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

3.2.4 · D4 · Physics › Orbital Mechanics & Astrodynamics › Orbit shape from eccentricity — circle (e=0), ellipse (0 - e

Yeh page ek self-test hai. Problem padho, solution cover karo, khud solve karo, PHIR collapsible kholo. Yahan jo bhi symbols use hue hain woh sab parent note the parent topic mein build kiye gaye hain; agar koi term unfamiliar lage, toh aage badhne se pehle wapas jaao.

Definition Paanch toolkit facts (neeche sab kuch sirf inhi se bana hai)

Poore note mein, $μ = GM$ (central mass times Newton's constant), $ε = E / m$ hai specific energy (orbiting body ki energy per kilogram), aur $h = L / m$ hai specific angular momentum (angular momentum per kilogram). Ellipse ki do purely geometric lengths bhi aati hain: semi-major axis $a$ (sabse lambe diameter ka aadha — orbit ka "average" radius), aur focal distance $c$ (har focus ellipse ke centre se kitna dur hai). Paanch facts:

Orbit shape: $r (θ) = \frac{p}{1 + e cos θ}$ , jisme semi-latus rectum $p = \frac{h ^{2}}{μ}$ .
Shape from energy: $e = 1 + \frac{2 ε h ^{2}}{μ ^{2}}$ .
Classification: $e = 0$ circle, $0 < e < 1$ ellipse, $e = 1$ parabola, $e > 1$ hyperbola.
Ellipse geometry: $p = a (1 - e^{2})$ , isliye $r_{m i n} = a (1 - e)$ , $r_{m a x} = a (1 + e)$ , aur $e = c / a$ (focal distance over semi-major axis).
$r \to \infty$ wahan hota hai jahan denominator zero ho jaaye: $cos θ = - 1/ e$ .

Inhe Conic Sections, Specific Orbital Energy, Angular Momentum in Orbits, aur Vis-viva Equation se anchor karo.

Level 1 — Recognition

L1·1 — Shape ka naam batao

Har eccentricity ke liye, shape batao aur yeh bhi batao ki orbit bound (wapas aati hai) hai ya unbound (escape kar jaati hai): (a) $e = 0$ , (b) $e = 0.6$ , (c) $e = 1$ , (d) $e = 1.8$ .

Recall Solution L1·1

Seedha Fact 3 se padho.

(a) $e = 0$ : circle, bound.
(b) $0 < e < 1$ : ellipse, bound.
(c) $e = 1$ : parabola, marginal escape (bound⇄unbound boundary).
(d) $e > 1$ : hyperbola, unbound. Yeh kaisa dikhta hai: neeche figure mein curves ki family dekho — jaise $e$ badhta hai, curve ek closed loop se ek open arm mein "khulta" jaata hai.

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

L1·2 — Length ya ratio?

Do ellipses: ek chhoti jiske foci $2 cm$ apart aur semi-major axis $5 cm$ hai, aur ek badi jiske foci $200, 000 km$ apart aur semi-major axis $500, 000 km$ hai. Kaun zyada eccentric hai?

Recall Solution L1·2

WHY a ratio: $e = c / a$ (Fact 4) dimensionless hai — units cancel ho jaate hain, isliye absolute size irrelevant hai. Dhyan do "foci apart" full separation $2 c$ hai, toh $c$ uska aadha hai.

Tiny: $c = 1 cm$ , $a = 5 cm$ , $e = 1/5 = 0.2$ .
Giant: $c = 100, 000 km$ , $a = 500, 000 km$ , $e = 100000/500000 = 0.2$ . Dono equally eccentric hain. Same shape, alag scale.

Level 2 — Application

L2·1 — Perihelion aur aphelion

Ek comet ka semi-major axis $a = 4 AU$ aur eccentricity $e = 0.75$ hai. Sun se uski closest ( $r_{m i n}$ ) aur farthest ( $r_{m a x}$ ) distance, AU mein nikalo.

Recall Solution L2·1

WHAT / WHY: Fact 4 seedha do turning points deta hai. $r_{m i n} = a (1 - e) = 4 (1 - 0.75) = 4 (0.25) = 1 AU$ $r_{m a x} = a (1 + e) = 4 (1 + 0.75) = 4 (1.75) = 7 AU$ Check: dono ka average $\frac{1 + 7}{2} = 4 = a$ . ✓ (semi-major axis hamesha do extremes ka mean hota hai — yahi "semi-major axis" hai.)

