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

3.2.3 · D2 · Physics › Orbital Mechanics & Astrodynamics › Orbit equation r = p - (1 + e·cos θ) — derivation from equat

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Step 1 — Ek dot, ek arrow, aur ek kheench

KYA. Socho ek tiny planet space mein kisi jagah par hai. Bade central mass (Sun) se planet ki taraf ek arrow banao. Woh arrow position vector $\vec{r}$ hai: yeh batata hai "kaunsi disha mein, aur kitni door." Iski length distance $r$ hai. Ek chhoti hat, $\hat{r}$, ka matlab hai "same direction mein lekin length 1 tak shrink kiya hua" — ek pure pointer.

KYUN. Pehle yeh bolne se pehle ki planet kaise move karta hai, humein woh kahan hai yeh naam dena padega. Iske baad sab kuch $\vec{r}$ aur $\hat{r}$ mein likha jaata hai, isliye pehle inhein earn karte hain.

PICTURE. Blue arrow $\vec{r}$ hai. Uske saath chhota orange arrow $\hat{r}$ hai (length 1). Gravity — red arrow — ulti taraf point karta hai, seedha Sun ki taraf wapas.

Newton's law, ek waqt mein ek symbol: $\underset{\text{arrow ki tip kaise accelerate karti hai}}{\underbrace{\ddot{\vec{r}}}}=\underset{\text{andar ki taraf}}{\underbrace{-}}\underset{\text{taakat, door jaane par kamzor}}{\underbrace{\frac{\mu }{r^2}}}\underset{\text{disha: Sun ki line ke saath}}{\underbrace{\hat{r}}}$

Minus sign gravity ki poori personality hai: yeh kabhi sideways nahi dhakelta, sirf centre ki taraf wapas kheenchta hai. Yeh ek fact hi agla step drive karta hai. Dekho Central Force Motion.

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Step 2 — Position ko ek angle se naam dena, aur uski do natural directions

Pehle "spin" ke baare mein ya "angle kitni tezi se sweep hota hai" ke baare mein baat karne se pehle, humein ek angle hona chahiye. Toh ab hum polar coordinates set up karte hain — yeh sabse pehli cheez hai jo humein chahiye.

KYA. Us flat plane mein jisme planet move karta hai, hum uski position do numbers se describe karte hain: Sun se distance $r$, aur ek chosen starting ray se measure kiya hua angle $\theta$. Yeh $\theta$ true anomaly hai. Planet ki location par hum do chhote unit arrows banate hain (har ek ki length 1):

• $\hat{r}$ — radial unit vector, $\vec{r}$ ke saath Sun se seedha baahir ki taraf point karta hai.
• $\hat{\theta }$ — tangential unit vector, sideways point karta hai, exactly $90^{\circ }$ us disha mein jahan angle $\theta$ badhta hai.

KYUN. Gravity purely $\hat{r}$ ke saath point karti hai. Agar hum har motion ko ek "$\hat{r}$ part" aur ek "$\hat{\theta }$ part" mein split karein, toh gravity unme se sirf ek mein aati hai aur doosra clean rehta hai. Yeh do arrows ek central force ke liye natural ruler hain, isliye inhe sabse pehle build karte hain.

PICTURE. Planet apne curve par, Sun origin par, starting ray se sweep hua angle $\theta$, aur do perpendicular unit arrows $\hat{r}$ (baahir) aur $\hat{\theta }$ (sideways-aage).

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Step 3 — Acceleration ko radial aur sideways parts mein split karna

KYA. Kyunki $\hat{r}$ aur $\hat{\theta }$ planet ke move karne par rotate karte hain, inhe acceleration likhne se chaar pieces milte hain (yeh standard polar decomposition hai). Direction ke hisaab se group kiya hua: $\vec{a}=\underset{\hat{r}\text{ ke saath}}{\underbrace{(\ddot{r}-r{\dot{\theta }}^2)}}\hat{r}+\underset{\hat{\theta }\text{ ke saath}}{\underbrace{(r\ddot{\theta }+2\dot{r}\dot{\theta })}}\hat{\theta }$

