Specific angular momentum h = √(GMp): Worked examples

3.2.12 · D3 · Physics › Orbital Mechanics & Astrodynamics › Specific angular momentum h = √(GMp)

Shuru karne se pehle, ek refresher taaki koi bhi symbol anjaan na rahe:

Scenario matrix

Is topic ka har problem inhi cells mein se kisi ek mein aata hai. Neeche ke examples ek-ek karke sab ko tick karte hain.

Cell	Kya special hai	Kaun sa tool use karo	Example
A Circle, $e = 0$	$p = a = r$ , speed constant	$h = GM r = v_{c i r c} r$	Ex 1
B Ellipse at an apside	$v ⊥ r$ , so $sin ϕ = 1$	$h = r_{p} v_{p} = r_{a} v_{a}$	Ex 2
C Ellipse at a general point	$v \neq ⊥ r$	$h = r v sin ϕ$	Ex 3
D Geometry $\to$ $h$	sirf $a, e$ diye hain	$p = a (1 - e^{2})$ , phir $h = GM p$	Ex 4
E Degenerate: parabola limit	$e \to 1$ , $a \to \infty$	$p$ finite rakho	Ex 5
F Degenerate: radial drop	$h = 0$ , straight-line fall	$sin ϕ = 0$	Ex 6
G Real-world word problem	GEO / apogee-kick style	B aur D mix karo	Ex 7
H Exam twist / trap	wrong-quadrant, unit trap	$sin ϕ$ dhyan se, vs $/2$	Ex 8

Signs & special values covered: $e = 0$ (A), $0 < e < 1$ (B,C,D,G), $e = 1$ (E), $h = 0$ (F); angle $ϕ = 9 0^{\circ}$ (B), $ϕ \neq = 9 0^{\circ}$ (C,H), $ϕ = 0^{\circ}$ (F). Constants: $G M_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ throughout.

Cell A — Circular orbit

Worked example Ex 1 —

$r = 7000 km$ par Circular orbit Ek satellite Earth ke around $r = 7 \times 1 0^{6} m$ radius par circle kar rahi hai. $h$ do tareekon se nikalo.

Forecast: circle mein har point "peri = apo" hai, isliye $h = r v$ with $v$ constant. Guess karo: kya $h$ $5 \times 1 0^{10}$ ke paas hai ya $5 \times 1 0^{12}$ ke?

Circle matlab $e = 0$ , so $p = a = r = 7 \times 1 0^{6} m$ . Yeh step kyun? $p = a (1 - e^{2})$ sirf $a$ ban jaata hai jab $e = 0$ , aur circle ke liye $a = r$ .
Master formula mein plug karo: $h = GM p = 3.986 \times 1 0^{14} \cdot 7 \times 1 0^{6} = 2.79 \times 1 0^{21} \approx 5.28 \times 1 0^{10} m^{2} /s .$ Yeh step kyun? $h = GM p$ hi poora topic hai; yahan $p$ bas radius hai.
Speed se cross-check: circular speed $v = GM / r = 3.986 \times 1 0^{14} /7 \times 1 0^{6} = 7546 m/s$ . Yeh step kyun? Circle mein $v ⊥ r$ har jagah hota hai, isliye seedha $h = r v$ .

Verify: $r v = 7 \times 1 0^{6} \cdot 7546 = 5.28 \times 1 0^{10} m^{2} /s$ ✓ step 2 se match karta hai. Units: $m^{3} / s^{2} \cdot m = m^{4} / s^{2} = m^{2} /s$ ✓.

Cell B — Ellipse at an apside ( $v ⊥ r$ )

Figure — Specific angular momentum h = √(GMp)

Worked example Ex 2 — Periapsis data

$\to$ $h, p, e$ Periapsis par ek probe ka $r_{p} = 8000 km = 8 \times 1 0^{6} m$ aur $v_{p} = 9.0 km/s = 9000 m/s$ hai. $h$ , $p$ , aur $e$ nikalo.

Forecast: periapsis par $v$ exactly tangential hoti hai (figure dekho: velocity arrow radius ke perpendicular hai). Isliye $h = r_{p} v_{p}$ bina kisi $sin$ loss ke. Guess karo kya $e$ $\sim 0.6$ aata hai ya $\sim 0.06$ .

$h = r_{p} v_{p} sin 9 0^{\circ} = r_{p} v_{p} = 8 \times 1 0^{6} \cdot 9000 = 7.2 \times 1 0^{10} m^{2} /s$ . Yeh step kyun? Apside par flight-path angle $ϕ = 9 0^{\circ}$ hota hai, isliye $sin ϕ = 1$ — yahi cell B ka poora trick hai.
$p = \frac{h ^{2}}{GM} = \frac{( 7.2 \times 1 0 ^{10} ) ^{2}}{3.986 \times 1 0 ^{14}} = 1.301 \times 1 0^{7} m \approx 13, 010 km$ . Yeh step kyun? $h = GM p$ ko invert karo.
$r_{p} = \frac{p}{1 + e}$ se: $e = \frac{p}{r _{p}} - 1 = \frac{13010}{8000} - 1 = 0.626$ . Yeh step kyun? Periapsis orbit equation mein $θ = 0$ hai, yahan se $e$ padhna sabse asaan hai.

