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

3.2.3 · D3 · Physics › Orbital Mechanics & Astrodynamics › Orbit equation r = p - (1 + e·cos θ) — derivation from equat

Shuru karne se pehle, formula ke paanch characters se milo — har ek plain words mein earned hai:

$r$ — focus se doori (Sun/planet wahin baith ta hai), hamesha $r > 0$ .
$θ$ — true anomaly, wo angle jo focus par measure hota hai, perihelion se shuru hota hai ( $θ = 0$ ).
$e$ — eccentricity, ek pure number $\geq 0$ jo shape decide karta hai.
$μ$ — gravitational parameter, $μ = GM$ : yeh gravitational constant $G$ ko central mass $M$ ke saath bundle karta hai, aur yeh ek hi number hai jo measure karta hai "pull kitna strong hai". Bada $μ$ matlab zyada strong gravity.
$h$ — specific angular momentum, $h = r^{2} \dot{θ}$ : orbiting body ka "spin amount per unit mass". Kyunki gravity seedha andar ki taraf kheenchti hai isliye $h$ kabhi change nahi kar sakti, toh $h$ poore orbit mein constant rehta hai (dekho Specific Angular Momentum h).

$μ$ aur $h$ ko naam de diya, toh orbit ki scale hai:

$p = \frac{h ^{2}}{μ}$ — semi-latus rectum, $r$ ki value jab $θ = 9 0^{\circ}$ ho (spin-squared divided by pull-strength).

Agar inme se koi cheez abhi bhi shaky lagti hai, toh derivation note for $r = p/(1+e\cos\theta)$ Newton's law se har ek cheez earn karta hai.

The scenario matrix

Is equation ke baare mein har sawaal inhi cells mein se kisi ek mein aata hai. Neeche ke examples tagged hain us cell ke saath jise woh cover karte hain.

Cell	Kya badlta hai	Kisko cover karta hai
A. Circle	$e = 0$ (degenerate: shape term vanish ho jaata hai)	Ex 1
B. Ellipse, given extremes	$0 < e < 1$ , $r_{m i n}, r_{m a x}$ se $e, p$ nikalo	Ex 2
C. Quadrant sweep	har quadrant mein $cos θ$ ka sign ( $θ = 0, 9 0^{\circ}, 18 0^{\circ}, 27 0^{\circ}$ )	Ex 3
D. Parabola limit	$e = 1$ (aphelion $\to \infty$ )	Ex 4
E. Hyperbola / escape	$e > 1$ (ek forbidden angle exist karta hai)	Ex 5
F. Recover $h$ or $μ$	$p = h^{2} / μ$ ko invert karo	Ex 6
G. Real-world word problem	messy units, physical sanity check	Ex 7
H. Exam twist	$r$ aur $θ$ diye hain, $θ$ ke liye solve karo (do answers!)	Ex 8

Worked examples

Ex 1 — Cell A: circle ( $e = 0$ )

Worked example

Ek weather satellite ka $p = 8000$ km aur $e = 0$ hai. $θ = 0^{\circ}$ , $9 0^{\circ}$ , aur $18 0^{\circ}$ par $r$ nikalo.

Forecast: padhne se pehle guess karo — kya $r$ badalta hai jab $θ$ ghoomta hai?

$e = 0$ ke saath equation likho: $r = \frac{p}{1 + 0 \cdot cos θ} = \frac{p}{1} = p$ . Yeh step kyun? Jab $e = 0$ hota hai toh poora $e cos θ$ term khatam ho jaata hai, isliye $θ$ bilkul bahar nikal jaata hai. Yeh degenerate case hai jahan shape term vanish ho jaata hai.
Isliye $r = 8000$ km har angle par. Yeh step kyun? Saare $θ$ ke liye constant $r$ circle ki definition hai — koi perihelion nahi, koi aphelion nahi, koi preferred direction nahi.

