Incompatibility of the Schwarzschild solution and Newton's Inverse Square Law

Newton proved, save claims to the contrary, that under an inverse square law force every orbit is a conic section. And the reverse is also true that every orbit which is a conic section corresponds to and inverse square law for the force. Newton showed this by using uniform circular motion in which the radius of curvature is constant.

The radius of the Schwarzschild metric is not constant, but, rather

$$\kappa=-\frac{\mathcal{R}}{r^3}\left(1-\frac{3\mathcal{R}}{4r}\right),$$

and so the Schwarzschild metric does not, nor cannot, lead to conical orbits. Contrary to this, Einstein showed that in the approximation $r^2\dot{\varphi}=L$ is constant then for a total negative energy, there exists an elliptical orbit

$$r=\frac{a(1-\epsilon^2)}{1+\epsilon\cos\varphi},$$

where $a$ is the semi-major  axis of the orbit of Mercury and $\epsilon$ is its eccentricity. The exact expression for the angular momentum,

$$\frac{r^2\dot{\varphi}}{1-\mathcal{R}/r}=L,$$

shows that the angular momentum is not conserved. And setting the denominator equal to unity is tantamount to neglecting the last term in the "perturbed" equation for the orbit,

$$\left(\frac{d\varrho}{d\varphi}\right)^2=A+B\varphi-\varrho^2+\mathcal{R}\varrho^3,$$

where $A$ and $B$ are constants and $\varrho=1/r$. Neglecting this term leads to the above equation for an elliptical orbit, and consequently no advance in the perihelion of Mercury.

In the early 1670's, Newton derived the following expression for the radius of curvature,

$$\kappa=\frac{r(1+z^2)^{3/2}}{1+z^2-dz/d\varphi},$$

where $z=d\ln r/\varphi$ is the slope of the curve. According to Proposition 1, Kepler's law of areas is

$$rv\sin\alpha=L,$$

where the angle $\alpha$ is the complement to the angle $\varphi$. Then Newton replaced the increment of motion along any general curve by one of uniform circular motion. Hence, the centripetal force is related to the force directed inward toward the center, $F_C$, by

$$F_{\varphi}=F_C\cos\varphi=F_C\sin\alpha=\frac{v^2}{\kappa},$$

where $v$ is the angular velocity. 

Newton explicitly employed  the assumption of uniform circular motion  in Proposition 7, Problem 2, Corollary 5. Inverting the above equation for the centripetal force, Newton obtained the curvature as

$$\kappa=v^2/F_C\sin\alpha=L^2/F_Cr^2\sin\alpha,$$

from which he could deduce

$$F_C=\frac{L^2}{r^2}\left(r^{-1}+\frac{d^2r^{-1}}{d\varphi^2}\right).$$

And with the equation of the elliptical orbit,

$$\frac{1}{r}=1+\frac{A}{B}\cos\varphi,$$

he obtained

$$F_C=\frac{L^2}{r^2}\left(1+\frac{A}{B}\cos\varphi-\frac{A}{B}\cos\varphi\right),$$

and concluded that all conical orbits result in inverse square force laws. Newton also proved that given the curvature of the force, the orbit is uniquely determined: Inverse square laws result in conical orbits. 

The assumption of uniform circular motion implies constant curvature. Metrics of non-constant curvature, like Schwarzschild's, are incompatible with Newton's inverse square law. Hence, it is also correct to assume that even asymptotically, that Newton's law should be obtained  in the metric coefficient of time,

$$g_{00}=1+2V,$$

with $V$ small. In Dirac's words, "[w]e get $g_{00}^{1/2}=1+V$ and $V$ becomes potential. It satisfies the Laplace equation, so that it can be identified with the Newtonian potential equal to $-m/r$ for a mass $m$ at the origin." Simply put, it can't because the radius of curvature, $\kappa$ isn't constant. Translated in terms of a metric, that implies that the metric be one of constant curvature. The inner solution of Schwarzschild's solution is, therefore, O.K., the outer one, not.