What is escape velocity from the center of the Earth?

Question

Escape velocity on the surface is 11.2 km/s.

If a hole is drilled to the center of the Earth, then what initial velocity should have the projectile launched from the center of the Earth in order to escape to the infinity?

Answer 1

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

Answer 2

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.

Answer 3

I agree with the analysis of rhsaunders. This is the equation I get when I put the analysis in math form (hope I haven't made a mistake):

V0² = 8πG/3·∫(rρ(r)dr) + Ve²

where:
V0 = required velocity at earth's center;
G = universal gravitational constant;
ρ(r) = density of earth as a function of distance r from center;
Ve = escape velocity at surface;
Limits of integration are from r=0 to r=(earth's radius)