Originally Posted by

**xyzt**
Originally Posted by

**grav-universe**
Does all of this look okay so far?

No, it doesn't, this is not the way GR deals with equations of motion. Shoehorning basic Newtonian equations is not the way to go.

All acceleration formulas will reduce to these when starting from rest and measuring over infinitesimal distance and time. If nothing else, we can define the coordinate acceleration in this way. Even the relativistic equations will reduce to the same thing in this case.

s = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1]

where t is infinitesimal, so

s = (c^2 / a) [(1 + (a dt / c)^2 / 2) - 1]

s = (c^2 / a) [a^2 dt^2 / (2 c^2)]

s = a dt^2 / 2

So if the object is initially at coordinates x_o=r, y_o=0, and travels an infinitesimal distance y = v t in the tangent direction, it will have travelled a distance of x = r - sqrt(r^2 - y^2) in the radial direction toward the mass

The stuff that you wrote is not correct even in Newtonian mechanics since you are mixing radial (linear) motion with circular motion. So, what you wrote is complete nonsense. If the infinitesimal motion is

in the tangential direction, then , in the radial direction you should have

, meaning that

.

For a circular orbit, we would have

X^2 + Y^2 = r^2, so X = sqrt(r^2 - Y^2). For infinitesimal y travelled, whereas y = Y, that gives for the distance x travelled in the radial direction

x = r - X = r - sqrt(r^2 - y^2) = y^2 / (2 r)