Orbit shape from eccentricity — circle (e=0), ellipse (0 - e - 1), parabola (e=1), hyperbola (e - 1): Worked examples

3.2.4 · D3 · Physics › Orbital Mechanics & Astrodynamics › Orbit shape from eccentricity — circle (e=0), ellipse (0 - e

Yeh page parent note ka ek hi idea drill karta hai: eccentricity $e$ shape decide karta hai. Yahan hum har woh case cover karte hain jo yeh topic throw kar sakta hai — energy ka har sign, har shape, degenerate limits, ek word problem, aur ek exam twist. Compute karne se pehle guess karo.

Shuru karne se pehle, teen symbols jinpe hum baar-baar rely karenge, sab parent note mein earn kiye gaye hain:

Recall Teen formulas jo hum reuse karte hain (sab parent se)

Orbit shape: $r (θ) = \frac{p}{1 + e cos θ}$ , jahan $p = \frac{h ^{2}}{μ}$ semi-latus rectum hai ( $θ = 9 0^{\circ}$ par $r$ ki distance).
Eccentricity from energy: $e = 1 + \frac{2 ε h ^{2}}{μ ^{2}}$ , jahan $ε = E / m$ specific energy hai (energy per kilogram) aur $h = L / m$ specific angular momentum hai (per kilogram). Dekho Specific Orbital Energy aur Angular Momentum in Orbits.
$μ = GM$ central mass ka gravitational parameter hai — ek number jo $G$ aur $M$ ko bundle karta hai taaki dono alag-alag carry na karne padein.

Upar ka har symbol parent mein build kiya gaya tha; yahan hum sirf unhe use karte hain. Agar koi naya tool aata hai (jaise kyun hum square root lete hain, kyun $cos$ ), woh spot par explain hoga.

Scenario matrix

Topic ko ek machine ki tarah socho: tum use $(ε, h)$ ya $(e)$ feed karte ho aur woh ek shape output karta hai. Neeche har "cell" us machine ka genuinely alag behaviour hai. Worked examples is tarah choose kiye gaye hain ki mil ke woh har cell ko light up karen.

Cell	Input character	Shape	Sign of $ε$	Covered by
A	$e = 0$ exactly (degenerate: foci merge)	circle	$ε < 0$	Ex 1
B	$0 < e < 1$ (generic bound)	ellipse	$ε < 0$	Ex 2
C	$e = 1$ exactly (knife-edge)	parabola	$ε = 0$	Ex 3
D	$e > 1$ (generic unbound)	hyperbola	$ε > 0$	Ex 4
E	Sign check: given $(ε, h)$ , classify	any	any	Ex 5
F	Angle jahan $r \to \infty$ (all $e$ )	para/hyper	$ε \geq 0$	Ex 6
G	Word problem: real body ( $e$ near 0)	near-circle	$ε < 0$	Ex 7
H	Limit: $e \to 1^{-}$ (ellipse becoming parabola)	ellipse→parabola	$ε \to 0^{-}$	Ex 8
I	Exam twist: same $e$ , different sizes	ellipse	$ε < 0$	Ex 9
J	Degenerate: $h = 0$ (radial fall, straight line)	$e = 1$ line	$ε < 0$ possible	Ex 10

Hum numbers simple rakhte hain $μ = 1$ ke units mein, jab tak koi real body naam se na di ho.

Neeche ki figure is poore page ka map hai. Yeh single focus se — origin par (kala dot = central mass) — char shapes draw karta hai, same $p = 1$ use karke. Dekho kaise ek dial $e$ ko $0$ se upar ghumane par ek circle (solid black) inflate hokar ellipse banta hai, phir — red dashed curve $e = 1$ par — loop khul jaata hai parabola mein, aur finally open hyperbola ( $e = 1.6$ , dotted) mein. Red curve woh knife-edge hai jahan har example baar-baar lauta hai.

Figure — Orbit shape from eccentricity — circle (e=0), ellipse (0 - e - 1), parabola (e=1), hyperbola (e - 1)

Worked examples

Cell A — circle ( $e = 0$ )

Worked example Ex 1 · Ek circular orbit

$μ = 1$ aur specific angular momentum $h = 2$ lo. Orbit ka $e = 0$ hai. $p$ nikalo, aur dikhao ki $r$ har angle par same hai. Woh constant radius kya hai?