Verify: $r_{p} = p / (1 + e) = 1.301 \times 1 0^{7} /1.626 = 8.0 \times 1 0^{6} m$ ✓ rebuild hota hai. Kyunki $0 < e < 1$ hai, orbit genuine ellipse hai ✓.

Cell C — Ellipse at a general point ( $v \neq ⊥ r$ )

Worked example Ex 3 — Mid-orbit snapshot se

$h$ Apsides ke beech kisi jagah ek spacecraft ka $r = 1.0 \times 1 0^{7} m$ , speed $v = 6500 m/s$ , aur $r$ aur $v$ ke beech angle $ϕ = 5 0^{\circ}$ hai. $h$ nikalo, phir $p$ .

Forecast: yahan $v$ tilted hai $r$ ke relative (figure dekho). Agar $sin ϕ$ bhool gaye toh overshoot ho jaoge. Guess karo: kya $h$ , $r v$ se bada hoga ya chhota?

General relation $h = r v sin ϕ$ use karo. Yeh step kyun? $h = ∣ r \times v ∣$ aur cross product ki magnitude $r v sin (angle between them)$ hoti hai. Sirf $v$ ka wo component jo $r$ ke perpendicular hai, area sweep karne mein contribute karta hai.
$sin 5 0^{\circ} = 0.766$ , isliye $h = 1.0 \times 1 0^{7} \cdot 6500 \cdot 0.766 = 4.98 \times 1 0^{10} m^{2} /s$ . Yeh step kyun? Numbers plug karo; note karo ki yeh $r v = 6.5 \times 1 0^{10}$ se kam hai — tilt ki wajah se kuch speed radially "waste" ho jaati hai.
$p = h^{2} / GM = (4.98 \times 1 0^{10})^{2} /3.986 \times 1 0^{14} = 6.22 \times 1 0^{6} m$ . Yeh step kyun? Wahi inversion jaise hamesha.

Verify: $h / (r v) = 4.98 \times 1 0^{10} /6.5 \times 1 0^{10} = 0.766 = sin 5 0^{\circ}$ ✓. Aur $h < r v$ jaisa forecast kiya tha kyunki $ϕ \neq = 9 0^{\circ}$ ✓.

Cell D — Pure geometry ( $a, e$ ) $\to h$

Worked example Ex 4 — Sirf ellipse ki shape diye ho toh

Ek Molniya-jaisi orbit ka semi-major axis $a = 2.66 \times 1 0^{7} m$ aur eccentricity $e = 0.74$ hai. $h$ nikalo.

Forecast: yahan $p = a$ mat use karna (woh sirf circles ke liye hai!). Hume $a$ ko $(1 - e^{2})$ se shrink karna hoga. Guess karo kya $h$ circular value $GM a$ se upar hoga ya neeche.

$p = a (1 - e^{2}) = 2.66 \times 1 0^{7} (1 - 0.7 4^{2}) = 2.66 \times 1 0^{7} \cdot 0.4524 = 1.203 \times 1 0^{7} m$ . Yeh step kyun? $p = a (1 - e^{2})$ "size" $a$ aur $h$ -carrying half-width $p$ ke beech ka bridge hai.
$h = GM p = 3.986 \times 1 0^{14} \cdot 1.203 \times 1 0^{7} = 4.795 \times 1 0^{21} = 6.93 \times 1 0^{10} m^{2} /s$ . Yeh step kyun? Master formula, ab jab $p$ pata hai.

Verify: "agar same $a$ ka circle hota" wala value $GM a = 3.986 \times 1 0^{14} \cdot 2.66 \times 1 0^{7} = 1.03 \times 1 0^{11}$ hota, aur sach mein $6.93 \times 1 0^{10} < 1.03 \times 1 0^{11}$ ✓ — eccentric orbits same $a$ ke liye kam $h$ carry karti hain (focus par wo zyada patli hoti hain). Units ✓.

Cell E — Degenerate limit: parabola ( $e \to 1$ )

Worked example Ex 5 — Escape (parabolic) trajectory

Ek comet parabolic ( $e = 1$ ) escape path par $r_{p} = 5.0 \times 1 0^{10} m$ par periapsis graze karta hai. $h$ nikalo. $a$ ka kya hota hai?

Forecast: parabola matlab "just barely unbound". Semi-major axis blow up kar jaata hai ( $a \to \infty$ ), lekin $p$ finite rehta hai. Kya phir bhi finite $h$ mil sakta hai?