KYUN yeh terms. Har piece ka ek seedha matlab hai jo feel ho sakta hai:

• $\ddot{r}$ — Sun ki line ke saath genuinely speed up ya slow down karna.
• $-r{\dot{\theta }}^2$ — curve par travel karne ka baahri "sling" (merry-go-round wala push). Yahan $\dot{\theta }$ ka matlab hai "angle kitni tezi se sweep hota hai"; yeh abhi define kiye $\theta$ ke change ki rate hai.
• $r\ddot{\theta }$ — angle ka speed up ya slow down hona.
• $2\dot{r}\dot{\theta }$ — Coriolis piece: rotate karte hue andar ya baahir move karna.

PICTURE. Planet par chaar contributions chhote arrows ke roop mein draw kiye gaye, do $\hat{r}$ (radial) terms aur do $\hat{\theta }$ (tangential) terms mein colour-split kiye gaye.

Ab Step 1 ka Newton's law apply karo. Gravity $-\frac{\mu }{r^2}\hat{r}$ hai — pure $\hat{r}$, zero $\hat{\theta }$. Do directions ko alag-alag match karte hue: $\underset{\text{radial: force hai}}{\underbrace{\ddot{r}-r{\dot{\theta }}^2=-\frac{\mu }{r^2}}},\underset{\text{tangential: force nahi}}{\underbrace{r\ddot{\theta }+2\dot{r}\dot{\theta }=0}}$

Sideways equation humein conserved "spin" free mein de deta hai. Isse $r$ se multiply karo: $r^2\ddot{\theta }+2r\dot{r}\dot{\theta }=0.$ Dhyan se dekho: left side exactly $r^2\dot{\theta }$ ka time-derivative hai (product rule se, $\frac{d}{dt}(r^2\dot{\theta })=2r\dot{r}\dot{\theta }+r^2\ddot{\theta }$). Toh: $\frac{d}{dt}\left(r^2\dot{\theta }\right)=0⟹h\equiv r^2\dot{\theta }=\text{constant}.$

Woh constant $h$ specific angular momentum hai. Humein koi 3D machinery nahi chahiye thi — no-sideways-force equation hi conservation law hai. Dekho Specific Angular Momentum h aur Kepler's Laws (yeh constant Kepler ka equal-area rule hai).

Recall "Specific" kyun, aur yeh

$|\vec{r}\times \dot{\vec{r}}|$ ke kyun barabar hai "Specific" = per unit mass (humne Step 1 mein hi $m$ divide kar diya tha). Cross product $\vec{r}\times \dot{\vec{r}}$ har second mein sweep hua area measure karta hai; polar form mein uski magnitude exactly $r^2\dot{\theta }$ nikhalti hai, wahi $h$. Dono raaste ek jagah milte hain.

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Step 4 — Jaadu ka swap $u=1/r$

Ab hamare paas do facts hain combine karne ke liye: radial equation $\ddot{r}-r{\dot{\theta }}^2=-\mu /r^2$, aur $h=r^2\dot{\theta }$. Lekin yeh $r$, $\theta$, aur time ko mix karti hain. Hum chahte hain $r$ ka ek clean equation sirf $\theta$ ke function ke roop mein.

KYA. Define karo $u=1/r$ (distance ki jagah "closeness" — bada $u$ = Sun ke paas). Pehle spin law ko rewrite karo: $h=r^2\dot{\theta }=\frac{\dot{\theta }}{u^2}$, toh $\dot{\theta }=h{u}^2.$

Ab $\dot{r}$ ko time-rate se $\theta$-rate mein chain rule use karke convert karo (neeche).

KYUN $d/dt\to d/d\theta$ swap. Chain rule kehta hai: yeh jaanne ke liye ki kuch cheez har second kitna change hoti hai, yeh multiply karo ki woh angle ke hisaab se kitna change hoti hai aur angle kitni tezi se ghoomta hai. Symbols mein, kisi bhi quantity $Q$ ke liye, $\frac{dQ}{dt}=\frac{dQ}{d\theta }\cdot \underset{=\dot{\theta }=h{u}^2}{\underbrace{\frac{d\theta }{dt}}}.$ Hum yeh isliye karte hain kyunki $\theta$ sirf badhta hai (planet ek hi taraf ghoomta rehta hai), toh yeh ek honest independent variable hai — $t$ se alag, jise hum hatana chahte hain.

