平均焦半径

hbghlyj

The true anomaly of point $P$ is the angle $f$. 用自变量 $f$ 表示距离$PF$：$$r = \frac{a(1-e^2)}{1+e\cos f}$$假设 $f$ 是均匀分布的，平均距离 $PF$ 是 $\frac{1}{2π} \int^{2π}_0 r df$

By residue theorem (two poles, one is inside unit circle, the other is outside it)\begin{aligned}
\int^{2π}_0\frac1{1+e\cos f}df&=\int_{\abs z=1}\frac1{1+\frac e2(z+z^{-1})}\frac{dz}{iz}\\
&=\frac1i\int_{\abs z=1}\frac2e\frac1{1+\frac 2ez+z^2}dz\\
&=2\pi\frac2e\frac1{2\sqrt{e^{-2}-1}}\\
&=\frac{2\pi}{\sqrt{1-e^2}}
\end{aligned}平均距离 $PF$ 是$\frac1{2\pi}a(1-e^2)\frac{2\pi}{\sqrt{1-e^2}}=a\sqrt{1-e^2}=b$

hbghlyj

另一方面，假设eccentric anomaly $E$ 是均匀分布的，用自变量 $E$ 表示距离$PF$：from MathWorld$$\tag9\label9
r=a(1-e\cos E)$$假设 $E$ 是均匀分布的，平均距离 $PF$ 是 $\frac{1}{2π}\int^{2π}_0 r dE=a$, 因为 cos 在 $[0,2\pi]$ 上的积分是 0.

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这可以如下解释。由于对称性，$PF_1$ 和 $PF_2$ 具有相同的平均值。但由于 $PF_1+PF_2$ 始终为 $2a$，$PF_1+PF_2$ 的平均值为 $2a$。因此，$PF_1$ 的平均值是 $a$。

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$E$ 均匀分布等价于 $P'$ 在圆上均匀分布。
如果 $P$ 均匀分布在椭圆上，因为对称性论证适用，$PF$的均值也是$a$。
参阅：
Mean distance from the focus of an ellipse in polar coordinates

When I try to do the arclength integral, I get something I do not know how to integrate. But here is the easy argument people have in mind, I think. You're averaging the distance from one focus. The distance from the other focus has the same average (by symmetry). But since the sum of the distances is always twice the semi-major axis, the average of their sum is twice the semi-major axis. Thus, the average of one of the distances is the semi-major axis.

The Average Distance of the Earth from the Sun

...It may not have been expected that the arc length representation would always produce the length of the semi-major axis, as it did, but this will be demonstrated below. As a consequence it is correct to describe an Astronomical Unit as the average relative to arc length of the earth's distance from the sun.
Invariance over eccentricity of the arc length based average In Figure 1 we offer a proof without words that the average distance using the arclength parameterization is always equal to the length of the semi-major axis.

hbghlyj

在上面我们证明了：
假设 $f$ 是均匀分布的，平均距离 $PF$ 是 $b$.
假设 $E$ 是均匀分布的，平均距离 $PF$ 是 $a$.
这种比较可以解释如下：
考虑点 $P$ 从点 $A$ 开始移动，其中 $f=E=0$.
如果 $E$ 均匀增加，$P$ 花费恰好一半时间在右半平面，平均距离 $PF$ 是 $a$ 是符合预期的。
如果 $f$ 均匀增加，当$E$为直角时，$PF$垂直于x轴，$P$仍在右半平面上，意味着 $P$ 花费多于一半时间在右半平面，但是当 $P$ 在右半平面上时$PF<a$，平均距离 $PF$ 是 $b$ 小于 $a$ 是符合预期的。

hbghlyj

在Revisiting Kepler’s measurements 第4页也证明了\eqref{9}:

Let's consider triangle $\triangle S P X$. $S X=e-x$. Also recall that $e^2=a^2-b^2$ and $\sin ^2 \beta+\cos ^2 \beta=1$. Using the Pythagorean theorem, Kepler calculated as follows :
$$
\begin{aligned}
r^2 & =y^2+(e-x)^2 \\
& =b^2 \sin ^2 \beta+(e-a \cos \beta)^2 \\
& =b^2 \sin ^2 \beta+a^2 \cos ^2 \beta-2 a e \cos \beta+e^2 \\
& =\left(a^2-e^2\right) \sin ^2 \beta+a^2 \cos ^2 \beta-2 a e \cos \beta+e^2 \\
& =a^2-e^2 \sin ^2 \beta-2 a e \cos \beta+e^2 \\
& =a^2+e^2\left(1-\sin ^2 \beta\right)-2 a e \cos \beta \\
& =a^2+e^2 \cos ^2 \beta-2 a e \cos \beta \\
& =(a-e \cos \beta)^2
\end{aligned}
$$
Since, $a>e \geq \cos \beta, a-e \cos \beta>0 . r>0$, so we can take the roots of both sides above to get $r=a-e \cos \beta$. We now factor the $a$ on the right side to get $r=a\left(1-\frac{e}{a} \cos \beta\right)$. Thus we have
$$
r=a(1-\varepsilon \cos \beta)
$$