Hi. I'm going to sticky/lock this topic, because it's a pretty frequently-asked question that demands an explanation that is too long for the FAQ.

If you measure the speed of the Earth, you will notice it to be always zero. That is, it's not getting any closer to you or farther from you, and it's not moving around in a circle. If you're standing on the Earth, the Earth appears to be quite stationary.

Similarly, if you are in an automobile that is driving around a sharp turn, you will feel yourself pressed to one wall (the wall opposite the turn). From your perspective, the automobile is stationary -- but, from your perspective, you can still tell that it is accelerating (or, equivalently, that it is generating a gravitational field).

On the other hand, as measured from somebody not in the car, the car is zipping around the turn -- i.e., moving *and* accelerating. Similarly, the Flat Earth can be observed by others not on the Earth to be accelerating and moving upwards. If you are worried that the Earth might eventually be going faster than the speed of light, you are only worried about it going faster than the speed of light as *those observers* measure it, not as *we on the Earth* measure it.

So, let's consider what happens from the perspective of one such observer -- call her Alice. In particular, let's suppose Alice jumps over the ice wall into empty space (or whatever's there). Since the Earth is no longer pushing her, she stops accelerating. From her perspective, she is now at rest, and at the instant she jumps, the Earth is also at rest. However, the Earth continues to accelerate upwards.

Let's say Bob stayed behind on Earth. He measures the Earth's acceleration to be g = 9.8 m/s^2 -- the same way you might measure the accleration of the automobile going around the curve. According to special relativity, Alice will measure a different acceleration -- in particular, she will measure it to be g/γ^3, where γ = 1/sqrt(1 - v^2/c^2). In other words, we have the following differential equation for the velocity v of the Earth: dv/dt = g/γ^3.

The solution to this equation is v = gt / sqrt( 1 + g^2t^2 / c^2 ), or v = g / sqrt(1/t^2 + g^2/c^2). We are interested in the infinite time limit of v -- i.e. what happens to v as we wait an arbitrarily long time. As t --> ∞, v --> g / sqrt(g^2/c^2) = c.

In other words, from Bob's perspective, the Earth is always stationary (we are Bob), but undergoing constant acceleration. From Alice's perspective, the Earth moves at an ever increasing rate, but the acceleration is not constant -- it decreases over time in such a way that the Earth never surpasses the speed of light.

For more information, especially on the derivation and solution of the differential equation I presented, see p. 37 of "Introducing Einstein's Relativity" by Ray D'Inverno, in particular section 3.8.