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Kepler's Second Law states that equal areas are swept in equal times. When calculating this area, why do we use the formula for the area of a triangle rather than the area of a sector?

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    $\begingroup$ If you are using a small section you can approximate the sector to be a triangle. However in general it actually isn't a triangle. Can you give more context to where you are seeing this? $\endgroup$ – Aaron Stevens Sep 3 '18 at 16:58
  • $\begingroup$ @AaronStevens I am seeing this in regard to ellipses. The area which a planet sweeps (with the focus acting like a "pivot") is a sector. So, I was wondering why we use the area of a triangle since it is really a circular sector. $\endgroup$ – Annabelle Sykes Sep 3 '18 at 17:00
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    $\begingroup$ ... and I always thought one used the area of the sector. Maybe you can edit your question to make it more explicit by including a figure or a specific reference? $\endgroup$ – ZeroTheHero Sep 3 '18 at 18:06
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Kepler's Second Law states that equal areas are swept in equal times.

So far so good.

When calculating this area, why do we use the formula for the area of a triangle rather than the area of a sector?

We don't. We use the area of the sector and we may use the sum of small triangles to approximate that.

The basic maths here is that :

$$dA = \frac 1 2 r^2d\theta$$

for some infinitesimal area (basic calculus).

In this expression the $r$ is the height of a triangle, the $rd\theta$ is the length of the arc swept out and we're using the well known formula for the area of a triangle (half the base by the perpendicular height).

Kepler's second law can be stated like this :

$$\int_{t_1}^{t_1+t_0}dA = \int_{t_2}^{t_2+t_0}dA$$

Now to find those values from pages of data you will use an approximation of some sort and the one you're going to use is a Riemann sum.

In the Riemann sum the integral can be approximated by summing small products :

$$\int_a^b f(x) dx \approx \sum_{k}f(x_k)(x_k-x_{k-1})$$

Assuming our numbers can be made accurate enough and we use small values for the $x_k-x_{k-1}$ intervals, this will be as accurate as we want.

In the case of Kepler, his work predated calculus so he would have been unaware of the formal maths, but would have used the approximation for his data (which was relatively inaccurate by modern standards and he had only discrete sets of points, not a continuous formula). Kepler had to use a triangle approximation, but was using it with the smallest intervals he could. Kepler was not even aware of the shape of orbits when he was working.

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