Ball dropped through a tunnel at Earth's center

[K.J.Healey]

The non generated one is just for :
G = 6.667428*10^-11 _m^3 _kg^-1 _s^-2

 

[Doc Al]

Heres how i did it in mathematica:
$$F= -G m M /r[t]{}^{\wedge}2$$
$$-\frac{G m M}{r(t)^2}$$

$$U=\text{Integrate}[F,r[t]]$$
$$\frac{G m M}{r(t)}$$

$$\rho = \text{ME}/((4/3)*\text{Pi}*\text{RE}{}^{\wedge}3)$$
$$\frac{3 \text{ME}}{4 \pi \text{RE}^3}$$

$$M=\rho *(4/3)*\text{Pi}*r[t]{}^{\wedge}3$$
$$\frac{\text{ME} r(t)^3}{\text{RE}^3}$$

$$T=(1/2)m (r'[t]){}^{\wedge}2$$
$$\frac{1}{2} m r'(t)^2$$

$$L=T-U$$
$$\frac{1}{2} m r'(t)^2-\frac{G m \text{ME} r(t)^2}{\text{RE}^3}$$

Since I don't use Mathematica I can't comment on your use of it, but your potential energy term is incorrect.

The force is:
$$F = -\frac{m M G}{R^3} r$$

So:
$$U = \frac{m M G}{2 R^3} r^2$$

 

[K.J.Healey]

Ah that's totally it, can't integrate F = GmM/r^2 to U when M is a function of R.