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I am currently reading Taylor's "Classical Mechanics", specifically the angular momentum chapter. In his definition of the conservation of angular momentum, Taylor states:

If the net external torque on an N particle system is 0, the system's total angular momentum $\vec{L} = \sum \vec{r}_\alpha \times \vec{p}_\alpha$

My problem is that when he discusses Kepler's Second Law (pg. 93), he shows that the area sweeped by a line connecting a planet to the sun is the same in any time interval $dt$:

$$ \frac{dA}{dt} = \frac{1}{2m} | \vec{r} \times \vec{p}| = \frac{1}{2m} |\vec{l}|$$

he states that because the angular momentum is conserved, $\frac{dA}{dt}$ is constant. This seems like it doesn't match with his principle as it is only the magnitude conserved and not the vector itself. So I'm confused as to which it is. Any clarification is appreciated.

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  • $\begingroup$ If a vector is constant then so is its magnitude. $\endgroup$ May 13 '20 at 8:11
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The angular momentum vector, which is perpendicular to the plane of motion, is conserved. So its magnitude is conserved, and hence the rate of sweeping out area.

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Treating the Sun as stationary, the planet's angular momentum about the centre of mass, O, of the Sun is conserved. So the vector $\vec r \times \vec p$ is conserved about O.

This implies that the direction of $\vec r \times \vec p$ stays constant, which implies that the planet's motion is confined to one plane, the plane containing $\vec r$ and $\vec p$ at any instant.

The conservation of $\vec L$ also implies that magnitude of $\vec r \times \vec p$ is constant, showing, as you have noted, that the rate of sweeping out of area by $\vec r$ is constant.

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