What is escape velocity from the center of the Earth?

An interesting question which I haven't encountered before.  I won't run the numbers, but I'll outline the solution: the total kinetic energy required will be that needed to reach infinity from the surface, plus that required to reach the surface from the center.  The former is easily determined from the stated escape velocity; the latter can be determined by integrating the gravitational force from zero to the surface.  Although the value of this depends on the earth's density at various depths, it is an adequate approximation to assume that the density is constant throughout, and in that case the gravitational pull is simply a linear function of the distance to the center, and the integral will be proportional to the square of that distance.

Brilliant question, simple answer (I think).

First dropping an apple towards the earth's center.  Its velocity should peak at the velocity of low earth orbit which is Escape Velocity/sqr(2).  (This assumes uniform density).

Therefore, using Conservation of Energy
Escape V center of Earth = sqr(Escape V^2 + (EscapeV/sqr(2))^2)
=sqr(1.5) * Escape V