Kepler's first law — orbits are conic sections: Worked examples

3.2.5 · D3 · Physics › Orbital Mechanics & Astrodynamics › Kepler's first law — orbits are conic sections

Yeh page Kepler's first law ka "no surprises" workbook hai. Parent note ne orbit equation prove ki thi: $r (θ) = \frac{p}{1 + e c o s θ}, p = \frac{h ^{2}}{μ}, e = 1 + \frac{2 E h ^{2}}{μ ^{2}}$ Yahan hum ise har tarah ke input se guzarenge: har eccentricity band, degenerate circle, zero-energy edge, negative-cosine back half of the orbit, ek real-world word problem, aur ek exam twist. Agar koi scenario exist karta hai, toh woh neeche ek row mein hai aur aage ek example mein bhi.

The scenario matrix

Neeche ka har cell kam se kam ek worked example se cover hota hai (uska number brackets mein hai).

Cell class	Specific case	Which example
$e = 0$ degenerate	circle, constant $r$	Ex 1
$0 < e < 1$ front half	perihelion, $θ = 0$ , $cos θ = + 1$	Ex 2
$0 < e < 1$ back half	aphelion, $θ = π$ , $cos θ = - 1$	Ex 2
$0 < e < 1$ side / intermediate $θ$	$θ = 9 0^{\circ}$ , $cos θ = 0$ (negative-cos region bhi probe ki gayi)	Ex 3
$e = 1$ edge	parabola — koi aphelion nahi, $r \to \infty$	Ex 4
$e > 1$ unbound	hyperbola — asymptote angle, forbidden angles	Ex 5
Inverse problem	do $r$ values diye, $e$ aur $p$ nikalo	Ex 6
Real-world word problem	satellite altitudes → orbit shape	Ex 7
Limiting behaviour	$e \to 1^{-}$ ellipse stretch hokar parabola banti hai	Ex 8
Exam twist	raw energy + $h$ se classify karo, sign traps	Ex 9

Definition Allowed

$θ$ -domain — formula valid kahan hai Yeh distinction poori page ka theme hai, toh koi bhi arithmetic karne se pehle isse pin down karo (neeche ka figure dekho). Sign convention se yaad karo ki negative $θ$ ka matlab sirf hai "perihelion se doosri taraf sweep kiya":

Ellipse / circle ( $0 \leq e < 1$ ): denominator $1 + e cos θ$ hamesha positive hai (kyunki $e cos θ \geq - e > - 1$ ), toh $r$ har angle ke liye finite aur positive hai. Domain: $0 \leq θ < 2 π$ rad ( $0^{\circ} \leq θ < 36 0^{\circ}$ ) — orbit khud par band ho jaati hai.
Parabola ( $e = 1$ ): denominator sirf $θ = π$ rad ( $18 0^{\circ}$ ) par $0$ hit karta hai, jahan $r \to \infty$ . Domain: $- π < θ < π$ rad ( $- 18 0^{\circ} < θ < 18 0^{\circ}$ ) — ek missing direction infinity ki taraf point karta hai.
Hyperbola ( $e > 1$ ): denominator $θ_{\infty} = arccos (- 1/ e)$ par $0$ hit karta hai. Domain: $- θ_{\infty} < θ < θ_{\infty}$ ; usse aage $r$ negative ho jaata — forbidden (Ex 5).

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

Upar ka figure chaaron cases ko ek hi focus (orange dot) se draw karta hai. Notice karo ki orange circle aur teal ellipse poora loop sweep karte hain, jabki plum parabola aur dark hyperbola open hote hain aur dashed directions ke along infinity ki taraf bhagte hain — yahi $θ$ -domain rule visible hai. Hum Ex 5 mein ise quantitatively dekhenge.

Ex 1 — Degenerate case: perfect circle ( $e = 0$ )

Worked example Ek moon fixed

$r = 4.0 \times 1 0^{8}$ m par orbit karta hai jahan $e = 0$ . $θ = 0$ , $9 0^{\circ}$ , $18 0^{\circ}$ par $r$ nikalo, aur $p$ bhi nikalo.

Forecast: padhne se pehle andaza lagao — kya $r$ ghumne se change hota hai? $p$ us fixed radius se compare mein kya hoga?