$e = 1$ ke liye: $r_{p} = \frac{p}{1 + e} = \frac{p}{2}$ , isliye $p = 2 r_{p} = 1.0 \times 1 0^{11} m$ . Yeh step kyun? Periapsis form kisi bhi $e$ ke liye valid hai; bas $e = 1$ set karo.
$a = \frac{p}{1 - e ^{2}} = \frac{p}{0} \to \infty$ — undefined/infinite. Yeh step kyun? $e = 1$ par $1 - e^{2} = 0$ ; ellipse formula degenerate ho jaata hai. Isliye hum limit mein $a$ nahin, $p$ use karte hain.
$h = GM p = 3.986 \times 1 0^{14} \cdot 1.0 \times 1 0^{11} = 3.986 \times 1 0^{25} = 6.31 \times 1 0^{12} m^{2} /s$ . Yeh step kyun? $p$ finite hai, isliye $h$ bilkul finite hai chahe $a$ na ho.

Verify: parabolic periapsis speed escape speed honi chahiye $v_{esc} = 2 GM / r_{p}$ . Check karo $h = r_{p} v_{esc} = r_{p} 2 GM / r_{p} = 2 GM r_{p} = 2 \cdot 3.986 \times 1 0^{14} \cdot 5 \times 1 0^{10} = 6.31 \times 1 0^{12}$ ✓ — match karta hai, aur $GM \cdot 2 r_{p} = GM p$ ✓ bhi.

Cell F — Degenerate limit: radial fall ( $h = 0$ )

Worked example Ex 6 — Centre ki taraf seedha girna

Ek patthar Earth ke upar se rest se release hota hai aur seedha neeche girta hai. Ek instant par $r = 1.0 \times 1 0^{7} m$ , $v = 3000 m/s$ , lekin $v$ seedha centre ki taraf point kar raha hai ( $ϕ = 0^{\circ}$ ). $h$ aur $p$ nikalo.

Forecast: agar velocity seedha focus ki taraf aim hai, toh koi area sweep ho raha hai? Guess karo $h = 0$ — lekin prove karo.

$h = r v sin ϕ = 1.0 \times 1 0^{7} \cdot 3000 \cdot sin 0^{\circ} = 0$ . Yeh step kyun? $r$ aur $v$ parallel (ya anti-parallel) hain, isliye unka cross product zero hai — koi perpendicular velocity component nahin, koi sweeping nahin.
$p = h^{2} / GM = 0$ . Yeh step kyun? Ek "line" orbit ki half-width zero hoti hai: conic ek radial segment ban gayi hai ( $e = 1$ aur $p = 0$ , ek degenerate straight line).

Verify: Kepler's Second Law kehta hai $d A / d t = h /2 = 0$ — radius koi area sweep nahin karta, jo radius ke saath seedhe motion ke saath consistent hai ✓. Units: $m^{2} /s$ , value $0$ ✓.

Cell G — Real-world word problem

Worked example Ex 7 — Geostationary transfer orbit (GTO)

Ek satellite ek transfer ellipse par hai jiska perigee radius $r_{p} = 6.578 \times 1 0^{6} m$ (200 km altitude) aur apogee radius $r_{a} = 4.216 \times 1 0^{7} m$ (GEO) hai. $h$ , $p$ , $e$ , aur perigee speed $v_{p}$ nikalo.

Forecast: do apside radii poori ellipse fix kar dete hain. Guess karo kya $v_{p}$ LEO circular speed ( $\approx 7.9 km/s$ ) se upar hogi ya neeche.

Semi-major axis $a = \frac{r _{p} + r _{a}}{2} = \frac{6.578 \times 1 0 ^{6} + 4.216 \times 1 0 ^{7}}{2} = 2.437 \times 1 0^{7} m$ . Yeh step kyun? Peri aur apo major axis ke do ends hain; unka average $a$ hai.
Eccentricity $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{3.558 \times 1 0 ^{7}}{4.874 \times 1 0 ^{7}} = 0.730$ . Yeh step kyun? $r_{a} = a (1 + e), r_{p} = a (1 - e)$ ; subtract karo aur divide karo.
$p = a (1 - e^{2}) = 2.437 \times 1 0^{7} (1 - 0.73 0^{2}) = 2.437 \times 1 0^{7} \cdot 0.4671 = 1.138 \times 1 0^{7} m$ . Yeh step kyun? Cell D bridge, ab derived $a, e$ ke saath.
$h = GM p = 3.986 \times 1 0^{14} \cdot 1.138 \times 1 0^{7} = 6.735 \times 1 0^{10} m^{2} /s$ .
Perigee speed: apside par $h = r_{p} v_{p} \Rightarrow v_{p} = h / r_{p} = 6.735 \times 1 0^{10} /6.578 \times 1 0^{6} = 10238 m/s \approx 10.2 km/s$ . Yeh step kyun? Cell B: perigee par $v ⊥ r$ hoti hai.

