# How does general relativity explain the impact force after radial inward fall toward Earth?

I understand that, in the context of the Schwarzschild spacetime (General Relativity), a radially inward falling observer follows a time-like geodesic with zero four-acceleration. There are no forces acting on this observer and he is thus in free fall. In addition, the (Schwarzschild) coordinate acceleration is nonzero, but this is interpreted as a result of a fictitious force (i.e. the gravitational "force").

If the Earth is at rest, how does general relativity explain the fact that the impact force (when the observer hits the ground) depends on the initial radial distance, even though the observer would have a zero four-acceleration regardless of how large the radial distance is?

If I resort to the equivalence principle, then I see why a free falling observer getting hit by an accelerating mass would result in an impact force that depends on the radial distance from the observer to the accelerating mass (a rocket, for example). The further away the observer, the more momentum is built up before impact, but in this case the mass is actually accelerating while the Earth is at rest.

• The surface of the Earth is not in free fall; an accelerometer at rest on the surface of the Earth registers non-zero acceleration. – Alfred Centauri Nov 7 '15 at 12:48

Imagine we have three massive objects: the Earth and two apples. Now, assume that apple A and apple B are initially held (by an external force) at radial distance $r_A$ and $r_B$ from the center of the Earth, respectively, with $r_A$ > $r_B$. It then follows that both apples exhibit nonzero four-acceleration. Next, release apple A. After a while, both apples will collide, since their worldlines will cross. The larger the radial separation between the apples, the stronger the collision, as apple B keeps being four-accelerated (apple A is free falling, i.e., it has zero four-acceleration). Note that apple B could play the role of the surface of the Earth.