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A particle of mass *m* and velocity * v* has linear momentum

L = r Ã— p (43)

Notice that angular momentum is always a vector perpendicular to the plane defined by the vectors * r* and

Figure: The angular momentum *L* of a particle traveling in a circular orbit.

*L = rp = mvr (44)*

The significance of angular momentum arises from its derivative with respect to time,* (45)*

where *p** *has been replaced by* m*v* *and the constant m has been factored out. Using* the *product rule of differential calculus,

(46)

In the first term on the right-hand side of equation (46), *d** r*/

(46)

(47)

Here, *d** v*/

(47)

(45)

(48)

Equation (48) means that any change in the angular momentum of a particle must be produced by a force that is not acting along the same direction as * r*. One particularly important application is the solar system. Each planet is held in its orbit by its gravitational attraction to the Sun, a force that acts along the vector from the Sun to the planet. Thus, the force of gravity cannot change the angular momentum of any planet with respect to the Sun. Therefore, each planet has constant angular momentum with respect to the Sun. This conclusion is correct even though the real orbits of the planets are not circles but ellipses.

(48)

The quantity * r* Ã—

(49)

Equation (49) means that if there is no torque acting on a particle, its angular momentum is constant, or conserved. Suppose, however, that some agent applies a force **F**_{a} to the particle resulting in a torque equal to * r*Ã—

(49)

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