Verify: $r$ ke do extremes wahan hote hain jahan $cos θ$ sabse extreme hota hai: kyunki $cos θ$ $[- 1, + 1]$ par range karta hai, sabse bada denominator (sabse chota $r$ ) $cos θ = + 1$ par hota hai, yaani $θ = 0$ , jisse $r_{m i n} = p / (1 + e)$ milta hai; sabse chota denominator (sabse bada $r$ ) $cos θ = - 1$ par hota hai, yaani $θ = 18 0^{\circ}$ , jisse $r_{m a x} = p / (1 - e)$ milta hai. Yahan $r_{m i n} = p / (1 + e) = 8000/1 = 8000$ aur $r_{m a x} = p / (1 - e) = 8000/1 = 8000$ . Dono agree karte hain — do "extremes" ek hi radius mein collapse ho jaate hain, bilkul waisa hi jaise ek circle demand karta hai. Units: poore mein km. ✓

Ex 2 — Cell B: apne do extremes se ellipse nikalna

Worked example

Ek comet ka perihelion $r_{p} = 0.6$ AU aur aphelion $r_{a} = 5.4$ AU hai. $e$ aur $p$ nikalo.

Forecast: kya $e$ zyada $0$ ke paas hoga (round) ya $1$ ke paas (bahut stretched)?

Extremes hain $r_{p} = \frac{p}{1 + e}$ ( $θ = 0$ par) aur $r_{a} = \frac{p}{1 - e}$ ( $θ = π$ par). Yeh step kyun? $cos θ$ sabse bada ( $+ 1$ ) $θ = 0$ par hota hai jisse sabse chota $r$ milta hai, aur sabse chota ( $- 1$ ) $θ = π$ par hota hai jisse sabse bada $r$ milta hai. Yeh sirf $e < 1$ ke liye valid hai; hum ise aakhir mein confirm karenge.
Dono ko divide karo: $\frac{r _{a}}{r _{p}} = \frac{1 + e}{1 - e}$ , toh $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{5.4 - 0.6}{5.4 + 0.6} = \frac{4.8}{6.0} = 0.8$ . Yeh step kyun? Divide karne se unknown $p$ cancel ho jaata hai, sirf $e$ mein ek equation bacha rehta hai.
Ab $p = r_{p} (1 + e) = 0.6 (1.8) = 1.08$ AU. Yeh step kyun? $e$ pata hone ke baad, koi bhi extreme equation seedha $p$ deta hai.

Verify: aphelion mein back-substitute karo: $p / (1 - e) = 1.08/ (1 - 0.8) = 1.08/0.2 = 5.4$ AU $= r_{a}$ . ✓ Aur $0 < e = 0.8 < 1$ confirm karta hai ki yeh sach mein ek (bahut stretched) ellipse hai, toh Step 1 legitimate tha.

Ex 3 — Cell C: charon quadrants mein sweep karna

Worked example

Ek orbit ka $p = 10000$ km aur $e = 0.5$ hai. $θ = 0^{\circ}, 9 0^{\circ}, 18 0^{\circ}, 27 0^{\circ}$ par $r$ tabulate karo aur motion describe karo.

Forecast: inme se kin chaar angles par body focus ke sabse paas hai? Sabse door?

Neeche ki figure bilkul yahi orbit plot karti hai: orange star focus hai, violet curve path hai, aur chaar dashed radii woh chaar angles hain jo hum compute karte hain. Dekho kaise radius arrow sabse chota daayein taraf point karta hai ( $θ = 0$ ) aur sabse bada baayein taraf ( $θ = 18 0^{\circ}$ ).

$θ = 0^{\circ}$ (perihelion): $cos 0 = + 1$ , toh $r = \frac{10000}{1 + 0.5 ( 1 )} = \frac{10000}{1.5} = 6666.7$ km — sabse chota. Yeh figure mein daayein point karne wala chota magenta arrow hai. Yeh step kyun? Poora $+ e$ add karne se denominator sabse bada ban jaata hai, isliye $r$ sabse chota hota hai. Yeh sabse paas ka point hai.
$θ = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0$ , toh $r = \frac{10000}{1 + 0} = 10000$ km $= p$ . Yeh $p$ length ka upar ki taraf orange arrow hai. Yeh step kyun? $9 0^{\circ}$ par cosine term vanish ho jaata hai aur $r$ semi-latus rectum ke barabar ho jaata hai — "sideways" reference distance. Step 4 ke hisaab se symmetry se $27 0^{\circ}$ par bhi same hoga.
$θ = 18 0^{\circ}$ (aphelion): $cos 18 0^{\circ} = - 1$ , toh $r = \frac{10000}{1 - 0.5} = \frac{10000}{0.5} = 20000$ km — sabse bada. Yeh baayein point karne wala lamba navy arrow hai. Yeh step kyun? Poora $e$ subtract karne se denominator sabse chota ban jaata hai, isliye $r$ sabse bada hota hai — sabse door ka point.
$θ = 27 0^{\circ}$ : $cos 27 0^{\circ} = 0$ , toh $r = 10000$ km phir se — neeche ki taraf orange arrow, Step 2 ka mirror. Yeh step kyun? Cosine perihelion axis ke baare mein even hai, isliye $θ$ aur $- θ$ (yaani $27 0^{\circ} = - 9 0^{\circ}$ ) identical $r$ dete hain. Orbit upar-neeche mirror-symmetric hai, jo tumhe figure mein seedha dikhtaa hai.

