\documentclass{jsarticle} \begin{document} \title{ITPASSレポート01} \author{佐藤緩奈} \date{2011年7月15日} \maketitle \section*{問題1} \section*{1.} $m_1$と$m_2$において運動方程式を立てると\\ \[ m_1\ddot{\mathbf{r_1}}=G\frac{m_1m_2}{r^2}\frac{\mathbf{r}}{r} \] \[ m_2\ddot{\mathbf{r_2}}=-G\frac{m_1m_2}{r^2}\frac{\mathbf{r}}{r} \]\\ 上式を整理して\\ \[ \ddot{\mathbf{r_1}}-\ddot{\mathbf{r_2}}=-G\frac{(m_1+m_2)}{r^3}\mathbf{r} \] \[ \ddot{\mathbf{r}}=\ddot{\mathbf{r_1}}-\ddot{\mathbf{r_2}}=-G\frac{(m_1+m_2)}{r^3}\mathbf{r} \]\\ \section*{2.} \[ (\dot{v_x},\dot{v_y})=-G\frac{m_1+m_2}{(x^2+y^2)^\frac{3}{2}}(x,y) \] これより \[ \dot{v_x}=-G\frac{m_1+m_2}{(x^2+y^2)^\frac{3}{2}}x \] \[ \dot{v_y}=-G\frac{m_1+m_2}{(x^2+y^2)^\frac{3}{2}}y \] \section*{問題2} \section*{1.} r=1,$\mu_1+\mu_2=1$,中心星、惑星がx軸上にあることから重心から中心星までの距離は \[1\times\frac{m_2}{m_1+m_2}=\frac{\frac{\mu_2}{G}}{\frac{\mu_1}{G}+\frac{\mu_2}{G}}=\frac{\mu_2}{\mu_1+\mu_2}=\mu_2\] 重心から惑星までの距離は \[1\times\frac{m_1}{m_1+m_2}=\frac{\frac{\mu_1}{G}}{\frac{\mu_1}{G}+\frac{\mu_2}{G}}=\frac{\mu_1}{\mu_1+\mu_2}=\mu_1\]\\ よって \[(x_1,y_1)=(-\mu_2,0)\] \[(x_2,y_2)=(\mu_1,0)\] \section*{2.} \[\theta=\omega{t}\] (ζ,η)を(x,y)に変換する回転行列を$\mathbf{R}$と置くと \begin{eqnarray*} \mathbf{R} &=& \left( \begin{array}{cc} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array} \right) \\ &=& \left( \begin{array}{cc} cos(\omega{t}) & -sin(\omega{t}) \\ sin(\omega{t}) & cos(\omega{t}) \end{array} \right)\\ \left( \begin{array}{c} \xi \\ \eta \end{array} \right) &=& \mathbf{R}\left( \begin{array}{c} x \\ y \end{array} \right) \end{eqnarray*} よって \begin{displaymath} \left\{ \begin{array}{l} \xi =xcos(\omega{t})-ysin(\omega{t}) \\ \eta =xsin(\omega{t})+ycos(\omega{t}) \end{array} \right. \end{displaymath} \section*{3.} 　2.2で求めた$\xi$と$\eta$をそれぞれ2階微分する \begin{eqnarray*} \dot{\eta}&=& \dot{x}sin(\omega{t})+\omega{x}cos(\omega{t})+\dot{y}cos(\omega{t})-\omega^2{y}sin(\omega{t})\\ \ddot{\eta}&=& \ddot{x}sin(\omega{t})+\omega\dot{x}cos(\omega{t})+\omega\dot{x}cos(\omega{t})-\omega^2{x}sin(\omega{t})\\ & & +\ddot{y}cos(\omega{t})-\omega\dot{y}sin(\omega{t})-(\omega\dot{y}sin(\omega{t})+\omega^2{y}cos(\omega{t}))\\ &=& (\ddot{y}+2\omega\dot{x}-\omega^2{y})cos(\omega{t})+(\ddot{x}-2\omega\dot{y}-\omega^2{x})sin(\omega{t}) \end{eqnarray*} \section*{4.} 問題文の$\ddot{\xi}$と$\ddot{\eta}$を2.3で求めた式に代入する。 \begin{eqnarray} \ddot{\xi}&=& \mu_1\frac{\xi_1-\xi}{r_1^3}+\mu_2\frac{\xi_2-\xi}{r_2^3}\nonumber \\ &=& (\ddot{x}-2\omega\dot{y}-\omega^2{y})cos(\omega{t})-(\ddot{y}+2\omega\dot{x}-\omega^2{y})sin(\omega{t})\\ \nonumber \\ \ddot{\eta}&=& \mu_1\frac{\eta_1-\eta}{r_1^3}+\mu_2\frac{\eta_2-\eta}{r_2^3}\nonumber \\ &=& (\ddot{y}+2\omega\dot{x}-\omega^2{y})cos(\omega{t})-(\ddot{x}-2\omega\dot{y}-\omega^2{x})sin(\omega{t}) \end{eqnarray}\\ (1)式に両辺$cos(\omega{t})$、(2)式に$sin(\omega{t})$をかけると \begin{eqnarray*} \left(\mu_1\frac{\xi_1-\xi}{r_1^3}+\mu_2\frac{\xi_2-\xi}{r_2^3}\right)cos(\omega{t})= \\ (\ddot{x}-2\omega\dot{y}-\omega^2{x})cos^2(\omega{t})-(\ddot{y}+2\omega\dot{x}-\omega^2{y})sin(\omega{t})cos(\omega{t})\\\\ \left(\mu_1\frac{\eta_1-\eta}{r_1^3}+\mu_2\frac{\eta_2-\eta}{r_2^3}\right)sin(\omega{t})= \\ (\ddot{y}+2\omega\dot{x}-\omega^2{y})cos(\omega{t})sin(\omega{t})-(\ddot{x}-2\omega\dot{y}-\omega^2{x})sin^2(\omega{t}) \end{eqnarray*}\\ 辺々足し合わせると \begin{eqnarray*} \ddot{x}-2\omega\dot{y}-\omega^2{x}&=& \mu_1\frac{\xi_1cos(\omega{t})-\xi{cos(\omega{t})}+\eta_1sin(\omega{t})-\eta{sin(\omega{t})}}{r_1^3}\\ \ &=& \mu_2\frac{\xi_2cos(\omega{t})-\xi{cos(\omega{t})}+\eta_2sin(\omega{t})-\eta{sin(\omega{t})}}{r_2^3}\\ \end{eqnarray*}\\ ここで、 \begin{eqnarray*} \left( \begin{array}{c} x \\ y \end{array} \right)&=& \left( \begin{array}{cc} cos(-\theta) & -sin(-\theta) \\ sin(-\theta)& cos(-\theta) \end{array} \right) \left( \begin{array}{c} \xi \\ \eta \end{array} \right)\\ &=& \left( \begin{array}{cc} cos\theta & sin\theta \\ -sin\theta& cos\theta \end{array} \right) \left( \begin{array}{c} \xi \\ \eta \end{array} \right)\\ \left\{ \begin{array}{l} \xi =xcos(\omega{t})-ysin(\omega{t}) \\ \eta =xsin(\omega{t})+ycos(\omega{t}) \end{array} \right. \end{eqnarray*} \begin{eqnarray*} \ddot{x}-2\omega\dot{y}-\omega^2{x}&=& \mu_1\frac{x_1-x}{r_1^3}+\mu_2\frac{x_2-x}{r_2^3}\\ &=& -\left[\mu_1\frac{x-x_1}{r_1^3}+\mu_2\frac{x-x_2}{r_2^3}\right]\\ &=& -\left[\mu_1\frac{x+\mu_2}{r_1^3}+\mu_2\frac{x-\mu_1}{r_2^3}\right] \end{eqnarray*} (∵1.1より$x_1=-\mu_2$、$x_2=\mu_1$)\\\\ 同様に(1)式に両辺$sin(\omega{t})$、(2)式に両辺$cos(\omega{t})$をかけて \begin{eqnarray*} \left(\mu_1\frac{\xi_1-\xi}{r_1^3}+\mu_2\frac{\xi_2-\xi}{r_2^3}\right)sin(\omega{t})= \\ (\ddot{x}-2\omega\dot{y}-\omega^2{x})cos(\omega{t})sin(\omega{t})-(\ddot{y}+2\omega\dot{x}-\omega^2{y})sin^2(\omega{t})\\ \left(\mu_1\frac{\eta_1-\eta}{r_1^3}+\mu_2\frac{\eta_2-\eta}{r_2^3}\right)cos(\omega{t})= \\ (\ddot{y}+2\omega\dot{x}-\omega^2{y})cos^2(\omega{t})+(\ddot{x}-2\omega\dot{y}-\omega^2{x})sin(\omega{t})cos(\omega{t}) \end{eqnarray*} これらを整理すると\\\\ \begin{eqnarray*} \ddot{y}-2\omega\dot{x}-\omega^2{y}&=& \mu_1\frac{-\xi_1sin(\omega{t})-\eta_1{cos(\omega{t})}+\xi{sin(\omega{t})}-\eta{sin(\omega{t})}}{r_1^3}\\ \ &=& \mu_2\frac{-\xi_2sin(\omega{t})+\eta_2{cos(\omega{t})}+\xi{sin(\omega{t})}-\eta{cos(\omega{t})}}{r_2^3}\\ &=& \mu_1\frac{y_1-y}{r_1^3}+\mu_2\frac{y_2-y}{r_2^3}\\ &=& -\left[\mu_1\frac{-y}{r_1^3}+\mu_2\frac{-y}{r_2^3}\right]\\ &=& -\left[\frac{\mu_1}{r_1^3}+\frac{\mu_2}{r_2^3}\right]y \end{eqnarray*} （∵2.1より$y_1=0$、$y_2=0$） よって導出できた。 \section*{5.} $P(x,y),\quad$$(x_1,y_1)=(-\mu_2,0),\quad$$(x_2,y_2)=(\mu_1,0)$より三平方の定理を用いて\\\\ \begin{eqnarray*} r_1 =\sqrt{(x+\mu_2)^2+y^2}\\ r_2 =\sqrt{(\mu_1-x)^2+y^2\\ \frac{\mu_1}{r_1}=\frac{\mu_1}{((x+\mu_2)^2+y^2))^\frac{1}{2}}\\ \end{eqnarray*} $\frac{\mu_1}{r_1}$を$x$で偏微分すると \begin{eqnarray*} \frac{\partial}{\partial x}\frac{\mu_1}{r_1}&=&-\frac{1}{2}\frac{\mu_1(2x+2\mu_2)}{\left\{(x+\mu_2)^2+y^2\right\}^\frac{3}{2}}\\ &=& -\mu_1\frac{(x+\mu_2)}{r_1^3} \end{eqnarray*}\\\\ 同様にして \begin{eqnarray*} \frac{\mu_2}{r_2}=\frac{\mu_2}{((\mu_1-x)^2+y^2)^\frac{1}{2}} \end{eqnarray*} $\frac{\mu_2}{r_2}$を$x$で偏微分すると \begin{eqnarray*} \frac{\partial}{\partial x}\frac{\mu_2}{r_2}&=&-\frac{1}{2}\frac{\mu_2(2x-2\mu_1)}{\left\{(\mu_1-x)^2+y^2\right\}^\frac{3}{2}}\\ &=& -\mu_2\frac{(x-\mu_1)}{r_2^3}\\\\ U &=& \frac{\omega^2}{2}(x^2+y^2)+\frac{\mu_1}{r_1}+\frac{\mu_2}{r_2}\\ \frac{\partial U}{\partial x}&=& \frac{\partial}{\partial x}\left(\frac{\omega^2}{2}(x^2+y^2)+\frac{\mu_1}{r_1}+\frac{\mu_2}{r_2}\right)\\ &=& \omega^2x-\left[\mu_1\frac{(x+\mu_2)}{r_1^3}+\mu_2\frac{(x-\mu_1)}{r_2^3}\right]\\ &=& \ddot{x}-2\omega\dot{y} \end{eqnarray*}\\\\ 同様に$y$でも偏微分する。 \begin{eqnarray*} \frac{\partial}{\partial y}\frac{\mu_2}{r_1}&=&-\frac{1}{2}\frac{\mu_1\cdot2y}{\left\{(x+\mu_2)^2+y^2\right\}^\frac{3}{2}}\\ &=& -\mu_1\frac{y}{r_1^3}\\ \frac{\partial}{\partial y}\frac{\mu_2}{r_2}&=&-\frac{1}{2}\frac{\mu_2\cdot2y}{\left\{(x+\mu_2)^2+y^2\right\}^\frac{3}{2}}\\ &=& -\mu_2\frac{y}{r_2^3}\\\\ \frac{\partial U}{\partial y}&=& \frac{\partial}{\partial y}\left(\frac{\omega^2}{2}(x^2+y^2)+\frac{\mu_1}{r_1}+\frac{\mu_2}{r_2}\right)\\ &=& \omega^2y-\left[\frac{\mu_1}{r_1^3}+\frac{\mu_2}{r_2^3}\right]y\\ &=& \ddot{y}-2\omega\dot{x}\\ \end{eqnarray*}\\\\ 以上より \begin{displaymath} \left\{ \begin{array}{l} \ddot{x}-2\omega\dot{y}=\frac{\partial}{\partial x}\\ \ddot{y}-2\omega\dot{x}=\frac{\partial}{\partial y} \end{array} \right. \end{displaymath}\\\\ \section*{6.} 2.5より \begin{eqnarray} \ddot{x}-2\omega\dot{y}=\frac{\partial}{\partial x}\\ \ddot{y}-2\omega\dot{x}=\frac{\partial}{\partial y} \end{eqnarray} (3)式に$\dot{x}$,(4)式に$\dot{y}$を両辺かけて辺々足し合わせると \begin{eqnarray*} \dot{x}\ddot{x}+\dot{y}\ddot{y}&=&\frac{\partial U}{\partial x}\dot{x}+\frac{\partial U}{\partial y}\dot{y}\\ &=& \left(\frac{\partial U}{\partial x}\frac{dx}{dt}+\frac{\partial U}{\partial y}\frac{dy}{dt}\right)\\ &=& \frac{dU}{dt} \end{eqnarray*}\\\\ 時間で積分する。積分定数を$C_J$と置くと\\ \begin{eqnarray*} \frac{1}{2}\dot{x}^2+\frac{1}{2}\dot{y}^2+C_J=U \end{eqnarray*}\\ よって \begin{eqnarray*} C_J=U-\frac{1}{2}(\dot{x}^2+\dot{y}^2) \end{eqnarray*} \end{document}