Kepler's Second Law

Suppose that we have an object $1$ in revolution around another object $2$.

Kepler's First Law states that the orbit is in the shape of an ellipse, with the other object at one focus.

Now, we talk about Kepler's Second Law: a line connecting the two objects sweeps out equal areas in equal times.

In other words, a line connecting the two objects sweeps out area at a constant rate. How may we show this?

Let the position vector of object $1$ relative to object $2$ be $\vec{r}$.

Suppose that over a small time interval $\Delta t$, the position vector changes by an amount $\Delta \vec{r}$.

For a sufficiently small time interval, the area swept out is roughly a triangle.

Using cross products, one can easily approximate $\Delta A$, the area swept out: $$\Delta A\approx\frac{1}{2}|\vec{r}\times\Delta \vec{r}|$$ Now, notice that: $$\Delta \vec{r}\approx\vec{v}\Delta t$$ where $\vec{v}$ is the velocity vector. This follows simply from the definition of velocity.

Now, substitute into the first equation: $$\Delta A\approx\frac{\Delta t}{2}|\vec{r}\times\vec{v}|$$ Now, divide by $\Delta t$ and take the limiting case: $$\frac{\Delta A}{\Delta t}\approx\frac{1}{2}|\vec{r}\times\vec{v}|$$ $$\frac{dA}{dt}=\frac{1}{2}|\vec{r}\times\vec{v}|$$ The Conservation of Angular Momentum guarantees that the following quantity is a constant: $$\vec{L}=m(\vec{r}\times\vec{v})$$ where $m$ is the mass of object $1$.

Thus, we can easily see that $\frac{dA}{dt}$ is a constant! We have completed a simple derivation of Kepler's Second Law.

Kepler's Second Law states that an object sweeps out equal areas in equal amounts of time. In other words, this can be written as $\frac{dA}{dt}$ is constant.

To prove this, we will consider how much area $dA$ an object sweeps up in a small time $dt$. If the object sweeps through an angle $d \theta$, we can approximate this area as a triangle with radius $r$ and base $r d \theta$. The area of this triangle is $$dA = \frac{1}{2} r^2 d \theta = \frac{1}{2} r^2 \dot \theta dt$$ Thus, $\frac{dA}{dt} = \frac{1}{2} r^2 \dot \theta = \frac{L}{2m}$. Since angular momentum is conserved for central forces, we have that $\frac{dA}{dt}$ is constant. Note that there were no assumptions about the nature of the gravitational force other than it is central. Therefore, this fact is true in general for central forces.