Question bank — Kepler's first law — orbits are conic sections

False. Ek ellipse ke do foci aur ek alag center hota hai; Sun ek focus par hota hai. Doosra focus aur center dono space mein empty points hain.

True. Focus–center distance $ae$ hai, toh focus–focus separation $2ae\to 0$ jab $e\to 0$; dono foci aur center coincide karte hain — woh limiting shape circle hai, isliye chote $e$ ke liye "Sun-at-center" ka illusion hota hai.

True. $r=p/(1+e\cos \theta )$ mein $e=0$ set karne se $r=p$ milta hai, jo constant hai — ek circle. Circle conic family ka ek genuine special case hai, koi alag law nahi.

True. Ek hi derivation ($u^{\prime \prime }+u=\mu /h^2$) inhe sab produce karta hai; kaun sa milega woh initial energy se fix hota hai, jisse $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$ milta hai.

False. Hyperbolic ($e>1$) trajectories real unbound flybys hain; interstellar object ʻOumuamua ka $e\approx 1.2$ tha. "Bound vs unbound" energy sign se set hota hai, $e$ ki kisi cap se nahi.

False. Sirf circle ($e=0$) mein constant speed hoti hai. $e>0$ ke liye body perihelion par sabse fast aur aphelion par sabse slow hoti hai — yeh Kepler's second law in action hai.

True. $\theta =\pm 90^{\circ }$ par, $\cos \theta =0$, toh exactly $r=p$ hai. Yahi $p$ ka geometric meaning hai: major axis ke perpendicular measure ki gayi focus-to-curve distance.

True. $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$ se, $\mathcal{E}<0$ ke saath, term $2\mathcal{E}{h}^2/{\mu }^2$ negative hai aur $h$ badhne par aur zyada negative hota jaata hai, toh $e$ decrease karke $0$ ki taraf jaata hai — orbit rounder hota jaata hai (circle given energy ke liye maximum-$h$ orbit hai).

Beech mein error hai: $\cos \pi =-1$ denominator ko sabse chota ($1-e$) banata hai, jo $r$ ko sabse bada banata hai. Conclusion (aphelion at $\theta =\pi$) sahi hai, lekin reason ulta hai — chota denominator, bada $r$.

Error: $r$ focus se measure kiya jaata hai (polar frame ka origin), kyunki wahi origin hai jahan attracting mass baithta hai. Center se measure karne par yeh clean single-focus equation nahi milegi.

Error: $e=1$ bound nahi hai. Yeh exact threshold $\mathcal{E}=0$ hai — object infinity tak pahunchta hai zero leftover speed ke saath aur kabhi nahi lautta. Bound orbits ke liye strictly $e<1$ chahiye.

Error, ulta hai: planet perihelion ke paas sabse fast move karta hai (equal areas in equal times), isliye wahan kam time spend karta hai. Woh aphelion ke paas ruka rehta hai jahan woh rengta hai.

Error: conservation ka strength se koi lena dena nahi. $h$ conserved hai kyunki gravity central hai ($\hat{r}$ ke along point karta hai), focus ke baare mein zero torque deta hai chahe kitna bhi strong ho.

Error: $u=1/r$ ek general polar-coordinate change of variables hai (jahan $u^{\prime \prime }$ ka matlab $d^{2}u/d{\theta }^2$ hai); yeh kisi bhi central-force radial equation ko $u$ mein ek form mein convert karta hai. Yeh $1/r^2$ ke liye especially clean hai (linear driven SHM deta hai), lekin yeh ek tool hai, gravity ka coincidence nahi.

Error: $\mathcal{E}>0$ se $e>1$ milta hai, ek hyperbola — ek open, unbound curve. Koi bhi positive-energy orbit kisi bhi ellipse mein close nahi hoti, chahe kitna bhi stretched ho.

Kyunki derivation ise force karta hai: $u^{\prime \prime }+u=\mu /h^2$ solve karne par $u=\mu /h^2+A\cos \theta$ milta hai, yaani $r=p/(1+e\cos \theta )$ — ek aisi form jiska polar origin (jahan se $r$ measure hota hai) attracting mass hai. Woh origin ek focus hai, center nahi, aur woh perihelion ($\theta =0$) side par $ae$ off-center baithta hai kyunki $\cos \theta$ term wahan $r$ ko sabse chota banata hai. Asymmetry equation mein built-in hai, sirf ek picture nahi.

Iska general solution $u=\mu /h^2+A\cos \theta$ hai (jahan $u^{\prime \prime }=d^{2}u/d{\theta }^2$), aur $r=1/u$ invert karne par exactly $r=p/(1+e\cos \theta )$ milta hai — conic ka polar form. Driven simple harmonic motion se koi aur function type nahi aa sakta.

Sirf $1/r^2$ (aur linear spring $\propto r$) aisi orbits deta hai jo precession ke bina ek revolution ke baad close hoti hain — yeh Bertrand's theorem hai; doosre powers rosette paths dete hain jo kabhi exactly repeat nahi hote. (Inverse-square law and Bertrand's theorem)

Kyunki $e$ aur $\mathcal{E}$ same information carry karte hain via $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$: $e<1\iff \mathcal{E}<0$ (bound), $e=1\iff \mathcal{E}=0$, $e>1\iff \mathcal{E}>0$. $e$ jaanna already energy ka sign fix kar deta hai.

Hum shape $r(\theta )$ chahte hain, timetable $r(t)$ nahi. Angular momentum $h=r^2\dot{\theta }$ time ko cleanly eliminate karne deta hai, orbit ki geometry seedhi deliver karta hai. (Conservation of angular momentum in central forces)

Negative energy ka matlab hai body gravitational potential well mein trapped hai — uske paas infinity tak pahunchne ki energy nahi hai, toh uski distance $r$ finite rehti hai aur orbit close hoti hai. (Escape velocity and orbital energy)

$r=p$ radius ka ek circle; perihelion aur aphelion coincide karte hain aur speed har jagah constant hoti hai. Yeh degenerate ellipse hai jisme dono foci center par hain.

$r\to \infty$: body ek asymptote ke along infinity ki taraf bhaag jaati hai. $e>1$ ke liye do aisi angles hain (hyperbola ki asymptotic directions); $e=1$ ke liye ek hai, $\theta =\pi$ par.

Nahi. Kyunki $e<1$ hai, denominator kam se kam $1-e>0$ hota hai ($\theta =\pi$ par), toh $r$ finite aur bounded rehta hai — orbit kabhi escape nahi karta.

Koi aphelion nahi hai. Jaise $\theta \to \pi$ hota hai denominator $1+\cos \theta \to 0$ aur $r\to \infty$; object hamesha ke liye door jaata hai, exactly zero speed ke saath infinity reach karta hai.

Formula degenerate ho jaata hai: $p=h^2/\mu =0$, toh "orbit" ek radial plunge seedhe center mein collapse ho jaati hai. Koi sideways motion nahi hai toh koi conic nahi hai — sirf ek line ke along free-fall.

Marginally unbound. Yeh exact dividing line hai ($\mathcal{E}=0$): body escape karti hai lekin infinity par zero remaining speed ke saath pahunchti hai — sabse slow possible escape.

Nahi. Ve same latus-rectum width share karte hain ($\theta =\pm 90^{\circ }$ par $r=p$), lekin alag $e$ bahut alag perihelion $p/(1+e)$ aur aphelion $p/(1-e)$ deta hai — isliye alag overall extent aur shape.