Verify: $r$ monotonically $6666.7 \to 10000 \to 20000$ rise karta hai jab $θ$ $0 \to 90 \to 180$ jaata hai, phir symmetrically wapas girta hai — bilkul ek closed ellipse ka smooth in-and-out. Bhi $r_{m i n}$ check: $p / (1 + e) = 10000/1.5 = 6666.7$ ✓.

Ex 4 — Cell D: parabola limit ( $e = 1$ )

Worked example

Ek probe ko parabolic escape trajectory par launch kiya jaata hai jisme $p = 12000$ km ( $e = 1$ ) hai. $θ = 0^{\circ}$ , $θ = 9 0^{\circ}$ par $r$ nikalo, aur describe karo kya hota hai jab $θ \to 18 0^{\circ}$ .

Forecast: kya probe kabhi wapas aayega? $r_{m a x}$ kya equal hoga?

Perihelion, $θ = 0$ : $r = \frac{12000}{1 + 1 ( 1 )} = \frac{12000}{2} = 6000$ km. Yeh step kyun? $e = 1$ par bhi perihelion formula $p / (1 + e)$ finite aur well-behaved hai — parabola ka ek closest point toh hota hi hai.
$θ = 9 0^{\circ}$ : $r = \frac{12000}{1 + 0} = 12000$ km $= p$ . Yeh step kyun? " $9 0^{\circ}$ par $r = p$ " ka rule har conic ke liye hold karta hai, kyunki yeh $cos 9 0^{\circ} = 0$ se aata hai, $e$ se independent.
Jab $θ \to 18 0^{\circ}$ : denominator $1 + cos θ \to 1 + (- 1) = 0$ , toh $r \to \frac{12000}{0 ^{+}} \to + \infty$ . Yeh step kyun? Exactly $e = 1$ ke saath, aphelion " $p / (1 - e)$ " zero se divide ho jaata — orbit kabhi close nahi hoti. Probe hamesha ke liye door jaata rehta hai, toh koi aphelion nahi hai. Yeh bound ( $e < 1$ ) aur unbound ( $e > 1$ ) ke beech ki borderline hai.

Verify: $r_{m i n} = p / (1 + e) = 12000/2 = 6000$ km Step 1 se match karta hai. Aur $lim_{θ \to π} p / (1 + cos θ) = \infty$ confirm karta hai ki path open hai, kabhi return nahi hoti. ✓

Ex 5 — Cell E: forbidden angle wala hyperbola ( $e > 1$ )

Worked example

Ek interstellar object Sun ke paas ek hyperbola par fly karta hai jisme $p = 2$ AU aur $e = 2$ hai. $θ = 0^{\circ}$ aur $θ = 6 0^{\circ}$ par $r$ nikalo, phir woh angle nikalo jiske aage $r$ negative ho jaata hai (physically unreachable).

Forecast: ek hyperbola ke liye, kya ek maximum angle hota hai jis par object kabhi bhi dekha ja sakta hai?

Figure mein open magenta hyperbola dikhta hai: orbit sirf do dotted navy asymptote rays ke beech exist karta hai. Jab object kisi bhi ray ki taraf race karta hai, uski doori $r$ infinity par jaati hai — wahi rays hain jinhe hum abhi compute karne wale hain.