Forecast: Guess karo — kya $r$ kabhi $θ$ par depend karega? Woh kis number par settle hoga?

$p = h^{2} / μ$ compute karo. $h = 2, μ = 1$ ke saath: $p = 4/1 = 4$ .
- Yeh step kyun? $p$ shape formula mein ekmaatra length hai; pehle ise nail karo.
$e = 0$ ko $r (θ) = \frac{p}{1 + e cos θ}$ mein plug karo. $cos θ$ term $e = 0$ se multiply hoti hai, toh woh vanish ho jaati hai: $r = \frac{p}{1 + 0} = p$ .
- Yeh step kyun? $e = 0$ set karna exactly "degenerate" cell hai — do foci merge ho jaate hain aur angle-dependence gayab ho jaati hai.
Radius read off karo. $r = p = 4$ har $θ$ ke liye.
- Yeh step kyun? Constant $r$ hi circle ki definition hai — yeh confirm karta hai ki $e = 0$ cell A deta hai.

Verify: $θ = 0, 9 0^{\circ}, 18 0^{\circ}$ test karo: denominator har baar $1$ hai, toh $r = 4$ hamesha. Units: $h^{2} / μ$ ki units length hain (jaisi honi chahiye, kyunki $r$ ek length hai). ✓

Cell B — generic ellipse ( $0 < e < 1$ )

Intuition

$r_{m i n}$ , $r_{m a x}$ aur $p = a (1 - e^{2})$ scratch se derive karna Ellipse ke endpoints ke baare mein sab kuch seedha shape formula $r (θ) = \frac{p}{1 + e cos θ}$ se nikalta hai.

Step 1 — near aur far points. $r$ wahan sabse chota hota hai jahan denominator sabse bada ho. $cos θ$ $+ 1$ par top out karta hai ( $θ = 0$ par), toh $r_{m i n} = \frac{p}{1 + e} .$ $r$ wahan sabse bada hota hai jahan denominator sabse chota ho: $cos θ = - 1$ ( $θ = 18 0^{\circ}$ par), toh $r_{m a x} = \frac{p}{1 - e} .$

Step 2 — size $a$ se connect karo. Hamari definition se $a = \frac{r _{m i n} + r _{m a x}}{2}$ . Dono fractions add karo: $r_{m i n} + r_{m a x} = \frac{p}{1 + e} + \frac{p}{1 - e} = \frac{p ( 1 - e ) + p ( 1 + e )}{( 1 + e ) ( 1 - e )} = \frac{2 p}{1 - e ^{2}} .$ Half karke, $a = \frac{p}{1 - e ^{2}}$ , yaani $p = a (1 - e^{2})$ .

Step 3 — endpoints $a, e$ ke terms mein. $p = a (1 - e^{2}) = a (1 + e) (1 - e)$ wapas substitute karo: $r_{m i n} = \frac{a ( 1 + e ) ( 1 - e )}{1 + e} = a (1 - e), r_{m a x} = \frac{a ( 1 + e ) ( 1 - e )}{1 - e} = a (1 + e) .$ Toh famous formulas $r_{m i n} = a (1 - e)$ aur $r_{m a x} = a (1 + e)$ derive kiye gaye hain, assume nahi kiye. Neeche figure mein red long-axis in do endpoints ko mark karta hai; red vertical segment $p$ hai, $θ = 9 0^{\circ}$ par sideways width.

Worked example Ex 2 · Ek ellipse ka perihelion aur aphelion

Ek orbit ka $e = 0.6$ aur semi-major axis $a = 10$ hai ( $μ = 1$ ke units mein). Sabse paas ki distance $r_{m i n}$ aur sabse door ki distance $r_{m a x}$ , aur semi-latus rectum $p$ nikalo.

Forecast: Guess karo — kya far point near point se zyada ya kam than twice hai?

