On a trip to the center

FullSimplify[∫_R^0 -1/v[r] r, AssumptionsR>0]

(π R^(3/2))/(2 (G M)^(1/2))

t0[R_, G_, M_] := π/2R^3/(G M)^(1/2)

Hence, for example, the time elapsed falling through the earth's center to the other side would be...

Convert[2 * t0[EarthRadius, GravitationalConstant, EarthMass], Second]

RowBox[{2534.49, , Second}]

In minutes, this is...

Convert[2 * t0[EarthRadius, GravitationalConstant, EarthMass], Minute]

RowBox[{42.2415, , Minute}]

It turns out (see the section on chords not through the center below) that π/2R^3/(G M)^(1/2)= 42.2415 Minutes is the elapsed time for all GraviTube frictionless surface-surface dives.  To determine the time elapsed partway into a radial dive, one integrates only to rf...

FullSimplify[∫_R^rf -1/v[r, R, G, M] r, Assumptions {R>0, rf>0, rf<R, G>0, M>0}]

(R^(3/2) (π - 2 ArcTan[rf/(R^2 - rf^2)^(1/2)]))/(2 (G M)^(1/2))

tRadial[rf_, R_, G_, M_] := R^(3/2)/(2 (G M)^(1/2)) (π - 2 ArcTan[rf/(R^2 - rf^2)^(1/2)])

Created by Mathematica  (June 17, 2004)