Perihelion, $θ = 0$ : $r = \frac{2}{1 + 2 ( 1 )} = \frac{2}{3} \approx 0.667$ AU (figure mein focus ke sabse paas violet dot). Yeh step kyun? Closest approach abhi bhi $p / (1 + e)$ use karta hai; bada $e$ object ko perihelion par aur tighter kheenchta hai.
$θ = 6 0^{\circ}$ : $cos 6 0^{\circ} = 0.5$ , toh $r = \frac{2}{1 + 2 ( 0.5 )} = \frac{2}{2} = 1$ AU. Yeh step kyun? Straightforward substitution; abhi bhi positive hai kyunki denominator abhi bhi positive hai.
Nikalo jahan denominator zero hit karta hai: $1 + e cos θ = 0 \Rightarrow cos θ = - \frac{1}{e} = - \frac{1}{2}$ , toh $θ = 12 0^{\circ}$ — figure mein exactly dotted asymptote ray. Yeh step kyun? Is angle ke aage $1 + e cos θ < 0$ ho jaata hai, jisse $r < 0$ banta hai — impossible. Toh object sirf $∣ θ ∣ < 12 0^{\circ}$ ke liye exist karta hai; do rays $θ = \pm 12 0^{\circ}$ hyperbola ke asymptotes hain. Yeh "forbidden angle" sirf tab exist karta hai jab $e > 1$ ho (kyunki humein $1/ e < 1$ chahiye).

Verify: $cos 12 0^{\circ} = - 0.5 = - 1/ e$ , aur $1 + 2 (- 0.5) = 0$ ✓ — denominator exactly wahin vanish hota hai. $e < 1$ ke liye equation $cos θ = - 1/ e$ ka koi solution nahi hai (kyunki $- 1/ e < - 1$ ), isliye ellipses mein koi forbidden angle nahi hota. Ex 2–3 ke saath consistent. ✓

Ex 6 — Cell F: geometry se $h$ recover karna

Worked example

Earth ke around ek satellite ka measured $p = 7000$ km hai. Earth ka $μ = 3.986 \times 1 0^{5} km^{3} / s^{2}$ hai. Specific angular momentum $h$ aur area-sweep rate nikalo.

Forecast: kya $h$ ke units km²/s honge ya km³/s²?

Definition $p = \frac{h ^{2}}{μ}$ ko invert karo: $h = μ p$ . Yeh step kyun? $p$ derivation mein $h^{2} / μ$ ke roop mein aaya tha; $h$ ke liye solve karne se orbit ki shape se hi physical constant of motion recover ho jaata hai.
Compute karo: $h = (3.986 \times 1 0^{5}) (7000) = 2.7902 \times 1 0^{9} \approx 5.282 \times 1 0^{4} km^{2} / s$ . Yeh step kyun? Units: $(km^{3} / s^{2}) (km) = km^{4} / s^{2} = km^{2} / s$ — ek area per time, jaisa $h$ hona chahiye.
Kepler's equal-areas rate hai $\frac{d A}{d t} = \frac{h}{2} \approx 2.641 \times 1 0^{4} km^{2} / s$ . Yeh step kyun? Radius vector dwara unit time mein sweep kiya gaya area exactly $h /2$ hai — yeh $h = r^{2} \dot{θ}$ se seedha nikalta Kepler's second law hai.

Verify: $h$ ko square karke wapas dekho: $h^{2} / μ = (5.282 \times 1 0^{4})^{2} / (3.986 \times 1 0^{5}) = 2.790 \times 1 0^{9} /3.986 \times 1 0^{5} = 7000$ km $= p$ ✓. Step 2 mein units check. Specific Angular Momentum h se links.

Ex 7 — Cell G: real-world word problem (Earth ki orbit)

Worked example

Earth ka perihelion $1.471 \times 1 0^{8}$ km aur aphelion $1.521 \times 1 0^{8}$ km hai. $e$ , $p$ nikalo, aur check karo ki $p$ lagbhag 1 AU hai ( $1 AU = 1.496 \times 1 0^{8}$ km).

Forecast: kya Earth ki orbit nearly circular hai ya noticeably squashed?

$e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{1.521 - 1.471}{1.521 + 1.471} = \frac{0.050}{2.992} \approx 0.01671$ . Yeh step kyun? Ex 2 wala same extreme-ratio trick; chota $e$ matlab nearly circular.
$p = r_{p} (1 + e) = 1.471 \times 1 0^{8} (1.01671) \approx 1.4956 \times 1 0^{8}$ km. Yeh step kyun? Scale paane ke liye $e$ ko perihelion equation mein plug karo.
1 AU se compare karo: $1.4956 \times 1 0^{8} /1.496 \times 1 0^{8} \approx 0.9997$ , essentially 1 AU. Yeh step kyun? Ek jaane-maane real number ke against sanity check — ek physicist hamesha poochta hai "kya yeh reality se match karta hai?"

