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

KYA: Hum ek element ko us quantity se match kar rahe hain jo use pin karti hai. ==Semi-major axis $a$ size set karta hai. Ise specific energy $\epsilon =v^2/2-\mu /r$== fix karta hai $a=-\mu /(2\epsilon )$ ke zariye. KYUN: Energy sirf is baat par depend karti hai ki ellipse kitna bada hai (vis-viva), body kahan hai is par nahi — isliye ek energy number se $a$ recover karna kaafi hai.

$\vec{n}=\hat{z}\times \vec{h}$. Kyunki $\vec{h}$ orbit plane ke perpendicular hai aur $\hat{z}$ reference frame ka vertical hai, unka cross product us line ke saath hota hai jahan orbit plane reference plane ko kaatta hai — line of nodes. Convention ke mutabiq $\vec{n}$ ascending node ki taraf point karta hai (jahan body upward cross karti hai). Agar $\vec{n}=\vec{0}$ toh orbit equatorial hai aur $\Omega$ undefined hai.

False. $\vec{e}=\frac{\vec{v}\times \vec{h}}{\mu }-\frac{\vec{r}}{r}$ points toward periapsis (closest approach). Iska magnitude scalar $e$ hai.

KYA: energy → size, ek plug-in. $\epsilon =\frac{v^2}{2}-\frac{\mu }{r}=\frac{7.8^2}{2}-\frac{398600}{6800}=30.42-58.62=-28.20{\text{km}}^2/{\text{s}}^2$ $a=-\frac{\mu }{2\epsilon }=-\frac{398600}{2(-28.20)}=7067\text{km}$ KYUN $a>0$: $\epsilon <0$ matlab bound (elliptical) orbit hai, aur minus signs cancel hokar positive size dete hain.

KYA: cross product phir ek cosine. $\vec{h}=\vec{r}\times \vec{v}=(0\cdot 3-0\cdot 6.5,0\cdot 0-8000\cdot 3,8000\cdot 6.5-0)=(0,-24000,52000)$ $h=\sqrt{0+24000^2+52000^2}=57271{\text{km}}^2/\text{s}$ $\cos i=\frac{h_z}{h}=\frac{52000}{57271}=0.9080\Rightarrow i=24.83^{\circ }$ KYUN: $\vec{h}$ orbit ka axle hai; $\hat{z}$ se uska tilt hi inclination hai.

$r_p=a(1-e)=7500(0.95)=7125\text{km}$; $r_a=a(1+e)=7500(1.05)=7875\text{km}$. KYUN: periapsis aur apoapsis major axis ke do endpoints hain; $a$ midpoint hai, $ae$ focus ka offset hai.

KYA: recipe se $\vec{e}$ banao, uski geometry padho. $r=7000$, $v^2=7.5^2+1.0^2=57.25$, $\vec{r}\cdot \vec{v}=0$. $\vec{e}=\frac{1}{\mu }[(v^2-\frac{\mu }{r})\vec{r}-(\vec{r}\cdot \vec{v})\vec{v}]=\frac{1}{398600}(57.25-56.943)(7000,0,0)$ $=\frac{0.30714\cdot 7000}{398600}(1,0,0)=(0.005393,0,0)$ Toh $e=0.00539$ — nearly circular. Sign check: $\vec{r}\cdot \vec{v}=0$ matlab body ek apse par hai (radial velocity zero). Kyunki $\vec{e}$ $+\hat{x}$ ki taraf point karta hai aur $\vec{r}$ bhi $+\hat{x}$ ki taraf hai, $\vec{r}$ aur $\vec{e}$ parallel hain → $\cos \nu =+1$ → body periapsis par baithe hai ($\nu =0$). KYUN parallel test: $\nu$ periapsis ($\vec{e}$) aur current position ($\vec{r}$) ke beech ka angle hai; parallel ⇒ zero angle ⇒ periapsis.

KYA: cosine ek raw angle deta hai; companion sign quadrant disambiguate karta hai. $n=\sqrt{3000^2+4000^2}=5000$. $\cos \Omega =n_x/n=-3000/5000=-0.6\Rightarrow {\Omega }_{\text{raw}}=126.87^{\circ }$. Quadrant fix: $n_y<0$ hai, isliye $\Omega =360^{\circ }-126.87^{\circ }=233.13^{\circ }$. KYUN: ${\cos }^{-1}$ sirf $[0^{\circ },180^{\circ }]$ return karta hai. Node yahan third quadrant mein point karta hai ($n_x<0,n_y<0$), jo $180^{\circ }$ ke baad hota hai; $n_y<0$ rule ise $x$-axis ke neeche reflect karta hai. Neeche ki figure raw aur corrected angles ko $x$-axis ke across mirror images ki tarah dikhati hai.

Ise aise padho: same $x$-projection, opposite $y$-sign — isliye cosine dono ko alag nahi kar sakta, lekin $n_y$ kar sakta hai.