$e = 0$ ke saath orbit equation likho. $r (θ) = \frac{p}{1 + 0 \cdot cos θ} = p$ . Yeh step kyun? Dial zero par set karne se $cos θ$ term bilkul khatam ho jaati hai, toh angle ka koi fark nahi — yeh circle ka algebraic fingerprint hai.
Har angle par evaluate karo. $r (0) = r (9 0^{\circ}) = r (18 0^{\circ}) = p$ . Yeh step kyun? Hume "all quadrants" check karne chahiye. Yahan sab equal hain — circle ka poora point yahi hai ki koi bhi direction special nahi hai.
$p$ identify karo. Kyunki $r$ constant hai aur $4.0 \times 1 0^{8}$ m ke barabar hai, hum padhte hain $p = 4.0 \times 1 0^{8}$ m. Yeh step kyun? $p$ define hota hai $r$ at $θ = 9 0^{\circ}$ ke roop mein; circle ke liye woh har jagah radius ke barabar hai.

Verify: perihelion $= a (1 - e) = a (1 - 0) = a$ aur aphelion $= a (1 + 0) = a$ same value par collapse ho jaate hain, confirming karta hai ki koi "close" ya "far" point nahi hai. Units: metres throughout. ✓

Ex 2 — Bound ellipse, front aur back halves ( $0 < e < 1$ )

Worked example Mars:

$e = 0.0934$ , semi-major axis $a = 2.279 \times 1 0^{11}$ m. Perihelion ( $θ = 0$ ) aur aphelion ( $θ = π$ ) distances nikalo.

Forecast: kaun sa bada hai, aur roughly $a$ ke kis fraction se?

$a$ aur $p$ ko relate karo. Recapped identity $p = a (1 - e^{2})$ use karo (upar ke [!formula] callout mein derive ki gayi). Yeh step kyun? Hume $a$ diya gaya hai, lekin orbit equation $p$ use karta hai; yeh identity unhe convert karta hai, aur hum ise pehle prove kar chuke hain, toh koi hand-waving nahi. $p = 2.279 \times 1 0^{11} (1 - 0.093 4^{2}) = 2.259 \times 1 0^{11}$ m.
Perihelion, $θ = 0$ (front half, $cos θ = + 1$ ). $r_{m i n} = \frac{p}{1 + e} = a (1 - e)$ . Yeh step kyun? Sabse bada $cos$ ⇒ sabse bada denominator ⇒ sabse chhota $r$ . Yeh matrix ka $cos θ = + 1$ corner hai. $r_{m i n} = 2.279 \times 1 0^{11} (1 - 0.0934) = 2.066 \times 1 0^{11}$ m.
Aphelion, $θ = π$ (back half, $cos θ = - 1$ ). $r_{m a x} = \frac{p}{1 - e} = a (1 + e)$ . Yeh step kyun? $cos π = - 1$ denominator ko sabse chhota banata hai, toh $r$ sabse bada hota hai — orbit ka negative-cosine region. $r_{m a x} = 2.279 \times 1 0^{11} (1 + 0.0934) = 2.492 \times 1 0^{11}$ m.

Verify: inка average $\frac{1}{2} (r_{m i n} + r_{m a x}) = \frac{1}{2} \cdot a [(1 - e) + (1 + e)] = a$ . Compute: $\frac{1}{2} (2.066 + 2.492) \times 1 0^{11} = 2.279 \times 1 0^{11}$ m $= a$ . ✓ Semi-major axis sach mein dono extremes ka mean hai.

Ex 3 — Intermediate angle aur negative-cosine region ( $0 < e < 1$ )

Worked example Ek orbit ke liye jahan

$p = 1.20 \times 1 0^{7}$ m aur $e = 0.50$ hai, $θ = 9 0^{\circ}$ aur $θ = 12 0^{\circ}$ par $r$ nikalo.

Forecast: $12 0^{\circ}$ par hum sideways se aage hain — kya $r$ , $9 0^{\circ}$ se bada hai ya chhota?