Verify: harmonic-mean check $\frac{2}{p} = \frac{1}{r _{p}} + \frac{1}{r _{a}}$ : RHS $= 1/6.578 \times 1 0^{6} + 1/4.216 \times 1 0^{7} = 1.7576 \times 1 0^{- 7}$ , aur $2/ p = 2/1.138 \times 1 0^{7} = 1.758 \times 1 0^{- 7}$ ✓. Aur $v_{p} = 10.2 > 7.9 km/s$ ✓ (apogee ko GEO tak uthane ke liye extra burn karna padta hai). Vis-viva equation $v_{p} = GM (2/ r_{p} - 1/ a)$ se cross-check karo (verify block dekho).

Cell H — Exam twist / trap

Worked example Ex 8 — Do-trap wala question

"Ek orbit ke kisi point par $r = 1.2 \times 1 0^{7} m$ , $v = 5000 m/s$ , aur velocity aur local horizontal (jo $r$ ke perpendicular hai) ke beech angle $γ = 3 0^{\circ}$ hai. Ek student likhta hai $h = GM \cdot r v$ . Do errors theek karo aur $h$ aur $p$ nikalo."

Forecast: do traps yahan chhupe hain — (i) $GM \cdot r v$ bakwaas hai ( $h = GM p$ hota hai, aur $r v$ $p$ nahin hai); (ii) diya gaya angle horizontal se measure kiya gaya hai, $r$ se nahin. Guess karo $sin$ mein kaun sa angle jaata hai.

$r$ aur $v$ ke beech angle $ϕ = 9 0^{\circ} - γ = 6 0^{\circ}$ hai, kyunki $γ$ (Flight-path angle) local horizontal se measure hota hai, jabki $ϕ$ radial line se measure hota hai. Yeh step kyun? $sin ϕ$ ke liye dono vectors ke beech ke angle ki zaroorat hai; horizontal aur radial $9 0^{\circ}$ se different hote hain.
$h = r v sin ϕ = 1.2 \times 1 0^{7} \cdot 5000 \cdot sin 6 0^{\circ} = 1.2 \times 1 0^{7} \cdot 5000 \cdot 0.8660 = 5.196 \times 1 0^{10} m^{2} /s$ . Yeh step kyun? Sahi general formula, sahi angle ke saath. (Equivalently $h = r v cos γ$ .)
$p = h^{2} / GM = (5.196 \times 1 0^{10})^{2} /3.986 \times 1 0^{14} = 6.774 \times 1 0^{6} m$ . Yeh step kyun? Master formula invert karo. Student ka $GM \cdot r v = 3.986 \times 1 0^{14} \cdot 6 \times 1 0^{10} = 1.55 \times 1 0^{13}$ lagbhag 300× off hota, aur dimensionally bhi galat hota ( $r v$ ke units $m^{2} /s$ hain, length nahin).

Verify: $h = r v cos γ = 6 \times 1 0^{10} \cdot cos 3 0^{\circ} = 6 \times 1 0^{10} \cdot 0.8660 = 5.196 \times 1 0^{10}$ ✓ (step 2 se match karta hai kyunki $sin 6 0^{\circ} = cos 3 0^{\circ}$ ). $p$ ke units: $(m^{2} /s)^{2} / (m^{3} / s^{2}) = m$ ✓.

Recall Scenario self-test

"Diye $a$ aur $e$ , nikalo $h$ " — kaun sa cell hai? ::: Cell D — pehle $p = a (1 - e^{2})$ compute karo. Kaun sa cell $h = 0$ deta hai aur kyun? ::: Cell F — radial fall, $v ∥ r$ isliye $sin ϕ = 0$ . Parabolic limit mein kya finite rehta hai, $a$ ya $p$ ? ::: $p$ finite rehta hai; $a \to \infty$ (Cell E). Horizontal se angle $γ$ diya ho toh $sin$ mein kya jaata hai? ::: $ϕ = 9 0^{\circ} - γ$ , yaani $h = r v cos γ$ (Cell H).

Connections

Parent: h = √(GMp) — woh formula jo har cell use karta hai.
Flight-path angle — angle $ϕ$ / $γ$ jo decide karta hai kya $h = r v$ hoga ya $h = r v sin ϕ$ .
Vis-viva equation — perigee speeds cross-check karta hai (Ex 7).
Kepler's Second Law — $d A / d t = h /2$ , jo Cell F ka $h = 0$ matlab "koi area sweep nahin" banata hai.
Two-body problem · Eccentricity vector · Specific orbital energy — jahan se $e$ , $a$ , $p$ aate hain.