It might also help to see it from the perspective of:
[tex]
3 \frac{GM}{rc^2} \frac{1}{1 - e^2} = 3 \frac{v^2}{c^2} \frac{1}{1 - e^2},[/tex]
[tex]
v = \sqrt{\frac{GM}{r}}.[/tex]
... where e is eccentricity, and r is the average orbit radius (semi-major axis).
From what we know about special relativity, kinematic time dilation, in relation to an orbiting body, is given by:
[tex]
\tau = t \sqrt{1 - \frac{v^2}{c^2}} = t \sqrt{1 - \frac{GM}{rc^2}}, [/tex]
From general relativity, gravitational time dilation is given by:
[tex]
\tau = t \sqrt{1 - \frac{2GM}{rc^2}}.[/tex]
As you can see, both forms of time dilation are quite similar when considering it from the perspective of an orbiting body.
For fun, we can obtain the Newtonian acceleration in a backwards sort of way, by differentiating the gravitational time dilation formula with respect to r:
[tex]
\tau \approx t \approx 1,[/tex]
[tex]
\frac{\partial \tau}{\partial r} \approx \frac{GM}{r^2 c^2},[/tex]
[tex]
a = \frac{\partial \tau}{\partial r} {c^2} = \frac{GM}{r^2},[/tex]
[tex]
v = \sqrt{a r} = \sqrt{\frac{GM}{r}}.[/tex]