I'm reading Theory of Orbit: The Restricted Problem of Three Bodies Victory Szebehely, and I'm having a problem to change the problem to dimensionless coordinates.
The original system is \begin{split} \frac{d^2X}{dt^{*2}}&=-\frac{m_1}{R_1^3}(X-b\cos{\omega t^*})-\frac{m_2}{R_2^3}(X+a\cos{\omega t^*})\\ \frac{d^2Y}{dt^{*2}}&=-\frac{m_1}{R_1^3}(Y-b\sin{\omega t^*})-\frac{m_2}{R_2^3}(Y+a\sin{\omega t^*}),\\ \end{split}
where $R_1=\sqrt{(X-b\cos{\omega t^*})^2+(Y-b\sin{\omega t^*})^2}$ and $R_2=\sqrt{(X+a\cos{\omega t^*})^2+(Y+a\sin{\omega t^*})^2}$
The change of variables that Victory Szebehely makes is: \begin{equation*} \begin{split} \xi = \frac{X}{R},\eta=\frac{Y}{R},t=\omega t^*,\mu_1=\frac{a}{R}=\frac{m_1}{m_1+m_2},\mu_2&=\frac{b}{R}=\frac{m_2}{m_1+m_2},\rho_1=\frac{R_1}{R},\rho_2=\frac{R_2}{R}, \end{split} \end{equation*} where $R$ is a constant, the distance between the primaries, that accordind to kepler third law satisfies the following relation: $\omega^2=\frac{m_1+m_2}{R^3}.$ I'm suppose to arrive to this: \begin{equation*} \begin{split} \frac{d^2\xi}{dt^{2}}&=-\frac{\mu_1}{\rho_1^3}(\xi-\mu_2\cos{t})-\frac{\mu_2}{\rho_2^3}(\xi+\mu_1\cos{t})\\ \frac{d^2\eta}{dt^{2}}&=-\frac{\mu_1}{\rho_1^3}(\eta-\mu_2\sin{t})-\frac{\mu_2}{\rho_2^3}(\eta+\mu_1\sin{t}).\\ \end{split} \end{equation*}
The main problem I'm having is that I don't know how to obtain the second derivative with respect to $t$.
Any suggestion?