\[ \begin{align} & x^{2}+y^{2}=R^{2} \\[6pt] & \dot{x}^{2}+\dot{y}^{2}=(\dot{R})^{2}+(R\dot{a})^{2} \\[6pt] & \ddot{x}^{2}+\ddot{y}^{2}=(\ddot{R}-R\dot{a}^{2})^{2}+(R\ddot{a}+2\dot{R}\dot{a})^{2} \\[6pt] & GM=C_{1} \\[6pt] & m\ddot{x}=m\frac{GM}{R^{2}}\frac{x}{R}=m\frac{C_{1}}{R^{2}}\frac{x}{R} \\[6pt] & m\ddot{y}=m\frac{GM}{R^{2}}\frac{y}{R}=m\frac{C_{1}}{R^{2}}\frac{y}{R} \\[6pt] & \ddot{x}=\frac{C_{1}}{R^{2}}\frac{x}{R} \\[6pt] & \ddot{y}=\frac{C_{1}}{R^{2}}\frac{y}{R} \\[6pt] & \ddot{x}^{2}+\ddot{y}^{2}=\frac{C_{1}}{R^{2}} \\[12pt] & \ddot{R}-R\dot{a}^{2}=\frac{C_{1}}{R^{2}} \\[6pt] & \ddot{R}\,dR=\bigl(R\dot{a}^{2}+\tfrac{C_{1}}{R^{2}}\bigr)\,dR \\[6pt] & d\!\bigl(-\tfrac{1}{2}\dot{R}^{2}\bigr)=d\!\bigl(\tfrac{1}{2}R^{2}\dot{a}^{2}-\tfrac{C_{1}}{R}\bigr) \\[6pt] & W=\tfrac{1}{2}\dot{R}^{2}-\tfrac{C_{1}}{R}+\tfrac{1}{2}R^{2}\dot{a}^{2} \\[12pt] & \tfrac{1}{R}\,\frac{d\bigl(R^{2}\dot{a}\bigr)}{dt}=0 \\[6pt] & \frac{da}{dt}=\dot{a}=\frac{C_{2}}{R^{2}} \\[6pt] & dt=\frac{R^{2}}{C_{2}}\,da \\[12pt] & 2W+2\frac{C_{1}}{R}-\frac{C_{2}^{2}}{R^{2}}=\dot{R}^{2}=\Bigl(\frac{dR}{dt}\Bigr)^{2}=\Bigl(\frac{dR}{R^{2}\,da/C_{2}}\Bigr)^{2} \\[12pt] & (da)^{2}=\frac{\bigl(\tfrac{C_{2}}{R^{2}}\,dR\bigr)^{2}}{2W+2\frac{C_{1}}{R}-\frac{C_{2}^{2}}{R^{2}}} \\[6pt] & da=\frac{\dfrac{C_{2}}{R^{2}}\,dR}{\sqrt{\,2W+2\dfrac{C_{1}}{R}-\dfrac{C_{2}^{2}}{R^{2}}\,}} \\[12pt] & da=\frac{\dfrac{C_{2}^{2}}{C_{1}R^{2}}\,dR}{\sqrt{\dfrac{2W C_{2}^{2}}{C_{1}^{2}}+\dfrac{2C_{2}^{2}}{C_{1}R}-\dfrac{C_{2}^{4}}{C_{1}^{2}R^{2}}}} \\[12pt] & da=\frac{\dfrac{C_{2}^{2}}{C_{1}R^{2}}\,dR}{\sqrt{\,1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}-\bigl(1-2\dfrac{C_{2}^{2}}{C_{1}R}+(\tfrac{C_{2}^{2}}{C_{1}R})^{2}\bigr)\,}} \\[12pt] & da=\frac{\dfrac{C_{2}^{2}}{C_{1}R^{2}}\,dR}{\sqrt{\,1-\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}-\dfrac{1}{\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}}\bigl(\tfrac{C_{2}^{2}}{C_{1}R}\bigr)\,}} \\[12pt] & d\Bigl(\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}-\dfrac{1}{\sqrt{\,1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}\,}}\bigl(\tfrac{C_{2}^{2}}{C_{1}R}\bigr)\Bigr)=da \\[12pt] & a+C_{3}=\arcsin\!\Bigl(\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}-\dfrac{1}{\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}}\bigl(\tfrac{C_{2}^{2}}{C_{1}R}\bigr)\Bigr) \\[12pt] & \sin(a+C_{3})=\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}-\dfrac{1}{\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}}\bigl(\tfrac{C_{2}^{2}}{C_{1}R}\bigr) \\[12pt] & \sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}\,\sin(a+C_{3})=\Bigl(1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}\Bigr)-\Bigl(\tfrac{C_{2}^{2}}{C_{1}R}\Bigr) \\[12pt] & R\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}\,\sin(a+C_{3})=R\Bigl(1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}\Bigr)-\Bigl(\tfrac{C_{2}^{2}}{C_{1}}\Bigr) \\[12pt] & R\sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}\,\bigl(\sin a\cos C_{3}+\sin C_{3}\cos a\bigr) =R\Bigl(1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}\Bigr)-\Bigl(\tfrac{C_{2}^{2}}{C_{1}}\Bigr) \\[12pt] & x=R\cos a \\[6pt] & y=R\sin a \\[12pt] & \sqrt{1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}}\,\bigl(y\cos C_{3}+x\sin C_{3}\bigr) =\sqrt{x^{2}+y^{2}}\;\Bigl(1+\dfrac{2W C_{2}^{2}}{C_{1}^{2}}\Bigr)-\dfrac{C_{2}^{2}}{C_{1}} \end{align} \]