Exercises — Orbit equation r = p - (1 + e·cos θ) — derivation from equations of motion

$r$ focus (central mass ki location, jaise Sun) se orbiting body tak ki distance hai — ellipse ke geometric centre se nahin. Yeh equation ki sabse zaroori reading hai. Focus vs. centre picture dekho:

Laal dot focus hai; har $r$ arrow waheen se shuru hota hai.

$e=0$ ke saath equation ban jaati hai $r=\frac{p}{1+0}=p$: ek constant. Focus se constant distance circle hoti hai. Kyunki $\cos \theta$ term $e=0$ se multiply hoti hai, angle bilkul nikal jaata hai — har direction mein same distance hai.

Mnemonic: Cats Eat Plump Herring jaise $e$ badhta hai $0\to >1$. Dekho Conic Sections.

Perihelion ($\theta =0$, $\cos 0=1$): $r_{\min }=\frac{p}{1+e}=\frac{8000}{1.25}=6400\text{km}.$ Aphelion ($\theta =180^{\circ }$, $\cos 180^{\circ }=-1$): $r_{\max }=\frac{p}{1-e}=\frac{8000}{0.75}\approx 10667\text{km}.$ Signs kyun? $\cos$ $\theta =0$ par $+1$ se ghoomke $\theta =180^{\circ }$ par $-1$ tak jaata hai, isliye denominator perihelion par sabse bada (→ sabse chhota $r$) aur aphelion par sabse chhota (→ sabse bada $r$) hota hai.

$\cos 90^{\circ }=0$, isliye $r=\frac{8000}{1+0.25(0)}=8000\text{km}=p.$ $90^{\circ }$ par tum focus se exactly ek semi-latus rectum door hote ho — yahi $p$ ka matlab hai.

$\cos 120^{\circ }=-\frac{1}{2}$, isliye $r=\frac{7000}{1+0.4(-0.5)}=\frac{7000}{1-0.2}=\frac{7000}{0.8}=8750\text{km}.$ Note karo ki $\theta =120^{\circ }$, $90^{\circ }$ se aage hai, isliye $\cos$ negative hai aur $r>p$ — tum orbit ke "door wale" half mein aphelion ki taraf ja rahe ho.

Use karo $r_p=\frac{p}{1+e}$ aur $r_a=\frac{p}{1-e}$. Divide karo: $\frac{r_a}{r_p}=\frac{1+e}{1-e}\Rightarrow e=\frac{r_a-r_p}{r_a+r_p}=\frac{0.425\times 10^8}{4.559\times 10^8}\approx \mathrm{0.0932.}$ Phir $p=r_p(1+e)=2.067\times 10^8(1.0932)\approx 2.259\times 10^8$ km. Divide kyun kiya? Divide karne se unknown $p$ cancel ho jaata hai aur $e$ ek clean step mein isolate ho jaata hai.

$p=h^2/\mu$ ko invert karo: $h=\sqrt{\mu p}=\sqrt{3.986\times 10^5\times 7000}\approx 5.28\times 10^4{\text{km}}^2/\text{s}.$ Yeh $h$ area-sweep rate $\frac{1}{2}h$ fix karta hai (Kepler's 2nd law — dekho Kepler's Laws aur Specific Angular Momentum h).

$p=a(1-e^2)=10000(1-0.36)=10000(0.64)=6400\text{km}.$ $\theta =90^{\circ }$ par: $r=p=6400$ km. $p=a(1-e^2)$ kyun? $p$ focus se guzarne wala half-chord hai, $a$ long axis ka aadha hai; yeh sirf tab coincide karte hain jab $e=0$ ho.

• $\theta =0$: $r=\frac{p}{1+1}=\frac{p}{2}=6.0\times 10^7$ km (perihelion).
• $\theta =90^{\circ }$: $r=p=1.2\times 10^8$ km.
• Jaise $\theta \to 180^{\circ }$: $\cos \theta \to -1$, denominator $1+e\cos \theta \to 1-1=0$, isliye $r\to \infty$.

