Kepler's laws
Kepler's second law
 16Kp3 Kepler

Kepler's second law.
We now proceed to directly address Kepler's second law, the one which states that a ray from the Sun to a planet sweeps out equal areas in equal times. This ray is simply the vector r that we've been using. (And we shall continue to use it; r, remember, is defined as the vector from the Sun to the planet.)

What we're looking for is the area that this vector sweeps out. Imagine the planet at some time t = 0, and then imagine at a short time afterward t = deltat. In that time, the vector has moved by a short displacement

deltar = r|(t = deltat) - r|(t = 0). equation 1
The three vectors r|(t = 0), deltar, and r|(t = deltat) form a triangle. The area of this triangle closely approximates the area swept out by the vector r during that short time deltat.

We can write this small area represented by this triangle, deltaA, as one-half of the parallelogram defined by the vectors r and deltar, or

deltaA = (1/2) |r cross deltar|. equation 2
We'll divide both sides of this equation by deltat, the short time involved. Because of this, and the associative properties of the cross product, we find:
deltaA/deltat = (1/2) (1/deltat) |r cross deltar| equation 3
deltaA/deltat = (1/2) |(r cross deltar)/deltat| equation 4
deltaA/deltat = (1/2) |r cross (deltar/deltat)|. equation 5
As we choose smaller and smaller values of deltat, we get better and better approximations of the area swept out by the ray. If we let deltat approach zero by taking the limit of both sides, the approximation approaches the real value and we find that
dA/dt = (1/2) |r cross dr/dt| equation 6
or
dA/dt = (1/2) |r cross v|. equation 7
Knowing
l = m |r cross v| equation 8
and dividing both sides by m, we find
l/m = |r cross v|. equation 9
We can substitute this into our expression for dA/dt and find that
dA/dt = l/(2 m). equation 10
That is, the instantaneous time rate of change of area is the magnitude of the angular momentum divided by twice the mass of the planet. But we know that the mass of the planet is constant, and we also know from our work earlier that the angular momentum vector is constant (and thus its magnitude certainly is). Therefore, the time derivative of area swept out by this ray is constant. In other words, no matter where on the orbit the planet is, its ray still sweeps out the same amount of area. This is Kepler's second law.
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