$\dot{r}$ ka step-by-step conversion (har arrow ek chhota move hai): $\dot{r}=\frac{dr}{dt}\stackrel{\text{chain rule}}{=}\frac{dr}{d\theta }\dot{\theta }\stackrel{r=1/u}{=}\frac{d}{d\theta }\left(\frac{1}{u}\right)(h{u}^2)=(-\frac{1}{u^2}\frac{du}{d\theta })(h{u}^2)=-h\frac{du}{d\theta }.$ Yahan $\frac{d}{d\theta }(1/u)=-\frac{1}{u^2}\frac{du}{d\theta }$ khud $1/u$ par chain rule hai; do $u^2$ phir cancel ho jaate hain.

Ab $\ddot{r}$ paane ke liye ek baar aur differentiate karo: $\ddot{r}=\frac{d}{dt}(-h\frac{du}{d\theta })\stackrel{\text{chain rule}}{=}\frac{d}{d\theta }(-h\frac{du}{d\theta })\dot{\theta }=(-h\frac{d^{2}u}{d{\theta }^2})(h{u}^2)=-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}.$

PICTURE. Side by side do curves: $r(\theta )$ perihelion par minimum tak dip karta hai (ek valley), jabki $u(\theta )=1/r$ wahan peak karta hai (ek hill). Swap valleys ko hills mein flip karta hai — aur, sabse zaroori, ek tedhi curve ko plain cosine wave mein badal deta hai, jo agla step trivial bana deta hai.

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Step 5 — Equation ek swing ban jaati hai

KYA. Radial equation lo aur har time-cheez ko $\theta$-cheez se replace karo. Humein do substitutions chahiye:

1. $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$ (abhi derive kiya).
2. $r{\dot{\theta }}^2$: kyunki $r=1/u$ aur $\dot{\theta }=h{u}^2$, hume milta hai $r{\dot{\theta }}^2=\frac{1}{u}(h{u}^2{)}^2=\frac{h^2{u}^4}{u}=h^2{u}^3$.

Dono ko $\ddot{r}-r{\dot{\theta }}^2=-\mu /r^2$ mein substitute karo, aur note karo right side par $1/r^2=u^2$ hai: $\underset{\ddot{r}}{\underbrace{-h^2{u}^2{u}^{\prime \prime }}}-\underset{r{\dot{\theta }}^2}{\underbrace{h^2{u}^3}}=-\mu u^2.$

Ab ek move at a time simplify karo. Har term ko $-h^2{u}^2$ se divide karo (allowed kyunki $u\ne 0$: planet kabhi infinitely door nahi hota): $\frac{-h^2{u}^2{u}^{\prime \prime }}{-h^2{u}^2}+\frac{-h^2{u}^3}{-h^2{u}^2}=\frac{-\mu u^2}{-h^2{u}^2}⟹u^{\prime \prime }+u=\frac{\mu }{h^2}.$ Pehla term $u^{\prime \prime }$ rakhta hai, doosra $u$ chodta hai, right side constant $\mu /h^2$ chodta hai. Clean: $\boxed{u^{\prime \prime }+u=\frac{\mu }{h^2}}$

KYUN yeh matter karta hai. Yeh physics mein sabse famous equation hai — ek simple harmonic oscillator (spring par mass, chhote swings ke liye pendulum) jisme right side par constant push $\mu /h^2$ hai. Humne $u$-substitution exactly isliye chase kiya taaki yahan pahunche, kyunki hum is equation ka answer pehle se jaante hain.

PICTURE. Spring par ek mass: kheecho, yeh constant force se offset rest position ke around oscillate karta hai. Yahan $u$ "oscillate" karta hai time mein nahi balki $\theta$ ke ghoomne ke saath — ek full swing per orbit. Rest position height $\mu /h^2$ par hai.

