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

Before anything, one refresher so no symbol is unearned:

The scenario matrix

Every problem on this topic lives in exactly one of these cells. The examples below tick off each one.

Cell	What makes it special	Which tool you reach for	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 $\to$ $h$	given $a, e$ only	$p = a (1 - e^{2})$ , then $h = GM p$	Ex 4
E Degenerate: parabola limit	$e \to 1$ , $a \to \infty$	keep $p$ finite	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	mix B and D	Ex 7
H Exam twist / trap	wrong-quadrant, unit trap	careful $sin ϕ$ , 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

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

Figure — Specific angular momentum h = √(GMp)

Worked example Ex 2 — Periapsis data

$\to$ $h, p, e$ At periapsis a probe has $r_{p} = 8000 km = 8 \times 1 0^{6} m$ and $v_{p} = 9.0 km/s = 9000 m/s$ . Find $h$ , $p$ , and $e$ .

Forecast: at periapsis $v$ is exactly tangential (see the figure: the velocity arrow is perpendicular to the radius). So $h = r_{p} v_{p}$ with no $sin$ loss. Guess whether $e$ comes out $\sim 0.6$ or $\sim 0.06$ .

$h = r_{p} v_{p} sin 9 0^{\circ} = r_{p} v_{p} = 8 \times 1 0^{6} \cdot 9000 = 7.2 \times 1 0^{10} m^{2} /s$ . Why this step? At an apside the flight-path angle $ϕ = 9 0^{\circ}$ , so $sin ϕ = 1$ — this is cell B's whole trick.
$p = \frac{h ^{2}}{GM} = \frac{( 7.2 \times 1 0 ^{10} ) ^{2}}{3.986 \times 1 0 ^{14}} = 1.301 \times 1 0^{7} m \approx 13, 010 km$ . Why this step? Invert $h = GM p$ .
From $r_{p} = \frac{p}{1 + e}$ : $e = \frac{p}{r _{p}} - 1 = \frac{13010}{8000} - 1 = 0.626$ . Why this step? Periapsis is $θ = 0$ in the orbit equation, the cleanest place to read $e$ .

Verify: rebuild $r_{p} = p / (1 + e) = 1.301 \times 1 0^{7} /1.626 = 8.0 \times 1 0^{6} m$ ✓. Since $0 < e < 1$ the orbit is a genuine ellipse ✓.

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

Worked example Ex 3 —

$h$ from a mid-orbit snapshot Somewhere between apsides a spacecraft has $r = 1.0 \times 1 0^{7} m$ , speed $v = 6500 m/s$ , and the angle between $r$ and $v$ is $ϕ = 5 0^{\circ}$ . Find $h$ , then $p$ .

Forecast: here $v$ is tilted relative to $r$ (figure). If you forget the $sin ϕ$ you'll overshoot. Guess: is $h$ bigger or smaller than $r v$ ?

Use the general relation $h = r v sin ϕ$ . Why this step? $h = ∣ r \times v ∣$ and a cross product's magnitude is $r v sin (angle between them)$ . Only the component of $v$ perpendicular to $r$ contributes to sweeping area.
$sin 5 0^{\circ} = 0.766$ , so $h = 1.0 \times 1 0^{7} \cdot 6500 \cdot 0.766 = 4.98 \times 1 0^{10} m^{2} /s$ . Why this step? Plug numbers; note it is less than $r v = 6.5 \times 1 0^{10}$ — the tilt "wastes" some speed radially.
$p = h^{2} / GM = (4.98 \times 1 0^{10})^{2} /3.986 \times 1 0^{14} = 6.22 \times 1 0^{6} m$ . Why this step? Same inversion as always.

Verify: $h / (r v) = 4.98 \times 1 0^{10} /6.5 \times 1 0^{10} = 0.766 = sin 5 0^{\circ}$ ✓. And $h < r v$ as forecast because $ϕ \neq = 9 0^{\circ}$ ✓.

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

Worked example Ex 4 — Given the ellipse's shape only

A Molniya-like orbit has semi-major axis $a = 2.66 \times 1 0^{7} m$ and eccentricity $e = 0.74$ . Find $h$ .

Forecast: never use $p = a$ here (that's only for circles!). We must shrink $a$ by $(1 - e^{2})$ . Guess whether $h$ is above or below the circular value $GM a$ .

$p = a (1 - e^{2}) = 2.66 \times 1 0^{7} (1 - 0.7 4^{2}) = 2.66 \times 1 0^{7} \cdot 0.4524 = 1.203 \times 1 0^{7} m$ . Why this step? $p = a (1 - e^{2})$ is the bridge between "size" $a$ and the $h$ -carrying half-width $p$ .
$h = GM p = 3.986 \times 1 0^{14} \cdot 1.203 \times 1 0^{7} = 4.795 \times 1 0^{21} = 6.93 \times 1 0^{10} m^{2} /s$ . Why this step? The master formula, now that $p$ is known.

