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开普勒第二定律告诉我们，时间与扫出的面积$A$成正比，我们可以计算这个面积作为我们的参数$$d A=\frac{1}{2} r^{2} d \theta$$计算$r$基于$A$的平均\begin{equation}\label1\operatorname{Avg}(r, A)=\frac{\int_{\theta=0}^{\theta=2 \pi} r d A}{\int_{\theta=0}^{\theta=2 \pi} d A}=a\left(1-e^{2}\right) \frac{\int_{0}^{2 \pi} \frac{d \theta}{(1+e \cos \theta)^{3}}}{\int_{0}^{2 \pi} \frac{d \theta}{(1+e \cos \theta)^{2}}}\end{equation}计算积分$$\int_{0}^{2 \pi} \frac{d \theta}{(1+e \cos \theta)^3}={π (e^2 + 2)\over(1 - e^2)^{5/2}}$$与$$\int_{0}^{2 \pi} \frac{d \theta}{(1+e \cos \theta)^2}={2 π\over(1 - f^2)^{3/2}}$$代入 \eqref{1}\[\operatorname{Avg}(r, A)=a(1 + \frac{1}{2}e^2)\]
这与天文学电子书 9.10: Mean Distance in an Elliptic Orbit中给出的结果一致（它不包含证明）。On the other hand, the mean distance averaged over the time is \(\frac{1}{2} P \int_0^{\frac{1}{2}P} r dt\). This one is slightly more tricky, but, following the hint for evaluating \(\frac{1}{\pi} \int^\pi_0 r dv\), you could try expressing \(r\) and \(v\) in terms of the eccentric anomaly. It will take you a moment or so, but you should eventually find that the mean distance averaged over the time is \(a(1 + \frac{1}{2}e^2)\). |
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