# At what rate does light 'bend' around the surface of the earth?

Since the g force of earth is 9.8 m/s*2 does that mean light 'drops' at that rate as it travels past earth? Or is general relativity a lot more complex than that?

If you assume that light "drops" at the gravitational acceleration $g$, and calculate the angle by which light is deflected using Newtonian mechanics, you end up with the formula:
$$\theta = \frac{2GM}{c^2r}$$
where $M$ is the mass of the deflecting object and $r$ is the distance of closest approach. A quick Google should find the calculation, or see this paper on the Arxiv. However when you do the calculation using general relativity you get the result:
$$\theta = \frac{4GM}{c^2r}$$