 # prove that the angular momentum is equal to twice the product of its mass and aerial velocity.

12 years ago

Dear charan,

The areal velocity vector is $\frac{d \vec{A}}{d t} = \lim_{\Delta t \rightarrow 0} {\Delta \vec{A} \over \Delta t} = \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times \vec{r}(t + \Delta t) \over 2 \Delta t}$ $= \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times [ \vec{r}(t) + \vec{r}\,'(t) \Delta t ] \over 2 \Delta t}$ $= \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times \vec{r}\,'(t) \over 2} \left( {\Delta t \over \Delta t} \right)$ $= {\vec{r}(t) \times \vec{r}\,'(t) \over 2}.$

But, $\vec{r}\,'(t)$ is the velocity vector $\vec{v}(t)$of the moving particle, so that $\frac{d \vec{A}}{d t} = {\vec{r} \times \vec{v} \over 2}.$

The areal velocity vector can be placed at the moving point B. As the particle moves along its path in space, it sweeps out a cone-shaped surface. The areal velocity vector is perpendicular to this surface, and, in general, varies in both magnitude and direction. In planar problems, such as the orbit of a planet about the sun, the direction of the areal velocity vector is perpendicular to the orbital plane. Kepler's 2nd law is a statement of the constancy of the rate at which the position vector of a planet sweeps out area, with the sun taken as origin. The path of the planet is an ellipse, with the sun at one focus.

The angular momentum of the particle is $\vec{L} = \vec{r} \times m \vec{v},$

and hence $\vec{L} = 2 m \frac{d \vec{A}}{d t}$.
The direction of the angular momentum vector L is always the same as that of the areal velocity vector. Angular momentum is conserved if and only if the areal velocity is a constant vector.

All the best.

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