Verify: the "if it were a circle of the same $a$ " value is $GM a = 3.986 \times 1 0^{14} \cdot 2.66 \times 1 0^{7} = 1.03 \times 1 0^{11}$ , and indeed $6.93 \times 1 0^{10} < 1.03 \times 1 0^{11}$ ✓ — eccentric orbits carry less $h$ for the same $a$ (they're skinnier at the focus). Units ✓.

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

Worked example Ex 5 — Escape (parabolic) trajectory

A comet grazes periapsis at $r_{p} = 5.0 \times 1 0^{10} m$ on a parabolic ( $e = 1$ ) escape path. Find $h$ . What happens to $a$ ?

Forecast: parabola means "just barely unbound". The semi-major axis blows up ( $a \to \infty$ ), yet $p$ stays finite. Can we still get a finite $h$ ?

For $e = 1$ : $r_{p} = \frac{p}{1 + e} = \frac{p}{2}$ , so $p = 2 r_{p} = 1.0 \times 1 0^{11} m$ . Why this step? The periapsis form still holds for any $e$ ; set $e = 1$ .
$a = \frac{p}{1 - e ^{2}} = \frac{p}{0} \to \infty$ — undefined/infinite. Why this step? $1 - e^{2} = 0$ at $e = 1$ ; the ellipse formula degenerates. This is why we work with $p$ , not $a$ , in the limit.
$h = GM p = 3.986 \times 1 0^{14} \cdot 1.0 \times 1 0^{11} = 3.986 \times 1 0^{25} = 6.31 \times 1 0^{12} m^{2} /s$ . Why this step? $p$ is finite, so $h$ is perfectly finite even though $a$ isn't.

Verify: parabolic periapsis speed should be escape speed $v_{esc} = 2 GM / r_{p}$ . Check $h = r_{p} v_{esc} = r_{p} 2 GM / r_{p} = 2 GM r_{p} = 2 \cdot 3.986 \times 1 0^{14} \cdot 5 \times 1 0^{10} = 6.31 \times 1 0^{12}$ ✓ — matches, and equals $GM \cdot 2 r_{p} = GM p$ ✓.

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

Worked example Ex 6 — Straight drop toward the centre

A stone is released from rest above the Earth and falls straight down. At one instant $r = 1.0 \times 1 0^{7} m$ , $v = 3000 m/s$ , but $v$ points directly toward the centre ( $ϕ = 0^{\circ}$ ). Find $h$ and $p$ .

Forecast: if the velocity is aimed straight at the focus, is any area being swept? Guess $h = 0$ — but prove it.

$h = r v sin ϕ = 1.0 \times 1 0^{7} \cdot 3000 \cdot sin 0^{\circ} = 0$ . Why this step? $r$ and $v$ are parallel (or anti-parallel), so their cross product vanishes — no perpendicular velocity component, no sweeping.
$p = h^{2} / GM = 0$ . Why this step? A "line" orbit has zero half-width: the conic has collapsed to a radial segment ( $e = 1$ and $p = 0$ , a degenerate straight line).

Verify: Kepler's Second Law says $d A / d t = h /2 = 0$ — the radius sweeps no area, consistent with motion straight along the radius ✓. Units: $m^{2} /s$ , value $0$ ✓.

Cell G — Real-world word problem

Worked example Ex 7 — Geostationary transfer orbit (GTO)

A satellite sits on a transfer ellipse with perigee radius $r_{p} = 6.578 \times 1 0^{6} m$ (200 km altitude) and apogee radius $r_{a} = 4.216 \times 1 0^{7} m$ (GEO). Find $h$ , $p$ , $e$ , and the perigee speed $v_{p}$ .

Forecast: two apside radii fix the whole ellipse. Guess whether $v_{p}$ is above or below LEO circular speed ( $\approx 7.9 km/s$ ).

Semi-major axis $a = \frac{r _{p} + r _{a}}{2} = \frac{6.578 \times 1 0 ^{6} + 4.216 \times 1 0 ^{7}}{2} = 2.437 \times 1 0^{7} m$ . Why this step? Peri and apo are the two ends of the major axis; their average is $a$ .
Eccentricity $e = \frac{r _{a} - r _{p}}{r _{a} + r _{p}} = \frac{3.558 \times 1 0 ^{7}}{4.874 \times 1 0 ^{7}} = 0.730$ . Why this step? $r_{a} = a (1 + e), r_{p} = a (1 - e)$ ; subtract & divide.
$p = a (1 - e^{2}) = 2.437 \times 1 0^{7} (1 - 0.73 0^{2}) = 2.437 \times 1 0^{7} \cdot 0.4671 = 1.138 \times 1 0^{7} m$ . Why this step? Cell D bridge, now with the derived $a, e$ .
$h = GM p = 3.986 \times 1 0^{14} \cdot 1.138 \times 1 0^{7} = 6.735 \times 1 0^{10} m^{2} /s$ .
Perigee speed: at an apside $h = r_{p} v_{p} \Rightarrow v_{p} = h / r_{p} = 6.735 \times 1 0^{10} /6.578 \times 1 0^{6} = 10238 m/s \approx 10.2 km/s$ . Why this step? Cell B: at perigee $v ⊥ r$ .

