The four coupled differential equations $$ \frac{dv_x}{dt}=-\frac{GM_{\odot}}{r^3}x, $$ $$ \frac{dx}{dt}=v_x, $$ $$ \frac{dv_y}{dt}=-\frac{GM_{\odot}}{r^3}y, $$ $$ \frac{dy}{dt}=v_y, $$ can be turned into dimensionless equations (as we did in project 2) or we can introduce astronomical units with \( 1 \) AU = \( 1.5\times 10^{11} \).

Using the equations from circular motion (with \( r =1\mathrm{AU} \)) $$ \frac{M_E v^2}{r} = F = \frac{GM_{\odot}M_E}{r^2}, $$ we have $$ GM_{\odot}=v^2r, $$ and using that the velocity of Earth (assuming circular motion) is \( v = 2\pi r/\mathrm{yr}=2\pi\mathrm{AU}/\mathrm{yr} \), we have $$ GM_{\odot}= v^2r = 4\pi^2 \frac{(\mathrm{AU})^3}{\mathrm{yr}^2}. $$