Exercises — Kepler's first law — orbits are conic sections

KYA check karte hain: $e=0.6$ eccentricity ruler par kahan fall karta hai.

• $e=0$ → circle
• $0<e<1$ → ellipse (bound)
• $e=1$ → parabola (marginally escapes)
• $e>1$ → hyperbola (unbound)

Kyunki $0<0.6<1$, yeh ek ellipse hai, aur yeh bound hai — body hamesha ke liye wapas aati rahti hai.

$r$ focus se measure hota hai, jo hamare polar coordinates ka origin hai. Isliye hi Sun (focus par) — empty geometric center nahi — reference point hai. Agar hum $r$ center se lete, toh ek alag, zyada complicated equation milti jisme $1/(1+e\cos \theta )$ form nahi hota.

Sabse tez perihelion par ($\theta =0$, sabse chhota $r$); sabse slow aphelion par ($\theta =\pi$, sabse bada $r$). Kepler's second law ke anusar, body equal times mein equal areas sweep karti hai, isliye jab woh paas hoti hai (chhota $r$) toh same area sweep karne ke liye tez chalti hai.

Step 1 — sabse clean point chuno. Perihelion par $\theta =0$ hai, isliye $\cos \theta =1$ aur equation collapse ho jaata hai: $r_{\min }=\frac{p}{1+e}.$ Yahan kyun? Hum $e$ solve karna chahte hain, lekin equation mein do unknowns hain, $e$ aur angle $\theta$. Perihelion ek aisa point hai jahan $\cos \theta$ ki known value ($+1$) hai, jo angle ko mita deta hai aur $e$ ko single unknown chhod deta hai — ek equation mein ek variable, jise hum hamesha invert kar sakte hain.

Step 2 — $e$ solve karo. $1+e=\frac{p}{r_{\min }}=\frac{8000}{6000}=1.3333,e=\mathrm{0.3333.}$ Orbit ek ellipse hai ($0<e<1$). ✓

Ek ellipse ke liye semi-major axis extremes se $r_{\min }=a(1-e)$ aur $r_{\max }=a(1+e)$ se relate hoti hai.

Yeh formulas kyun? Perihelion aur aphelion major axis ke opposite ends par hain, focus se $a(1-e)$ aur $a(1+e)$ distance par; unka average $a$ hai (center-to-vertex distance) aur unka difference $2ae$ hai (focus offset ka double).

Perihelion: $r_{\min }=1.496\times 10^{11}(1-0.0167)=1.4710\times 10^{11}$ m. Aphelion: $r_{\max }=1.496\times 10^{11}(1+0.0167)=1.5210\times 10^{11}$ m.

Difference sirf $a$ ka lagbhag $1.7\%$ hai — Earth ka orbit almost circular hai.

$\theta =90^{\circ }$ par: $\cos 90^{\circ }=0$, isliye $r=\frac{p}{1+e\cdot 0}=p=8000\text{ km}.$ Yeh angle special kyun hai: $90^{\circ }$ par $\cos$ term bilkul band ho jaata hai, isliye $r$ equal hota hai $p$ ke — eccentricity chahe jo bhi ho. Isliye $p$ ko semi-latus rectum kehte hain — yeh literally orbital radius hai jab tum perihelion se $90^{\circ }$ par, focus ke sidha "upar" khade ho.

$\theta =180^{\circ }$ (aphelion) par: $\cos 180^{\circ }=-1$, isliye $r_{\max }=\frac{p}{1-e}=\frac{8000}{1-0.333}=\frac{8000}{0.667}=11994\text{ km}\approx 1.199\times 10^4\text{ km}.$ Yahan $1-e$ kyun: $\cos 180^{\circ }=-1$ denominator ko $1+e(-1)=1-e$ banata hai, iska sabse chhota possible value, isliye $r$ sabse bada hai — exactly aphelion, jaise expect kiya tha.

(a) Definition ke anusar $p=h^2/\mu$. Yeh relation kyun? Yeh seedha orbit derivation (parent note, Step 6) se aata hai: linearized equation $u^{\prime \prime }+u=\mu /h^2$ mein constant driving term size scale set karta hai, aur $u=1/r$ invert karne par $p=h^2/\mu$ natural length ban jaata hai. Isliye $h$ aur $\mu$ jaanne par $p$ milta hai bina kisi extra data ke. $p=\frac{(4.0\times 10^{15}{)}^2}{1.327\times 10^{20}}=\frac{1.6\times 10^{31}}{1.327\times 10^{20}}=1.2057\times 10^{11}\text{m}.$

(b) Perihelion par, $r_{\min }=p/(1+e)$. Phir perihelion kyun? Wohi reason jaise Exercise 2.1 mein: yeh aika point hai jahan $\cos \theta$ known hai ($+1$), isliye do-unknown orbit equation ek solvable equation ban jaata hai $e$ ke liye. $1+e=\frac{p}{r_{\min }}=\frac{1.2057\times 10^{11}}{8.0\times 10^{10}}=1.5071,e=\mathrm{0.5071.}$ Ek ellipse — yeh comet bound aur periodic hai (ek "returning" comet).

