The gravitational time dilation for a static "shell" observer in the Schwarzschild metric observed from infinity is
$$ d\tau_{\rm shell} = dt\left( 1 - \frac{r_s}{r}\right)^{1/2}\ .$$
We could label this as the "General Relativistic time dilation".

The speed of an object in a circular orbit around a central mass, as measured by the stationary shell observer at radius $r$, is
$$ v_{\rm shell} = c\left(\frac{r_s}{2(r-r_s)}\right)^{1/2}\ .$$

The time dilation due to the orbital motion with respect to the local inertial frame of the shell observer is
$$d\tau = \frac{dt_{\rm shell}}{\gamma_{\rm shell}}\ ,$$
where $\tau$ here is proper time measured by the orbiting observer.
We could label this as the "Special Relativistic time dilation".

In a local inertial frame, $dt_{\rm shell} = d\tau_{\rm shell}$, so
$$d \tau = \gamma_{\rm shell}^{-1}\ d\tau_{\rm shell} = \left( 1 - \frac{r_s}{r}\right)^{1/2} \gamma_{\rm shell}^{-1}\ dt\ ,$$
and we are effectively multiplying the "General Relativistic" and "Special Relativistic" time dilations to get$^{*}$
$$ d\tau = \left[ \left(1 - \frac{r_s}{r}\right)\left(1 - \frac{r_s}{2(r-r_s)}\right) \right]^{1/2} = \left(1 - \frac{3r_s}{2r}\right)^{1/2}\ .$$
**No approximations have been made, this is an exact result.**

**EDIT: To address the last part of the question.**

In the more "holistic" version of the calculation - directly using the Schwarzschild metric - the question arises as to why one uses the "Newtonian" value of $v = rd\phi/dt = (GM/r)^{1/2}$? Note that this is not an assumption or an approximation, it is exactly true for the Schwarzschild metric.

Apologies, but this is a lengthy derivation. To see this we have to go back to geodesics in the Schwarzschild metric, which are defined in terms of the two constants of motion that are often labelled $E/mc^2$ and $L/m$, the coordinate specific energy and coordinate specific angular momentum respectively.

In terms of these quantities, the rate of change of radial coordinate with respect to proper time can be written
$$ \frac{dr}{d\tau} = \pm c\left[ \left(\frac{E}{mc^2}\right)^2 - \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{m^2r^2c^2}\right) \right]^{1/2}\ , \tag*{(1)} $$
where
$$ \left(1 - \frac{r_s}{r}\right)\frac{dt}{d\tau} = \frac{E}{mc^2}\ , $$
$$ r^2 \frac{d\phi}{d\tau} = \frac{L}{m}\ . $$

We can now express the circular velocity as seen by a distant observer, $v_{\rm circ} = r d\phi/dt$ as :
$$ v_{\rm circ}^2 = r^2\left(\frac{d\phi}{dt}\right)^2 = r^2\left(\frac{d\phi}{d\tau}\right)^2 \left(\frac{d\tau}{dt}\right)^2 = \frac{L^2}{m^2r^2} \left(1 - \frac{r_s}{r}\right)^2 \left(\frac{mc^2}{E}\right)^2\, . $$
But for a circular orbit $dr/d\tau=0$ and eqn. (1) gives us a relation between $E$ and $L$ --
$$ \left(\frac{E^2}{mc^2}\right)^2 = \left(1 - \frac{r_s}{r}\right)\left(1 + \frac{L^2}{m^2r^2c^2}\right) \ ; $$
and substituting this into the equation for $v_{\rm circ}^2$, we get
$$ v_{\rm circ}^2 = \frac{L^2}{m^2 r^2} \left(1 - \frac{r_s}{r}\right) \left(1 + \frac{L^2}{m^2 r^2 c^2}\right)^{-1} . \tag*{(2)} $$