Closest approach $r_{m i n} = a (1 - e)$ . $= 10 (1 - 0.6) = 10 (0.4) = 4$ .
- Yeh step kyun? $θ = 0$ se $cos θ = + 1$ milta hai, sabse bada denominator $1 + e$ , isliye sabse chota $r$ — woh formula jo humne abhi derive kiya.
Farthest point $r_{m a x} = a (1 + e)$ . $= 10 (1.6) = 16$ .
- Yeh step kyun? $θ = 18 0^{\circ}$ se $cos θ = - 1$ milta hai, sabse chota denominator $1 - e$ , sabse bada $r$ .
Semi-latus rectum $p = a (1 - e^{2})$ . $= 10 (1 - 0.36) = 10 (0.64) = 6.4$ .
- Yeh step kyun? Upar derive ki gayi identity use karke ( $p = a (1 + e) (1 - e)$ ); yeh do views (shape formula vs $a, e$ ) ko link karta hai, aur Ex 6 mein zaruurat padegi.

Verify: $r_{m i n} + r_{m a x} = 4 + 16 = 20 = 2 a$ ✓ (major axis = near + far). Aur $p$ ko $θ = 9 0^{\circ}$ par $r$ ke barabar hona chahiye: $p / (1 + 0.6 cos 9 0^{\circ}) = 6.4/1 = 6.4 = p$ ✓. Dekho Kepler's First Law.

Cell C — parabola ( $e = 1$ )

Worked example Ex 3 · Ek parabolic comet

Ek comet parabola par hai, $e = 1$ , $p = 2$ ke saath (toh $μ = 1, h = 2$ ). Perihelion par $r$ nikalo ( $θ = 0$ ) aur describe karo kya hota hai jab $θ \to 18 0^{\circ}$ .

Forecast: Guess karo — kya $r$ ek finite maximum tak pahunchta hai, ya infinity tak jaata hai?

Perihelion, $θ = 0$ . $r = \frac{p}{1 + 1 \cdot cos 0} = \frac{2}{1 + 1} = \frac{2}{2} = 1$ .
- Yeh step kyun? Sabse paas ka point hamesha $θ = 0$ par hota hai; yeh woh ek distance hai jo hum exactly pin down kar sakte hain.
$θ = 18 0^{\circ}$ approach karo. Denominator $1 + cos 18 0^{\circ} = 1 - 1 = 0$ , toh $r \to \infty$ .
- Yeh step kyun? Yeh zero-denominator $e = 1$ ki signature hai: orbit khul jaati hai. Ellipse ke ulat yeh kabhi close nahi hoti.
Energy interpret karo. $e = 1$ ke liye, relation $e = 1 + 2 ε h^{2} / μ^{2}$ force karta hai $2 ε h^{2} / μ^{2} = 0$ , yaani $ε = 0$ .
- Yeh step kyun? Zero specific energy = barely escape karta hai, infinity par zero speed se pahunchta hai. Yeh Escape Velocity ki boundary hai.

Verify: $μ = 1$ ke saath, $p = h^{2} / μ = h^{2} \Rightarrow h = 2 \Rightarrow h^{2} = 2 = p$ ✓. Aur $ε = 0$ deta hai $e = 1 + 0 = 1$ ✓, jo assumed shape se match karta hai.

Cell D — generic hyperbola ( $e > 1$ )

Worked example Ex 4 · Ek hyperbolic flyby

Ek interstellar object ka $e = 2$ aur $p = 3$ hai. Perihelion $r_{m i n}$ aur woh angle nikalo jahan $r \to \infty$ hai (uska asymptote).

Forecast: Guess karo — kya "escape angle" parabola ke $18 0^{\circ}$ se zyada hoga ya kam?

Perihelion, $θ = 0$ . $r_{m i n} = \frac{p}{1 + e} = \frac{3}{1 + 2} = \frac{3}{3} = 1$ .
- Yeh step kyun? Same closest-point logic; $cos 0 = 1$ denominator ko maximize karta hai.
$r \to \infty$ kahan? Denominator $= 0$ chahiye: $1 + 2 cos θ = 0 \Rightarrow cos θ = - \frac{1}{2}$ .
- Yeh step kyun? Yeh "forbidden angle" hai — iske baad formula negative $r$ deta hai, jo unphysical hai. Yeh woh mark karta hai jahan body infinity tak fly off hoti hai.
$cos θ = - 0.5$ solve karo. $θ = 12 0^{\circ}$ (aur symmetry se $- 12 0^{\circ}$ aane ke raaste par).
- Yeh step kyun? $cos$ woh tool hai kyunki shape formula angle ko cosine ke andar pack karta hai; hum poochte hain "kis angle ka yeh cosine hai?" — woh inverse question $arccos$ hai.

