Schwarzschild metric:
$d\tau^2 = -(1 - \frac{R}{r}) dt^2 + (1 - \frac{R}{r})^{-1} dr^2 + r^2 d\Omega^2$
(treating $c = 1$)

Within a spherical body:
$R = R_0 \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi r_0^3} = R_0 \frac{r^3}{r_0^3}$
for $R_0$ the Schwarzschild radius
for $r_0$ the body radius (with constant density)

For purely timelike geodesics:
$d\tau^2 = -(1 - \frac{R}{r}) dt^2$
substitute:
$\frac{d\tau^2}{dt^2} = -(1 - \frac{R_0}{r_0^3} r^2)$

Circular orbit in Minkowski spacetime:
$\vec{c} = \pmatrix{c_t \\ c_x \\ c_y} = \pmatrix{\mu \\ \alpha cos (\beta \mu) \\ \alpha sin (\beta \mu) }$
$\frac{\partial \vec{c}}{\partial \mu} = \pmatrix{1 \\ -\alpha \beta sin (\beta \mu) \\ \alpha \beta cos (\beta \mu) }$
Using a Minkowski metric:
$|\frac{\partial \vec{c}}{\partial \mu}|^2 = \frac{\partial c_\mu}{\partial \mu} \frac{\partial c_\nu}{\partial \mu} \eta_{\mu\nu} = -(1 - \alpha^2 \beta^2)$

Do we need to reparameterize by arclength? Probably not. Matching derivative magnitudes enough.
$\sqrt{\left| |\frac{\partial \vec{c}}{\partial \mu}|^2 \right|} = \sqrt{1 - \alpha^2 \beta^2}$
Arclength of circular orbit in Minkowski spacetime:
$\tau = \int_0^\mu \sqrt{\left| |\frac{\partial \vec{c}}{\partial \mu}|^2 \right|} d\mu$ $ = \int_0^\mu \left( \sqrt{1 - \alpha^2 \beta^2} \right) d\mu$ $ = \mu \sqrt{1 - \alpha^2 \beta^2}$
Reparameterization by arclength:
$\mu = \frac{\tau}{\sqrt{1 - \alpha^2 \beta^2}}$
$\vec{c} = \pmatrix{c_t \\ c_x \\ c_y} = \pmatrix{ \frac{\tau}{\sqrt{1 - \alpha^2 \beta^2}} \\ \alpha cos (\frac{\tau \beta}{\sqrt{1 - \alpha^2 \beta^2}}) \\ \alpha sin (\frac{\tau \beta}{\sqrt{1 - \alpha^2 \beta^2}}) }$

Equate $\frac{d\tau^2}{dt^2} = |\frac{\partial \vec{c}}{\partial \mu}|^2$:
$\frac{R_0}{r_0^3} r^2 = \alpha^2 \beta^2$
$r \sqrt{ \frac{R_0}{r_0^3} } = \alpha \beta$ for arbitrary $\alpha \beta$
$R_0 = 2 M$, $\rho = \frac{M}{\frac{4}{3} \pi r_0^3}$, $M = \frac{4}{3} \pi r_0^3 \rho$
$\alpha \beta = r \sqrt{ \frac{2 \cdot \frac{4}{3} \pi r_0^3 \rho}{r_0^3} }$
$\alpha \beta = r \sqrt{ \frac{8}{3} \pi \rho }$

Kepler's 1-2-3 law:
$m' = \omega^2 r'^3$ (treating $G = 1$)
Match the period of the Minkowski spacetime path spiral with the period of the Kepler 1-2-3 law:
Let $\beta = \omega$
Then $\alpha \beta = \alpha \sqrt{\frac{m'}{r'^3}} = r \sqrt{\frac{R_0}{r_0^3}}$
This also equates the Kepler law's body' mass $m'$ with the Schwarzschild radius $R_0$
and the Kepler law's orbit distance $r'$ with the spherical body size radius $r_0$
So this all works if...
$\alpha = r$
$\beta = \omega = \sqrt{\frac{m'}{r'^3}}$
$r' = r_0$
$m' = R_0$
This makes the Minkowski metric parameterized spiral path equatl to:
$\vec{c} = \pmatrix{c_t \\ c_x \\ c_y} = \pmatrix{\tau \\ r cos (\omega \tau) \\ r sin (\omega \tau) }$
$\omega = \sqrt{\frac{R_0}{r_0^3}}$
Notice that $r$ is the distance from the center of the body being measured by our Schwarzschild metric, not the distance from the center of our Minkowski metric spiral (which is $\sqrt{(c_x)^2 + (c_y)^2}$).

Alternatively, we could match $\alpha$ to the radius of the particle, or $\beta$ to the angular velocity