Verify: harmonic-mean check $\frac{2}{p} = \frac{1}{r _{p}} + \frac{1}{r _{a}}$ : RHS $= 1/6.578 \times 1 0^{6} + 1/4.216 \times 1 0^{7} = 1.7576 \times 1 0^{- 7}$ , and $2/ p = 2/1.138 \times 1 0^{7} = 1.758 \times 1 0^{- 7}$ ✓. And $v_{p} = 10.2 > 7.9 km/s$ ✓ (you burn extra to raise apogee to GEO). Cross-check with Vis-viva equation $v_{p} = GM (2/ r_{p} - 1/ a)$ (see verify block).

Cell H — Exam twist / trap

Worked example Ex 8 — The two-trap question

"At a point on an orbit, $r = 1.2 \times 1 0^{7} m$ , $v = 5000 m/s$ , and the angle between velocity and the local horizontal (perpendicular to $r$ ) is $γ = 3 0^{\circ}$ . A student writes $h = GM \cdot r v$ . Fix the two errors and find $h$ and $p$ ."

Forecast: two traps hide here — (i) $GM \cdot r v$ is nonsense ( $h = GM p$ , and $r v$ isn't $p$ ); (ii) the given angle is measured from the horizontal, not from $r$ . Guess which angle goes into $sin$ .

The angle between $r$ and $v$ is $ϕ = 9 0^{\circ} - γ = 6 0^{\circ}$ , because $γ$ (the Flight-path angle) is measured off the local horizontal, while $ϕ$ is measured off the radial line. Why this step? $sin ϕ$ needs the angle between the two vectors themselves; horizontal and radial differ by $9 0^{\circ}$ .
$h = r v sin ϕ = 1.2 \times 1 0^{7} \cdot 5000 \cdot sin 6 0^{\circ} = 1.2 \times 1 0^{7} \cdot 5000 \cdot 0.8660 = 5.196 \times 1 0^{10} m^{2} /s$ . Why this step? The correct general formula, with the correct angle. (Equivalently $h = r v cos γ$ .)
$p = h^{2} / GM = (5.196 \times 1 0^{10})^{2} /3.986 \times 1 0^{14} = 6.774 \times 1 0^{6} m$ . Why this step? Invert the master formula. The student's $GM \cdot r v = 3.986 \times 1 0^{14} \cdot 6 \times 1 0^{10} = 1.55 \times 1 0^{13}$ would be off by ~300×, and dimensionally wrong ( $r v$ has units $m^{2} /s$ , not length).

Verify: $h = r v cos γ = 6 \times 1 0^{10} \cdot cos 3 0^{\circ} = 6 \times 1 0^{10} \cdot 0.8660 = 5.196 \times 1 0^{10}$ ✓ (matches step 2 since $sin 6 0^{\circ} = cos 3 0^{\circ}$ ). Units of $p$ : $(m^{2} /s)^{2} / (m^{3} / s^{2}) = m$ ✓.

Recall Scenario self-test

Which cell is "given $a$ and $e$ , find $h$ "? ::: Cell D — compute $p = a (1 - e^{2})$ first. Which cell gives $h = 0$ and why? ::: Cell F — radial fall, $v ∥ r$ so $sin ϕ = 0$ . In the parabolic limit what stays finite, $a$ or $p$ ? ::: $p$ stays finite; $a \to \infty$ (Cell E). Given angle $γ$ from the horizontal, what enters $sin$ ? ::: $ϕ = 9 0^{\circ} - γ$ , i.e. $h = r v cos γ$ (Cell H).

Connections

Parent: h = √(GMp) — the formula every cell uses.
Flight-path angle — the angle $ϕ$ / $γ$ that decides whether $h = r v$ or $h = r v sin ϕ$ .
Vis-viva equation — cross-checks perigee speeds (Ex 7).
Kepler's Second Law — $d A / d t = h /2$ , which makes Cell F's $h = 0$ mean "no area swept".
Two-body problem · Eccentricity vector · Specific orbital energy — where $e$ , $a$ , $p$ come from.