Verify: $cos 12 0^{\circ} = - 0.5$ toh $1 + 2 (- 0.5) = 0$ ✓. Object sirf $- 12 0^{\circ} < θ < 12 0^{\circ}$ ke liye exist karta hai, ek $24 0^{\circ}$ arc — ek open curve, jaisi hyperbola honi chahiye. Energy: $e > 1 \Rightarrow ε > 0$ , unbound. ✓

Cell E — $(ε, h)$ se classify karo

Worked example Ex 5 · Kaun si shape? Teen inputs

$μ = 1$ ke saath, har ek ke liye orbit classify karo: (a) $ε = - 0.5, h = 1$ ; (b) $ε = 0, h = 1$ ; (c) $ε = + 0.5, h = 1$ .

Forecast: Guess karo — compute karne se pehle, kaun sa unbound hai?

Case (a). $e = 1 + 2 (- 0.5) (1)^{2} /1 = 1 - 1 = 0 = 0$ .
- Yeh step kyun? $ε$ ka sign ek square root ke neeche $1 + (woh)$ ke roop mein rehta hai. Yahan woh poori cheez $0$ tak drive karta hai → circle (cell A).
Case (b). $e = 1 + 2 (0) (1) /1 = 1 = 1$ → parabola.
- Yeh step kyun? $ε = 0$ knife-edge hai; $h$ chahe kuch bhi ho, hamesha $e = 1$ .
Case (c). $e = 1 + 2 (0.5) (1) /1 = 2 \approx 1.414$ → hyperbola (unbound).
- Yeh step kyun? Positive energy root ke neeche ki term ko $1$ se upar push karta hai, toh $e > 1$ .

Verify: Signs shape ke saath step mein chalta hai: $ε < 0 \Rightarrow e < 1$ , $ε = 0 \Rightarrow e = 1$ , $ε > 0 \Rightarrow e > 1$ — exactly parent ki table. Numerically $2 \approx 1.41421$ ✓. Dekho Vis-viva Equation ki $ε$ speed se kaise measure hoti hai.

Cell F — woh angle jahan distance blow up hoti hai

Worked example Ex 6 · Ellipse ke liye koi blow-up nahi

Ex 2 ki ellipse ke liye ( $e = 0.6, p = 6.4$ ), dikhao ki koi angle nahi hai jahan $r \to \infty$ . $e = 1$ aur $e = 1.5$ se contrast karo.

Forecast: Guess karo — kya koi $θ$ hai jo denominator ko zero karta hai jab $e = 0.6$ ?

Denominator zero set karo. $1 + e cos θ = 0 \Rightarrow cos θ = - 1/ e$ .
- Yeh step kyun? $r = p / (denominator)$ sirf wahan explode ho sakta hai jahan denominator vanish ho.
$e = 0.6$ plug karo. $cos θ = - 1/0.6 \approx - 1.667$ .
- Yeh step kyun? Cosine sirf $[- 1, 1]$ mein values le sakta hai. Yahan hamein $- 1.667$ chahiye, jo impossible hai — toh aisa koi $θ$ exist nahi karta. Ellipse har jagah finite rehti hai (cell B, bounded).
Contrast. $e = 1 \Rightarrow cos θ = - 1$ (reachable $18 0^{\circ}$ par, exactly ek direction). $e = 1.5 \Rightarrow cos θ = - 0.667$ (reachable $θ \approx 131. 8^{\circ}$ par).
- Yeh step kyun? Yeh single inequality $∣ - 1/ e ∣ \leq 1$ poori bound-vs-unbound dividing line hai: yeh sirf tab hold karti hai jab $e \geq 1$ .

Verify: $∣ - 1/ e ∣ \leq 1$ require karta hai $e \geq 1$ . Check $e = 1.5$ : $arccos (- 0.667) \approx 131.8 1^{\circ}$ , aur $cos 131.8 1^{\circ} \approx - 0.667$ ✓.

Cell G — real-world word problem

Worked example Ex 7 · Earth ki near-circular orbit

Earth ki orbit ka $e = 0.0167$ hai. Perihelion par Earth Sun se aphelion ke comparison mein kitne percent closer hai? $r_{m a x} / r_{m i n} = (1 + e) / (1 - e)$ use karo.

