Exercises — Physical meaning of each orbital element

Dono poles ke upar se guzarna matlab orbit plane equator se poori ek quarter-turn tilted hai. Woh tilt hi inclination $i$ hai. $i=90^{\circ }$ $i$ kyun aur $\Omega$ kyun nahi? $\Omega$ sirf yeh batata hai ki tilt kis compass direction mein face karta hai; $i$ batata hai ki woh kitna steep hai. Polar = maximum steepness = $90^{\circ }$.

Sirf ==$\nu$== (true anomaly) change hota hai. Baaki paanchon constant hain kyunki energy, orbit plane, aur ellipse ki shape/orientation — yeh sab pure two-body problem mein conserved hain. (Real perturbations — dekho Orbital perturbations — baaki ko dhire dhire drift karte hain, lekin woh ideal case se pare hai.)

$e=\frac{r_a-r_p}{r_a+r_p}=\frac{0}{14400}=0$ Yeh ek circle hai. Aur kyunki $a=(r_p+r_a)/2=7200$ km constant radius ke barabar hai, yahan (aur sirf yahan) $a$ literally satellite ki distance hai.

Altitudes ko focal distances mein convert karo (Earth ki radius add karo, kyunki $r$ Earth ke centre = focus se measure hoti hai): $r_p=500+6378=6878\text{km},r_a=5500+6378=11878\text{km}$ $a=\frac{r_p+r_a}{2}=\frac{18756}{2}=9378\text{km}$ $e=\frac{r_a-r_p}{r_a+r_p}=\frac{5000}{18756}=0.2666$ $R_{\oplus }$ kyun add karte hain? Altitude surface se measure hoti hai, lekin har orbital formula focus (Earth ke centre) se distance use karta hai.

Kepler's third law (Kepler's Laws): $T=2\pi \sqrt{\frac{a^3}{\mu }}=2\pi \sqrt{\frac{9378^3}{398600}}$ $a^3=8.248\times 10^{11}{\text{km}}^3,\frac{a^3}{\mu }=2.0693\times 10^6{\text{s}}^2$ $T=2\pi \sqrt{2.0693\times 10^6}=2\pi (1438.5)=9038\text{s}\approx 150.6\text{min}$ Sirf $a$ kyun aata hai? Period energy par depend karti hai, aur energy sirf $a$ par depend karti hai — eccentricity $e$ kabhi $T$ mein enter nahi karta.

Vis-viva equation kisi bhi radius par speed deta hai: $v=\sqrt{\mu \left(\frac{2}{r}-\frac{1}{a}\right)}=\sqrt{398600\left(\frac{2}{6878}-\frac{1}{9378}\right)}$ $\frac{2}{6878}=2.9078\times 10^{-4},\frac{1}{9378}=1.0663\times 10^{-4}$ $v=\sqrt{398600\times 1.8415\times 10^{-4}}=\sqrt{73.40}=8.567\text{km/s}$ Yahan sabse fast kyun? Periapsis sabse paas ka point hai, isliye kinetic energy sabse zyada hoti hai (Kepler's Laws, equal-areas → paas hone par fast).

Specific angular momentum $\vec{h}=\vec{r}\times \vec{v}$ compute karo (Angular momentum in orbits): $\vec{h}=\left|\begin{array}{ccc}\hat{I}& \hat{J}& \hat{K}\\ 7000& 0& 0\\ 0& 6& 4\end{array}\right|=(0\cdot 4-0\cdot 6,0\cdot 0-7000\cdot 4,7000\cdot 6-0)$ $\vec{h}=(0,-28000,42000){\text{km}}^2/\text{s}$ Magnitude: $h=\sqrt{0^2+28000^2+42000^2}=\sqrt{2.548\times 10^9}=50478{\text{km}}^2/\text{s}$. Inclination $\vec{h}$ ke $z$-component ko uski length se use karta hai: $\cos i=\frac{h_z}{h}=\frac{42000}{50478}=0.8320\Rightarrow i=33.69^{\circ }$ $h_z/h$ kyun? $\vec{h}$ orbit plane ke perpendicular hai; $\hat{K}$ equator ke perpendicular hai. Dono normals ke beech ka angle hi dono planes ke beech ka angle hai.

