Kepler's first law — orbits are conic sections

3.2.5 · Physics › Orbital Mechanics & Astrodynamics

The law (ek line mein): Kisi planet ki orbit ek ellipse hoti hai, jisme Sun ek focus par hota hai (center par nahi!). Zyada broadly bolen to, kisi bhi do-body gravitational orbit ki shape ek conic section hoti hai — ellipse, parabola, ya hyperbola.

Yeh law exist hi kyun karta hai?

KYA chahiye: prove karna ki orbit ki shape $r(\theta )$ ek conic hai. KYUN matter karta hai: yeh batata hai ki planet/satellite kahan hai, trajectories classify karne deta hai (bound vs escape), aur poori astrodynamics ki neenv hai. KAISE pahunche: angular momentum conservation + Newton's gravity combine karo, aur ek clever substitution se radial equation solve karo.

Building blocks

Derivation scratch se (note ka dil)

Humare paas mass $m$ ka ek chhota body hai jo origin par rakhe bade mass $M$ ke around orbit kar raha hai. Maan lo $\mu =GM$.

Step 1 — Equation of motion

Newton's second law gravity ke saath: $\ddot{\vec{r}}=-\frac{\mu }{r^2}\hat{r}$ Yeh step kyun? Gravity $m$ se $M$ ki taraf (origin ki taraf) point karti hai, magnitude per unit mass $GM/r^2$ hai.

Step 2 — Angular momentum conservation use karo

Kyunki force central hai, $h=r^2\dot{\theta }$ constant hai. Isse hum variable change karenge — time $t$ se angle $\theta$ par. Kyun? Time awkward hai; hum shape $r(\theta )$ chahte hain, timetable nahi. $t$ eliminate karne se seedha geometry milti hai.

Step 3 — Magic substitution $u=1/r$

Maan lo $u=1/r$. Tab: $\dot{\theta }=\frac{h}{r^2}=h{u}^2$ $\dot{r}=\frac{dr}{d\theta }\dot{\theta }=\frac{d}{d\theta }\left(\frac{1}{u}\right)h{u}^2=-\frac{1}{u^2}\frac{du}{d\theta }h{u}^2=-h\frac{du}{d\theta }$ $\ddot{r}=\frac{d}{dt}\left(-h\frac{du}{d\theta }\right)=-h\frac{d^{2}u}{d{\theta }^2}\dot{\theta }=-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}$ Yeh step kyun? $u=1/r$ ugly nonlinear $1/r^2$ equation ko ek clean linear equation mein badal deta hai — yahi woh famous trick hai jo poori cheez haath se solve karne layak banati hai.

Step 4 — Radial equation of motion

Radial acceleration (polar coordinates mein) $\ddot{r}-r{\dot{\theta }}^2$ hai. Ise radial force per mass ke barabar set karo: $\ddot{r}-r{\dot{\theta }}^2=-\frac{\mu }{r^2}$ Substitute karo $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$, $r=1/u$, ${\dot{\theta }}^2=h^2{u}^4$: $-h^2{u}^2{u}^{\prime \prime }-\frac{1}{u}{h}^2{u}^4=-\mu u^2$ $-h^2{u}^2$ se divide karo: $u^{\prime \prime }+u=\frac{\mu }{h^2}$ Yeh step kyun? Dekho yeh — yeh sirf simple harmonic motion hai constant driving ke saath! Hum pehle se jaante hain ise kaise solve karein.

Step 5 — Linear ODE solve karo

$u(\theta )=\frac{\mu }{h^2}+A\cos (\theta -{\theta }_0)$ Angle origin choose karo taki ${\theta }_0=0$ ho (perihelion se $\theta$ measure karo). Phir $r=1/u$ par wapas jao: $\boxed{r(\theta )=\frac{h^2/\mu }{1+(A{h}^2/\mu )\cos \theta }}$

Step 6 — Conic padhlo

Conic $r=\frac{p}{1+e\cos \theta }$ se compare karo: $p=\frac{h^2}{\mu },e=\frac{A{h}^2}{\mu }$ Yeh IS ek conic section hai. Kepler's first law proved. ∎

Eccentricity ko energy se connect karna (taki pata chale kaun si conic hai)

Total energy per unit mass $\mathcal{E}=\frac{1}{2}{v}^2-\mu /r$ hai. Ise work through karne par (using $v^2={\dot{r}}^2+r^2{\dot{\theta }}^2$ aur upar ki solution) yeh khoobsurat relation milti hai: $e=\sqrt{1+\frac{2\mathcal{E}{h}^2}{{\mu }^2}}$

Worked examples

Common mistakes (Steel-manned)

Flashcards

Connections

• Kepler's second law — equal areas in equal times (conservation of $h$, same ingredient)
• Kepler's third law — period vs semi-major axis (ellipse geometry se follow karta hai)
• Conservation of angular momentum in central forces
• Vis-viva equation ($v^2=\mu (2/r-1/a)$ — is orbit ka energy version)
• Conic sections — geometry of ellipse, parabola, hyperbola
• Two-body problem and reduced mass
• Escape velocity and orbital energy
• Inverse-square law and Bertrand's theorem (kyun sirf $1/r^2$ orbits close karta hai)

Concept Map