km/s ke saath, kya yeh bound hai ya escape kar raha hai? $ε$ ka sign guess karo.

Energy. $v = 12$ , $ε = \frac{144}{2} - \frac{398600}{6000} = 72 - 66.433 = + 5.567$ . Ye step kyun? Positive energy = unbound = hyperbola. Formula $a = - μ /2 ε$ abhi bhi laagu hota hai.
Semi-major axis. $a = - \frac{398600}{2 \times 5.567} = - 35802$ km. Ye step kyun? Negative $a$ ek hyperbola ke liye correct aur normal hai — yeh koi "fix karne wali" galti nahi hai.
Eccentricity. $r \cdot v = 0$ , $v^{2} - μ / r = 144 - 66.433 = 77.567$ , $e = \frac{77.567}{398600} (6000, 0, 0) = (1.1676, 0, 0)$ , $e = 1.168 > 1$ . Ye step kyun? $e > 1$ ek hyperbola ka algebraic signature hai, $ε > 0$ ke consistent.

Verify: $p = h^{2} / μ$ aur $e = 1 + 2 ε h^{2} / μ^{2}$ se cross-check: $h = 6000 \times 12 = 72000$ , $e = 1 + \frac{2 ( 5.567 ) ( 7200 0 ^{2} )}{39860 0 ^{2}} = 1.168$ ✓. Dono routes agree karte hain, aur $a (1 - e^{2}) = p$ $a < 0$ ke saath bhi hold karta hai. ✓

Ex 6 — Elements → State off periapsis (cell C7)

Forecast: $ν = 9 0^{\circ}$ pe satellite periapsis se quarter-orbit aage hai. Guess karo kya $r$ , periapsis distance $a (1 - e) = 7200$ km se bada hai ya chhota.

Figure — Converting between orbital elements and state vectors (r, v)

Semi-latus rectum & radius. $p = a (1 - e^{2}) = 8000 (0.99) = 7920$ ; $r = \frac{p}{1 + e cos 9 0 ^{\circ}} = \frac{7920}{1 + 0} = 7920$ km. Ye step kyun? Orbit equation kisi bhi $ν$ ke liye distance deta hai; $9 0^{\circ}$ pe $cos ν = 0$ toh $r = p$ exactly (isliye $p$ ko orbit ka "width scale" kehte hain). Figure mein: yeh radius yellow arrow ki length hai.
Perifocal position — figure se seedha padho. $r_{pq w} = r (cos ν, sin ν, 0) = 7920 (cos 9 0^{\circ}, sin 9 0^{\circ}, 0) = 7920 (0, 1, 0) = (0, 7920, 0)$ km. Ye step kyun? Perifocal frame mein periapsis $\overset{p}{^}$ axis pe baitha hai (figure mein green dot). Kyunki $ν = 9 0^{\circ}$ hai, yellow arrow $r_{pq w}$ $\overset{p}{^}$ se quarter-turn ghoom ke seedha $\overset{q}{^}$ axis ki taraf point karta hai — isliye sirf middle ( $q$ ) component non-zero hai. Aise $(a, e, ν)$ concrete vector $(0, 7920, 0)$ bana.
Inertial mein rotate karo. $\tilde{R}$ laagoo karo (upar define hua, jahan $c i = cos i$ , $s i = sin i$ ) $c Ω = cos 4 0^{\circ} = 0.766$ , $s Ω = sin 4 0^{\circ} = 0.643$ , $c ω = cos 6 0^{\circ} = 0.5$ , $s ω = sin 6 0^{\circ} = 0.866$ , $c i = cos 3 0^{\circ} = 0.866$ , $s i = sin 3 0^{\circ} = 0.5$ use karte hue. Kyunki $r_{pq w} = (0, 7920, 0)$ mein sirf $q$ -component hai, yeh $\tilde{R}$ ki middle column select karta hai: $r_{x} = 7920 (- c Ω s ω - s Ω c ω c i) = 7920 (- 0.766 \cdot 0.866 - 0.643 \cdot 0.5 \cdot 0.866) = - 7459$ km; $r_{y} = 7920 (- s Ω s ω + c Ω c ω c i) = 7920 (- 0.643 \cdot 0.866 + 0.766 \cdot 0.5 \cdot 0.866) = - 1783$ km; $r_{z} = 7920 (c ω s i) = 7920 (0.5 \cdot 0.5) = 1980$ km. Ye step kyun? Hum perifocal se inertial mein transform kar rahe hain, jo $\tilde{R}$ mein bake-in negative-angle rotations use karta hai. Figure mein: laal arrow yeh tilted-and-swivelled result hai — same length yellow arrow jaisi, naya direction.