$\vec{n}=\hat{K}\times \vec{h}=(h\text{-based})=(-h_y,h_x,0)=(28000,0,0)$ $|\vec{n}|=28000$. $\cos \Omega =\frac{n_x}{|\vec{n}|}=\frac{28000}{28000}=1\Rightarrow \Omega =0^{\circ }\text{ or }360^{\circ }$ Quadrant check: $n_y=0$ ($<0$ nahi), isliye $\Omega =0^{\circ }$. $n_y$ ka sign kyun? $\cos$ $\Omega$ aur $360^{\circ }-\Omega$ ke liye same value deta hai; sirf $y$-sign batata hai ki node vernal-equinox axis ke kis side par hai.

$\epsilon =\frac{v^2}{2}-\frac{\mu }{r}=\frac{100}{2}-\frac{398600}{8000}=50-49.825=+0.175{\text{km}}^2/{\text{s}}^2$ $\epsilon >0$, isliye orbit hyperbolic hai (unbound — yeh escape kar raha hai). Is radius par escape speed check karo: $v_{esc}=\sqrt{2\mu /r}=\sqrt{2\cdot 398600/8000}=\sqrt{99.65}=9.982$ km/s. Kyunki $10.0>9.982$, object (barely) escape kar raha hai. ✓

Orbit equation $r=\frac{a(1-e^2)}{1+e\cos \nu }$ ko $\cos \nu$ ke liye rearrange karo: $1+e\cos \nu =\frac{a(1-e^2)}{r}\Rightarrow \cos \nu =\frac{1}{e}\left(\frac{a(1-e^2)}{r}-1\right)$ Semi-latus rectum compute karo $p=a(1-e^2)=9378(1-0.07108)=8711.4$ km. $\cos \nu =\frac{1}{0.2666}\left(\frac{8711.4}{8000}-1\right)=\frac{1}{0.2666}(0.08893)=0.3336$ $\nu =\arccos (0.3336)=70.51^{\circ }\text{or}289.49^{\circ }$ Kyunki satellite baahir ki taraf move kar raha hai ($\vec{r}\cdot \vec{v}>0$), woh periapsis aur apoapsis ke beech ascending half par hai, isliye $0<\nu <180^{\circ }$: $\boxed{\nu =70.51^{\circ }}$ Direction kyun matter karta hai: same $r$ ek orbit mein do baar hoti hai (ek baar baahir jaate waqt, ek baar andar aate waqt). Sirf radial-velocity sign unhe distinguish karta hai.

$\cos \omega =\frac{\vec{n}\cdot \vec{e}}{|\vec{n}||\vec{e}|}$ $\vec{n}\cdot \vec{e}=1(0.2)+0(0.2)+0(0.1)=0.2$. $|\vec{e}|=\sqrt{0.2^2+0.2^2+0.1^2}=\sqrt{0.09}=0.3$. $\cos \omega =\frac{0.2}{1\times 0.3}=0.6667\Rightarrow \omega =48.19^{\circ }$ Quadrant check: $e_z=0.1>0$, isliye periapsis reference plane ke upar hai aur $\omega <180^{\circ }$. $\omega =48.19^{\circ }$ rakho. $\vec{e}$ kyun? Uski direction hi periapsis ki taraf line hai aur uska magnitude hi $e$ hai — toh yeh dono shape aur in-plane pointing ek hi vector mein carry karta hai (Eccentricity vector).