Sideways point, $θ = 9 0^{\circ}$ ( $cos = 0$ ). $r = \frac{p}{1 + e \cdot 0} = p = 1.20 \times 1 0^{7}$ m. Yeh step kyun? Yeh $p$ ka defining point hai; negative-cosine zone mein push karne se pehle ek clean anchor.
Past sideways, $θ = 12 0^{\circ}$ ( $cos 12 0^{\circ} = - 0.5$ ). $r = \frac{p}{1 + 0.50 ( - 0.5 )} = \frac{1.20 \times 1 0 ^{7}}{0.75} = 1.60 \times 1 0^{7}$ m. Yeh step kyun? Jab $cos θ$ negative ho jaata hai toh denominator 1 se neeche girta hai, toh $r$ badhta hai — body aphelion ki taraf ja rahi hai. Yeh "intermediate $θ$ with negative cosine" cell cover karta hai.

Verify: dono extremes in numbers ko bracket karne chahiye: $r_{m i n} = p /1.5 = 8.0 \times 1 0^{6}$ m aur $r_{m a x} = p /0.5 = 2.4 \times 1 0^{7}$ m. Dono $1.20 \times 1 0^{7}$ aur $1.60 \times 1 0^{7}$ , $[8.0 \times 1 0^{6}, 2.4 \times 1 0^{7}]$ ke andar hain. ✓ Aur $r (12 0^{\circ}) > r (9 0^{\circ})$ forecast ke mutabiq hai, kyunki hum aphelion ki taraf gaye.

Ex 4 — Parabolic edge ( $e = 1$ )

Worked example Ek comet ka

$e = 1.000$ aur $p = 2.0 \times 1 0^{11}$ m hai. Perihelion distance nikalo, aur describe karo ki $θ \to 18 0^{\circ}$ jaane par $r$ kya hota hai.

Forecast: kya yeh comet kabhi aphelion reach karta hai?

Perihelion, $θ = 0$ . $r_{m i n} = \frac{p}{1 + 1} = \frac{p}{2} = 1.0 \times 1 0^{11}$ m. Yeh step kyun? Perihelion abhi bhi normally behave karta hai — fark sirf far side par dikhta hai.
$θ \to 18 0^{\circ}$ ( $π$ rad) ke paas jaana. Denominator $1 + cos θ \to 1 + (- 1) = 0$ , toh $r \to \infty$ . Yeh step kyun? $e = 1$ ke liye denominator zero hit kar sakta hai. Yeh naya phenomenon hai jo ellipses mein absent tha: curve kabhi close nahi hota, yeh infinity ki taraf escape karta hai jab woh perihelion ke opposite line up karta hai.
Conclusion: koi aphelion exist nahi karta; comet ek baar aata hai aur chala jaata hai. Yeh step kyun? Yeh exactly "wapas aata hai" ( $e < 1$ ) aur "hamesha ke liye chala gaya" ke beech ki boundary hai.

Verify: ellipse formula $r_{m a x} = p / (1 - e)$ ; $e \to 1$ plug karne par $p /0 = \infty$ milta hai, limit se match karta hai — parabola ek stretched ellipse ka continuous limit hai. (Note karo $p = a (1 - e^{2})$ bhi $a$ -picture ko blow up karta hai: $a = p / (1 - e^{2}) \to \infty$ , toh parabola infinite semi-major axis wala ellipse hai.) Units: $r_{m i n}$ metres mein. ✓

Ex 5 — Hyperbolic flyby ( $e > 1$ ) aur forbidden angles

Worked example Ek interstellar object ka

$e = 1.5$ , $p = 3.0 \times 1 0^{11}$ m hai. Perihelion nikalo, aur woh asymptote angle nikalo jiske aage koi orbit exist nahi karta.

Forecast: hyperbola ke liye, kya saare angles $θ$ physically reachable hain?

Perihelion, $θ = 0$ . $r_{m i n} = \frac{p}{1 + e} = \frac{3.0 \times 1 0 ^{11}}{2.5} = 1.2 \times 1 0^{11}$ m. Yeh step kyun? Same closest-approach logic; unbound orbits ka bhi ek well-defined nearest point hota hai.
$r \to \infty$ kahan? Denominator zero set karo. $1 + e cos θ_{\infty} = 0 \Rightarrow cos θ_{\infty} = - \frac{1}{e} = - \frac{1}{1.5} = - 0.6667$ . Yeh step kyun? Kyunki $e > 1$ hai, value $- 1/ e$ $[- 1, 1]$ ke andar hai, toh ek real angle exist karta hai jahan $r$ blow up hota hai. Yeh asymptote direction hai. $θ_{\infty} = arccos (- 0.6667) = 131. 8^{\circ}$ .
Forbidden region. $θ > θ_{\infty}$ ke liye (up to $18 0^{\circ}$ tak), $1 + e cos θ < 0$ , negative $r$ milta hai — unphysical. Body physically sirf $- 131. 8^{\circ} < θ < 131. 8^{\circ}$ mein exist karti hai (sign convention se yaad karo: woh range perihelion ke kisi bhi taraf $131. 8^{\circ}$ cover karta hai). Yeh step kyun? Yahi woh case coverage hai jise parent ne skip kiya: hyperbolas mein angular no-go zones hain, ellipses ki tarah nahi jo poora $36 0^{\circ}$ sweep karte hain.

