Specific angular momentum h = √(GMp): Question bank

True or false — justify

Each line: read the claim, decide, and — most importantly — say why.

T/F:

h

is measured in

kg m^{2} /s

False.

h

is angular momentum per unit mass,

h = L / m

, so the kg divides out — units are

m^{2} /s

. Being mass-free is the entire point.

T/F: For a fixed central mass, a wider orbit (larger

p

) always has larger

h

True.

h = GM p

is monotonic in

p

, so more half-width means more spin. This is why

p

and

h

carry the same information.

T/F:

h = r v

holds at every point of an elliptical orbit.

False. In general

h = r v sin ϕ

; only at periapsis and apoapsis is

ϕ = 9 0^{\circ}

sin ϕ = 1

. Everywhere else

v

has a radial component that contributes nothing to

r \times v

T/F: Two orbits with the same

h

must have the same shape.

False. Same

h

fixes only

p = h^{2} / GM

; you still need

e

(or the energy) to pin down whether it's a fat ellipse, a circle, or a hyperbola sharing that half-width.

T/F: If gravity suddenly doubled (

GM \to 2 GM

) while

r, v

stayed the same,

h

would double.

False.

h = ∣ r \times v ∣

depends only on the state vectors, not on

GM

; it wouldn't budge at that instant. The orbit's shape would change afterward, but

h

at that moment is unchanged.

T/F: For a circular orbit

p = a = r

True. Circle means

e = 0

, and

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

; a circle's radius equals its semi-major axis, so all three coincide.

T/F:

h

points along the direction of motion.

False.

h = r \times v

is perpendicular to the orbital plane (perpendicular to both

r

and

v

). The direction of motion lies in the plane;

h

sticks out of it.

T/F: Conservation of

h

is what forces the orbit to stay in one flat plane.

True. The motion is always perpendicular to the fixed vector

h

; a set of points all perpendicular to one fixed direction is a plane through the focus.

T/F:

h = GM a

for any ellipse.

False. That drops the

(1 - e^{2})

factor. The correct form is

h = GM a (1 - e^{2})

; it reduces to

GM a

only when

e = 0

Why questions

Why do we divide angular momentum by mass to define

h

In the Two-body problem the orbiting mass cancels from

\ddot{r} = - \frac{GM}{r ^{2}} \overset{r}{^}

, so the dynamics don't care about it — the natural conserved quantities should be mass-free too, letting one formula serve any orbiter.

Why is

v \times v = 0

used in the conservation proof?

Any vector is parallel to itself, and the cross product of parallel vectors is zero. This kills one term in the product rule so only

r \times a

survives — which then also vanishes because

a ∥ r

Why does the semi-latus rectum, and not periapsis distance, appear in

h = GM p

p

falls straight out of the orbit equation as

p = h^{2} / GM

, making it the natural geometric partner of

h

. Periapsis

r_{p} = p / (1 + e)

mixes in

e

, so it's a less clean handle on the spin.

Why does flight-path angle

ϕ

enter the general relation

h = r v sin ϕ

Only the component of

v

perpendicular to

r

produces angular momentum;

sin ϕ

extracts exactly that perpendicular component. See Flight-path angle for how

ϕ

varies around the orbit.

Why does a circular orbit give the simplest

h = GM r

With

e = 0

every point is a "peri = apo"-like point:

v ⊥ r

always and

r

never changes, so

p = a = r

and the formula collapses to a single radius.

Why can we read off

p = h^{2} / GM

just by comparing two written equations?

Both

r = \frac{h ^{2} / GM}{1 + e c o s θ}

(derived) and

r = \frac{p}{1 + e c o s θ}

(definition) hold for all

θ

; two fractions equal for all

θ

must have equal numerators, so

p = h^{2} / GM

Why doesn't

h

tell you whether the orbit is bound or unbound?

h

fixes only the half-width

p

; a parabola and a slim ellipse can share the same

p

. You need the sign of the energy (or

e

via the Eccentricity vector) to decide bound vs unbound.

Edge cases

Edge: What is

h

for a purely radial fall (straight toward the star, no sideways motion)?

v ∥ r

r \times v = 0

h = 0

. A zero-

h

"orbit" is a degenerate line through the centre, not an area-sweeping ellipse.

Edge: Can

h

be zero for an orbit that still has nonzero speed?

Yes — if

v

is exactly along

r

(radial), speed is large but

sin ϕ = 0

makes

h = 0

. Nonzero speed ≠ nonzero spin.

Edge: What happens to

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

e \to 1

(parabola)?

a \to \infty

while

a (1 - e^{2})

stays finite, so

p

remains a well-defined nonzero number even though

a

blows up — which is why

p

, not

a

, is the safe parameter near parabolic orbits.

Edge: For a hyperbolic orbit (

e > 1

), is

h

still real and conserved?

Yes.

p = h^{2} / GM > 0

regardless of

e

, so

h = GM p

is real and central-force conservation still gives

d h / d t = 0

; only

a

turns negative.

Edge: If two orbits have the same

r

and

v

magnitudes at some instant but different

ϕ

, do they share

h

No.

h = r v sin ϕ

depends on the angle between the arrows, so different flight-path angles give different

h

— same speeds are not enough.

Edge: As

e \to 0

from above, does the periapsis-based formula

h = r_{p} v_{p}

break?

No — it stays valid and smoothly becomes

h = r v_{c i r c}

, since a circle can be regarded as an ellipse whose "periapsis" is every point. The formula degenerates gracefully.

Edge: At the exact instant a planet crosses

θ = 9 0^{\circ}

(where

r = p

), is anything special about

h

Nothing special about

h

— it's constant everywhere. What's special is geometry:

r

equals the semi-latus rectum there, which is how we physically interpret

p

, not a change in the spin.

Recall Quick self-audit

If you nailed every "why" and every "edge" above, you understand that $h$ is (a) mass-free, (b) constant, (c) perpendicular to the plane, and (d) tied to shape through $p$ but not to energy. Any miss points to exactly which pillar to reread.

Connections

Parent — full derivation & examples
Flight-path angle — the $ϕ$ in $h = r v sin ϕ$ .
Kepler's Second Law — the $d A / d t = h /2$ trap.
Specific orbital energy — why $h$ alone can't fix the orbit.
Eccentricity vector — the missing $e$ .
Vis-viva equation — pairs speed with radius using energy.

Question bank — Specific angular momentum h = √(GMp)

True or false — justify

Spot the error

Why questions

Edge cases

Connections