Yeh parabola ki pehchaan hai: iska koi aphelion nahin hota — comet ek open arc mein infinity tak escape kar jaata hai. $\theta =180^{\circ }$ direction tak pohonchna possible nahin (uske liye infinite $r$ chahiye hoga). Limiting-case figure dekho:

$r\to \infty$ tab hota hai jab denominator zero ho jaaye: $1+e\cos {\theta }_{\infty }=0\Rightarrow \cos {\theta }_{\infty }=-\frac{1}{e}=-\frac{1}{1.5}=-\mathrm{0.6667.}$ ${\theta }_{\infty }=\arccos (-0.6667)\approx 131.8^{\circ }(\text{aur symmetry se }-131.8^{\circ }).$ $\pm 131.8^{\circ }$ se aage ke angles par denominator negative ho jaata → unphysical negative $r$. Isliye body sirf $|\theta |<131.8^{\circ }$ mein rehti hai: "swoosh" ki do branches. Isi liye hyperbolas open hote hain.

$\cos \theta =-1/e$ ka solution sirf $e>1$ par kyun hota hai? Kyunki $-1/e$, $[-1,1]$ mein hona chahiye; uske liye $e\ge 1$ zaroori hai. Ellipse ke liye ($e<1$) denominator kabhi zero nahin hota, isliye $r$ har jagah finite rehta hai — ek closed orbit.

$e=0$ ke saath: saare $\theta$ ke liye $r=p=h^2/\mu$ — ek constant, isliye orbit radius $a=p$ ka ek circle hai. $r=\frac{h^2}{\mu }=\frac{(6.0\times 10^4{)}^2}{3.986\times 10^5}=\frac{3.6\times 10^9}{3.986\times 10^5}\approx 9032\text{km}.$

$r_{\min }=\frac{p}{1+e}=\frac{a(1-e^2)}{1+e}=\frac{a(1-e)(1+e)}{1+e}=a(1-e).$ $r_{\max }=\frac{p}{1-e}=\frac{a(1-e^2)}{1-e}=a(1+e).$ Add karo: $r_{\min }+r_{\max }=a(1-e)+a(1+e)=2a.$ Yeh exactly ellipse ka long axis hai — confirm karta hai ki $a$ semi-major axis hai. Note karo ki factorisation $1-e^2=(1-e)(1+e)$ hi cheez hai jo ise itna clean banata hai.

Equation ko $\frac{1}{r}=\frac{1}{p}(1+e\cos \theta )$ likhो. Do data points do linear equations dete hain unknowns $x=1/p$ aur $y=e/p$ mein: $\frac{1}{r}=x+y\cos \theta .$

• $\theta =60^{\circ },\cos =0.5$: $\frac{1}{7500}=x+0.5y$.
• $\theta =120^{\circ },\cos =-0.5$: $\frac{1}{9000}=x-0.5y$.

Dono add karo: $2x=\frac{1}{7500}+\frac{1}{9000}=1.3333\times 10^{-4}+1.1111\times 10^{-4}=2.4444\times 10^{-4}$, isliye $x=1.2222\times 10^{-4}\Rightarrow p=\frac{1}{x}\approx 8182\text{km}.$ Subtract karo: $y=\frac{1}{0.5}(\frac{1}{7500}-x)=2(1.3333\times 10^{-4}-1.2222\times 10^{-4})=2.2222\times 10^{-5}$, aur $e=y\cdot p=y/x$: $e=\frac{2.2222\times 10^{-5}}{1.2222\times 10^{-4}}\approx \mathrm{0.1818.}$ Toh $p\approx 8182$ km, $e\approx 0.182$. $1/r$ mein linearise kyun kiya? Kyunki equation $x$ aur $y$ mein linear hai — ek mushkil fit ko simple simultaneous equations mein badal deta hai.

$p$ ke dono expressions ko equal rakho: $\frac{h^2}{\mu }=a(1-e^2)\Rightarrow h=\sqrt{\mu a(1-e^2)}.$ Plug in karo: $h=\sqrt{3.986\times 10^5\times 10000\times (1-0.04)}=\sqrt{3.986\times 10^5\times 9600}.$ $h=\sqrt{3.8266\times 10^9}\approx 6.186\times 10^4{\text{km}}^2/\text{s}.$ Yeh geometry ($a,e$) ko directly conserved $h$ se link karta hai — Vis-viva Equation aur Eccentricity and Orbital Energy ka bridge hai.

• $\theta =90^{\circ }$ par $r$? ::: $r=p$.
• Ellipse ke liye $r_{\min }+r_{\max }$? ::: $2a$.
• Hyperbola par $r\to \infty$ kahan hota hai? ::: $\cos \theta =-1/e$.
• Relation $h(a,e)$? ::: $h=\sqrt{\mu a(1-e^2)}$.