# Specific relative angular momentum

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See also: Classical central-force problem

In celestial mechanics, the specific relative angular momentum (h) of two orbiting bodies is the vector product of the relative position and the relative velocity. Equivalently, it is the total angular momentum divided by the reduced mass.[1] Specific relative angular momentum plays a pivotal role in the analysis of the two-body problem.

## Definition

Specific relative angular momentum, represented by the symbol ${\displaystyle \mathbf {h} \,\!}$, is defined as the cross product of the relative position vector ${\displaystyle \mathbf {r} \,\!}$ and the relative velocity vector ${\displaystyle \mathbf {v} \,\!}$.

${\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} ={\mathbf {L} \over \mu }}$

where:

The units of ${\displaystyle \mathbf {h} \,\!}$ are m2s−1.

For unperturbed orbits the ${\displaystyle \mathbf {h} \,\!}$ vector is always perpendicular to the fixed orbital plane. However, for perturbed orbits the ${\displaystyle \mathbf {h} \,\!}$ vector is generally not perpendicular to the osculating orbital plane

As usual in physics, the magnitude of the vector quantity ${\displaystyle \mathbf {h} \,\!}$ is denoted by ${\displaystyle h\,\!}$:

${\displaystyle h=\left\|\mathbf {h} \right\|}$

## Elliptical orbit

In an elliptical orbit, the specific relative angular momentum is twice the area per unit time swept out by a chord from the primary to the secondary: this area is referred to by Kepler's second law of planetary motion.

Since the area of the entire orbital ellipse is swept out in one orbital period, ${\displaystyle h\,\!}$ is equal to twice the area of the ellipse divided by the orbital period, as represented by the equation

${\displaystyle h={\frac {2\pi ab}{2\pi {\sqrt {\frac {a^{3}}{G(M\!+\!m)}}}}}=b{\sqrt {\frac {G(M\!+\!m)}{a}}}={\sqrt {a(1-e^{2})G(M\!+\!m)}}={\sqrt {pG(M\!+\!m)}}}$.

where