From One of My Unpublished Papers $$\frac{d^2 x^{\alpha}}{d\tau^2}=-\Gamma^{\alpha}_{\beta \gamma}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau} \tag{1}$$
For radial motion in Schwarzschild’s Geometry we have,
$$\frac{d^2 r}{d \tau^2}=-\frac{M}{r^2}\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2+\frac{M}{r^2}\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2\tag{2}$$
Again from radial motion, we have from Schwarzchild’s metric:
$$d\tau^2=\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2\tag{3}$$
Dividing both sides of (3) by $d\tau^2$ we have,
$$1=\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2-\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2\tag{4}$$
Using relation (4) in (2), after factoring out $M/r^2$ from the RHS of (2), we obtain:
$$ \frac{d^2 r}{d \tau^2}=-\frac{M}{r^2}\tag{5}$$
The inverse square law should hold accurately if proper time is used. Here $r$ represents the coordinate distance along the radius.
One may use the relations:
$$M ~\rightarrow~ GM/c^2\qquad\text{and}\qquad\tau ~\rightarrow~ c\tau,$$
to obtain the exact "form" of the law of Gravitation.
Query: Is equation (5) indicative of the fact that Gauss law may be used in the same classical "form" in GR?
[We may introduce a symbol $F=m\frac{d^2 r}{d\tau^2}$.]
