A New Earth!

First of all, from laws of gravitation, we know, $\begin{aligned}a(r) &= \dfrac{G}{r^2}M(r)\\&=\dfrac{4\rho G \pi r^3}{3r^2}\\a&=\dfrac{4\rho G \pi}{3}r\end{aligned}$

Note one thing that the above equation represents that of a Simple Harmonic Motion. So, $\omega = \sqrt{\dfrac{4\rho G \pi}3}$ and Amplitude $A = R$ $\begin{aligned}T_1 &= \dfrac{T}{4}=\dfrac{2\pi}{4\omega}\\&=\dfrac{1}{4}\sqrt{\dfrac{3\pi}{\rho G}}\\v_1 &= \omega A\\&=2R\sqrt{\dfrac{\rho G \pi}3}\end{aligned}$

Now comes the tricky part. Let's investigate...

Take the centre of the cavity to be $C$ and the vector from $C$ to $O$ be $\vec c$ and the radius vector to any arbitrary point be $\vec r$ so that the radius vector of that point to the centre of the cavity becomes $\vec c+\vec r$ .

Now, since the cavity is removed from the whole mass, we can take the negative mass system corresponding to the cavity. Also, the accelerations will be proportional to the radius vectors and the constant of proportionality is $\omega^2$ . Thus, $\begin{aligned}a &= \omega^2 (-\vec r) + -\left[\omega^2 (-\{\vec r + \vec c\})\right]\\&=\omega^2 \vec c\\&=\omega^2\dfrac{R}{2}\end{aligned}$

The acceleration here is thus a constant.

So, $\begin{aligned}T_2 &= \dfrac2{ \omega}\\&=\sqrt{\dfrac 3{\rho G \pi}}\\v_2 &= \omega R\\&=2R\sqrt{\dfrac{\rho G \pi}3}\end{aligned}$

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Thus, $x=\dfrac{T_1}{T_2} = \dfrac{\dfrac{1}{4}\sqrt{\dfrac{3\pi}{\rho G}}}{\sqrt{\dfrac 3{\rho G \pi}}}=\dfrac{\pi}{4}$

And, $y=\dfrac{v_1}{v_2} = \dfrac{2R\sqrt{\dfrac{\rho G \pi}3}}{2R\sqrt{\dfrac{\rho G \pi}3}}=1$

$\Huge \therefore \boxed{\dfrac{x}{y} = \dfrac{\pi}4}\large\approx 0.7854$