Verify: ellipse ( $e < 1$ ) se contrast karo: wahan $- 1/ e < - 1$ jo $[- 1, 1]$ ke bahar hai, toh denominator kabhi zero nahi reach karta, $r$ finite rehta hai, aur har angle allowed hai — exactly yahi wajah hai ki ellipses close hote hain. $e = 1.5$ ke liye, $θ_{\infty}$ $9 0^{\circ}$ se zyada aur $18 0^{\circ}$ se kam hona chahiye: $131. 8^{\circ}$ ✓.

Figure mein plum hyperbola sirf allowed wedge $- 131. 8^{\circ} < θ < 131. 8^{\circ}$ trace karta dikhta hai. Dashed teal lines $θ_{\infty}$ par asymptotes hain; unke peeche shaded grey wedges forbidden directions hain jahan formula negative $r$ return karta. Is truncated fan ko Ex 1–3 ke full loops se compare karo: woh visual contrast hi matrix section ka $θ$ -domain rule hai.

Ex 6 — Inverse problem: do radii tumhe shape dete hain

Worked example Ek probe track kiya gaya:

$θ = 0$ par uski distance $r_{1} = 7.0 \times 1 0^{6}$ m hai; $θ = 18 0^{\circ}$ par $r_{2} = 2.1 \times 1 0^{7}$ m hai. $e$ , $p$ , aur $a$ nikalo.

Forecast: $r_{2} = 3 r_{1}$ ke saath, kya yeh nearly circular hai ya quite eccentric?

Dono extremes likho. $r_{1} = \frac{p}{1 + e}$ , $r_{2} = \frac{p}{1 - e}$ . Yeh step kyun? Do data points, do unknowns ( $p, e$ ) — perihelion aur aphelion sabse clean equations hain kyunki cosines $\pm 1$ hain.
$p$ hatane ke liye divide karo. $\frac{r _{2}}{r _{1}} = \frac{1 + e}{1 - e} = \frac{2.1 \times 1 0 ^{7}}{7.0 \times 1 0 ^{6}} = 3$ . Yeh step kyun? Ratios size eliminate karte hain, shape parameter $e$ isolate karte hain.
$e$ ke liye linear equation solve karo. Cross-multiply $\frac{1 + e}{1 - e} = 3$ karo taaki $3 (1 - e) = 1 + e$ mile, phir expand aur collect karo: $3 - 3 e = 1 + e \Rightarrow 2 = 4 e \Rightarrow e = 0.5$ . Yeh step kyun? Cross-multiplying fraction clear karta hai toh $e$ linearly appear karta hai; $e$ -terms ko ek taraf gather karna step 2 ke baad ek hi bache unknown ko isolate karne ka ek hi tarika hai.
$p$ ke liye back-substitute karo. $p = r_{1} (1 + e) = 7.0 \times 1 0^{6} (1.5) = 1.05 \times 1 0^{7}$ m. Yeh step kyun? $e$ ke jaante hi, koi bhi extreme equation $p$ directly yield karta hai; perihelion sabse simple hai.
Semi-major axis. $a = \frac{1}{2} (r_{1} + r_{2}) = \frac{1}{2} (7.0 \times 1 0^{6} + 2.1 \times 1 0^{7}) = 1.4 \times 1 0^{7}$ m. Yeh step kyun? $a$ extremes ka mean hai (Ex 2 ke verify mein prove kiya gaya).