Verify: rotation se length preserved hai: $∣ r ∣ = 745 9^{2} + 178 3^{2} + 198 0^{2} = 7920$ km $= r$ ✓ (rotations kabhi stretch nahi karte). Units sab km. ✓

Ex 7 — Real-world: ISS-jaisi circular orbit (cell C8)

Forecast: ISS roughly kitna fast jaata hai — 3 km/s ke kareeb ya 8 km/s ke? Guess karo.

Circular speed via vis-viva. $e = 0$ ke liye, $r = a$ , toh $v = μ / r = 398600/6798 = 7.657$ km/s. Ye step kyun? Vis-viva $v^{2} = μ (2/ r - 1/ a)$ $a = r$ hone par $μ / r$ ban jaata hai.
Period. $T = 2 π a^{3} / μ = 2 π 679 8^{3} /398600 = 5578$ s $= 92.97$ min. Ye step kyun? Kepler's third law poori ellipse sweep karne ka time deta hai; dekho Kepler's equation and time-of-flight.
Ek consistent velocity. $r$ ke $+ x$ along circular equatorial orbit ke liye, velocity perpendicular aur in-plane hai: $v = (0, 7.657, 0)$ km/s. Ye step kyun? Circular ⇒ velocity purely tangential (koi radial part nahi), aur equatorial ⇒ koi $z$ -component nahi.

Verify: $92.97$ min real ISS period se match karta hai ( $\approx 93$ min) ✓. Aur $r \cdot v = 0$ zero radial speed confirm karta hai, exactly woh jo "circular" demand karta hai. ✓

Ex 8 — Exam twist: parabolic limit (cell C9)

Forecast: parabola ke liye $ε$ kya hota hai? Aur $a = - μ /2 ε$ kya karta hai jab $ε = 0$ ?

Escape speed. $v_{esc} = 2 μ / r = 2 \times 398600/10000 = 8.929$ km/s. Ye step kyun? Parabola = exactly zero total energy = bas barely escaping.
Energy. $ε = \frac{v _{esc}^{2}}{2} - \frac{μ}{r} = \frac{79.72}{2} - 39.86 = 39.86 - 39.86 = 0$ . Ye step kyun? Parabola ki defining property $ε = 0$ confirm karta hai.
Kyun $a$ toot ta hai. $a = - μ / (2 \times 0) = - \infty$ . Ek parabola ka koi finite semi-major axis nahi hota — uske dono "ends" infinity tak jaate hain. Ye step kyun? Dikhata hai kyun textbooks open orbits ko $p$ (ya perigee $r_{p}$ ) se parametrise karte hain, $a$ se kabhi nahi.
$p$ use karo. $r \cdot v = 0$ (perigee pe) ke saath, $h = r v = 10000 \times 8.929 = 89290$ , $p = h^{2} / μ = 8929 0^{2} /398600 = 20000$ km. Aur indeed $r_{p} = p / (1 + e) = 20000/2 = 10000$ km $= r$ ✓. Ye step kyun? $p$ finite rehta hai aur parabola ko fully describe karta hai; $e = 1$ ke liye perigee radius $= p /2$ .

Verify: $ε = 0$ exactly ✓; $r_{p} = 10000$ km di gayi position recover karta hai ✓. Parabola ellipses ( $ε < 0$ , Ex 3–4) aur hyperbolas ( $ε > 0$ , Ex 5) ke beech ka clean boundary hai. ✓

Recall Which companion sign fixes which angle?

$Ω$ ke liye, $n_{y}$ check karo ::: agar $n_{y} < 0$ toh $Ω = 36 0^{\circ} - Ω_{raw}$ . $ω$ ke liye, $e_{z}$ check karo ::: agar $e_{z} < 0$ toh $ω = 36 0^{\circ} - ω_{raw}$ . $ν$ ke liye, $r \cdot v$ check karo ::: agar $r \cdot v < 0$ toh $ν = 36 0^{\circ} - ν_{raw}$ .

Flashcards

Semi-major axis

a

negative kab hota hai?

Hyperbolas ke liye, jahan

ε > 0

, kyunki

a = - μ /2 ε

Circular non-equatorial orbit ke liye

ω

ki jagah kya aata hai?

Argument of latitude

u = ω + ν

, ascending node se satellite tak measure kiya hua.

Parabola ke liye

a

kyun useless hai?

Kyunki

ε = 0

a = - μ / (2 ε) \to \infty

ho jaata hai; semi-latus rectum

p = h^{2} / μ

use karo.

Which companion sign tells you

ν

is past

18 0^{\circ}

A negative

r \cdot v

(radius shrinking, falling toward periapsis).