Estimate the eccentricity of the Earth's orbit using equinoxes I was reading the wikipedia article on Kepler's laws (https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion) and I came across the following.
"The eccentricity of the orbit of the Earth makes the time from the March equinox to the September equinox, around 186 days, unequal to the time from the September equinox to the March equinox, around 179 days. A diameter would cut the orbit into equal parts, but the plane through the Sun parallel to the equator of the Earth cuts the orbit into two parts with areas in a 186 to 179 ratio, so the eccentricity of the orbit of the Earth is approximately."
This makes sense, but what follows is not immediately obvious to me. Using the data above, the eccentricity ($e$) of the Earths orbit is given by
\begin{equation}
e \approx \frac{\pi}{4}\frac{186 - 179}{186 + 179}.
\end{equation}
Just for curiosity, can anyone kindly explain how this approximation comes about?
 A: 
Let's assume that the Earth's orbit is an ellipse with eccentricity $\varepsilon$, semi-major axis $a$, and semi-minor axis $b=a\sqrt{1-\varepsilon^2}$, as in the figure on the left. First, we let the perihelion $A$ coincide with the December solstice; in reality, the perihelion occurs in early January, so this is a reasonable approximation. Consequently, the points $B$ and $D$ correspond with the March and September equinoxes, respectively.
The line $BD$ divides the ellipse into two areas, $S_1$ and $S_2$, which are swept out in 179 and 186 days, respectively. In other words,
$$
S_1 = 179K,\\
S_2 = 186K,
$$
for some constant $K$. To calculate the area $S_1$, we could use elliptic integrals, but it is easier to first calculate the area of the corresponding segment $S'_1$ in a circle with radius $a$ (see the figure on the right). This area is equal to the area of the sector enclosed by the arc $B'A'D'$ (blue area + grey area) minus the area of the triangle $\Delta OB'D'$. We find
$$
S'_1 = \eta\,a^2 - \varepsilon ab,
$$
where
$$
\cos\eta = \varepsilon.
$$
By re-scaling the vertical axis by a factor $b/a$, we can transform the circle into our ellipse; the areas scale accordingly, and thus
$$
S_1 = \frac{b}{a}S'_1 = \eta\,ab - \varepsilon b^2,
$$
and
$$
S_2 = \pi ab - S_1 = \pi ab - \eta\,ab + \varepsilon b^2.
$$
From this we obtain
$$
\begin{align}
S_2-S_1 &= \pi ab - 2\eta\,ab + 2\varepsilon b^2= (186-179)K,\\
S_2+S_1 &= \pi ab = (186+179)K,
\end{align}
$$
so that
$$
\pi\frac{S_2-S_1}{S_2+S_1} = \pi - 2\eta + 2\varepsilon\frac{b}{a} = \pi\frac{186-179}{186+179}.
$$
Since $\varepsilon$ is small, we can use the additional approximations
$$
\varepsilon = \cos\eta = \sin(\pi/2-\eta) \approx \pi/2-\eta,\\
2\varepsilon\frac{b}{a} = 2\varepsilon\sqrt{1-\varepsilon^2} \approx 2\varepsilon,
$$
and the equation finally simplifies to
$$
4\varepsilon \approx \pi\frac{186-179}{186+179}.
$$
A: We can also derive this approximation via Kepler's equation. Using the same variable names as Pulsar, that equation is
$$M = \eta - \varepsilon\sin\eta$$
where $M$ is the mean anomaly, $\varepsilon$ is the eccentricity, and $\eta$ is the eccentric anomaly.
We're assuming that the perihelion and northern summer solstice coincide, at $M=\eta=0$, and that the March equinox occurs when $M=\frac{179\pi}{186+179}$.
The coordinates of the Earth (using the Sun as the origin) are:
$$\begin{align}
x &= a(\cos\eta - \varepsilon)\\
y &= b\sin\eta
\end{align}
$$
At the equinox, $x=0$, so
$$\cos\eta = \varepsilon$$
We know that the eccentricity of Earth's orbit is fairly small ($\approx\frac1{60}$), so at the equinox $\eta$ must be close to a right angle, and therefore so must $M$.
For this problem, it's convenient to work with the complements of the angles. Let $\bar\theta$ denote the complementary angle of $\theta$. That is, $\theta+\bar\theta=\frac\pi2$. Both $\bar M$ and $\bar\eta$ are close to zero.
Rewriting the Kepler equation in terms of the complementary angles, and substituting $\varepsilon=\cos\eta=\sin\bar\eta$,  we get
$$\bar M = \bar\eta + \sin\bar\eta\cos\bar\eta$$
$$\bar M -\bar\eta = \frac12\sin2\bar\eta$$
Using the small angle approximation, $sin\theta\approx\theta$, we get
$$\bar M -\bar\eta \approx \bar\eta$$
That is,
$$\bar\eta \approx \frac{\bar M}2$$
Now
$$\begin{align}
\bar M &= \frac\pi2 - M\\
&= \frac\pi2 - \frac{179\pi}{186+179}\\
&= \frac\pi2\left(\frac{186+179-2\times179}{186+179}\right)\\
\bar M &= \frac\pi2\left(\frac{186-179}{186+179}\right)
\end{align}$$
And from the small angle approximation,
$$\bar\eta\approx\sin\bar\eta=\varepsilon$$
Hence
$$\varepsilon \approx  \frac\pi4\left(\frac{186-179}{186+179}\right)$$