Verify: check karo $p = a (1 - e^{2}) = 1.4 \times 1 0^{7} (1 - 0.25) = 1.05 \times 1 0^{7}$ m ✓, aur $r_{2} = p / (1 - e) = 1.05 \times 1 0^{7} /0.5 = 2.1 \times 1 0^{7}$ m input se match karta hai. ✓

Ex 7 — Real-world word problem: satellite altitudes se

Worked example Earth (radius

$R_{E} = 6.371 \times 1 0^{6}$ m) ke around ek satellite ka perigee altitude $500$ km aur apogee altitude $5000$ km hai. $e$ aur $p$ nikalo. (Altitude = surface se upar ki height.)

Forecast: classic trap — kya hum orbit equation mein altitudes use karte hain ya distances-from-centre?

Altitudes ko focus-distances mein convert karo. Focus Earth ka centre hai, toh $R_{E}$ add karo: $r_{m i n} = 6.371 \times 1 0^{6} + 5.00 \times 1 0^{5} = 6.871 \times 1 0^{6}$ m, $r_{m a x} = 6.371 \times 1 0^{6} + 5.00 \times 1 0^{6} = 1.1371 \times 1 0^{7}$ m. Yeh step kyun? $r = p / (1 + e cos θ)$ mein $r$ focus se measure hota hai, kabhi surface se nahi — $R_{E}$ bhoolna #1 satellite-exam error hai.
Ratio se eccentricity. $\frac{r _{m a x}}{r _{m i n}} = \frac{1 + e}{1 - e} = \frac{1.1371 \times 1 0 ^{7}}{6.871 \times 1 0 ^{6}} = 1.6549$ . Solve: $e = \frac{1.6549 - 1}{1.6549 + 1} = 0.2467$ . Yeh step kyun? Same divide-out trick jaise Ex 6 mein; ratio size strip kar leta hai aur Earth ka radius offset already true $r$ values mein baked in hai.
Semi-latus rectum. $p = r_{m i n} (1 + e) = 6.871 \times 1 0^{6} (1.2467) = 8.568 \times 1 0^{6}$ m. Yeh step kyun? $e$ jaante hi, perihelion $p$ ke liye invert karne ka sabse clean equation hai.

Verify: $r_{m a x} = p / (1 - e) = 8.568 \times 1 0^{6} /0.7533 = 1.137 \times 1 0^{7}$ m, step 1 se match karta hai ✓. Sanity: $e = 0.25$ ek mild ellipse hai, ek real transfer-style orbit ke liye reasonable. Units: metres throughout. ✓

Ex 8 — Limiting behaviour: ellipse ko parabola ki taraf stretch karna

Worked example Perihelion

$r_{m i n} = 1.0 \times 1 0^{11}$ m par fixed rakho jabki $e$ ko $0.9, 0.99, 0.999$ se badhao. Aphelion aur ratio $r_{m a x} / r_{m i n}$ track karo.

Forecast: jaise jaise $e \to 1$ hota hai, kya $r_{m a x}$ finite value approach karta hai ya bhaag jaata hai?

Aphelion ko perihelion se express karo. $r_{m a x} = r_{m i n} \cdot \frac{1 + e}{1 - e}$ . Yeh step kyun? $r_{m i n}$ fixed rakhne se akele far side par $e$ ka effect isolate hota hai.
Tabulate karo.
- $e = 0.9$ : ratio $= 1.9/0.1 = 19$ , toh $r_{m a x} = 1.9 \times 1 0^{12}$ m.
- $e = 0.99$ : ratio $= 1.99/0.01 = 199$ , toh $r_{m a x} = 1.99 \times 1 0^{13}$ m.
- $e = 0.999$ : ratio $= 1.999/0.001 = 1999$ , toh $r_{m a x} = 1.999 \times 1 0^{14}$ m. Yeh step kyun? $e = 1$ ke paas har factor-of-10 aphelion ko ~10x multiply karta hai — ellipse bina bound ke elongate hota jaata hai.
Limit lo. Jaise jaise $e \to 1^{-}$ hota hai, $r_{m a x} \to \infty$ : closed ellipse Ex 4 ki parabola mein open ho jaata hai. Yeh step kyun? Yeh poore conic family ki continuity hai — matrix rows ek dial hain, alag alag universes nahi.

Verify: $e = 0.999$ par, $r_{m a x} / r_{m i n} = 1999$ aur sach mein $(1 + e) / (1 - e) = 1.999/0.001 = 1999$ ✓. Trend monotonic hai aur diverge karta hai, parabolic limit confirm karta hai.