L2·2 — $h$ se Semi-latus rectum

Earth ke around ek orbit ka specific angular momentum $h = 7.0 \times 1 0^{10} m^{2} / s$ hai, aur $μ_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ . Semi-latus rectum $p$ metres mein nikalo.

Recall Solution L2·2

WHY $p = h^{2} / μ$ : $p$ wo value hai jo $r$ ki $θ = 9 0^{\circ}$ par hoti hai, jahan $cos θ = 0$ toh $r = p$ . Yeh purely angular momentum se fix hoti hai (Fact 1), energy se independent. $p = \frac{h ^{2}}{μ} = \frac{( 7.0 \times 1 0 ^{10} ) ^{2}}{3.986 \times 1 0 ^{14}} = \frac{4.9 \times 1 0 ^{21}}{3.986 \times 1 0 ^{14}} \approx 1.229 \times 1 0^{7} m$ Lagbhag $12, 290 km$ — Earth ki surface se comfortably upar, toh yeh ek real orbit hai.

L2·3 — Ek given angle par distance

Parent-note ke comet ke liye jisme $e = 1$ aur semi-latus rectum $p$ hai, $θ = 9 0^{\circ}$ aur $θ = 12 0^{\circ}$ par $r$ nikalo, $p$ ke multiples mein.

Recall Solution L2·3

Fact 1 use karo $e = 1$ ke saath: $r = \frac{p}{1 + cos θ}$ (yaad raho $θ = 0$ perihelion hai).

$θ = 9 0^{\circ}$ par: $cos 9 0^{\circ} = 0$ , toh $r = \frac{p}{1 + 0} = p$ .
$θ = 12 0^{\circ}$ par: $cos 12 0^{\circ} = - \frac{1}{2}$ , toh $r = \frac{p}{1 - \frac{1}{2}} = \frac{p}{1/2} = 2 p$ . Kaisa dikhta hai: comet $12 0^{\circ}$ par $9 0^{\circ}$ ke mukable mein already do guna dur hai, aur infinity ki taraf ja raha hai jaise $θ \to 18 0^{\circ}$ (jahan $1 + cos θ \to 0$ ).

Level 3 — Analysis

L3·1 — Energy aur momentum se shape

Ek planet ( $μ = 4.0 \times 1 0^{14} m^{3} / s^{2}$ ) ke around ek probe ka specific energy $ε = - 2.0 \times 1 0^{6} J/kg$ aur $h = 6.0 \times 1 0^{10} m^{2} / s$ hai. $e$ nikalo aur shape ka naam batao.

Recall Solution L3·1

WHY Fact 2: eccentricity energy aur angular momentum ke combination se fix hoti hai — yeh formula dono ko package karta hai. $\frac{2 ε h ^{2}}{μ ^{2}} = \frac{2 ( - 2.0 \times 1 0 ^{6} ) ( 6.0 \times 1 0 ^{10} ) ^{2}}{( 4.0 \times 1 0 ^{14} ) ^{2}}$ Numerator: $2 \times (- 2.0 \times 1 0^{6}) \times 3.6 \times 1 0^{21} = - 1.44 \times 1 0^{28}$ . Denominator: $1.6 \times 1 0^{29}$ . Ratio $= - 0.09$ . Phir $e = 1 + (- 0.09) = 0.91 \approx 0.954.$ Kyunki $0 < e < 1$ , yeh ek ellipse hai (bound) — $ε < 0$ ke saath consistent. Highly eccentric lekin phir bhi closed.

L3·2 — Escape ka knife-edge

Dikhao ki agar $ε = 0$ toh $e = 1$ exactly hoga, kisi bhi nonzero $h$ ke liye. Physically explain karo kyun $h$ ka bahar nikalna sense deta hai.

Recall Solution L3·2

Algebra: Fact 2 mein $ε = 0$ daalo: $e = 1 + \frac{2 ( 0 ) h ^{2}}{μ ^{2}} = 1 + 0 = 1.$ $h$ term ko $ε = 0$ se multiply kiya jaata hai, toh woh har value of $h$ ke liye vanish ho jaata hai. Physics — WHY $h$ yahan irrelevant hai: $ε = 0$ matlab body infinity tak exactly zero speed ke saath pahunchti hai (saari kinetic energy well se bahar nikalte nikalte khatam ho jaati hai). Woh "barely escapes" condition sirf energy ke baare mein ek statement hai; kitna sideways swing ( $h$ ) hai woh sirf parabola ki width badalta hai (uska $p = h^{2} / μ$ ), yeh nahi ki woh parabola hai ya nahi. Har zero-energy orbit ek parabola hai. Dekho Escape Velocity.

