Areal velocity is closely related to
angular momentum. Any object has an orbital angular momentum about an origin, and this turns out to be, up to a multiplicative scalar constant, equal to the areal velocity of the object about the same origin. A crucial property of angular momentum is that it is conserved under the action of central forces (i.e. forces acting radially toward or away from the origin). Historically, the law of conservation of angular momentum was stated entirely in terms of areal velocity. A special case of this is
Kepler's second law, which states that the areal velocity of a planet, with the sun taken as origin, is constant with time. Because the gravitational force acting on a planet is approximately a central force (since the mass of the planet is small in comparison to that of the sun), the angular momentum of the planet (and hence the areal velocity) must remain (approximately) constant.
Isaac Newton was the first scientist to recognize the dynamical significance of Kepler's second law. With the aid of his laws of motion, he proved in 1684 that any planet that is attracted to a fixed center sweeps out equal areas in equal intervals of time. For this reason, the law of conservation of angular momentum was historically called the "principle of equal areas". The law of conservation of angular momentum was later expanded and generalized to more complicated situations not easily describable via the concept of areal velocity. Since the modern form of the law of conservation of angular momentum includes much more than just Kepler's second law, the designation "principle of equal areas" has been dropped in modern works.
Derivation of the connection with angular momentum In the situation of the first figure, the area swept out during time period Δ
t by the particle is approximately equal to the area of triangle
ABC. As Δ
t approaches zero this near-equality becomes exact as a
limit. Let the point
D be the fourth corner of parallelogram
ABDC shown in the figure, so that the vectors
AB and
AC add up by the
parallelogram rule to vector
AD. Then the area of triangle
ABC is half the area of parallelogram
ABDC, and the area of
ABDC is equal to the magnitude of the
cross product of vectors
AB and
AC. This area can also be viewed as a (pseudo)vector with this magnitude, and pointing in a direction perpendicular to the parallelogram (following the
right hand rule); this vector is the cross product itself: \text{vector area of parallelogram }ABCD = \mathbf{r}(t) \times \mathbf{r}(t + \Delta t). Hence \text{vector area of triangle }ABC = \frac{\mathbf{r}(t) \times \mathbf{r}(t + \Delta t)}{2}. The areal velocity is this vector area divided by Δ
t in the limit that Δ
t becomes vanishingly small: \begin{align} \text{areal velocity} &= \lim_{\Delta t \rightarrow 0} \frac{\mathbf{r}(t) \times \mathbf{r}(t + \Delta t)}{2 \Delta t} \\ &= \lim_{\Delta t \rightarrow 0} \frac{\mathbf{r}(t) \times \bigl( \mathbf{r}(t) + \mathbf{r}\,'(t) \Delta t \bigr)}{2 \Delta t} \\ &= \lim_{\Delta t \rightarrow 0} \frac{\mathbf{r}(t) \times \mathbf{r}\,'(t)}{2} \left( {\Delta t \over \Delta t} \right) \\ &= \frac{\mathbf{r}(t) \times \mathbf{r}\,'(t)}{2}. \end{align} But, \mathbf{r}\,'(t) is the velocity vector \mathbf{v}(t) of the moving particle, so that \frac{d \mathbf{A}}{d t} = \frac{\mathbf{r} \times \mathbf{v}}{2}. On the other hand, the angular momentum of the particle is \mathbf{L} = \mathbf{r} \times m \mathbf{v}, and hence the angular momentum equals 2
m times the areal velocity. ==Relationship with magnetic dipoles==