Step 1 — magnitudes. KYUN pehle: har downstream formula ($\epsilon$, $\cos i$, $\vec{e}$) ko scalar lengths $r=|\vec{r}|$ aur $v=|\vec{v}|$ chahiye, isliye inhe ek baar pehle compute kar lete hain. $r=\sqrt{6524^2+6862^2+6449^2}=\sqrt{42562576+47087044+41589601}=\sqrt{131239221}=11455.1\text{km}$ $v=\sqrt{4.902^2+5.533^2+1.976^2}=\sqrt{24.0296+30.6141+3.9046}=\sqrt{58.5483}=7.6517\text{km/s}$ Step 2 — energy → $a$. KYUN energy: specific energy $\epsilon$ sirf orbit size par depend karta hai, isliye yeh sabse clean single quantity hai jo humein $a$ deta hai ($a=-\mu /2\epsilon$ ke zariye) bina kisi angle ke. $\epsilon =\frac{7.6517^2}{2}-\frac{398600}{11455.1}=29.274-34.797=-5.523{\text{km}}^2/{\text{s}}^2$ $a=-\frac{398600}{2(-5.523)}=36090\text{km}$ Step 3 — $\vec{h}$ aur $i$. KYUN $\vec{h}$: angular-momentum vector orbit plane ke perpendicular axle hai, isliye $\hat{z}$ se uska tilt hi inclination hai — ek cosine se padh lo. $\vec{h}=\vec{r}\times \vec{v}=(6862\cdot (-1.976)-6449\cdot 5.533,6449\cdot 4.902-6524\cdot (-1.976),6524\cdot 5.533-6862\cdot 4.902)$ $=(-49246.8,44510.8,2469.5){\text{km}}^2/\text{s},h=\sqrt{49246.8^2+44510.8^2+2469.5^2}=66436{\text{km}}^2/\text{s}$ $\cos i=\frac{h_z}{h}=\frac{2469.5}{66436}=0.03717\Rightarrow i=87.87^{\circ }(\text{near-polar})$ Step 4 — $\vec{e}$ phir $e$. KYUN $\vec{e}$: eccentricity vector woh akela quantity hai jo shape ($e=|\vec{e}|$) aur periapsis direction dono ek saath carry karta hai, isliye yeh conversion ka shape wala aadha khatam karta hai (aur $\omega ,\nu$ ke liye bhi kaam aata agar pucha jaata). Pehle hume $\vec{r}\cdot \vec{v}$ chahiye, jo radial-motion indicator hai: $\vec{r}\cdot \vec{v}=6524(4.902)+6862(5.533)+6449(-1.976)=31980.6+37967.5-12743.2=57204.9$. $\vec{e}=\frac{1}{\mu }[(v^2-\frac{\mu }{r})\vec{r}-(\vec{r}\cdot \vec{v})\vec{v}]$ $v^2-\mu /r=58.548-34.797=23.751$. Phir: $\vec{e}=\frac{1}{398600}[23.751(6524,6862,6449)-57204.9(4.902,5.533,-1.976)]$ $=\frac{1}{398600}[(154951,162978,153170)-(280418,316505,-113037)]=\frac{(-125467,-153527,266207)}{398600}$ $\vec{e}=(-0.3148,-0.3852,0.6679),e=|\vec{e}|=0.8329$ Toh yeh ek highly eccentric, near-polar orbit hai jisme $a\approx 36090$ km, $e\approx 0.833$, $i\approx 87.9^{\circ }$.

Step 1 — perifocal position. KYUN semi-latus rectum aur orbit equation: perifocal frame mein periapsis $\hat{p}$ axis par hota hai, isliye position trivially $r(\cos \nu ,\sin \nu ,0)$ hai — lekin pehle hume is true anomaly par scalar distance $r$ chahiye. Orbit equation $r=p/(1+e\cos \nu )$ exactly woh conic hai jo focus se likhi gayi hai, aur $p$ uska natural size-scale hai ($\nu =90^{\circ }$ par width). $p=a(1-e^2)$ use karke humara given size element $a$ us scale mein convert hota hai: $p=a(1-e^2)=8000(0.99)=7920\text{km}$ $r=\frac{p}{1+e\cos \nu }=\frac{7920}{1+0.1\cos 30^{\circ }}=\frac{7920}{1.08660}=7288.8\text{km}$ ${\vec{r}}_{pqw}=r(\cos \nu ,\sin \nu ,0)=7288.8(0.86603,0.5,0)=(6312.3,3644.4,0)\text{km}$ Step 2 — PQW ko ECI mein rotate karo. Is page ke upar display ki gayi composite matrix $\tilde{R}$ lagao. Kyunki $z_{pqw}=0$ hai, sirf pehle do columns kaam karte hain, jo dete hain (jahan $c=\cos ,s=\sin$): $x=(c\Omega c\omega -s\Omega s\omega ci)x_{pqw}+(-c\Omega s\omega -s\Omega c\omega ci)y_{pqw}$ KYUN yeh exact cosine–sine combinations? $\tilde{R}$ teen stacked rotations ka product hai, innermost→outermost padho:

• $R_3(-\omega )$ pehle periapsis ($\hat{p}$ axis) ko orbit plane ke andar $\omega$ se aage swing karta hai — isliye $\cos \omega ,\sin \omega$ in-plane coordinates ke saath paired aate hain.
• $R_1(-i)$ phir poore plane ko node line ke baare mein inclination se tilt karta hai — isliye har term jo "plane se bahar jaati hai" woh $\cos i$ ya $\sin i$ se multiply hoti hai ($ci$, $si$ factors).
• $R_3(-\Omega )$ finally node line ko $\hat{z}$ ke baare mein uske sky longitude tak swivel karta hai — yahi $\cos \Omega ,\sin \Omega$ factors ka source hai. Har entry har rotation se ek ek factor ka product hai, isliye $s\Omega c\omega ci$ jaisa term literally padha jaata hai "swivelled by $\Omega$, spun by $\omega$, tilted by $i$." Negatives isliye hain kyunki hum components ko PQW se ECI mein transform kar rahe hain, frame rotate karne ka inverse.