Step 1 — energy → $a$. $r=8000$, $v=7.5$. $\epsilon =\frac{v^2}{2}-\frac{\mu }{r}=\frac{56.25}{2}-\frac{398600}{8000}=28.125-49.825=-21.70{\text{km}}^2/{\text{s}}^2$ Negative ⇒ bound. Phir $a=-\frac{\mu }{2\epsilon }=-\frac{398600}{2(-21.70)}=9184\text{km}$.

Step 2 — eccentricity vector. $\vec{e}=\frac{1}{\mu }\left[\left(v^2-\frac{\mu }{r}\right)\vec{r}-(\vec{r}\cdot \vec{v})\vec{v}\right]$ Yahan $\vec{r}\cdot \vec{v}=8000(0)+0(7.5)+0=0$ (velocity purely tangential hai ⇒ hum ek apse par hain!). $v^2-\mu /r=56.25-49.825=6.425$. $\vec{e}=\frac{1}{398600}\left[6.425(8000,0,0)-0\right]=\frac{(51400,0,0)}{398600}=(0.1290,0,0)$ Toh $e=0.1290$.

Step 3 — true anomaly. $\vec{e}$ $+x$ ki taraf point karta hai, aur $\vec{r}$ bhi $+x$ ki taraf point karta hai, isliye $\vec{r}$ periapsis par hai: $\cos \nu =\frac{\vec{e}\cdot \vec{r}}{|\vec{e}|r}=\frac{0.1290\cdot 8000}{0.1290\cdot 8000}=1\Rightarrow \nu =0^{\circ }$ $\vec{r}\cdot \vec{v}=0$ se confirmed, speed high hai ⇒ periapsis. Answers: $a=9184$ km, $e=0.129$, $\nu =0^{\circ }$, bound. ✓

$a$ ke liye Kepler III invert karo: $T=2\pi \sqrt{\frac{a^3}{\mu }}\Rightarrow a^3=\mu {\left(\frac{T}{2\pi }\right)}^2$ $\frac{T}{2\pi }=\frac{7200}{6.28319}=1145.92\text{s},{\left(\frac{T}{2\pi }\right)}^2=1.31313\times 10^6$ $a^3=398600\times 1.31313\times 10^6=5.2341\times 10^{11}{\text{km}}^3$ $a=(5.2341\times 10^{11}{)}^{1/3}=8059\text{km}$ Altitude $=a-R_{\oplus }=8059-6378=1681\text{km}$. Kepler III kyun invert karo? Period design goal hai; $a$ woh knob hai jo ise set karta hai, aur circle ke liye altitude directly milti hai kyunki $r=a$.

$a_A=\frac{7000+11000}{2}=9000\text{km},a_B=\frac{8000+10000}{2}=9000\text{km}$ Equal $a$ ⇒ equal energy ⇒ equal period: $T=2\pi \sqrt{\frac{9000^3}{398600}}=2\pi \sqrt{1.8290\times 10^6}=8497\text{s}\approx 141.6\text{min (dono ke liye)}$ Eccentricities alag hain: $e_A=\frac{11000-7000}{18000}=0.2222,e_B=\frac{10000-8000}{18000}=0.1111$ Takeaway: identical $a$ clock fix karta hai; $e$ sirf us fixed time mein satellite ke path ko reshape karta hai. Yeh "energy sirf $a$ par depend karti hai" ka concrete payoff hai.

Kaunsa element "lap kitna lamba hai" ka jawab deta hai? ::: $a$ ($T=2\pi \sqrt{a^3/\mu }$ ke zariye) Kaunsa sign test $\nu$ ko disambiguate karta hai? ::: $\vec{r}\cdot \vec{v}$ ka sign (outward $>0$ ⇒ $\nu <180^{\circ }$) Orbit equation ka numerator kya hai? ::: semi-latus rectum $p=a(1-e^2)$, $a$ nahi Positive specific energy $\epsilon$ matlab kaunsi orbit? ::: hyperbolic (unbound / escaping) Do orbits, same $a$, alag $e$ — same ya alag period? ::: same period (energy sirf $a$ par depend karti hai)