Term by term: $\underset{\text{u-wave ki curvature}}{\underbrace{u^{\prime \prime }}}+\underset{\text{centre ki taraf restoring pull}}{\underbrace{u}}=\underset{\text{constant offset (rest line)}}{\underbrace{\frac{\mu }{h^2}}}$

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Step 6 — Solution padhna, piece by piece

KYA. Equation $u^{\prime \prime }+u=\mu /h^2$ linear hai, toh iska poora answer hai ek particular solution plus general free wobble:

• Particular: ek constant jo isse satisfy kare. Try karo $u=C$; phir $u^{\prime \prime }=0$, toh $0+C=\mu /h^2$, jo deta hai $C=\mu /h^2$. Yeh flat "rest line" hai.
• Homogeneous (free wobble): $u^{\prime \prime }+u=0$ solve karo. Woh functions jinki second derivative minus themselves hoti hai exactly $\cos$ aur $\sin$ hain, toh general wobble hai $A\cos (\theta -{\theta }_0)$ constants $A,{\theta }_0$ ke liye.

Inhe add karo: $u(\theta )=\underset{\text{rest line}}{\underbrace{\frac{\mu }{h^2}}}+\underset{\text{free wobble}}{\underbrace{A\cos (\theta -{\theta }_0)}}.$

KYUN constants. $A$ aur ${\theta }_0$ planet launch hone ke tarike se fix hote hain. ${\theta }_0$ ka apna free choice hum ${\theta }_0=0$ set karke use karte hain — yeh sirf decide karta hai ki hum angle kahan se count karna shuru karein. Hum ise wobble ke peak par rakhte hain (sabse bada $u$ = sabse chhota $r$ = sabse nazdik point = perihelion).

PICTURE. Ek orbit mein solution: $\mu /h^2$ par flat rest line, uske upar amplitude $A$ ki cosine swaari kar rahi hai. Peak = perihelion $\theta =0$ par; trough = aphelion $\theta =\pi$ par.

Ab $u$ ko $r$ mein wapas flip karo, ek algebraic move at a time. ${\theta }_0=0$ ke saath: $\frac{1}{r}=\frac{\mu }{h^2}+A\cos \theta .$ Right side se $\mu /h^2$ factor karo: $\frac{1}{r}=\frac{\mu }{h^2}(1+\underset{\text{isko e kaho}}{\underbrace{\frac{A{h}^2}{\mu }}}\cos \theta ).$ Us dimensionless bundle ko naam do $e\equiv A{h}^2/\mu$ (eccentricity). Dono sides ka reciprocal lo: $\boxed{r=\frac{h^2/\mu }{1+e\cos \theta }=\frac{p}{1+e\cos \theta }},p\equiv \frac{h^2}{\mu }.$ Scale $p=h^2/\mu$ aur shape number $e$ apne aap nikal aate hain. Bada $A$ ⇒ bada $e$ ⇒ zyada lopsided orbit. Dekho Eccentricity and Orbital Energy.

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Step 7 — Har shape, har case

KYA. Ek formula secretly chaar alag curves contain karta hai, jo poori tarah $e$ se decide hote hain (Conic Sections se):

$e$	denominator $1+e\cos \theta$	kya hota hai	shape
$0$	hamesha $1$	$r=p$ har jagah	circle
$0<e<1$	kabhi $0$ nahi hota	$r$ finite rehta hai, closed loop	ellipse
$1$	$\theta \to \pi$ par $\to 0$	$r\to \infty$: escape	parabola
$>1$	kisi $\theta <\pi$ par $0$ hota hai	$r\to \infty$ jaldi: ud jaata hai	hyperbola

KYUN. Humein sabhi cases check karne chahiye, sirf tidy ellipse nahi. Key sawaal: kya denominator zero tak pahunch sakta hai? Jahan yeh hota hai, $r$ infinity tak blow up hota hai aur orbit open hai (body kabhi wapas nahi aati).