Forecast: Guess karo — kuch percent, ya tens of percent?

Ratio banao. $\frac{r _{m a x}}{r _{m i n}} = \frac{1 + 0.0167}{1 - 0.0167} = \frac{1.0167}{0.9833}$ .
- Yeh step kyun? Ratio mein $a$ cancel ho jaata hai, toh Earth ki actual distance ki zaruurat hi nahi — pure geometry from $e$ .
Divide karo. $\frac{1.0167}{0.9833} \approx 1.0340$ .
- Yeh step kyun? Yeh kehta hai aphelion perihelion se lagbhag $3.4%$ door hai.
Interpret karo. $\sim 3.4%$ distance swing — tiny. Seasons axial tilt se aate hain, isse nahi.
- Yeh step kyun? Number ko physical reality se jodate hain, jo ek word problem ka point hai.

Verify: $(1.0167/0.9833 - 1) \times 100 \approx 3.40%$ ✓. Chota $e$ → near-circle (cell A limit), $ε < 0$ deeply bound ke saath consistent.

Cell H — limit $e \to 1^{-}$

Worked example Ex 8 · Ek ellipse parabola ki taraf stretch karti hui

$p = 2$ fix karo aur $e$ badhate jao: $e = 0.9, 0.99, 0.999$ . $r_{m a x} = \frac{p}{1 - e}$ (aphelion) track karo. $e \to 1^{-}$ limit mein kya hota hai?

Forecast: Guess karo — kya $r_{m a x}$ ek finite ceiling approach karta hai ya run away karta hai?

$e = 0.9$ : $r_{m a x} = \frac{2}{1 - 0.9} = \frac{2}{0.1} = 20$ .
- Yeh step kyun? Aphelion $θ = 18 0^{\circ}$ par hota hai jahan denominator $1 - e$ hai; jaise $e \to 1$ yeh shrink hota hai.
$e = 0.99$ : $r_{m a x} = \frac{2}{0.01} = 200$ . $e = 0.999$ : $r_{m a x} = \frac{2}{0.001} = 2000$ .
- Yeh step kyun? Har extra "9" far distance ko $\sim 10$ se multiply karta hai — yeh without bound grow karta hai.
Limit. Jaise $e \to 1^{-}$ , $1 - e \to 0^{+}$ toh $r_{m a x} \to \infty$ : ellipse ka far end infinity tak run off karta hai aur closed loop khulkar Ex 3 ki parabola ban jaata hai.
- Yeh step kyun? Isliye $e = 1$ genuinely ek change of kind hai, na ki "ek thinner ellipse" — woh mistake jo parent warn karta hai.

Verify: $2/0.1 = 20$ , $2/0.01 = 200$ , $2/0.001 = 2000$ ✓ — har baar $10 \times$ jaise $e$ ek nine gain karta hai.

Cell I — exam twist: same shape, different size

Worked example Ex 9 · Do ellipses, ek eccentricity

Orbit P ka $a = 5$ hai; orbit Q ka $a = 50$ hai; dono ka $e = 0.4$ hai. Dikhao ki dono ki same shape hai par different sizes hain, $r_{m i n} / a$ aur $r_{m a x} / r_{m i n}$ compare karke.

Forecast: Guess karo — kya dono ratios P aur Q mein differ karte hain?

Ratio $r_{m i n} / a = (1 - e)$ . Dono ke liye: $1 - 0.4 = 0.6$ .
- Yeh step kyun? $a$ se divide karne par size strip ho jaati hai, sirf shape info bachti hai (jo sirf $e$ par depend karti hai).
Ratio $r_{m a x} / r_{m i n} = (1 + e) / (1 - e) = 1.4/0.6 \approx 2.333$ . Dono ke liye same.
- Yeh step kyun? Phir $a$ cancel ho jaata hai; identical $e$ → identical proportions.
Absolute sizes differ karti hain. P: $r_{m i n} = 5 (0.6) = 3$ . Q: $r_{m i n} = 50 (0.6) = 30$ .
- Yeh step kyun? Yeh parent ki warning expose karta hai — $e$ ek ratio hai, length nahi; Q 10× bada hai phir bhi equally "squashed" hai.

Verify: $1.4/0.6 \approx 2.3333$ ✓; $5 \times 0.6 = 3$ , $50 \times 0.6 = 30$ ✓. Same $e$ , same shape, sizes $a$ ke saath scale karti hain.