Step 1 — correction term compute karo. $\frac{2\mathcal{E}{h}^2}{{\mu }^2}=\frac{2(5.0\times 10^6)(3.0\times 10^{15}{)}^2}{(1.327\times 10^{20}{)}^2}.$ Numerator: $2\times 5.0\times 10^6\times 9.0\times 10^{30}=9.0\times 10^{37}$. Denominator: $(1.327\times 10^{20}{)}^2=1.7609\times 10^{40}$. Ratio: $9.0\times 10^{37}/1.7609\times 10^{40}=5.111\times 10^{-3}$.

Step 2 — root lo. $e=\sqrt{1+5.111\times 10^{-3}}=\sqrt{1.005111}=\mathrm{1.00255.}$

$\mathcal{E}$ ka sign kyun matter karta hai: upar ki derivation se, $e^2-1=2\mathcal{E}{h}^2/{\mu }^2$, isliye positive energy force karta hai $e^2>1$, yaani $e>1$. Yeh ek hyperbola hai — ek unbound flyby jo leftover speed ke saath escape karta hai. Barely, lekin unmistakably.

Step 1 — $p$ aur $a$ relate karo. Ek ellipse ke liye, $p=a(1-e^2)$. Yeh relation picture par kyun dikhta hai? Solution figure s01 dekho: major axis perihelion se aphelion tak failti hai, total length $r_{\min }+r_{\max }=2a$. Dono extreme distances add karo: $r_{\min }+r_{\max }=\frac{p}{1+e}+\frac{p}{1-e}=p\frac{(1-e)+(1+e)}{(1+e)(1-e)}=\frac{2p}{1-e^2}=2a.$ Isliye $a=\frac{p}{1-e^2}$, equivalently $p=a(1-e^2)$. Yeh bas "dono vertices ka average $a$ hai," algebra mein convert kiya — kuch mysterious nahi. $a=\frac{p}{1-e^2}=\frac{1.2057\times 10^{11}}{1-0.5071^2}=\frac{1.2057\times 10^{11}}{1-0.2571}=\frac{1.2057\times 10^{11}}{0.7429}=1.6229\times 10^{11}\text{m}.$

Step 2 — aphelion. $r_{\max }=a(1+e)=1.6229\times 10^{11}(1.5071)=2.4459\times 10^{11}\text{m}.$ Check: $r_{\max }$ yeh bhi equal hona chahiye $p/(1-e)=1.2057\times 10^{11}/0.4929=2.446\times 10^{11}$ m. ✓

(a) Perihelion par velocity purely tangential hai (radial rate $\dot{r}=0$ wahan, kyunki $r$ minimum par hai). Isliye $h=r_{\min }{v}_{\min }$: $h=(7.0\times 10^6)(9.0\times 10^3)=6.30\times 10^{10}{\text{m}}^2/\text{s}.$ $\dot{r}=0$ kyun? Closest point par, $r$ shrink hona momentarily band kar deta hai aur badhna shuru karta hai — ek minimum — isliye uska time-derivative zero hai, sirf sideways (tangential) motion bachti hai.

(b) $p=h^2/\mu =(6.30\times 10^{10}{)}^2/3.986\times 10^{14}=3.969\times 10^{21}/3.986\times 10^{14}=9.9573\times 10^6\text{m}.$

(c) Perihelion par $r_{\min }=p/(1+e)$: $1+e=\frac{p}{r_{\min }}=\frac{9.9573\times 10^6}{7.0\times 10^6}=1.4225,e=\mathrm{0.4225.}$

(d) $0<e<1$ → ellipse, ek bound satellite orbit. ✓

Step 1 — perihelion par energy. $\frac{1}{2}{v}_{\min }^2=\frac{1}{2}(9.0\times 10^3{)}^2=4.05\times 10^7\text{J/kg}.$ $\frac{\mu }{r_{\min }}=\frac{3.986\times 10^{14}}{7.0\times 10^6}=5.694\times 10^7\text{J/kg}.$ $\mathcal{E}=4.05\times 10^7-5.694\times 10^7=-1.644\times 10^7\text{J/kg}.$ Negative → bound → ellipse. Consistent. ✓

Step 2 — $e$ recover karo. $\frac{2\mathcal{E}{h}^2}{{\mu }^2}=\frac{2(-1.644\times 10^7)(6.30\times 10^{10}{)}^2}{(3.986\times 10^{14}{)}^2}.$ Numerator: $2\times (-1.644\times 10^7)\times 3.969\times 10^{21}=-1.3050\times 10^{29}$. Denominator: $(3.986\times 10^{14}{)}^2=1.5888\times 10^{29}$. Ratio: $-0.8214$. $e=\sqrt{1-0.8214}=\sqrt{0.1786}=\mathrm{0.4226.}$ Part (c) ke $e=0.4225$ se match karta hai (rounding). Do independent routes, same answer — ek strong check. Is ke peeche general speed–radius relation ke liye Vis-viva equation dekho.

