初投稿（楕円軌道上を等速度運動する点について！！？？）

質量を$m$, 速さを$v$とし, 楕円軌道を定数$a,b$と媒介変数$\theta$で$(x,y)=(a\cos \theta ,b\sin \theta )$とおく. $\theta$に対応する速度ベクトル, 加速度ベクトル, 力ベクトルをそれぞれ$\mathit{v}(\theta ),\mathit{\alpha }(\theta ),\mathit{F}(\theta )$とおく. ${\displaystyle \frac{dx}{d\theta }}=-a\sin \theta ,{\displaystyle \frac{dy}{d\theta }}=b\cos \theta$より$$L=\int \sqrt{(dx{)}^2+(dy{)}^2}={\int }_0^{2\pi }\sqrt{(a\sin \theta {)}^2+(b\cos \theta {)}^2}d\theta$$とできる. ここで$h(x)=\sqrt{(a\sin x{)}^2+(b\cos x{)}^2}$とおき, ${\displaystyle H(x)={\int }_0^{x}h(u)du}$を定義する.
任意に一つ$\theta$をとり, $\mathit{\Delta }t$の時間に$\theta \to \mathit{\Delta }\theta$と変化したとする. このとき$\mathit{\Delta }t$と$\mathit{\Delta }\theta$の間には${\displaystyle v\cdot \mathit{\Delta }t={\int }_{\theta }^{\theta +\mathit{\Delta }\theta }h(u)du=H(\theta +\mathit{\Delta }\theta )-H(\theta )}$という関係が成立している.
運動方程式より$\mathit{F}(\theta )=m\cdot \mathit{\alpha }(\theta )=m\cdot (\mathit{v}(\theta +\mathit{\Delta }\theta )-\mathit{v}(\theta ))\cdot {\displaystyle \frac{1}{\mathit{\Delta }t}}$であり, $\mathit{v}(\theta )=v\cdot ({\displaystyle \frac{dx}{d\theta }}\cdot {\displaystyle \frac{1}{h(\theta )}},{\displaystyle \frac{dy}{d\theta }}\cdot {\displaystyle \frac{1}{h(\theta )}})$なので$\mathit{F}(\theta )$の$x$成分について考えると$$\begin{array}{rl}(\mathit{F}(\theta ){)}_x& =mv\cdot {\displaystyle \frac{-\frac{a\sin (\theta +\mathit{\Delta }\theta )}{h(\theta +\mathit{\Delta }\theta )}+\frac{a\sin \theta }{h(\theta )}}{\frac{1}{v}\cdot (H(\theta +\mathit{\Delta }\theta )-H(\theta ))}}\\ & =-am{v}^2\cdot {\displaystyle \frac{\frac{\sin (\theta +\mathit{\Delta }\theta )}{h(\theta +\mathit{\Delta }\theta )}-\frac{\sin \theta }{h(\theta )}}{\mathit{\Delta }\theta }}\cdot {\displaystyle \frac{\mathit{\Delta }\theta }{H(\theta +\mathit{\Delta }\theta )-H(\theta )}}\\ & =-am{v}^2\cdot {\displaystyle \frac{\cos \theta \cdot h(\theta )-\sin \theta \cdot h^{\prime }(\theta )}{h(\theta {)}^2}}\cdot {\displaystyle \frac{1}{h(\theta )}}(商の微分より)\\ & =-{\displaystyle \frac{abm{v}^2}{h(\theta {)}^4}}\cdot b\cos \theta \end{array}$$
$y$成分も同様に計算することで$\mathit{F}(\theta )={\displaystyle \frac{abm{v}^2}{h(\theta {)}^4}}\cdot (-b\cos \theta ,-a\sin \theta )$とわかる.
初めの時刻を$0$として時刻$t$のときを考える. $t$と$\theta$の関係は$vt=H(\theta )⟺\theta =H^{-1}(vt)$とできるので力の大きさの平均$\bar{F}$は${\displaystyle \bar{F}={\displaystyle \frac{v}{L}}{\int }_0^{\frac{v}{L}}\Vert \mathit{F}(H^{-1}(vt))\Vert dt}$と表すことができる. $t={\displaystyle \frac{H(\theta )}{v}}$と置換すると$$\bar{F}={\displaystyle \frac{v}{L}}{\int }_0^{2\pi }\Vert \mathit{F}(\theta )\Vert \cdot {\displaystyle \frac{h(\theta )}{v}}d\theta ={\displaystyle \frac{abm{v}^2}{L}}{\int }_0^{2\pi }{\displaystyle \frac{d\theta }{(a\sin \theta {)}^2+(b\cos \theta {)}^2}}$$ここで一般に$$\int {\displaystyle \frac{d\theta }{(a\sin \theta {)}^2+(b\cos \theta {)}^2}}={\displaystyle \frac{1}{ab}}{\tan }^{-1}({\displaystyle \frac{a}{b}}\tan \theta )+C(Cは積分定数)$$であるので,$$\bar{F}={\displaystyle \frac{abm{v}^2}{L}}\cdot 4\cdot {\displaystyle \frac{1}{ab}}(\underset{\theta \to \frac{\pi }{2}-0}{lim}{\tan }^{-1}({\displaystyle \frac{a}{b}}\tan \theta ))={\displaystyle \frac{2\pi m{v}^2}{L}}$$