Cell J — degenerate: zero angular momentum (radial fall)

Worked example Ex 10 · Seedha neeche girna,

$h = 0$ Ek body rest se drop ki jaati hai (koi sideways motion nahi), toh uski specific angular momentum $h = 0$ hai. $μ = 1$ aur specific energy $ε = - 0.5$ ke saath, $e$ compute karo aur "orbit" interpret karo.

Forecast: Guess karo — koi sideways swing nahi, toh kya yeh ellipse hai, line hai, ya kuch aur?

$h = 0$ ko $e = 1 + 2 ε h^{2} / μ^{2}$ mein plug karo. $h^{2}$ poori doosri term ko kill kar deta hai: $e = 1 + 0 = 1$ .
- Yeh step kyun? Energy sign chahe kuch bhi ho, $h = 0$ force karta hai $e = 1$ kyunki energy term $h^{2}$ se multiply hoti hai.
Shape formula check karo. $p = h^{2} / μ = 0$ , toh $r = p / (1 + cos θ)$ ka numerator $0$ hai: "conic" collapse hokar focus se gujarti straight line ban gayi hai (ek degenerate parabola).
- Yeh step kyun? $p = 0$ matlab curve ki koi sideways width nahi; yeh limiting radial line hai. Body seedha andar girta hai.
Physics interpret karo. Bhale hi $ε = - 0.5 < 0$ (bound, toh energetically ise ellipse hona chahiye), path ek fat ellipse nahi hai — yeh ek line segment hai jis par body centre ki taraf plunge karta hai aur wapas bahar swing aata. Toh shortcut " $ε < 0 \Rightarrow$ ellipse" ko caveat chahiye "aur $h \neq = 0$ ."
- Yeh step kyun? Yeh woh sach mein degenerate corner hai jo shape table chupaata hai: akela eccentricity ( $e = 1$ yahan) poori story nahi batata jab $h = 0$ ho.

Verify: $e = 1 + 2 (- 0.5) (0) /1 = 1 = 1$ ✓, $ε$ se independent. Aur $p = 0^{2} /1 = 0$ ✓ → ek degenerate radial line, area-filling conic nahi. Consistency check: ek genuine ellipse ko $h \neq = 0$ chahiye taaki $p > 0$ mile; yahan $p = 0$ sahi se degeneracy flag karta hai.

Active recall

Which cell

e = 0.6, p = 6.4

ko kabhi infinity tak nahi pahunchne deta, aur kyun?

Cell F/B —

cos θ = - 1/0.6 \approx - 1.667

chahiye, jo

[- 1, 1]

se bahar hai, impossible.

e = 2, p = 3

ke liye, perihelion aur asymptote angle kya hai?

r_{m i n} = 1

;

r \to \infty

θ = 12 0^{\circ}

h = 0

eccentricity ke saath kya karta hai, energy chahe kuch bhi ho?

Forces

e = 1

(energy term

\times h^{2} = 0

) aur

p = 0

: ek degenerate radial line.

e \to 1^{-}

hone par fixed

p

ke saath aphelion ka kya hota hai?

r_{m a x} = p / (1 - e) \to \infty

; ellipse khulkar parabola ban jaati hai.

Do orbits same

e

, different

a

— same ya different shape?

Same shape (sab ratios sirf

e

par depend karte hain); different absolute size (

a

ke saath scale hota hai).

p = a (1 - e^{2})

kyun hai?

Kyunki

r_{m i n} + r_{m a x} = 2 p / (1 - e^{2}) = 2 a

, toh

a = p / (1 - e^{2})

, yaani

p = a (1 - e^{2})

Semi-major axis

a

kya hai?

Lambe axis ka aadha,

a = (r_{m i n} + r_{m a x}) /2

— orbit ki size.

Connections

Parent topic
Conic Sections — har cell ke peeche ki geometry
Vis-viva Equation — $ε$ kahan se aata hai (kisi bhi $r$ par speed)
Specific Orbital Energy — woh sign jo shape pick karta hai
Angular Momentum in Orbits — $h$ sets $p$ aur $h = 0$ degenerate case
Kepler's First Law — nature mein ellipse cell
Escape Velocity — Ex 3 ki $e = 1$ boundary