• $e<1$: kyunki $e\cos \theta \ge -e>-1$, denominator hamesha positive hai — $r$ hamesha finite hai — orbit close hoti hai.
• $e=1$: $\theta =\pi$ par, $\cos \theta =-1$, denominator $=1-1=0$ — $r\to \infty$ exactly peeche. Borderline escape (parabola).
• $e>1$: denominator $\cos \theta =-1/e$ par $0$ hota hai ($\pi$ se pehle ka ek angle). Uske baad $r$ negative ho jaata — meaningless — toh real path sirf us arc par exist karta hai jahan denominator positive rehta hai. Woh arc ek hyperbola hai jisme do "asymptote" directions hain.

PICTURE. Charon curves same focus (Sun) aur same $p$ share karte hue, colour-coded, parabola aur hyperbola ke runaway arms infinity ki taraf shoot karte hue.

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Ek-picture summary

Yeh raha poora safar ek canvas par: gravity ka inward pull → $\hat{r}$ aur $\hat{\theta }$ mein split → sideways part deta hai $h=$const → radial part + $u=1/r$ swap → spring equation → uska cosine solution → conic. Arrows ko left se right follow karo.

Recall Feynman retelling — poora walkthrough simple words mein

Ek fact se shuru karo: Sun planet ko seedha andar kheenchta hai, kabhi sideways nahi. Iske liye hum do chhote arrows set up karte hain jo planet ke saath ride karte hain — ek baahir point karta hai ($\hat{r}$), ek sideways ($\hat{\theta }$). Hum planet ki acceleration ko in dono directions mein split karte hain. Gravity poori tarah outward-inward hai, toh sideways equation kehti hai "koi sideways force nahi" — aur woh ek line, clean ki gayi, humein batati hai ki $r^2\dot{\theta }$ kabhi change nahi ho sakta. Yeh conserved "spin" hai, aur yeh poore dance ko ek plane mein bhi flatten karta hai.

Bacha hua outward-inward equation awkward hai kyunki yeh distance, angle aur time ko mix karti hai. Hum ek clever trick khelein: distance ki jagah closeness $u=1/r$ track karo, aur "har second kahan hai?" ki jagah "har angle par kahan hai?" poochho. Chain rule translating karta hai. Woh double swap gravity ke awkward $1/r^2$ ko magically cancel kar deta hai aur humein physics ki sabse famous equation deta hai: ek spring par mass. Spring ka answer ek rest position plus ek gentle cosine wobble hai.

Swap undo karo, tidily factor karo, aur woh cosine ban jaata hai $r=p/(1+e\cos \theta )$. Size $p$ aur shape $e$ apne aap pop out karte hain. Chhota $e$: circle ya gentle oval. $e=1$: planet barely ek parabola par escape karta hai. $e>1$: yeh ek visitor hai, ek baar hyperbola par swing karke aata hai aur kabhi wapas nahi aata. Ek equation, chaar destinies — sab "gravity sirf andar kheenchti hai" se.

Recall Rapid self-test

• Kaun si equation $h=$const deti hai, aur kaise? ::: Tangential (sideways) wali, $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$; $r$ se multiply karne par yeh $\frac{d}{dt}(r^2\dot{\theta })=0$ hai.
• Kaun sa tool $d/dt$ ko $d/d\theta$ mein translate karta hai? ::: Chain rule, $\frac{dQ}{dt}=\frac{dQ}{d\theta }\dot{\theta }$ jahaan $\dot{\theta }=h{u}^2$.
• $u^{\prime \prime }$ ka kya matlab hai? ::: $d^{2}u/d{\theta }^2$ — $\theta$ ke respect mein do baar differentiate karo.
• $u=1/r$ kya faayda deta hai? ::: Yeh $1/r^2$ force ko cancel karta hai aur ODE ko $u^{\prime \prime }+u=\mu /h^2$, ek spring equation, mein badal deta hai.
• Orbit infinity par kab open hoti hai? ::: Jab denominator $1+e\cos \theta$ $0$ tak pahunch sakta hai, yaani $e\ge 1$.