I wanted to add the actual GR math. The outcome of this is exactly as everyone who has any experience in GR said. Indeed, from first principles it could be no other way. But I had time yesterday to play around with this. All equations are using geometrized units where ##c=G=1##.
We start with the weak field metric in cylindrical coordinates: $$ds^2 = (1 - 2 U) dr^2 + (-1 - 2 U) dt^2 + (1 - 2 U) dz^2 + (r^2 - 2 r^2 U) d\phi^2 $$ with the standard gravitational potential in cylindrical coordinates $$ U=-\frac{M}{\sqrt{r^2+z^2}} $$
Now, an object in orbit is in free-fall, so the worldline of the planet is a geodesic. To calculate the orbit of the earth we therefore calculate the geodesic using the equations described
here. When we do so, we get the following equations: $$0 = \left(
\begin{array}{c}
0 \\
\frac{2 r^2 \left(z^2-2 M^2\right)
\ddot r+r \left(-z^2 \left(z^2-4
M^2\right) \ \dot \phi ^2-M \dot r^2 \left(2 M+3
\sqrt{r^2+z^2}\right)+M \dot z^2
\left(\sqrt{r^2+z^2}-2 M\right)+M
\left(\sqrt{r^2+z^2}-2
M\right)\right)+z \left(z
\left(z^2-4 M^2\right) \ddot r-4 M
\sqrt{r^2+z^2} \dot r
\dot z\right)+r^3 \ \dot phi^2 \left(M
\left(2 M+\sqrt{r^2+z^2}\right)-2
z^2\right)+r^4 \ddot r+r^5
\left(-\dot \phi
^2\right)}{\left(r^2+z^2\right)
\left(-4 M^2+r^2+z^2\right)} \\
\frac{2 \dot r \dot \phi \left(-4 M^2+r^2
\left(1-\frac{2
M}{\sqrt{r^2+z^2}}\right)+z^2\right)
+r \left(\left(r^2-4 M^2\right) \ddot \phi
-\frac{4 M z \dot z \dot \phi
}{\sqrt{r^2+z^2}}+z^2 \ddot \phi
\right)}{r \left(-4
M^2+r^2+z^2\right)} \\
\frac{r \left(r \left(r^2-4 M^2\right)
\ddot z-4 M \sqrt{r^2+z^2} \dot r
\dot z\right)+2 z^2 \left(r^2-2
M^2\right) \ddot z+M z \left(\dot r^2
\left(\sqrt{r^2+z^2}-2 M\right)-\dot z^2
\left(2 M+3
\sqrt{r^2+z^2}\right)+\left(\sqrt{r^
2+z^2}-2 M\right) \left(r^2 \dot \phi
^2+1\right)\right)+z^4
\dot z}{\left(r^2+z^2\right) \left(-4
M^2+r^2+z^2\right)} \\
\end{array}
\right) $$
To specifically find a circular orbit we can set ##z=0## and ##r=R## and ##\phi = d\phi \ t##. That simplifies the geodesic equation to: $$ 0=\left(
\begin{array}{c}
0 \\
\frac{-\text{d$\phi $}^2 M R^2-\text{d$\phi
$}^2 R^3+M}{2 M R+R^2} \\
0 \\
0 \\
\end{array}
\right) $$ so $$ {d\phi}=\frac{\sqrt{M}}{\sqrt{M
R^2+R^3}} $$
Solving for ##\phi = 2\pi## we get $$ t_{2\pi}=\frac{2 \pi \sqrt{R^2 (M+R)}}{\sqrt{M}} $$
Evaluating proper time along the worldline of the earth we get $$ \frac{{d\tau}}{{dt}}=\frac{\sqrt{-4
M^2-2 M R+R^2}}{\sqrt{M R+R^2}} $$ which we can integrate to get Big Ben's time over one year to get $$\tau_{2\pi} =\int_0^{t_{2\pi}} \frac{d\tau}{dt} dt = 2 \pi R \sqrt{-\frac{4 M}{R}+\frac{R}{M}-2} $$ Plugging in the mass of the sun ##M=1480## in geometrized units (meters) and the orbital radius of the Earth ##R=1.496 \ 10^{11}## we get the proper time ##\tau_{2\pi} = 9.45 \ 10^{15}## which is one year in geometrized units.
I will post the analysis in a moving frame in a separate post.