L3·3 — Hyperbola ke forbidden angles

Ek interstellar object ka $e = 2.5$ hai. (a) Kis angle par $r \to \infty$ ? (b) $θ$ ki kaun si range mein object actually apni orbit par present hai?

Recall Solution L3·3

WHY Fact 5: object orbit par sirf wahan hai jahan $r$ positive aur finite hai; $r$ wahan blow up hota hai jahan denominator zero ho jaaye. (a) $cos θ = - 1/ e = - 1/2.5 = - 0.4$ . Toh $θ = arccos (- 0.4) \approx 113. 6^{\circ}$ (aur symmetry se $- 113. 6^{\circ}$ ). (b) Physical arc hai $- 113. 6^{\circ} < θ < 113. 6^{\circ}$ . Iske bahar, $1 + e cos θ$ negative ho jaata hai → $r$ negative → unphysical. Woh do limiting angles asymptotes hain: woh straight-line directions jin par object aata hai aur jaata hai. Kaisa dikhta hai: figure mein deflected flyby path — ek single open swing, kabhi loop nahi.

Level 4 — Synthesis

L4·1 — $r_{m i n}, r_{m a x}$ se sab kuch nikalna

Ek satellite ka $r_{m i n} = 7000 km$ aur $r_{m a x} = 21000 km$ hai. (a) $a$ , (b) $e$ , (c) $p$ nikalo, aur (d) shape confirm karo.

Recall Solution L4·1

WHY $a$ se shuru karo: yeh extremes ka average hai, toh Fact 4 cleanly invert hota hai. (a) Fact 4 ki do relations add karo: $r_{m i n} + r_{m a x} = a (1 - e) + a (1 + e) = a [(1 - e) + (1 + e)] = 2 a$ . $\pm e$ cancel ho jaata hai, toh $a = \frac{r _{m i n} + r _{m a x}}{2} = \frac{7000 + 21000}{2} = 14000 km .$ (b) Unhe subtract karo: $r_{m a x} - r_{m i n} = a (1 + e) - a (1 - e) = a [(1 + e) - (1 - e)] = 2 a e$ . Constant parts cancel ho jaate hain, sirf $e$ term bachta hai: $e = \frac{r _{m a x} - r _{m i n}}{2 a} = \frac{r _{m a x} - r _{m i n}}{r _{m a x} + r _{m i n}} = \frac{21000 - 7000}{28000} = \frac{14000}{28000} = 0.5.$ (c) $p = a (1 - e^{2}) = 14000 (1 - 0.25) = 14000 (0.75) = 10500 km$ . (d) $0 < e < 1$ → ellipse, bound. ✓ Cross-check: $r_{m i n} = a (1 - e) = 14000 (0.5) = 7000$ ✓ aur $r_{m a x} = a (1 + e) = 14000 (1.5) = 21000$ ✓.

L4·2 — Ek parabolic capture design karo

Tum chahte ho ki ek probe ek planet ( $μ = 3.0 \times 1 0^{13} m^{3} / s^{2}$ ) par parabolic orbit ( $e = 1$ ) mein perihelion distance $r_{p} = 5.0 \times 1 0^{6} m$ ke saath pahunche. Uske paas kaun sa specific angular momentum $h$ hona chahiye?

Recall Solution L4·2

WHY $r_{p}$ ko $p$ se link karo: ek parabola ( $e = 1$ ) ke liye, Fact 1 $θ = 0$ (perihelion) par $r_{p} = \frac{p}{1 + 1} = \frac{p}{2}$ deta hai, toh $p = 2 r_{p}$ . Phir $p = h^{2} / μ \Rightarrow h = μ p = μ \cdot 2 r_{p}$ . $h = (3.0 \times 1 0^{13}) (2) (5.0 \times 1 0^{6}) = 3.0 \times 1 0^{20} \approx 1.732 \times 1 0^{10} m^{2} / s .$ Sanity: bada required perihelion ⇒ bada $h$ ⇒ zyada sideways swing, jo geometry se match karta hai.

Level 5 — Mastery

L5·1 — Classification threshold prove karo

$e = 1 + 2 ε h^{2} / μ^{2}$ use karke prove karo ki $ε$ ka sign exactly bound vs unbound determine karta hai, aur $ε$ ki woh single value identify karo (in terms of $μ, h$ ) jo $e$ ko uski minimum possible value $e = 0$ (a circle) banati hai.