Verify: $e \approx 0.0167$ famously Earth ki eccentricity hai (nearly circular, jaise expected). Aphelion check: $p / (1 - e) = 1.4956 \times 1 0^{8} / (0.98329) \approx 1.521 \times 1 0^{8}$ km $= r_{a}$ ✓.

Ex 8 — Cell H: exam twist — angle ke liye solve karo (do answers)

Worked example

Ek orbit ka $p = 9000$ km, $e = 0.3$ hai. Kis true anomaly $θ$ par $r = 10000$ km hai?

Forecast: $[0^{\circ}, 36 0^{\circ})$ mein kitne angles yeh satisfy karte hain — ek, ya do?

$r = \frac{p}{1 + e cos θ}$ se shuru karo aur $cos θ$ ke liye solve karo: $1 + e cos θ = \frac{p}{r}$ , toh $cos θ = \frac{1}{e} (\frac{p}{r} - 1)$ . Yeh step kyun? Hum formula invert kar rahe hain — "angle in, distance out" ki jagah hum "distance in, angle out" ja rahe hain.
Plug in karo: $cos θ = \frac{1}{0.3} (\frac{9000}{10000} - 1) = \frac{1}{0.3} (0.9 - 1) = \frac{- 0.1}{0.3} = - 0.3333$ . Yeh step kyun? Numbers substitute kiye; negative value batati hai ki hum $9 0^{\circ}$ ke aage hain (semi-latus rectum se aage, aphelion ki taraf badhte hue).
$θ = arccos (- 0.3333) \approx 109.4 7^{\circ}$ — aur bhi $θ = 36 0^{\circ} - 109.4 7^{\circ} = 250.5 3^{\circ}$ . Yeh step kyun? Cosine $θ$ aur $- θ$ ke liye same value deta hai, isliye orbit par do points is doori par hain — ek baahir jaate waqt (aphelion ki taraf) aur ek wapas aate waqt. Doosri root miss karna classic exam slip hai.

Verify: $θ = 109.4 7^{\circ}$ par: $r = 9000/ (1 + 0.3 (- 0.3333)) = 9000/ (1 - 0.1) = 9000/0.9 = 10000$ km ✓. $θ = 250.5 3^{\circ}$ par: $cos 250.5 3^{\circ} = - 0.3333$ bhi, toh $r = 9000/ (1 + 0.3 (- 0.3333)) = 9000/0.9 = 10000$ km phir ✓. Dono angles valid physical positions hain. Range check ke liye, $r = 10000$ km ko $r_{m i n} = p / (1 + e) = 9000/1.3 = 6923$ km aur $r_{m a x} = p / (1 - e) = 9000/0.7 = 12857$ km ke beech hona chahiye — aur hai. Conclusion: satellite exactly do true anomalies par $r = 10000$ km par hai, $θ \approx 109.4 7^{\circ}$ (outbound) aur $θ \approx 250.5 3^{\circ}$ (inbound). ✓

Active recall

Recall Har fact kis cell se aata hai?

$e = 0$ $θ$ ko vanish karta hai, constant $r$ deta hai → Cell A (circle).
Wo angle jahan $1 + e cos θ = 0$ sirf tab exist karta hai jab $e > 1$ → Cell E (hyperbola asymptote).
$e = 1$ $θ = 18 0^{\circ}$ par $r \to \infty$ bhejna → Cell D (parabola, koi aphelion nahi).
Diye hue $r$ se $θ$ solve karne par do answers milte hain → Cell H.

Recall Self-test

$r = r_{m i n}$ observed diya hai, kaunsa $θ$ ? ::: $θ = 0$ (perihelion). $e = 2$ ke liye, forbidden angle satisfy karta hai? ::: $cos θ = - 1/ e = - 0.5$ , yaani $θ = 12 0^{\circ}$ . Distance-to-angle problems mein do solutions kyun hote hain? ::: Kyunki $cos θ = cos (- θ)$ , isliye $θ$ aur $36 0^{\circ} - θ$ same $r$ share karte hain.

Inhe bhi dekho Conic Sections, Vis-viva Equation, Eccentricity and Orbital Energy, aur Central Force Motion — jahan se yeh numbers aate hain.