Ex 9 — Exam twist: energy aur angular momentum se seedha classify karo

Worked example Sun (

$μ = 1.327 \times 1 0^{20} m^{3} / s^{2}$ ) ke around ek body ka specific energy $E = + 4.0 \times 1 0^{8} J/kg$ aur $h = 4.5 \times 1 0^{15} m^{2} / s$ hai. Orbit classify karo aur $e$ aur $p$ nikalo.

Forecast: positive energy — bound hai ya unbound? Compute karne se pehle conic guess karo.

Energy par sign check. $E > 0$ ⇒ unbound. Predict karo $e > 1$ , ek hyperbola. Yeh step kyun? $E$ ka sign kisi bhi arithmetic se pehle conic decide karta hai — sabse fast exam move aur ek common trap (log needlessly compute karte hain).
$e$ compute karo. $e = 1 + \frac{2 E h ^{2}}{μ ^{2}}$ . Andar: $\frac{2 ( 4.0 \times 1 0 ^{8} ) ( 4.5 \times 1 0 ^{15} ) ^{2}}{( 1.327 \times 1 0 ^{20} ) ^{2}} = \frac{2 ( 4.0 \times 1 0 ^{8} ) ( 2.025 \times 1 0 ^{31} )}{1.7609 \times 1 0 ^{40}} = \frac{1.620 \times 1 0 ^{40}}{1.7609 \times 1 0 ^{40}} = 0.9200$ . Toh $e = 1.9200 = 1.386$ . Yeh step kyun? Sign-prediction quantitatively confirm karta hai; $e > 1$ ⇒ hyperbola. ✓
Semi-latus rectum. $p = \frac{h ^{2}}{μ} = \frac{( 4.5 \times 1 0 ^{15} ) ^{2}}{1.327 \times 1 0 ^{20}} = \frac{2.025 \times 1 0 ^{31}}{1.327 \times 1 0 ^{20}} = 1.526 \times 1 0^{11}$ m. Yeh step kyun? $p$ sirf $h$ aur $μ$ par depend karta hai (parent ka Step 6), energy se independent.
Reality check ke roop mein closest approach. $r_{m i n} = \frac{p}{1 + e} = \frac{1.526 \times 1 0 ^{11}}{2.386} = 6.40 \times 1 0^{10}$ m. Yeh step kyun? Escape flyby ka bhi ek well-defined perihelion hota hai; ise compute karna confirm karta hai ki hyperbola ek physical trajectory hai jo Sun ko graze karta hai aur chala jaata hai.

Verify: $E > 0 \Leftrightarrow e > 1$ ; humne $e = 1.386 > 1$ paya — consistent. Units of $p$ : $(m^{2} / s)^{2} / (m^{3} / s^{2}) = m$ ✓. Asymptote angle $arccos (- 1/1.386) = arccos (- 0.7215) = 136. 2^{\circ}$ jo $(9 0^{\circ}, 18 0^{\circ})$ mein hai, jaise har hyperbola mein hona chahiye.

Recall check

Recall Kaun sa cosine value perihelion deta hai, kaun sa aphelion?

$cos θ = + 1$ (at $θ = 0$ ) → perihelion (sabse chhota $r$ ); $cos θ = - 1$ (at $θ = π$ rad $= 18 0^{\circ}$ ) → aphelion (sabse bada $r$ ). Aphelion sirf tab exist karta hai jab $e < 1$ .

Recall Hyperbola (

$e > 1$ ) ke liye kaun sa angle asymptote mark karta hai? $cos θ_{\infty} = - 1/ e$ ; usse aage denominator negative ho jaata hai aur koi orbit exist nahi karta.

Recall Ek satellite problem altitudes deta hai. Pehla move?

Planet ka radius add karo: $r = altitude + R_{E}$ , kyunki $r$ focus (centre) se measure hota hai.

Recall Kin conics ke liye har angle allowed hai?

Sirf circle aur ellipse ( $0 \leq e < 1$ ) poore $0 \leq θ < 2 π$ rad allow karte hain. Parabola $θ = π$ rad kho deta hai; hyperbola $- arccos (- 1/ e) < θ < arccos (- 1/ e)$ tak confined rehta hai.

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