Derivation. $r_{\min }=\frac{p}{1+e},r_{\max }=\frac{p}{1-e}.$ Multiply karo: $r_{\min }{r}_{\max }=\frac{p}{1+e}\cdot \frac{p}{1-e}=\frac{p^2}{(1+e)(1-e)}=\frac{p^2}{1-e^2}.$ Ab $p=a(1-e^2)$ ek baar substitute karo: $\frac{p^2}{1-e^2}=\frac{p\cdot a(1-e^2)}{1-e^2}=ap$. Aur kyunki $p=a(1-e^2)$, yeh $a\cdot a(1-e^2)=a^2(1-e^2)$ bhi equal hota hai. Teeno forms agree karte hain. ∎

Numeric check. $r_{\min }=p/(1+e)=1.2057\times 10^{11}/1.5071=8.000\times 10^{10}$ m aur $r_{\max }=2.4459\times 10^{11}$ m use karke: $r_{\min }{r}_{\max }=(8.000\times 10^{10})(2.4459\times 10^{11})=1.9567\times 10^{22}{\text{m}}^2.$ Compare karo $ap=(1.6229\times 10^{11})(1.2057\times 10^{11})=1.9567\times 10^{22}{\text{m}}^2$ se. ✓

(a) $p=r_{\min }(1+e)=3.0\times 10^{10}(1+1.20)=3.0\times 10^{10}\times 2.20=6.60\times 10^{10}$ m.

(b) $r\to \infty$ matlab denominator $1+e\cos \theta \to 0$, yaani $\cos {\theta }_{\infty }=-\frac{1}{e}=-\frac{1}{1.20}=-\mathrm{0.8333.}$ Principal value hai ${\theta }_{\infty }=\arccos (-0.8333)=146.44^{\circ }=2.5559\text{rad}.$ $\arccos$ kyun? Hum escape angle ka cosine jaante hain aur angle chahiye — $\arccos$ exactly "kaun sa angle is cosine ke saath hai?" ka undo-operation hai, aur yeh $[0^{\circ },180^{\circ }]$ mein single value return karta hai. Lekin cosine even hota hai: $\cos (-\theta )=\cos \theta$, isliye $\theta =-146.44^{\circ }$ bhi same equation satisfy karta hai. Isliye ek hyperbola ke do asymptote angles hote hain, ${\theta }_{\infty }=\pm 146.44^{\circ }$ — incoming aur outgoing directions jis par object travel karta hai. Physically yeh sirf wedge $-146.44^{\circ }<\theta <+146.44^{\circ }$ sweep karta hai; us se aage denominator $1+e\cos \theta$ negative ho jaata hai (forbidden, jaise upar edge-case callout mein bataya gaya).

(c) Geometric reason (figure s02 dekho). Ellipse ke liye $e<1$, isliye $-1/e<-1$. Lekin $\cos \theta$ kabhi $-1$ se neeche nahi jaata — yeh $[-1,+1]$ mein rehta hai. Isliye equation $\cos \theta =-1/e$ ka koi solution nahi: denominator $1+e\cos \theta$ har angle ke liye $\ge 1-e>0$ rehta hai, $r$ hamesha finite hota hai, aur curve khud par loop back karta hai — closed orbit jisme koi escape direction nahi. Hyperbola ke liye $e>1$, isliye $-1/e$ inside $[-1,0)$ hai, ek achievable cosine. Us angle par denominator zero ho jaata hai aur $r$ blow up karta hai; orbit infinity ki taraf do straight-line asymptotes $\pm {\theta }_{\infty }$ par chala jaata hai. Short mein: infinity ek real angle par reachable hai ya nahi yeh bilkul isi baat se decide hota hai ki $1/e\le 1$ hai ya nahi, yaani ki $e\ge 1$ hai ya nahi.

$e$ nikalne ke liye sabse clean inversion point ::: Perihelion, jahan $\cos \theta =1$ deta hai $r_{\min }=p/(1+e)$. Perihelion data ke terms mein $h$ ::: $h=r_{\min }{v}_{\min }$ (velocity wahan purely tangential hoti hai). Hyperbola ke escape (asymptote) angles ::: $\cos {\theta }_{\infty }=-1/e$, deta hai $\theta =\pm {\theta }_{\infty }$; sirf tab exist karta hai jab $e\ge 1$. Bound ellipse ke liye energy ka sign ::: $\mathcal{E}<0$. Perihelion aur aphelion distances ka product ::: $r_{\min }{r}_{\max }=ap=a^2(1-e^2)$. $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$ kahan se aata hai ::: Perihelion par $\mathcal{E}=\frac{1}{2}{v}^2-\mu /r$ evaluate karo; yeh rearrange hota hai $e^2-1=2\mathcal{E}{h}^2/{\mu }^2$ mein.