Numbers: $c\Omega =0.76604,s\Omega =0.64279,c\omega =0.5,s\omega =0.86603,ci=0.86603,si=0.5$. Row 1 coefficients: $0.76604(0.5)-0.64279(0.86603)(0.86603)=0.38302-0.48209=-0.09907$; $-0.76604(0.86603)-0.64279(0.5)(0.86603)=-0.66342-0.27835=-0.94177$. $x=-0.09907(6312.3)+(-0.94177)(3644.4)=-625.4-3432.3=-4057.7\text{km}$ Row 2 coefficients: $s\Omega c\omega +c\Omega s\omega ci=0.64279(0.5)+0.76604(0.86603)(0.86603)=0.32140+0.57451=0.89590$; $-s\Omega s\omega +c\Omega c\omega ci=-0.64279(0.86603)+0.76604(0.5)(0.86603)=-0.55668+0.33172=-0.22496$. $y=0.89590(6312.3)+(-0.22496)(3644.4)=5655.2-819.9=4835.3\text{km}$ Row 3 coefficients: $s\omega si=0.86603(0.5)=0.43301$; $c\omega si=0.5(0.5)=0.25$. $z=0.43301(6312.3)+0.25(3644.4)=2733.3+911.1=3644.4\text{km}$ Answer: ${\vec{r}}_{\text{ECI}}=(-4058,4835,3644)\text{km}$. Check: $|\vec{r}|=\sqrt{4058^2+4835^2+3644^2}=7289\text{km}$ = $r$ ✓ (rotations length preserve karti hain).

KYA toot ta hai: $i=0$ ke saath orbit plane reference plane hi hai, toh woh usse "upar se cross" kabhi nahi karti — line of nodes gayab ho jaati hai: $\vec{n}=\hat{z}\times \vec{h}=\vec{0}$. Phir $\cos \Omega =n_x/n$ matlab $0/0$ → $\Omega$ undefined hai, aur kyunki $\omega$ (missing) node se measure hota hai, $\omega$ bhi undefined ho jaata hai. Fix: eccentricity direction abhi bhi perfectly good hai ($e\ne 0$), isliye true longitude of periapsis $\varpi =\Omega +\omega$ use karo jo seedha $\hat{x}$ axis se measure hoti hai, $\vec{e}$ se compute ki jaati hai: $\cos \varpi =e_x/e$ (quadrant $e_y$ se). Zyada generally, equinoctial elements sabhi aisi $0/0$ cases hatate hain. KYUN yeh kaam karta hai: combination $\Omega +\omega$ tab bhi observable rehti hai jab har piece individually nahi hoti — ambiguity sirf angle ko kahan split karein mein hai, total mein nahi.

KYA toot ta hai: $e=0$ ke saath eccentricity vector collapse ho jaata hai, $\vec{e}=\vec{0}$. Ab koi periapsis direction nahi hai, isliye:

• $\omega$ (node se periapsis tak measure hota hai) undefined hai — $\cos \omega =\frac{\vec{n}\cdot \vec{e}}{ne}$ matlab $0/0$.
• $\nu$ (periapsis se measure hota hai) bhi undefined hai — $\cos \nu =\frac{\vec{e}\cdot \vec{r}}{er}$ matlab $0/0$. Node abhi bhi exist karta hai ($i\ne 0$ ⇒ $\vec{n}\ne \vec{0}$), isliye $\Omega$ theek hai. Fix: position ko missing periapsis ki jagah node se measure karo. Argument of latitude $u=\omega +\nu$ use karo, jo seedha node aur position vectors se compute hota hai: $\cos u=\frac{\vec{n}\cdot \vec{r}}{nr}(\text{quadrant: agar }{r}_z<0,u=2\pi -u)$ KYUN yeh kaam karta hai: sum $\omega +\nu$ woh angle hai node se body tak, jo perfectly well-defined hai chahe periapsis "kahan hota" is baat se beparwah. Sirf $\omega$ aur $\nu$ mein split karna ambiguous hai. (Agar orbit dono circular aur equatorial ho, toh $\Omega$ bhi mar jaata hai aur tum true longitude $\ell =\Omega +\omega +\nu$ par fall back karte ho, ya equinoctial elements par.)

Har quadrant-ambiguous element ke liye companion sign batao. $\Omega :::n_y$ $\omega :::e_z$ $\nu :::\vec{r}\cdot \vec{v}$