If the Earth is modeled as a constant density, non-rotating sphere, then Schwarzschild's interior solution can be used. When G=c=1,
<br />
d\tau^{2}=\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}dt^{2}-\left( 1-\frac{2Mr^{2}}{R^{3}}\right) ^{-1}dr^{2}-r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right),<br />
where R is the r coordinate at the surface of the Earth.
If an observer on the Earth's surface uses a telescope to look down a tunnel to a clock at the Earth's centre, he will see his clock running faster than the clock at the Earth's centre.
Consider two dentical clocks, one moving around the Earth once a day on the Earth's surface at the equator (\theta = \pi/2) and one at the Earth's centre. Both clocks have constant r values, so dr=0 for both clocks, and, after factoring out a dt^2, the above equation becomes
<br />
\left( \frac{d\tau }{dt}\right) ^{2}=\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}-v^{2},<br />
where v=rd\phi/dt is, approximately, the speed of something moving along a circular path. At the centre, v=r=0, and, on the surface, r = R and v = 1.544 \times 10^{-6}, which is one Earth circumference in one day.
Then, with G and c restored,
<br />
\frac{d\tau_{centre}}{d\tau_{surf}}=\left( \frac{d\tau_{centre}}{dt}\right) \left( \frac{d\tau_{surface}}{dt}\right)^{-1} =\frac{\frac{3}{2}\sqrt{1-\frac{2GM}{c^{2}R}}-\frac{1}{2}}{\sqrt{1-\frac{2GM}{c^{2}R}-v^{2}}}<br />.
Running, the numbers, I get
<br />
\frac{d\tau_{centre}}{d\tau_{surf}} = 1 - 3.5 \times 10^{-10}.<br />
Lots of places errors could have crept in, though.
