It might be instructive to work out the problem using GR in the Earth centered frame, using the Schwarzschild metric, for a circular orbit at the equator. This will also show that the usual SR velocity time dilation formula is only an approximation.
It might be helpful to recap how we get the equation for time dilation in SR, first.
The time dilation that we wish to solve for is just
<br />
\frac{dt}{d\tau}<br />
In SR, we know the flat space-time metric is
c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
dividing both sides by dt^2, we get
c^2 \left( \frac{d\tau}{dt} \right) ^2 = c^2 - \left( \frac{dx}{dt} \right)^2 - \left( \frac{dy}{dt} \right)^2 - \left( \frac{dz}{dt} \right)^2
Solving for d\tau / dt we get
\frac{d\tau}{dt} = \sqrt{1 - \frac{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 }{c^2} }<br />
We recognize this as
\frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}}
but since we want the reciprocal, dt / d\tau, we get the usual relation
\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
Now, we just need to repeat this for the case of a gravitating body. We'll use Schwarzschild coordinates to get the curved space-time metric, rather than the flat metric we used in SR.
Note that in Schwarzschild coordinates, \theta is Pi at the equator, and \phi varies from 0 to 2Pi as it sweeps out the orbit. r is constant for a circular orbit.
So we write
<br />
c^2 d\tau^2 = g_{tt} dt^2 + g_{rr} dr^2 + g_{\theta\theta} d\theta^2 + g_{\phi\phi} d\phi^2<br />
dividing both sides of the equation by dt^2, and dropping some terms that we know to be zero, such as \frac{dr}{dt} and \frac{d\theta}{dt}, we get
<br />
c^2 \left( \frac{d\tau}{dt} \right)^2 = g_{tt} + g_{\phi\phi}\left(\frac{d\phi}{dt}\right)^2<br />
We know that for the schwarzschild metric
http://en.wikipedia.org/w/index.php?title=Schwarzschild_metric&oldid=325313722
g_{tt} = \left(1-\frac{r_s}{r}\right)^2\,c^2
and we know that
g_{\phi\phi} = -r^2
as sin\phi is one.
Putting this together we get
<br />
c^2 \left( \frac{d\tau}{dt} \right)^2 = c^2 \, \left(1-\frac{r_s}{r}\right)^2\ - r^2 \left(\frac{d\phi}{dt}\right)^2<br />
This can be rewritten as
<br />
\frac{d\tau}{dt}= \sqrt{ g_{tt} } \sqrt{1 - \frac{1}{g_{tt}} \left( \frac{r \frac{d\theta}{dt} }{c} \right)^2<br />
This is almost in the form of the product of the GR and SR time dilation but not quite exactly.
Note that if we set d\phi / dt to zero, we see that the time dilation is just the gravitational time dilation
<br />
\frac{dt}{d\tau}= \frac{1}{\sqrt{g_{tt}}}
But because the gravitational time dilation is so nearly unity, it provides only a tiny correction to the velocity in the SR formula to multiply it by g_tt so it's approximately correct to multiply the SR time dilation by the gravitational time dilation for a non-moving object to get the total time dilation.