\documentstyle{article}
\oddsidemargin -15.0mm
\evensidemargin -15.0mm
\textwidth 180mm
\textheight 270mm
\topmargin -40mm
\columnsep 5mm
\pagestyle{plain}
\pagenumbering{arabic}
\title{ITPASS�½������ͷ׻��½����ꤽ�Σ�}
\author{joho04  ������}
\date{2010ǯ7��15�� ���}
\begin{document}
\maketitle
\section{�����Ϥˤ����롢�濴���������֤α�ư������}

�濴���μ��̤�$m_{1}$�������μ��̤�$m_{2}$���濴���ΰ��֥٥��ȥ��$\mbox{\boldmath $r_{1}$}$��\\
�����ΰ��֥٥��ȥ��$\mbox{\boldmath $r_{2}$}$�����Х٥��ȥ��$\mbox{\boldmath $r$}=\mbox{\boldmath $r_{2}$}-\mbox{\boldmath $r_{1}$}$,
��ͭ���������G�Ȥ��롣\\
���ΤȤ������Ϥ�ڤܤ�������ŷ�Τα�ư�Τߤ�ͤ���Τ���ͭ���Ϥ�ͤ���Ф褤��\\
�������äơ��濴���������Ϥ��줾��˴ؤ�������Ω�ı�ư��������
\begin{eqnarray}
m_{1}\frac{d^{2}\mbox{\boldmath $r_{1}$}}{dt^{2}}= G\frac{m_{1}m_{2}}{r^{3}}\mbox{\boldmath $r$}
\end{eqnarray}
\begin{eqnarray}
m_{2}\frac{d^{2}\mbox{\boldmath $r_{2}$}}{dt^{2}}= -G\frac{m_{1}m_{2}}{r^{3}}\mbox{\boldmath $r$}
\end{eqnarray}
�Ǥ��롣\\
��äơ�1�˼���$m_{1}$����2�˼���$m_{2}$�ǳ�ꡢ��2�ˤ����1�ˤ�����Ȥ���ȡ�
\begin{eqnarray}
\frac{d^{2}\mbox{\boldmath $r$}}{dt^{2}} &=& \frac{d^{2}\mbox{\boldmath $r_{2}$}}{dt^{2}}-\frac{d^{2}\mbox{\boldmath $r_{1}$}}{dt^{2}} \\
&=& -G\frac{m_{1}}{r^{3}}\mbox{\boldmath $r$}-G\frac{m_{2}}{r^{3}}\mbox{\boldmath $r$}
\end{eqnarray}
�������ä�
\begin{eqnarray}
\frac{d^{2}\mbox{\boldmath $r$}}{dt^{2}} = -G\frac{m_{1}+m_{2}}{r^{3}}\mbox{\boldmath $r$} 
\end{eqnarray}
�Ȥʤꡢ���뼰��Ƴ�Ф���롣
�ޤ������μ�����ŷ�Τ����Х٥��ȥ�μ��ȤʤäƤ��뤳�Ȥ��顢\\
1 �Ĥμ��������Ȥ�����ɸ�ϤǤ��濴�ϤΤߤ�Ư���褦�ˤʤäƤ��ꡢ
�����Ϥˤ����Ƥ���ŷ�Τνſ����濴�Ȥ�����®�߱�ư�򤷤Ƥ��뤳�Ȥ򼨤��Ƥ��롣
\section{��ʬ��ʬ��}
1����ʬx��y��ʬ�򤹤롣\\
��5�˼���$\mbox{\boldmath$r$}=(x,y)$��������$\mbox{\boldmath$v$}=(v_{x},v_{y})$�Ǥ��뤳�Ȥ���
\begin{eqnarray}
\frac{dv_{x}}{dt}=\frac{d^{2}x}{dt^{2}}=-G\frac{m_{1}+m_{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}x\\
\frac{dv_{y}}{dt}=\frac{d^{2}y}{dt^{2}}=-G\frac{m_{1}+m_{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}y
\end{eqnarray}
�Ȥʤ롣
\end{document}i