The last part of the puzzle is to find an expression for $L/m$ in terms of the radial coordinate for a circular orbit. This is done by writing an expression for the effective potential in the Schwarzschild metric and finding where it is a minimum. Thus:
$$ \frac{V_{\rm eff}}{mc^2} = -\frac{r_s}{2r} +\frac{L^2}{2m^2r^2c^2} - \frac{r_sL^2}{2m^2 r^3 c^2}\ . $$
Differentiating and equating to zero, we get
$$ \left(\frac{L}{m}\right)^2 = \frac{c^2\,r^2\,r_s}{2r - 3r_s}\ . \tag*{(3)} $$

Replacing $L^2/m^2$ in eqn. (2) using eqn. (3),
\begin{eqnarray}
v_{\rm circ}^2 & = & c^2\left(\frac{2r_s}{2r- 3r_s}\right) \left(1 - \frac{r_s}{r}\right) \left(1 + \frac{r_s}{2r-3r_s}\right)^{-1} \nonumber \\
& = & c^2 \left(\frac{2r_s}{2r- 3r_s}\right) \left(\frac{r - r_s}{r}\right)\left(\frac{2r - 3r_s}{2r -2r_s}\right) \nonumber \\
& = & c^2\, \frac{r_s}{2r} = \frac{GM}{r}\, .
\end{eqnarray}
Thus the coordinate speed according to a distant observer is $\sqrt{GM/r}$, **the same as the Newtonian result!**

I called the function f(x), so $x$ instead of $\Delta r$ and all divided by $t$.

$f(x) = \sqrt{1-\frac{2GM}{r_Ac^2}+\frac{2GM x}{r_A^2c^2}}$

$f'(x) = \frac{1}{2} \frac{1}{\sqrt{1-\frac{2GM}{r_Ac^2}+\frac{2GM x}{r_A^2c^2}}} \frac{2GM }{r_A^2c^2}$

With normal Taylor expansion around $x_0 = 0 $ to first order:

$f(x) \approx \sqrt{1-\frac{2GM}{r_Ac^2}} + \frac{1}{ \sqrt{1-\frac{2GM}{r_Ac^2}}}\frac{GM x}{r_A^2c^2}$

$ = \sqrt{1-\frac{2GM}{r_Ac^2}} (1+\frac{1}{1-\frac{2GM}{r_Ac^2}} \frac{GM x}{r_A^2c^2})$

If you say $\frac{1}{1-\frac{2GM}{r_Ac^2}} \approx 1$ you recover your result.
But as for now, doing so does not make sense to me.

## Best Answer

This is how to calculate the time dilation for an object moving at velocity $v$ in a radial direction towards or away from the black hole.

Because the object is moving radially $d\theta = d\phi = 0$ and the Schwarzschild metric simplifies to:

$$ c^2d\tau^2 = c^2\left(1-\frac{r_s}{r}\right)dt^2 - \frac{dr^2}{1-r_s/r} \tag{1}$$

$d\tau$ is the proper time, and this corresponds to the time shown on the falling objects clock. $dt$ and $dr$ and the time and radial displacement measured by the distant observer. The time dilation is $d\tau/dt$, and to calculate this we have to note that if the velocity measured by the Schwarzschild observer is $v$ then $dr = vdt$. Substituting this into equation (1) we get:

$$ c^2d\tau^2 = c^2\left(1-\frac{r_s}{r}\right)dt^2 - \frac{v^2dt^2}{1-r_s/r} $$

And rearranging this gives:

$$ \left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{r_s}{r} - \frac{v^2}{c^2}\frac{1}{1-r_s/r} \tag{2} $$

I've left $v$ in the equation. To eliminate $v$ you need to use the expression relating $v$ to $r$ for an object free-falling from infinity:

$$ \frac{v}{c} = - \left( 1 - \frac{r_s}{r} \right) \left( \frac{r_s}{r} \right)^{1/2} $$

I'll leave the working as an exercise for the reader. The rather surprising result after we've done the substitution is:

$$ \frac{d\tau}{dt} = 1 - \frac{r_s}{r} \tag{3} $$