Recall Solution L5·1

Part 1 — $ε$ ka sign: Root ke andar quantity hai $1 + \frac{2 ε h ^{2}}{μ ^{2}}$ . Kyunki $h^{2} / μ^{2} > 0$ :

$ε < 0$ ⇒ added term negative ⇒ root $< 1$ ⇒ $e < 1$ (bound: circle/ellipse).
$ε = 0$ ⇒ root $= 1 = 1$ ⇒ $e = 1$ (parabola, the boundary).
$ε > 0$ ⇒ root $> 1$ ⇒ $e > 1$ (hyperbola, unbound). Teen cases saare real $ε$ exhaust karte hain, toh classification complete aur gap-free hai.

Part 2 — circle condition: $e = 0$ require karta hai ki poora square root $0$ ho: $1 + \frac{2 ε h ^{2}}{μ ^{2}} = 0 \Rightarrow ε = - \frac{μ ^{2}}{2 h ^{2}} .$ Yeh given $h$ ke liye sabse zyada negative $ε$ hai: fixed angular momentum ke liye, sabse deep (most bound) energy sabse round orbit deti hai. Koi bhi kam-negative $ε$ ise ellipse mein stretch kar deta hai. Kaisa dikhta hai: L1 figure mein innermost, perfectly round loop — shape ka ground state.

L5·2 — Do orbits, same $p$ , alag energy

Do bodies ka identical semi-latus rectum $p$ hai (same $h$ ), same $μ$ ke around. Body A ka $e = 0.5$ hai, Body B ka $e = 2$ . (a) Kaun bound hai? (b) Har ek ke liye $θ = 6 0^{\circ}$ par $r$ ko $p$ ke multiple ke roop mein nikalo. (c) Denominator se explain karo kyun B ki distances $θ$ ke saath zyada tezi se badhti hain.

Recall Solution L5·2

(a) $e = 0.5 < 1$ → A ek ellipse (bound) hai; $e = 2 > 1$ → B ek hyperbola (unbound) hai. (b) Fact 1 ke saath $cos 6 0^{\circ} = 0.5$ (angles perihelion se measure hote hain, $θ = 0$ ):

A: $r_{A} = \frac{p}{1 + 0.5 ( 0.5 )} = \frac{p}{1.25} = 0.8 p$ .
B: $r_{B} = \frac{p}{1 + 2 ( 0.5 )} = \frac{p}{2} = 0.5 p$ . (Interesting: $6 0^{\circ}$ par hyperbola actually zyada paas hai — uska perihelion tight hai kyunki same $p$ ke saath bada $e$ matlab chhota $r_{p} = p / (1 + e)$ hota hai.) (c) Jaise $θ$ $9 0^{\circ}$ ke baad badhta hai, $cos θ$ negative ho jaata hai. B ke liye term $e cos θ = 2 cos θ$ do guna hard swing karta hai, toh denominator zero ki taraf race karta hai aur $cos θ = - 0.5$ ( $θ = 12 0^{\circ}$ ) par hit karta hai — $r$ diverge hota hai. A ke liye, $0.5 cos θ$ kabhi denominator ko zero tak nahi kheench sakta (uska sabse negative value $1 - 0.5 = 0.5 > 0$ hai), toh $r$ hamesha finite rehta hai. Yahi exactly reason hai ki $e \geq 1$ escape karta hai aur $e < 1$ nahi karta — poora bound/unbound distinction usi ek denominator mein rehta hai.

Active recall

Which factor,

(1 - e)

(1 + e)

, gives perihelion?

(1 - e)

— chhota number, kyunki

r_{m i n}

sabse bade denominator ke saath pair hota hai.

Parabola ke liye, perihelion ka

p

se kya relation hai?

r_{p} = p /2

(denominator

1 + cos 0 = 2

Fixed

h

par perfect circle banane ke liye kaun sa

ε

chahiye?

ε = - μ^{2} / (2 h^{2})

r_{m i n}, r_{m a x}

diye hon toh

e

kaise nikaalte hain?

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

cos θ = - 1/ e

par

r

ke saath kya ho raha hai?

Woh infinity ki taraf diverge karta hai — woh angle asymptote hai (sirf tabhi reachable jab

e \geq 1

ho).

Orbit par

θ = 0

kahan hai?

Perihelion par (closest approach), jahan

cos θ = + 1

r

ko sabse chhota banata hai.

Connections

Parent topic
Conic Sections · Vis-viva Equation · Specific Orbital Energy · Angular Momentum in Orbits · Kepler's First Law · Escape Velocity