Kepler's Second Law can be stated in several equivalent ways:

If we draw a line from the sun to the planet in question (a radius), then as
the planet moves around in its orbit it will sweep out some area $A_1$ in time
$t$. If we consider the planet elsewhere on its orbit, then over the same time
interval $t$ its radius will sweep out another area, $A_2$. Kepler's Second
Law states that $A_1 = A_2$. This law is often referred to as the "law of
equal areas."

Alternatively, any two radial lines between the sun and the elliptical
orbit of a planet form some area (for convenience let us again call this $A_1$).
The points where these radii intersect the orbit are labeled $p_1$ and $q_1$.
We then choose two more radial lines that form another area $A_2$ that is equal
in size to $A_1$ and label the points where these radii intersect $p_2$ and
$q_2$. Then Kepler's Second Law tells us that the time taken for the
planet to pass between points $p_1$ and $q_1$ is equal to the time taken to pass
between points $p_2$ and $q_2$.

Keplers Second Law means that the closer a planet is to the sun, the faster it
must be moving on its orbit. When the planet is far away from the sun, it only
has to move a relatively small distance in order to sweep out a large area.
However, when the planet is close to the sun it must move a lot further in order
to sweep out an equal area. This can be seen most clearly in .

Kepler's Second Law and Conservation of Angular Momentum

Kepler's Second Law is an example of the principle of conservation of
angular momentum for
planetary
systems. We can make a geometrical argument to show how this works.

Consider two points $P$ and $Q$ on the orbit of a planet, separated by avery
small distance. Suppose that it takes a small time $dt$ for the planet to move
from $P$ to $Q$. Because the line segment $\vec{PQ}$ is small, we can make the
approximation that it is a straight line. Then $\vec{PQ}$, being the
infinitesimal distance $dx$ over which the planet moved in time $dt$, represents
the average velocity of the planet over that small range. That is $\vec{PQ} =
\vec{v}$. Now consider the area swept out in this time $dt$. It is given by
the area of the triangle $SPQ$, which has height $PP'$ and base $r$. But it is
also clear from that $PP' = |PQ|\sin\theta$. Thus the area
swept out per time $dt$ is given by:
\begin{equation}
\frac{dA}{dt} = \frac{1}{2}\times r \times |PQ| \times \sin\theta =
\frac{rv\sin\theta}{2}
\end{equation}
But Kepler's Second Law asserts that equal areas must be swept out in equal
time intervals or, expressed differently, area is swept out at a constant rate
($k$). Mathematically:
\begin{equation} \frac{dA}{dt} = k \end{equation}
But we just this value:
\begin{equation} \frac{dA}{dt} = k = \frac{rv\sin\theta}{2}
\end{equation}
Angular momentum is given by the expression:
\begin{equation} \vec{L} = m(\vec{v} \times \vec{r}) =
mvr\hat{n}\sin\theta \end{equation}
where $m$ is the mass being considered. The magnitude of the angular
momentum is clearly $mvr\sin\theta$ where we
are now
considering the magnitudes of $\vec{v}$ and $\vec{r}$. Kepler's Second Law
has demonstrated that $ k = \frac{rv\sin\theta}{2}$, and thus:
\begin{equation} 2km = mvr\sin\theta = |\vec{L}| \end{equation}
Since the mass of any planet remains constant around the orbit, we have
shown that the magnitude of the angular momentum is equal to a constant. Thus
Kepler's Second Law demonstrates that angular momentum is conserved for an
orbiting planet.